1

THE ACADEMY CORNERNo. 22

Bruce Shawyer

All communications about this column should be sent to Bruce

Shawyer, Department of Mathematics and Statistics, Memorial University

of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

APICS Mathematics Competition 1998

held at Saint Mary's University, Halifax, Nova Scotia,

Rules:

� Teams of two are to work in cooperation and to submit one set ofanswers each.

� No notes, calculators, or other such aids are permitted.

� You may not communicate with noncontestants (except invigilators) orother teams.

� There are nine questions.

1. Fred and Cathy play the following game. They are given the polynomialf(x) = ax3 + bx2 + cx + d. They take turns, Cathy �rst, in replacinga, then b, then c and �nally d with positive integers. Fred wins if theresulting polynomial has at least two distinct roots. Who should winand what is the winning strategy?

2. De�ne the integer sequence fTng by T0 = 0, T1 = 1, T2 = 2 andTn+1 = Tn + Tn�1 + Tn�2 (n � 2). Compute

S :=

1Xn=0

Tn

2n.

3. Let X1; X2; : : : ; Xn be independent, integer-valued random variableswith p =probabilityfXk is eveng. Form the sum Sn of the randomvariables. Show that the probability that the sum is even is

[1 + (2p� 1)n]=2 .

2

4. Show that there do not exist four points in the Euclidean plane suchthat the pairwise distances between them are all odd integers.

5. If fang is a sequence of positive integers such that

limn!1

an

a1 + a2 + � � �+ an= 0 ,

show that there is a sequence fbng of positive integers such that forevery positive integer n � 2

bn

b1 + b2 + � � �+ bn� 1

3,

and for some positive integer N we have an = bn for all n � N .

6. For a > 1 evaluate Za

0

xa(�bloga xc)dx ,

where btc denotes the greatest integer less than or equal to t.

7. Let ABCD be a cyclic quadrilateral, inscribed in a circle !. Let A0, B0,C0, D0 be the points where the tangents at A and B, at B and C, atC and D and at D and A, respectively, intersect. Prove that the linesAC, BD, A0C0 and B0D0 are concurrent; that is, they intersect at onepoint.

8. The expression

(: : : (((x� 2)2 � 2)2 � 2)2 � : : :� 2)2| {z }n� times

is multiplied out and coe�cients of equal powers are collected. Findthe coe�cient of x2.

9. Let f(n) = 2n2 + 14n+ 25. We see that f(0) = 25 = 52. Find twopositive integers n such that f(n) is a perfect square.

The winning teams were:

1. Ian Caines and Alex Fraser | Dalhousie University;

2. Dave Morgan and Shannon Sullivan | Memorial University;

3. Tara Small and Kit Yan Wong | University of New Brunswick.

All six students received a free subscription to CRUX with MAYHEM, in ad-dition to some other prizes.

We will publish solutions later this year. Will your name be attachedto a solution? Send them to me as soon as you can!

Thanks to Karl Dilcher, Dalhousie University, Halifax, Nova Scotia, forsending me the LATEX �le.

3

THE OLYMPIAD CORNERNo. 195

R.E. Woodrow

All communications about this column should be sent to Professor R.E.

Woodrow, Department of Mathematics and Statistics, University of Calgary,

Calgary, Alberta, Canada. T2N 1N4.

Here it is, the start of another year. How time ies. It is time torecognize all the individual contributions which go to make up the Olympiad

Corner. First I would like to thank Joanne Longworthwho somehowmanagesto translate my o�erings into LATEX, and to keep me nearly on schedule too.My thanks, too, for the excellent �gures prepared by the Editorial O�ce.

The Corner could not exist without its readers who supply me withgood Olympiad materials as well as interesting solutions, generalizations,and corrections. I hope that we have missed no one in the following list ofcontributors for 1998:

Mohammed Aassila Yin LeiMiguel Amengual Covas Pavlos MaragoudakisDavid Arthur Vasiliou MeletisJamie Batuwantudawe Richard NowakowskiFrancisco Bellot Rosado Bob PrielippAdrian Birka Bill SandsMansur Boase Sree SanyalPierre Bornsztein Heinz-J �urgen Sei�ertAdrian Chan Toshio SeimiyaSonny Chan Michael SelbyKeon Choi Zun ShanJimmy Chui D.J. SmeenkFilip Crnogorac Daryl TingleyF.J. Flanigan Jim TottenChen He Panos E. TsaoussoglouWalther Janous Enrique Valeriano CubaMurray Klamkin Aliya WaljiMarcin Kuczma Edward T.H. WangMichael Lebedinsky

Thanks, and all the best for good problem solving in 1999.

4

As a �rst sample of Olympiad materials we give the Second Round 1995and First Round 1996 Problems of the Bundeswettbewerb Mathematik (theFederal Contest in Mathematics in Germany). My thanks go to Ravi Vakil,Canadian Team Leader to the 37 th IMO in Mumbai, India, for collectingthem.

BUNDESWETTBEWERB MATHEMATIKFederal Contest in Mathematics (Germany)

Second Round 1995

1. Starting in (1=1), a stone is moved according to the following rules:

(i) From any point (a=b), it can be moved to (2a=b) or (a=2b).

(ii) From any point (a=b), if a > b it can be moved to (a� b=b), and if a < b

it can be moved to (a=b� a).

Determine a necessary and su�cient relation between x and y so that thestone can reach (x=y) after some moves.

2. In a segment of unit length, a �nite number of mutually disjointsubsegments are coloured such that no two points with distance 0:1 are bothcoloured. Prove that the total length of the coloured subsegments is notgreater than 0:5.

3. Every diagonal of a given pentagon is parallel to one side of the pen-tagon. Prove that the ratio of the lengths of a diagonal and its correspondingside is the same for each of the �ve pairs. Determine the value of this ratio.

4. Prove that every integer k, (k > 1) has a multiple which is lessthan k4 and which can be written in decimal representation with at mostfour di�erent digits.

First Round 1996

1. Is it possible to cover a square of length 5 completely with threesquares of length 4?

2. The cells of an n�n-board are numbered according to the exampleshown for n = 5. You may choose n cells, not more than one from each rowand each column, and add the numbers in the cells chosen. Which are thepossible values of this sum?

1 2 3 4 5

6 7 8 9 10

11 12 13 14 15

16 17 18 19 20

21 22 23 24 25

5

3. There are four straight lines in the plane, each three of them deter-mining a triangle. One of these straight lines is parallel to one of the mediansof the triangle formed by the other three lines. Prove that each of the otherstraight lines has the same property.

4. Determine the set of all positive integers n for which n � 2n�1 is aperfect square.

As a second set this issue, we give the problems of the Grade XII levelof the XLV Lithuanian Mathematical Olympiad (1996). My thanks again goto Ravi Vakil for collecting them when he was Canadian Team Leader at the37 th IMO at Mumbai, India.

XLV LITHUANIANMATHEMATICAL OLYMPIAD1996

XII Grade

1. Solve the following equation in positive integers:

x3 � y3 = xy + 61 .

2. Sequences a1, : : : , an, : : : and b1, : : : , bn, : : : are such that a1 > 0,b1 > 0, and

an+1 = an +1

bn, bn+1 = bn +

1

an, n 2 N .

Prove thata25 + b25 > 10

p2 .

3. Two pupils are playing the following game. In the system8<:

�x+ �y+ �z = 0 ,�x+ �y+ �z = 0 ,�x+ �y+ �z = 0 ,

they alternately replace the asterisks by any numbers. The �rst player winsif the �nal system has a non-zero solution. Can the �rst player always win?

4. How many sides has the polygon inscribed in a given circle and suchthat the sum of the squares of its sides is the largest one?

5. Given ten numbers 2, 3, 5, 6, 7, 8, 10, 11, 12 ,13, one must crossout several of them so that the total of any of the remaining numbers wouldnot be an exact square (that is, the sum of any two, three, four,: : : , and of allthe remaining numbers would not be an exact square). At most, how manynumbers can remain?

6

And now, shifting to the other side of the globe, we give the problemsof Category A and Category B of the Vietnamese Mathematical Olympiad3/1996. Again our thanks go to Ravi Vakil, Canadian Team Leader at the37 th IMO in Mumbai, India, for collecting the contest materials.

VIETNAMESE MATHEMATICAL OLYMPIAD 3/1996Category A

First Day | Time: 3 hours

1. Solve the system of equations:8<:p3x�1 + 1

x+y

�= 2 ,

p7y�1� 1

x+y

�= 4

p2 .

2. Let Sxyz be a trihedron (a �gure determined by the intersectionof three planes). A plane (P ), not passing through S, cuts the rays Sx,Sy, Sz respectively at A, B, C. In the plane (P ), construct three trianglesDAB, EBC, FCA such that each has no interior point of triangle ABCand 4DAB = 4SAB, 4EBC = 4SBC, 4FCA = 4SCA. Considerthe sphere (T ) satisfying simultaneously two conditions:

(i) (T ) touches the planes (SAB), (SBC), (SCA), (ABC);

(ii) (T ) is inside the trihedron Sxyz and is outside the tetrahedron SABC.

Prove that the circumcentre of triangle DEF is the point where (T ) touches(P ).

3. Given two positive integers k and n, 0 < k � n, �nd the number ofthe k-arrangements (a1; a2; : : : ; ak) of the �rst n positive integers 1, 2, : : : ,n such that each k-arrangement (a1; a2; : : : ; ak) satis�es at least one of thetwo conditions:

(i) there exist s, t 2 f1, 2, : : : , kg such that s < t and as > at;

(ii) there exists s 2 f1, 2, : : : , kg such that (as � s) is not divisible by 2.

Second Day | Time: 3 hours

4. Determine all functions f : N� ! N� satisfying:

f(n) + f(n+ 1) = f(n+ 2)f(n+ 3)� 1996

for every n 2 N�, (N� is the set of positive integers).

5. Consider the triangle ABC, the measure of side BC (which is 1)and the measure of angle BAC (which is a given number �

�� > �

3

�). For

triangle ABC, �nd the distance from the incentre to the centre of gravity ofABC which attains the least value. Calculate this least value in terms of �.

7

Let f(�) be the least value. When � varies in the interval��

3; ��, at

which value of � does the function f(�) attain its greatest value?

6. We are given four non-negative real numbers a, b, c, d satisfyingthe condition:

2(ab+ ac+ ad+ bc+ bd+ cd) + abc+ abd+ acd + bcd = 16 .

Prove that:

a+ b+ c+ d � 2

3(ab+ ac + ad + bc+ bd+ cd) .

When does equality occur?

Category BFirst Day | Time: 3 hours

1. Determine the number of real solutions of the system of equationswith unknowns x, y: �

x3y � y4 = a2 ,x2y + 2xy2 + y3 = b2 ,

where a, b are real parameters.

2. Let ABCD be a tetrahedron with AB = AC = AD, inscribed ina sphere with centre O. Let G be the centre of gravity of triangle ACD, Ebe the mid-point of BG and F be the mid-point of AE. Prove that OF isperpendicular to BG if and only if OD is perpendicular to AC.

3. Let us be given n (n � 4) numbers a1, a2, : : : , an, distinct fromone another. Determine the number of permutations of these n numberssuch that in each permutation, no three of the four numbers a1, a2, a3, a4lie in three consecutive positions.

Second Day | Time: 3hours

4. Determine all functions f : Z ! Z satisfying simultaneously twoconditions:

(i) f(1995) = 1996

(ii) for every n 2 Z, if f(n) = m then f(m) = n and f(m+ 3) = n� 3,(Z is the set of integers).

5. Consider trianglesABC, with sideBC of length 1, and angleBACmeasuring �

�� > �

3

�. For triangles ABC, does the distance from the in-

centre to the centre of gravity of ABC attain the least value? Calculate thisleast value in terms of �.

8

6. Let x, y, z be three non-negative real numbers satisfying the con-dition:

xy + yz+ zx+ xyz = 4 .

Prove that:

x+ y+ z � xy + yz+ zx .

When does equality occur?

Last number we began giving readers' solutions to problems proposedto the jury but not used at the 37 th IMO at Mumbai, India [1997: 450]. Wenow continue with readers' solutions.

13. Let ABC be an acute-angled triangle with circumcentre O andcircumradius R. Let AO meet the circle BOC again in A0, let BO meet thecircle COA again in B0 and let CO meet the circle AOB again in C0. Provethat

OA0 �OB0 � OC0 � 8R3 .

When does equality hold?

Solution by Toshio Seimiya, Kawasaki, Japan.

A

B C

A0

B0

C0

D

EFf

Let D, E, F be the intersections of AO, BO, CO with BC, CA, AB,respectively. We denote the area of 4PQR by [PQR]. SinceAD

OD=

[ABC]

[OBC], we get

OA

OD=

AD

OD� 1 =

[ABC]� [OBC]

[OBC]=

[OCA] + [OAB]

[OBC].

We put [OBC] = x, [OCA] = y and [OAB] = z. Then we haveOA

OD=y + z

x; that is

R

OD=y+ z

x.

9

Similarly we haveR

OE=z + x

yand

R

OF=x+ y

z. Multiplying these

three equations, we have

R3

OD �DE � OF =(y+ z)(z+ x)(x+ y)

xyz. (1)

Since y + z � 2pyz, z + x � 2

pzx and x + y � 2

pxy, we have

(y + z)(z + x)(x+ y)� 8xyz, with equality holding when x = y = z.

Hence from (1),

R3

OD � OE � OF� 8 . (2)

Since O, B, A0, C are concyclic we get

\OA0C = \OBC = \OCB = \OCD ,

so that we get OD � OA0 = OC2 = R2. Thus we have OA0 = R2

OD.

Similarly, we have OB0 = R2

OE, and OC0 = R

2

OF. Multiplying these

three equations we get

OA0 � OB0 �OC0 =R6

OD �DE � OF . (3)

Therefore, we obtain from (2) and (3)

OA0 �OB0 � OC0 � 8R3 .

Equality holds when [OBC] = [OCA] = [OAB]; that is when 4ABC isequilateral.

16. Four integers are marked on a circle. On each step we simulta-neously replace each number by the di�erence between this number and thenext number on the circle, moving in a clockwise direction; that is, the num-bers a, b, c, d are replaced by a� b, b� c, c� d, d� a. Is it possible after1996 such steps to have numbers a, b, c, d such that the numbers jbc� adj,jac � bdj, jab� cdj are primes?

Solution by Pierre Bornsztein, Courdimanche, France.La r �eponse est non, apr �es 1996 �etapes tout comme apr �es n'importe

quel nombre d' �etapes.

En e�et, si apr �es k �etapes (k 2 N�) on obtient a, b, c, d ; on appellem, n, p, q, les entiers pr �ec �edents avec a = m � n, b = n � p, c = p � q,d = q �m, et donc �d = a+ b+ c.

Mais alors,

bc� ad = bc+ a(a+ b+ c) = (a+ b)(a+ c)

ac� bd = ac+ b(a+ b+ c) = (a+ b)(b+ c)

ab� cd = ab+ c(a+ b+ c) = (a+ c)(b+ c) ,

10

d'o �u

jbc� adj jac � bdj jab� cdj = [(a+ b)(b+ c)(c+ a)]2 .

Or, le produit de trois nombres premiers ne peut pas etre un carr �e, d'o �ul'a�rmation du d �epart.

18. Find all positive integers a and b for which�a2

b

�+

�b2

a

�=

�a2 + b2

ab

�+ ab ,

where, as usual, [t] refers to the greatest integer which is � t.

Solution by Pierre Bornsztein, Courdimanche, France.Soient a, b dans N� qui v �eri�ent�

a2

b

�+

�b2

a

�=

�a2 + b2

ab

�+ ab . (1)

Par sym�etrie des roles on peut imposer a � b.

Dans le cas o �u a = b, (1) s' �ecrit a+ a = 2+ a2 ; c. �a.d. (a� 1)2+1 = 0, quiest impossible.

Si a > b, on pose x =ha2

b

iet y =

hb2

a

i, x; y 2 N. Remarquons que x > 1

car a2

b> a > b � 1, a et b entiers. De plus x � a

2

bet y � b

2

a, d'o �u xy � ab.

Et alors, d'apr �es (1)

x+ y =

�a

b+b

a

�+ ab � 2 + ab

�car t+

1

t� 2

�� 2 + xy ,

d'o �u (x � 1)(y � 1) � �1 avec x > 1 et y � 0, y entier, et donc y = 0.C'est �a dire a > b2.

On pose a = b2 + c avec c 2 N�.Cons �equemment, (1) s' �ecrit

b3 + 2bc+

�c2

b

�= b+

�c

b+

b

b2 + c

�+ b3 + bc ;

c. �a.d.

b(c� 1) +

�c2

b

�=

�c

b+

b

b2 + c

�. (2)

Puisque �c2

b

�>

c2

b� 1 >

c

b� 1

11

et �c

b+

b

b2 + c

�� c

b+

b

b2 + c� c

b+ 1 ,

nous avons �a partir de (2) b(c� 1) + c

b� 1 < c

b+1, donc b(c� 1) < 2 avec

b � 1 et c � 1 et alors c � 2.

Lorsque c = 2, b = 1 et donc a = b2 + c = 3. Or pour a = 3, b = 1,ha2

b

i+hb2

a

i= 9 et

�a

b+ b

a

�+ ab = 3+ 3 = 6, donc (1) n'est pas v �eri� �ee.

En�n si c = 1, a = b2 + 1, alors�a2

b

�+

�b2

a

�=

�a2

b

�= b3 + 2b+

�1

b

�, et donc�

a

b+b

a

�+ ab = b+

�1

b+

b

b2 + 1

�+ b3 + b .

Pour b = 1�1

b

�=h1

b+ b

b2+1

i= 1, d'o �u (1) est vraie.

Pour b > 1 1

b� 1

2et b

b2+1< 1

2, donc

�1

b

�= 0 =

h1

b+ b

b2+1

i, et

donc (1) est v �eri� �ee.

Par suite, les couples-solutions sont les couples de la forme (b2+1; b)

ou (b; b2 + 1) avec b 2 N�.

Another solution has arrived to a problem of the Irish MathematicalOlympiad 1994 given in the November 1997Corner [1997: 388] and for whichsome solutions appeared last issue [1998: 456].

4. Consider the set of m� n matrices with every entry either 0 or 1.Determine the number of such matrices with the property that the numberof \1"s in each row and in each column is even.

Solution by Felipe Gago, Universidade de Santiago de Compostela,

Spain.Given anym� n matrix whose entries are all 0 or 1, we can construct

an m� (n+ 1) whose entries are also 0 or 1 and whose rows have an evennumber of 1's.

Indeed, in each row we must enter a 1 if the number of 1's was odd, and a 0otherwise. If we choose the number of the new column, say the last one,this process is unique.

12

Similarly we can construct an (m+1)�nmatrix whose entries are also0 or 1 and whose columns have each an even number of 1's.

If we combine both processes, given an m � n matrix, we obtain acon�guration like the one depicted above in which all complete rows andcolumns have an even number of 1's. Therefore, them� (n+1)matrix andthe (m+1)�nmatrix contain both an even number of 1's, and so the numberof 1's in the new column and in the new row have the same parity. By addinga 1, if they are odd, or a 0, if they are even, we obtain an (m+ 1)� (n+1)

matrix with the required property. The uniqueness of the processes tells usthat the number of such (m + 1) � (n+ 1) matrices equals the number ofm � n matrices whose entries are 0 or 1; that is, equals 2m�n, and so thesolution to our problem is 2(m�1)�(n�1).

We next turn to solutions by our readers to problems of the CroatianNational Mathematics Competition (4th Class), May 13, 1994 [1997: 454].

1. One member of an in�nite arithmetic sequence in the set of naturalnumbers is a perfect square. Show that there are in�nitely many membersof this sequence having this property.

Solutions by Murray S. Klamkin, University of Alberta, Edmonton,

Alberta; by Bob Prielipp, University of Wisconsin{Oshkosh,Wisconsin, USA;

and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario.

We give Wang's solution and comment.Let the arithmetic sequence be fa + nd : n = 0, 1, 2, : : : , g where

a, d 2 Z, a, d > 0. Suppose a+ kd = q2, for some k, q 2 Z, k � 0, q > 0.Then

a+ (k+ 2q2 + dq2)d = a+ kd+ (d2 + 2d)q2

= q2 + (d2 + 2d)q2

= ((d+ 1)q)2 ,

showing that a + (k+ 2q2 + dq2)d is also a perfect square. The conclusionthen follows by iteration.

Remark: The iteration described above by no means exhausts all the perfectsquares in the sequence. For example, if d � 2 then by the same method, wesee that if a+kd = q2, then a+(k�2q2+dq2)d = ((d�1)q)2. For example,

13

if the arithmetic sequence is f1+3n : n = 0, 1, 2, : : : , g, then starting with1, the �rst perfect square, we would get, using the iteration given in oursolution, the sequence of perfect squares 12, 42, 162, 642, : : : . If we use thesecond iteration mentioned above, then we would get the \re�nement" 12,22, 42, 82, : : : . Note, however, that neither of these two sequences includesthe perfect square 49 which is a member of the given arithmetic sequence.

Solution and Generalization of Klamkin.If the square term is a2 and the common di�erence is d, then the terms

of the progression a2 + d(2an + n2d) for all natural n are perfect squares(a+ nd)2.

More generally, if there is a term of the form P (a) in the arithmeticprogression, where P is an integral polynomial, (e.g. 2a7 + 3a6 � 1), then

the terms P (a) + d�P (a+nd)�P (a)

d

�are of the same form for all natural n.

2. For a complex number z let w = f(z) =2

3� z.

(a) Determine the set fw : z = 2+ iy; y 2 Rg in the complex plane.

(b) Show that the function w can be written in the form

w� 1

w� 2= �

z � 1

z � 2.

(c) Let z0 =1

2and the sequence (zn) be de�ned recursively by

zn =2

3� zn�1, n � 1 .

Using the property (b) calculate the limit of the sequence (zn).

Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,

Ontario.(a) Let w = u+ iv, where u; v 2 R. Then from w = 2

1�iy =2(1+iy)

1+y2we get

u = 2

1+y2and v = 2y

1+y2. Eliminating y, we get u2 + v2 = 4

1+y2= 2u, or

(u� 1)2+ v2 = 1. Hence fw : z = 2� iy; y 2 Rg is the set of all points onthe circle centred at (1; 0) with radius 1.

(b) Straightforward computations show that

w� 1

w� 2=

�z � 1

3� z

���2z � 4

3� z

�=

1

2

�z � 1

z � 2

�;

that is, � = 1

2.

(c) Using the equation in (b), we �nd, by iteration, that

zn � 1

zn � 2=

1

2

�zn�1 � 1

zn�1 � 2

�=

�1

2

�n �z0 � 1

z0 � 2

�=

1

3

�1

2

�n.

14

Since (12)n ! 0 as n!1, we �nd limn!1 zn = 1.

Remark: One could also solve for zn from the equation in (b) and obtain1 + 1

zn�2 = 1

3(12)n or zn = 2 + (1

3(12)n � 1)�1. This explicit expression for

zn can also be proved directly by induction.

3. Determine all polynomials P (x) with real coe�cients such that forsome n 2 N we have xP (x� n) = (x� 1)P (x), for all x 2 R.

Solutions by Murray S. Klamkin, University of Alberta, Edmonton,

Alberta; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,

Ontario. We give the solution of Wang.The only such polynomials are P (x) = cx, where c is an arbitrary con-

stant. Clearly P (x) � 0 satis�es the given equation. So assumeP (x) 6� 0. Setting x = 0 in xP (x � n) = (x � 1)P (x), we get P (0) = 0

and thus (n� 1)P (n) = 0. If n 6= 1, then P (n) = 0 and a straightforwardinduction shows that P (kn) = 0 for all k 2 N, which is impossible sinceP (x) 6� 0. Hence n = 1 and we have xP (x� 1) = (x� 1)P (x) which im-plies P (2) = 2P (1). Suppose that P (m) = mP (1) for some m � 2. Thenfrom (m + 1)P (m) = mP (m + 1), we get P (m + 1) = (m + 1)P (1).Hence P (m) = mP (1) = mc for all m 2 N , where c = P (1). LetQ(x) = P (x)� cx. Then Q(m) = P (m)� cm = 0 for all m 2 N and soQ(x) � 0 from which P (x) � cx. Noting that c = 0 if and only if P (x) � 0,the conclusion follows.

4. In the plane �ve points P1, P2, P3, P4, P5 are chosen having integercoordinates. Show that there is at least one pair (Pi; Pj), for i 6= j such thatthe line PiPj contains a point Q, with integer coordinates, and is strictlybetween Pi and Pj.

Solution by Murray S. Klamkin, University of Alberta, Edmonton,

Alberta.With respect to parity, there are only 4 kinds of points (x; y). They are

(E;E), (E;O), (O;E) and (O;O), where E denotes an even integer, and Odenotes an odd integer. Since there are 5 points, there must be at least two(x1; y1) and (x2; y2) such that x1 and x2 have the same parity and also y1and y2. Hence Q can be taken as the mid-point of these two points.

We now give a solution to one of the problems of theAdditional Compe-

tition for the Olympiad of the Croatian National Mathematical Competition,May 14, 1994 [1997: 454].

2. Construct a triangle ABC if the lengths jAOj, jAU j and radius r ofincircle are given, where O is orthocentre and U the centre of the incircle.

Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;

and by Murray S. Klamkin, University of Alberta, Edmonton, Alberta. We

give the solution by Amengual Covas.

15

Denote the orthocentre, the incentre and the circumcentre of triangleABC byH, I, and O, respectively.

As� a

r

\A=2

I

Figure 1

O

A0

V

HI

CB

A

Figure 2

O

A0

V

H

I

CB

A

Figure 3

A

B

C

r

r

y

�

s

9

R

s

Figure 4Analysis. Imagine the construction completed (Figures 2 and 3), with A0 themid-point of sideBC and V the midpoint of arc BC. It is a standard exercise(for example #1.8 and #3.4 in Trajan Lalesco, La G �eom�etrie du Triangle) toshow that

OA0 =1

2AH and BV = V I .

Geometrical Construction. SinceAI and r are given, \A and s�a are known(Figure 1).

If \A = 90� (or equivalently, AI = rp2), then A andH coincide and

there are an in�nite number of triangles circumscribing the r circle.

16

Suppose \A 6= 90�. In isosceles triangle BOC we then know

\BOC =

8>><>>:

2 � \A if \A < 90�

(or, equivalently AI > rp2)

2 � (180� � \A); if \A > 90�

(or, equivalently AI < rp2) ,

and altitude OA0 (= 1

2AH), whence4BOC and the circumcircle of4ABC

can be constructed. Then,

Solution 1. TriangleBOC and the circumcircle of4ABC being constructed,draw XY parallel to BC and a distance r from it (on the same side as O if\A < 90� and on the opposite side if \A > 90�). Bisect the arc BC at Vand let the circle, centre V , radius V B, cut XY at I and I0. These pointswill be the incentres for the two (symmetrical) solutions. Let V I meet thecircumcircle again at A. Then ABC is the required triangle.

Solution 2. Triangle BOC being constructed, we then know \A, r and s

(= a+ (s� a)) where s is the semiperimeter of4ABC and a = BC.

These elements su�ce to determine the incircle and the excircle lyingin \A, which are readily constructed.

The common internal tangents of these circles intersect the sides of theangle A at B and C to form the required triangle (Figure 4).

The details of the construction may be omitted since they are obvious.

We �nish this number of the Corner with a more direct solution thanthe one published in the [1997: 460]. There are also some generalizations.

18. Show that in a non-obtuse triangle the perimeter of the triangleis always greater than two times the diameter of the circumcircle.

Solution by Murray S. Klamkin, University of Alberta, Edmonton,

Alberta.Here we give a more direct solution and some generalizations.

If we also allow degenerate triangles, then the inequality here is

a+ b+ c � 4R ,

and since a = 2R sinA, etc., we equivalently have the known inequality

sinA+ sinB + sinC � 2 (1)

for non-obtuse triangles. Also, by using the pedal angle transformation,A = (� � A0)=2, etc., (1) becomes

cosA0=2 + cosB02 + cosC0=2 � 2 (10)

17

for all triangles A0B0C0.

The following generalization will include (1). We consider the sameproblem where the maximum angle of the triangle is given as � and we havetwo cases:

(i) � > �=2,

(ii) � � �=2.

To establish our inequalities, we use the powerfulmajorization inequal-ity. Here we say a vector (x; y; z) majorizes another vector (x0; y0; z0) ifx � y � z, x0 � y0 � z0 and x � x0, x+y � x0+y0, x+y+z = x0+y0+z0

and we write this as (x; y; z) � (x0; y0; z0). Then for any convex function F ,we have the inequality

F (x) + F (y)+ F (z) � F (x0) + F (y0) + F (z0) ,

and for a concave function like sin t, the inequality is reversed.

Case (i). Here (�; � � �; 0) � (A;B; C) � (�=3; �=3; �=3), so that

3 sin�=3 � sinA+ sinB + sinC � 2 sin � .

Case (ii). Here (�; �; � � 2�) � (A;B;C) � (�=3; �=3; �=3), so that

3 sin�=3 � sinA+ sinB + sinC � 2 sin � + sin2� .

If we now let � = �=2, we recapture (1) and there is equality only for thedegenerate triangle of angles �=2, �=2, 0.

We now consider the same original problem where the minimum angleof the triangle is given as '. Then since

(� � 2'; '; ') � (A;B;C) � (f�� 'g=2; f� � 'g=2; ') ,

2 cos'=2 + sin' � sinA+ sinB + sinC � sin 2'+ 2 sin' .

What is done above in restricting one of the angles can be applied togeneralize other triangle inequalities; for example,

3=2 � sinA=2 + sinB=2 + sinC=2 � 1 ,

3p3=8 � sinA sinB sinC � 0 ,

2p3=9 � sinA sinB sinC=2 � 0 ,

3p3=8 � cosA=2 cosB=2 cosC=2 � 0 ,

tanA=2 + tanB=2 + tanC=2 �p3 ,

cotA=2 + cotB=2 + cotC � 3p3 .

That concludes theOlympiadCorner for this issue. Sendme your OlympiadContests and your nice solutions.

18

BOOK REVIEWS

ALAN LAW

First, we must introduce our new Book Reviews Editor, Alan Law.

Alan is a Professor in the Department of Computer Science, Faculty ofMathematics, University of Waterloo. He was previously Dean of Scienceat Memorial University, St. John's, Newfoundland, and before that, Headof the Department of Computer Science at the University of Regina, Regina,Saskatchewan. He has also served on the Canadian Mathematical OlympiadCommittee.

During his 37 years as a full-time academic, he has authored or co-authored over 45 papers in refereed journals and over 30 refereed conferenceproceedings papers, as well as having edited or co-edited three books. Hehas given lectures all over the world at universities and research institutes.He was also instrumental in persuading the present Editor-in-Chief to takeon the position.

Since Andy Liu has, as usual, been very e�cient in getting books re-viewed, the next few issues will, in fact, be from him.

Over and Over Again, by Genghe Chang and Thomas Sederberg,published by the Mathematical Associatlon of America, 1997,P.O. Box 91112, Washington DC 20090-1112.ISBN # 0-88385-641-7, softcover, 309+ pages, $31.50.Reviewed by Murray S. Klamkin, University of Alberta, Edmonton,

Alberta.

This book is #39 in the excellent well edited New Mathematical Libraryseries of the MAA. It consists of 32 chapters dealing mainly with varied re-peated transformations (maps) over and over again; that is, iterations of var-ious functions.

The �rst eighteen chapters require secondary school mathematics witha few exceptions. Also included here are geometry, complex numbers, vec-tors, and inequalities. Most of the problems at the end of these chapterscome from mathematical olympiads from various countries, and there arehints and solutions at the end of the book for them. This part of the bookshould be interesting and very suitable for students studying for mathemati-cal competitions. The following is a small sample of some of these problems.

1. We start a list of numbers with 1 and 2. The operation O consists oftaking all pairs a and b of di�erent numbers already listed and addingthe numbers a + b + ab to the list. Give a simple description of thenumbers which will be in the list if we repeat O in�nitely often.

19

2. Let d(n) be the number of 1's in the base 2 representation of n. Showthat the number of odd binomial coe�cients

�n

i

�equals d(n).

3. Five points A, B, C, U and V lie in the same plane. If trianglesAUV ,UBV and UV C are directly similar to each other, show that triangleABC is directly similar to each of them.

4. Show that the equation zn+1� zn � 1 = 0 (n a positive integer) has asolution on the unit circle in the complex plane if and only if 6j(n+2).

5. Show that a circular diskD can never be covered by two smaller circulardisks D1 and D2.

The last fourteen chapters are more advanced and so are more suitablefor college students. They deal with functional iteration, Chebyshev polyno-mials, chaos, Bernstein polynomials, Bezier curves, surfaces and splines. Asbefore, there are also problems at the end of these chapters with hints andsolutions at the end of the book. This part of the book is a good supplementfor study in calculus, numerical analysis, and computer-aided geometric de-sign. The following is a small sample of the problems from these chapters.

6. Determine limn!1an where an+1 =

a2

n

an�1 , n = 0; 1; : : : and a0 = 2.

(The solution given indicates how to determine an explicitly.)

7. Let fn+1 = f(fn), where f0 = x and f(x) = 4�x� 1

2

�2. Show that

for any positive integer p, there is an x in [0,1] such that the sequencefn is periodic of least period p.

8. Show that if faig, i = 0; 1; : : : is an arithmetic progression, thennXi=1

ai

�n

i

�xi(1� x)n�i is either a constant or a polynomial of degree 1

for any positive integer n.

9. Show that the parametric equations x = P1(t) and y = P2(t), whereP1 and P2 are polynomials, cannot represent the arc of a circle exactly.Find a pair of rational functions which does represent an arc of a circle.

10. Show that a polynomial P (x) of degree at most m can be representeduniquely in the form P (x) = a0x

n + a1(x+ 1)n + � � �+ an(x+ n)n,where n � m.

20

Generalisations of a Four-Square Theorem

Hiroshi Okumura and John F. Rigby

In the 17 th{19 th centuries, Japanese people often wrote their mathe-matical results on a wooden board and dedicated it to a shrine or a temple.Then the board was hung under the roof there. Such a board is called a san-gaku. In this paper, we shall discuss generalisations of a sangaku probleminvolving four squares, which we state as the following theorem (see Fig-ure 1.). [Ed. see CRUX problem 1496 [1989: 298; 1998: 56].]

Theorem 1. [1, p.47]Let CAEA0, CBGB0, ADFB, A0D0F 0B0 be squares

as in Figure 1. Then D, E, D0 are collinear if and only if CAEA0 is half thesize of CBGB0.

rr

rr

r

rr

r

r

r

r

r

r

r

r

r

D0 E D

F 0

B0

G

B

FA

C

A0

Figure 1.

The angleCBA, denoted by \CBA, is the angle between�!BC and

�!BA;

it is positive if the direction of rotation from�!BC to

�!BA is anticlockwise, and

negative otherwise. The following lemma is the key of our generalizations(see Figure 2).

Lemma 1. If the vertices of two triangles ABC and AED lie anticlockwise

in this order, and if they satisfy

BC � CA

AB=

AE

DA; \BAC + \EAD = � ; (1)

then 2\DEA = � � \ACB.

Proof. Let f be a dilative rotation with centre A mappingB into D. Ifwe denote the ratio of magni�cation by �, then

� =DA

AB=

AE

BC � CA

Research supported in part by NFS grant DMS-9300657.

Copyright c 1999 Canadian Mathematical Society

21

D

P

B

C

E

A

Figure 2.

by the �rst part of (1). Let P = f(C). Then A lies on the segment PE since\DAP + \EAD = � by the second part of (1), while

PE = PA+ AE = �CA+ AE =CA � AEBC � CA

+AE

=BC � AEBC � CA

= �BC = DP :

Hence EPD is an isosceles triangle with summit P . Therefore, we get2\DEA = � � \APD = � � \ACB.

Re ecting Figure 2 in a line, we get a similar lemma with opposite ori-entations: If the vertices of two triangles ABC and AED lie clockwise in

this order, and if they satisfy

BC �CA

AB=

AE

DA; \BAC + \EAD = �� ;

then 2\DEA = �� � \ACB.

Lemma 2. If \EAC = ��=2 in Lemma 1, then the angle bisector of \BCA

is parallel toDE.

Proof. In this situation, the triangleADP is obtained from the triangleABC by a dilative rotation with centre A and angle of rotation ��=2.

In the following examples, we shall agree thatACA0E,CBGB0,ADFBand A0B0F 0D0 have anti-clockwise orientations. Complete proofs will begiven later.

Example 1. (Figure 3) The ratio of the sides of the two squares ACA0Eand CBGB0 is n : n+1 (n is an integer). The two rectangles consist ofn squares as in the �gure. Let AE = na; then BC = (n + 1)a. Hence(BC � CA)=AB = a=AB = (na)=(nAB) = AE=DA. And we can showthat D, E, D0 are collinear and that this line is parallel to the angle bisectorsof \ACB and \A0CB0.

Example 2. (Figure 4) Four similar rectangles where the ratio of the sides ofCAEA0 and CBGB0 is 1 : 2. Then D, E, D0 are collinear, and this line isparallel to the angle bisectors of \ACB and \A0CB0 (this is a special caseof Example 4 below).

22

r

rr

r

r r

rrr

r

r

r

rr

rr

D0 E D

F

B

GB0

F 0

A

C

A0

Figure 3. CA : CB = n : (n+1) (n = 2).

r

r

r

rr

rr

r

r

rr

rr

rr

r

D0 E D

FB

G

B0

F 0

AA0

C

Figure 4. CA : CB = 1 : 2

Example 3. (Figure 5) Four similar rectangles where CA : AE = s : a andCA : BC = s2 : s2+1. We get BC = (s2 + 1)CA=s2 = (s2 + 1)CA0=s,and

BC � CA

AB=

(s2 + 1)CA=s2� CA

DA=s=

CA=s

DA=

AE

DA;

as well as

B0C � CA0

A0B0 =sBC �CA0

A0B0 =(s2 + 1)CA0 � CA0

sD0A0 =sCA0

D0A0 =A0E

D0A0 :

And we can show that D, E, D0 are collinear, and that this line is parallel tothe angle bisectors of \ACB and \A0CB0.

Example 4. (Figure 6) Four similar cyclic quadrilaterals. The cyclic quadrilat-erals CAEA0, CBGB0, BADF , B0F 0D0A0 are similar, with\CAE = �=2 and BC = 2CA. In the �gure, we get (BC � CA)=AB =

CA=AB = AE=DA. Further, we can show that D, E,D0 are collinear, andthis line is parallel to the angle bisectors of \ACB and \A0CB0.

r

rr

rr

rr

r

r

rrr

rr

rr

D0 E D

F

B

G

B0

F 0

A0

C

A

Figure 5.

r

r

r

r

r

r

rr

rr

r

r

r

r r

rD0 E D

FB

G

B0F 0

C

AA0

-

-

-�

Figure 6.

Example 5. (Figure 7) Four similar harmonic cyclic quadrilaterals: CAEA0 isa cyclic quadrilateral with the property CA�EA0 = AE �A0C. The quadrilat-erals CBGB0 BFDA, B0A0D0F 0 are similar to CAEA0, and

23

BC = 2CA. In the �gure,

BC � CA

AB=

CA

AB=

AE � A0C

EA0FD

DA � BF=

AE

DA

(A0C=EA0)

(BF=FD)

=AE

DA

(CA=AE)

(BF=FD)=

AE

DA;

and we can show that D, E, D0 are collinear.

In example 4, the quadrilaterals CBGB0, BADF , B0F 0D0A0 are di-rectly similar, and CAEA0 is oppositely similar to the other three, whereas,in example 5, it is CBGB0 that is oppositely similar to the other threequadrilaterals.

In all the above examples, the trianglesABC andAED satisfy the hy-potheses of Lemma 1 (see Figure 8). Also, the triangles A0B0C andA0ED0 satisfy the hypotheses of the re ected lemma. Thus we get\DEA = �=2 � \ACB=2, and \D0EA0 = ��=2 � \A0CB0=2. Hencewe get

\DED0 = \DEA+ \AEA0 + \A0ED0

= 1

2(� � \ACB) + (� � \A0CA) + 1

2(� + \A0CB0)

= 2� � 1

2(\ACB + 2\A0CA+ \B0CA0) :

But since \A0CA = \BCB0, we have

1

2(\ACB + 2\A0CA+ \B0CA0)

= 1

2(\ACB + \BCB0 + \B0CA0 + \A0CA) � 0 (mod �) :

Hence \DED0 � 0 (mod �). This implies that D, E, D0 are collinear.Also, examples 1, 2, 3, and 4 satisfy the hypotheses of Lemma 2, andDD0 isparallel to the angle bisector of \ACB in all four examples.

r

r

r

r

r

rr

r

r

r rr r r

r

r

D0 E D

F 0

B0

G

F

BC

A0

A

�

��

�

Figure 7.

D0 E D

B

B0

AC

A0

Figure 8.

24

Example 6. (Figures 9A and 9B) Four regular 2n{gons. The smallest oneis half the size of the one above. The collinearity described in the �gure isobtained from example 4.

rrr

r

r

r rr

r

rr

r

r

r

rr

r

r

r

r

r

r

r

rr

r

r

r

rr

r

r

r

rr r

r

r

rr

Figure 9A (n = 5).

rr

r

r

r

r r

r

r

r

rr

r

r

rr r

r

r

r

r

r

r

rr

r

r

r

r rr rrr

rrrrrr

Figure 9B (n = 5).

Reference

[1.] H. Fukagawa and Pedoe, D., Japanese Temple Geometry Problems,Charles Babbage Research Centre, Winnipeg, Canada, 1989.

Hiroshi Okumura John F. Rigby

Maebashi Institute of Technology School of Mathematics

460-1 Kamisadori Maebashi University of Wales Cardi�

Gunma 371, Japan Cardi� CF2 4AG, Wales, UK

25

THE SKOLIAD CORNERNo. 35

R.E. Woodrow

We begin this issue with the problems of the �rst round of the OldMutual Mathematical Olympiad 1992. Thanks go to John Grant McLoughlinof the Faculty of Education, Memorial University of Newfoundland, forcollecting this contest and forwarding it for use in the Corner.

OLD MUTUAL MATHEMATICAL OLYMPIAD 1992Time: 1 hour

1. (0:4)2� (0:1)2 equals:

(a) 0:04 (b) 1:5 (c) 0:15 (d) 0:6 (e) 0:06

2. The angles of a triangle are in the ratio 2 : 3 : 4. The size of thelargest angle in degrees is:

(a) 40� (b) 80� (c) 45� (d) 90� (e) 72�

3. Reduced to the lowest terms a2�b2ab

� ab�b2ab�a2 is equal to:

(a) a2�2b2ab

(b) a� 2b (c) a2 (d) ab

(e) none of these

4. If the radius of a circle is increased by 100%, the area is increasedby:

(a) 100% (b) 200% (c) 300% (d) 400% (e) 10000%

5. The area of the largest triangle that can be inscribed in a semicircleof radius r is:

(a) r2 (b) r3 (c) 2r2 (d) 2r3 (e) 1

2r2

6. A manufacturer built a machine which will address 500 envelopesin 8minutes. He wishes to build another machine so that when both are op-erating together they will address 500 envelopes in 2minutes. The equationused to �nd how many minutes x, it would require the second machine toaddress 500 envelopes alone is:

(a) 8� x = 2 (b) 500

8+ 500

x= 500 (c) 1

8+ 1

x= 1

2

(d) x2+ x

8= 1 (e) none of these.

7. The expressionq

4

3�q

3

4is equal to:

(a)q

7

12(b)p�1 (c) 1

2p3

(d)q

1

7(e) �1

26

8. Two circles of equal size are contained in a rectangle as shown.

If the radius of each circle is 1 cm, then the area of the shaded portion in cm2

is:

(a) � � 4 (b) 4� 2� (c) 8� � (d) 8� 2� (e) 4

9. The yearly changes in the population of a town for four consecu-tive years are respectively 10% increase, 10% increase, 10% decrease, 10%decrease. The net change over four years to the nearest percent is:

(a) �2 (b) �1 (c) 0 (d) 1 (e) 12

10. If log10 2 = a and log10 3 = b, then log10 12 equals:

(a) a2 + b (b) 2a+ b (c) 4b (d) 2ab (e) a2b

11. Of the following, which is the best approximation to the positivesquare root of 1992

10000:

(a) 0:0045 (b) 0:014 (c) 0:0446 (d) 0:1411 (e) 0:4463

12. A circle and a square have the same perimeter. Then:

(a) their areas are equal

(b) the area of the circle is 4

�times that of the square

(c) the area of the circle is 2

�times that of the square

(d) the area of the circle is � times the area of the square

(e) none of these

13. The square of an integer is called a perfect square; for example,4, 9, 16, 25 are all perfect squares. If n is a perfect square, then the nextperfect square greater than n is:

(a) n+ 1 (b) n2 + 1 (c) n2 + 2n+ 1 (d) n2 + n (e) n+ 2pn+ 1

14. The last digit of 71992 is:

(a) 1 (b) 2 (c) 6 (d) 7 (e) 9

15. The �gure below shows a triangle PQR, where P bQR > Q bPR.The altitude to the base PQ, divides P bRQ into two parts R1 and R2.

R

PQ

R1 R2

27

Then:

(a) R1 +R2 = bP + bQ (b) R1 � R2 = bQ� bP (c) R1 � R2 = bP � bQ(d) R1 + R2 = bQ� bP (e) R1 � R2 = bQ+ bP

16. A cylinder is such that the area of its curved surface is twice itsvolume. Then its radius is:

(a) 1

2(b) 2 (c) �

2(d)q

2

�(e) 1

17.��1

8

��1=3is equal to:

(a) �2 (b) 2 (c) �1

2(d) 8 (e) 83

18. A man wishes to travel 1992 kilometres at an average speed of100 kmh�1. He travels the �rst half of this distance at 50 kmh�1. How fastmust he go over the remaining half?

(a) 150 kmh�1 (b) 200 kmh�1 (c) 400 kmh�1

(d) 496 kmh�1 (e) none of these

19. Six numbers are in arithmetic progression. The sum of the �rstand last is 5. Then the sum of the third and fourth is:

(a) 5 (b) 6 (c) 7 (d) 12 (e) impossible to determine

20. If n means nn, so that 3 = 33 = 27, then 2 is:

(a) 2 (b) 4 (c) 16 (d) 64 (e) 256

Last issue we gave the problems of the Final Round of the Old Mu-tual Mathematical Olympiad 1991. My thanks go to John Grant McLoughlinof the Faculty of Education, Memorial University of Newfoundland for for-warding the questions and \o�cial" solutions to the questions of paper 2,and problem 4 of paper 1.

OLD MUTUAL MATHEMATICAL OLYMPIAD 1991Final Paper 1

Solutions (Time: 2 hours)

1. In the �gure shown ABC and AEB are semicircles and F is themid-point of AC and AF = 1 cm. Find the area of the shaded region.

CA F

B

D

E

28

Solution. AB is the diameter of the semicircle AEB and thehypotenuse of isosceles right triangle AFB, so AB =

p2, and the area of

semicircle AEB is 1

2�(

p2

2)2 = �

4. Now semicircle ABC has area 1

2�, while

right triangle AFB has area 1

2. Hence the area of the circular sector ADB

and the triangle AFB is is �

4� 1

2, and the shaded area is �

4� (�

4� 1

2) = 1

2.

2. What is the value ofp17� 12

p2 +

p17 + 12

p2 in its simplest

form?

Solution. Let x =p17� 12

p2 +

p17 + 12

p2. Now

x2 = 17� 12p2 + 17 + 12

p2 + 2

q17� 12

p2

q17 + 12

p2

= 34 + 2

q(17� 12

p2)(17 + 12

p2)

= 34 + 2

q(17)2 � (12)2(

p2)2

= 34 + 2p289� 288 = 34 + 2 = 36 .

So x = �6, or x = 6. Since x > 0, x = 6.

3. In a certain mathematics examination, the average grade of thestudents passing was x%, while the average of those failing was y%. Theaverage of all students taking the examination was z%. Find, in terms of x,y and z, the percentage who fail.

Solution. The total of the scores for those passing the exam is x

100� p,

where p is the number of students who pass. Similarly, with f the numberwho fail y

100f is the total of the failing scores. The total of all students scores

is z(p+f)

100= xp

100+ yf

100so zp+ zf = xp+ yf ,

(z � x)p = (y � z)f and p =y � z

z � xf .

Now

p+ f =y � z

z � xf + f =

�y � z

z � x+ 1

�f

=y � x

z � xf .

The percentage who fail is

f

p+ f� 100 =

fy�xz�xf

� 100 =z � x

y � x� 100 .

29

4. In the �gure shownAB = AD =p130 cm andBEDC is a square.

A

D

B

CE

Also the area of 4AEB = area of square BEDC.Find the area of BEDC.

Solution. Let BOD be a diagonal of the square, with O its centre.Now triangles 4BEA and BCE have the same altitude, from B, namelyBO. Since the area of square BEDC equals the area of triangle 4AEB,and the area of the square is twice the area of4BCA, we have AE = 2EC.Clearly BO = EO = 1

2EC, so AE = 4EO.

By Pythagoras, AO2+ BO

2= AB

2= 130. Thus,

(AE + EO)2 + EO2= 130

or

(4EO+EO)2 + EO2

= 130

(25 + 1)EO2

= 130

EO2

=130

26= 5 .

But the area of the square is just twice the area of the square with side lengthEO, so the area is 2� 5 = 10.

Final Paper 2Solutions (Time: 2 hours)

1. If the pattern below of dot-�gures is continued, how many dots willthere be in the 100 th �gure?

w w

w

w

w

w

w

w w

w

w

w

w

w

w w

w

w

w

w

w

w

w

w

Solution. By inspection the number of dots in the n th �gure is 2n+1.So the 100 th has 201 dots.

30

2. It is required to place a small circle in the space left by a large circleas shown. If the radius of the large one is a and that of the small one is b,�nd the ratio a=b.

Solution. Let O be the centre of the larger circle and let O1 be thecentre of the small circle. The distance of O from P , the point of intersectionof the two tangents shown, is

p2a, by Pythagoras, and similarly the distance

from O1 to P isp2 b. But OO1 = a + b. So

p2 b+ a + b =

p2 a, giving

a

b=

p2+1p2�1 = 3+ 2

p2.

3. Find all solutions to the simultaneous equations

x+ y = 2 ,

xy � z2 = 1 ,

and prove that there are no other solutions.

Solution. We �nd all real solutions. The �rst observation is that

xy � 1 (1)

in order that xy� z2 = 1. Thus x and y are both positive or both negative.But for x + y = 2, x and y must both be positive. Now for any x; y > 0,pxy � x+y

2(this is just the Arithmetic-Geometric Mean Inequality). Thusp

xy � 2

2= 1, and xy � 1. From (1) we have xy = 1 and z = 0. But the

AM/GM inequality is strict unless x = y so x = y = 1 and z = 0 is the onlysolution.

4. If a, b, c and d are numbers such that

a+ b < c+ d ,

b+ c < d+ e ,

c+ d < e+ a ,

and d+ e < a+ b .

Prove that the largest number is a and the smallest is b.

31

Solution.From d+ e < a+ b and a+ b < c+ d we see d+ e < c+ d so e < c. (1)

From a+ b < c+ d and c+ d < e+ a we get b < e. (2)

From b+ c < d+ e and d+ e < a+ b we get c < a. (3)

Now from d+ e < a+ b we read d� a < b� e < 0 by (2), so d < a.From (1), (2) and (3) b < e < c < a, so with d < a, a is the largest number.

Similarly from a + b < c + d we see 0 < a � c < d � b from (3) andb < d. This makes b the smallest number.

5. The diagram below [rotated through 90�] shows a container whoselower part is a hemisphere and whose upper part is a cylinder.

6?

20cm

-�220cm

The cylindrical part has internal diameter of 20 cm and is 220 cm long. Wateris poured into it and rises to a height of 20 cm in the cylindrical part. Thetop is then sealed with a at cover and the container is turned upside down.The water is now 200 cm high in the cylindrical part.

(i) Calculate the volume of the hemisphere in terms of �.(ii) Find the total height of the container.

[Note: The volume of a sphere of radius R is 4

3�R3.]

Solution. (Diagrams are not to scale!)

r

20

220

20

6

??6

r

20

220200

?

6

?

6

The volume V of water is V = � �102 �200, from the information aboutthe upside down �gure.

But V = 1

2

4

3�r3 + � � 102 � 20 from the initial con�guration.

32

So the volume of the hemisphere VH is

VH =1

2

4

3�r3 = � � 102 � 200� � � 202 � 20

= � � 18000 .

Now

2

3�r3 = � � 18000

r3 = 27000

r = 30 .

The total height is 220 + 30 = 250 cm

That completes the Skoliad Corner for this issue. Please send me con-test materials and suggestions for other features of the Corner.

Challenge

What is the 10 th term in the following sequence, and why?

n xn

0 0

1 116

�p30 +

p10 �p

6 �p2 � 2

�p3� 1

�q5 +

p5

�

2 18

�q30� 6

p5 �p

5 � 1

�

3 18

�p10 +

p2� 2

q5 �p

5

�

4 18

�q10 + 2

p5 �p

15 +p3

�

5 14

�p6 �

p2�

6 14

�p5 � 1

�

7 116

�2�p

3� 1�q

5�p5�p

30 +p10 �p

6 +p2

�

8 18

�p15 +

p3�

q10 � 2

p5

�

9 18

�2

q5 +

p5�p

10 +p2

�

10 ?

33

MATHEMATICAL MAYHEM

Mathematical Mayhem began in 1988 as a Mathematical Journal for and byHigh School and University Students. It continues, with the same emphasis,as an integral part of Crux Mathematicorum with Mathematical Mayhem.

All material intended for inclusion in this section should be sent to theMayhem Editor, Naoki Sato, Department of Mathematics, Yale University,PO Box 208283 Yale Station, New Haven, CT 06520{8283 USA. The electronicaddress is still

mayhem@math.toronto.edu

The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).The rest of the sta� consists of Adrian Chan (Upper Canada College), DonnyCheung (University of Waterloo), Jimmy Chui (Earl Haig Secondary School),David Savitt (Harvard University) and Wai Ling Yee (University of Waterloo).

A new year brings change, and this is especially true for 1999. We are pleased

to announce two new appointments. Adrian Chan, a high school student in his grad-

uating year at Upper Canada College, is the new High School Problems Editor, and

Donny Cheung, a 3 rd year undergraduate student at the University of Waterloo, is

the new Advanced Problems Editor. Both bring their knowledge and experience to

MAYHEM, including the IMO. We also introduce a new column, \Problem of the

Month", which will hopefully appeal to young problem solvers. We hope these

changes better serve you, our reader, which brings me to my second point.

MAYHEM was founded on the principle that it be a journal produced for stu-

dents, by students (people still sending mail to Professor Naoki Sato may take par-

ticular note), and it is a principle that we have been committed to, for we take our

audience as seriously as we take our content. We encourage our readership, in par-

ticular students, to take an active role in what goes into MAYHEM, and there are

many ways one can do this. First, you can comment on our material. Do you think

it is too easy or too hard? Are there any particular topics you would like to see? We

are a small organization, so this kind of feedback is important to us. So, write to us,

and remember, it does not hurt to be speci�c. Second, you can have a stronger say

by writing an article yourself. It does not have to be deep or profound (that is for

CRUX), just something that appeals to students. Finally, if you think you have what it

takes, then consider joining our \illustrious" sta�. (By contract, we must always use

the word \illustrious" when describing our sta�.) Candidates must be enthusiastic

about mathematics, dedicated to their work, knowledgeable with word processing,

and naturally, students. We cannot promise prestige or glamour. We cannot promise

large sums of money, or even small sums of money. But, we do o�er a chance to

be a part of a unique Canadian mathematics journal. I am very serious about these

o�ers, because MAYHEM has always been strongly driven by its readership, and I

want to make this absolutely clear. That being said, I feel very optimistic about the

coming year, and am con�dent that our cause will prosper. I would like to thank all

the sta� who have worked hard to make MAYHEM what it is today, and the CMS,

and in particular Bruce Shawyer for his patience and guidance. I will see you all in

1999!

Your Editor, Naoki Sato

34

Mayhem Problems

The Mayhem Problems editors are:

Adrian Chan Mayhem High School Problems Editor,Donny Cheung Mayhem Advanced Problems Editor,David Savitt Mayhem Challenge Board Problems Editor.

Note that all correspondence should be sent to the appropriate editor |see the relevant section. In this issue, you will �nd only problems | thenext issue will feature only solutions.

We warmly welcome proposals for problems and solutions. With thenew schedule of eight issues per year, we request that solutions from thisissue be submitted by 1 June 1999, for publication in issue 2 of volume 26.

High School Problems

Editor: Adrian Chan, 229 Old Yonge Street, Toronto, Ontario, Canada.M2P 1R5 <all238@ipoline.com>

H249. For a certain positive composite integer x, when the fraction(60 � x)=120 is reduced to lowest terms, the sum of the numerator anddenominator exceeds 120. Determine x.

H250. Let ABCD be a unit square, and let E, F , G, and H bepoints onAB, BC, CD, andDA respectively such thatAE = BF = CG =

DH = 1999=2000. Construct the triangles AGB, BHC, CED, and DFA,and let S be the area of the region that is common to all four triangles. Showthat

S =1

19992 + 20002.

H251. We say that an arithmetic sequence is astonishing if it satis�esthe following conditions:

(a) Every term in the sequence is an integer.

(b) No term in the sequence is greater than 10000.

(c) There are at least three terms in the sequence.

(d) The sum of the terms is 1999.

For example, the arithmetic sequence �998, �997, : : : , 999, 1000 is aston-ishing. How many astonishing arithmetic sequences are there?

35

H252. Find a solution (a; b) in rational numbers to the followingsystem:

9a2 + 16b2 = 25 ,

a2 + b2 <25

16+

1

10.

(Query: Can you determine an in�nite set of rational solutions to thissystem?)

Advanced Problems

Editor: Donny Cheung, c/o Conrad Grebel College, University of Wa-terloo, Waterloo, Ontario, Canada. N2L 3G6 <dccheung@uwaterloo.ca>

A225. In an acute-angled triangle, ABC, label its orthocentre H andits circumcentre O. LineBO is extended to meet the circumcircle atD. Showthat ADCH is a parallelogram.

A226. Proposed by Naoki Sato.

Let n be a positive integer. A 2�n array is �lled with the entries 1, 2,: : : , 2n, using each exactly once, such that the entries increase reading leftto right in each row, and top to bottom in each column.

For example, for n = 5, we could have the following array:

1 2 4 5 9

3 6 7 8 10

Find the number of possible such arrays in terms of n.

A227. Proposed by Mohammed Aassila, Universit �e Louis Pasteur,

Strasbourg, France.

Let f , g : N ! N be functions such that f is surjective, g is injective,and f(n)� g(n) for all n 2 N. Prove that f = g.

A228. Given a sequence a1, a2, a3, : : : of positive integers in whichevery positive integer occurs exactly once. Prove that there exist integersk < ` < m, such that ak + am = 2a`.

(1997 Baltic Way)

36

Challenge Board Problems

Editor: David Savitt, Department of Mathematics, Harvard University,1 Oxford Street, Cambridge, MA, USA 02138 <dsavitt@math.harvard.edu>

C83. Proposed by Dima Arinkin, graduate student, Harvard Univer-

sity.Let f be a function from the plane R2 to the reals. Given a polygon

P in the plane, let f(P) denote the sum of the values of f at each of thevertices of P. Suppose there exists a convex polygon Q in the plane suchthat for every polygon P similar to Q, we �nd that f(P) = 0. Show that fis identically zero.

C84. Proposed by Christopher Long, graduate student, Rutgers Uni-

versity.

Let A(x) =P1

m=0 amxm be a formal power series, with each am

either 0 or 1 and with in�nitely many of the am non-zero. Give a necessaryand su�cient condition on the am for there to exist a formal power seriesB(x) =

P1n=0 bnx

�n with each bn either 0 or 1, with in�nitely many of thebn non-zero, and such that the formal product A(x)B(x) exists.

Problem of the Month

Jimmy Chui, student, Earl Haig S.S.

We introduce a new section, originally titled \Problem of the Month"(yes, this is the most creative name we could think of). As you might guess,we plan to exhibit a problem and solution each issue, and not just for itselegance, but for its simplicity. For example, the last problems on a contestor olympiad always seem di�cult, but the o�cial solutions are inevitably afew lines long, and usually easy to understand. We would like to illustratethe same insight here. We begin with a problem from the 1997 Asian Paci�cMathematical Olympiad.

Problem. (1997 APMO)Let ABC be a triangle inscribed in a circle and let

la =ma

Ma

, lb =mb

Mb

, lc =mc

Mc

,

where ma, mb, mc are the lengths of the angle bisectors (internal to thetriangle) and Ma, Mb, Mc are the lengths of the angle bisectors extendeduntil they meet the circle. Prove that

la

sin2A+

lb

sin2B+

lc

sin2C� 3 ,

37

and that equality holds if and only if ABC is an equilateral triangle.

Solution.Let the altitudes from A to BC, B to CA, and C to AB be ha, hb,

and hc respectively. Let the circumradius of triangle ABC be R. Notethat ha = b sinC, hb = c sinA, and hc = a sinB, so that hahbhc =

abc sinA sinB sinC. We can see that ma � ha, mb � hb, mc � hc,Ma � 2R, Mb � 2R, andMc � 2R. Hence,

la

sin2A+

lb

sin2B+

lc

sin2C

=ma

Ma sin2A

+mb

Mb sin2B

+mc

Mc sin2C

� ha

2R sin2A+

hb

2R sin2B+

hc

2R sin2C

=ha

a sinA+

hb

b sinB+

hc

c sinC(Extended Sine Law)

� 33

rha

a sinA� hb

b sinB� hc

c sinC(AM-GM)

= 3 .

J.I.R. McKnight Problems Contest 1986

1. A bookshelf whose width lies between 30 and 50 cm holds exactly 5

poetry books each p cm thick, 3 history books each h cm thick, and 5

dictionaries each d cm thick (p, h, and d are unequal integers). It couldinstead hold exactly 4 poetry books, 5 history books and 4 dictionaries.If instead 7 poetry and 4 history books are placed on it, how manydictionaries must then be added to �ll the shelf exactly? Also �nd thethickness of each book and the width of the shelf.

2. ABCDEF is a regular hexagon, G is the mid-point of AB, and GEmeets AC at P . Find the ratio into which P divides AC.

3. Prove that the sum of the squares of the �rst n even natural numbersexceeds the sum of the squares of the �rst n odd natural numbers byn(2n+1). Hence, or otherwise, �nd the sum of the squares of the �rstn odd natural numbers.

4. (a) Andre, Barbara, Carole and Donato, whose ages were respectively19, 17, 15 and 13 years, inherited 13 750 dollars which was so di-vided that their respective shares, at 10 percent per annum simpleinterest, amounted to equal sums when they arrived at the age of21 years. What was the share that each received when they reachedthe age of 21?

38

(b) Prove that in any acute triangle ABC,

cotA+ cotB + cotC =a2 + b2 + c2

4K,

where K is the area of triangle ABC.

5. The point P (t2; t3) lies on the curve y2 = x3. Q is another point onthe curve such that PQ subtends a right angle at the origin O. Provethat the tangents at P and Q intersect on the curve 4y2 = 3x� 1.

[Ed. Unfortunately, we have lost the last page with questions 6, 7and 8. Anyone with a spare copy, please send to the editor. Thankyou.]

Contest Dates

Here are some upcoming contest dates to mark on your calendar.

Contest Grade Date

Gauss Grades 7 & 8 May 14, 1999

Pascal Grade 9Cayley Grade 10 February 24, 1999Fermat Grade 11Euclid Grade 12 April 20, 1999Descartes Grades 12 & 13 April 21, 1999

AHSME High School February 9, 1999

AIME High School March 16, 1999USAMO High School April 27, 1999

OPEN High School November 24, 1999

CMO High School March 31, 1999IMO High School July 10{22, 1999

More information can be found at the following sites:

<http://camel.math.ca/CMS/Competitions>

Canadian Mathematical Society Competitions

<http://math.uwaterloo.ca/CMC/CMCHome.html>

Canadian Mathematics Competitions

<http://www.unl.edu/amc/>

American Mathematics Competitions

39

Counting Snakes, Di�erentiating the

Tangent Function, and Investigating the

Bernoulli-Euler Triangle

Harold Reiter

In this paper we will examine three apparently unrelated mathematicalobjects. One is a family of permutations of �nite sets of integers, calledsnakes. The second is the tangent and secant functions from trigonometry,and their higher derivatives. The third is a Pascal-like triangular array ofnumbers called the Bernoulli-Euler triangle.

Snakes and Up-Down Permutations.

What are \snakes" and what are \up-down" permutations? Before wecan answer these questions, we must discuss permutations of �nite sets andestablish some notation. We denote by [n] the set f1; 2; 3; 4; : : : ; ng.

De�nition. A permutation � of the set [n] is a one-to-one function from[n] to itself: �

1 2 3 : : : n

�(1) �(2) �(3) : : : �(n)

�;

or simply (�(1); �(2); �(3); : : : ; �(n)). For example, the permutation � of[4] given by �(1) = 3; �(2) = 1; �(3) = 2 and �(4) = 4 is denoted(3; 1; 2; 4).

De�nition. A permutation � is called

A. an up-down permutation if �(1) < �(2); �(2) > �(3); �(3) < �(4); : : : ;

and

B. a down-up permutation if �(1) > �(2); �(2) < �(3); �(3) > �(4); : : : .

A permutation which is either up-down or down-up is called a snake.Our main goal here is to �nd a formula for the number of snakes on [n]. Letus list the up-down and the down-up permutations of [n] for small values ofn.

1. The up-down permutations of [1]: (1)The down-up permutations of [1]: (1)

2. The up-down permutations of [2]: (1; 2)The down-up permutations of [2]: (2; 1)

Copyright c 1999 Canadian Mathematical Society

40

3. The up-down permutations of [3]: (1; 3; 2), (2; 3; 1)The down-up permutations of [3]: (2; 1; 3), (3; 1; 2)

4. The up-down permutations of [4]: (1; 3; 2; 4), (2;3; 1; 4), (1; 4; 2; 3),(2; 4; 1; 3), (3; 4; 1; 2)The down-up permutations of [4]: (4; 2; 3; 1), (4;1; 3; 2), (3; 2; 4; 1),(3; 1; 4; 2), (2; 1; 4; 3)

Let An andBn denote the number of up-down and down-up permuta-tions of [n], n � 1. For convenience, let A0 = 1 and B0 = 1. Note from theabove that A1 = B1 = 1, A2 = B2 = 1, A3 = B3 = 2 and A4 = B4 = 5.Let us �nd A5.

Each up-down permutation on [5] has the form

i i i

i i

�(1) �(2) �(3) �(4) �(5)

Clearly there are just two places for the 5. That is 5 must be �(2) or�(4):

A:

5

�(1) �(2) �(3) �(4) �(5)

i i i

i i

and B:

5

�(1) �(2) �(3) �(4) �(5)

i i i

i i

Do you see a one-to-one correspondence between those of type A andthose of type B? Because of this correspondence, we will concentrate only onthose of type A. Notice that �(1) can be any of 1, 2, 3, or 4 and the otherthree can become �(3), �(4), �(5) in two ways (�(4) is the largest of thethree, and �(3) could be either of the other two). Thus, there are 4� 2 = 8

of type A, and so there are 16 up-down perms of [5]. Our next step is to �nda general formula for An. Before we do this, let us consider the relationshipbetween the An and the Bn.

Theorem. For all n;An = Bn.

Proof. Let � = (�(1);�(2); : : : ; �(n)) be an up-down permutation of[n]. Consider the function� = (n+1��(1);n+1��(2); : : : ; n+1��(n)).For example if � = (1; 3; 2; 5; 4), then � = (6�1;6�3;6�2;6�5; 6�4) =

(5; 3; 4; 1; 2). Note that � is a function from [n] to [n], that � is one-to-one(hence a permutation), and that � is a down-up permutation. Notice alsothat if �1 6= �2, then �1 6= �2. This means that the mapping � ! � isone-to-one. It is also onto; that is, if � is any down-up permutation, thereis an up-down permutation �0 such that the mapping above takes � to �0.Try working out �0 from �. Thus An = Bn for all n.

In what follows, we will �nd a formula for the numbers An + Bn =

2An. Fix n. Consider the possible positions for n in a down-up or an up-down permutation. Of course, n must appear in an up position, which wedenote by k.

41

r

r

r

r

r

r

r

r

r

r

r�(k)

: : :

�(k � 1)

: : :

�(k + 1)

Notice that � is an up-down permutation if and only if k is even. Inother words, the even positions are up positions if � is an up-down permu-tation and down positions if � is a down-up permutation. The numbers inpositions 1 through k � 1 are in up-down or down-up order, and those inpositions k + 1 through n are in up-down order. There are

�n�1k�1

�ways to

select the numbers for positions 1 through k. Having made that selection,there are Ak�1 ways of arranging them and An�k ways to arrange the rest

of the numbers. Thus, there are

�n� 1

k � 1

�Ak�1An�k up-down or down-up

permutations with the number n in the kth position. (Our choice of A0 = 1

comes into play here.) Since n can appear in any of the n positions, we have

2An = An +Bn =

nXk=1

�n� 1

k � 1

�Ak�1An�k . (2)

To illustrate for n = 5:

5i i i

i i there are

�4

0

�A0A4 = 1 � 1 � 5 = 5;

5

i i i

i i

there are

�4

1

�A1A3 = 4 � 1 � 2 = 8;

5i i i

i i there are

�4

2

�A2A2 = 6 � 1 � 1 = 6;

5i i i

i i

there are

�4

3

�A3A1 = 4 � 2 � 1 = 8;

5i i i

i i there are

�4

4

�A4A0 = 1 � 5 � 1 = 5.

Hence, A5 + B5 = 5 + 8 + 6 + 8 + 5 = 32 and A5 = 16.

42

Let ak = Ak=k!. Thus a0 = 1=0! = 1, a1 = 1=1! = 1, a2 = 1=2! = 1=2,etc. Writing out (2) and using the factorial form of

�n�1k�1

�, we get

2An =

nXk=1

�n� 1

k� 1

�Ak�1An�k =

nXk=1

(n� 1)!Ak�1An�k(k� 1)!(n� k)!

= (n� 1)!

nXk=1

Ak�1(k� 1)!

An�k(n� k)!

= (n� 1)!

nXk=1

ak�1an�k .

We may conclude from this that

2nAn

n(n� 1)!=

nXk=1

ak�1an�k ,

which is equivalent to

2nan =

nXk=1

ak�1an�k . (3)

Let f be the ordinary generating function of the an. That is,

f(t) =

1Xk=1

aktk .

When expressed in terms of the An, f is the exponential generating function.That is

f(t) =

1Xk=1

Ak

k!tk .

Next, compute f2(t) using equation (3).

f2(t) = (a0 + a1t+ a2t2 + : : : )(a0 + a1t+ a2t

2 + : : : )

= a20 + (a0a1 + a1a0)t+ (a0a2 + a1a1 + a2a0)t2 + : : :

+

nXk=1

(ak�1an�k)tn�1 : : :

= a20 + 2 � 2 � a2t+ 2 � 3 � a3t2 + � � �+ 2nantn�1 + : : :

= a20 + 2

" 1Xk=2

kaktk�1

#= a20 + 2

" 1Xk=1

kaktk�1 � a1

#

= a20 � 2a1 + 2f 0(t) = 1� 2 + 2f 0(t) .

43

Hence f satis�es the di�erential equation

y2 + 1 = 2y0 .

To solve this, write 1=2 = y0=(y2+1) and anti-di�erentiate both sides withrespect to t to get

t

2+ C = arctan y .

Take the tangent of both sides to get

tan

�t

2+ C

�= tan(arctan(f(t))) = f(t)

for some constant C. At t = 0, we have a0 = f(0) = 1 = tan(C), soC = �=4 will do. Then

f(t) = tan

�t

2+�

4

�= a0 + a1t+ a2t

2 + : : : ,

and

f(�t) = tan

�� t

2+�

4

�= a0 � a1t+ a2t

2 � : : : .

It can be shown that

tan

�t

2+�

4

�+ tan

�� t

2+�

4

�= 2= cos t

and that

tan

�t

2+�

4

�� tan

�� t

2+�

4

�= 2 tan t

for all t 2���

2; �2

�. Thus f(t) + f(�t) = 2= cos t, or

sec t = a0 + a2t2 + a4t

4 + : : : = A0 +A2

2!t2 +

A4

4!t4 + : : : ,

and similarly

tan t = a1 + a3t3 + a5t

5 + : : : = A1 +A3

3!t3 +

A5

5!t5 + : : : .

Therefore, to �nd An, for odd n, we �nd the n th derivative of tan t, andevaluate at t = 0. That is,

An =dn

dtntan t

�t=0

and for even n, we �nd the n th derivative of sec t, and evaluate at t = 0.That is,

An =dn

dtnsec t

�t=0

.

44

Di�erentiating the Tangent Function.

We seek to di�erentiate the tangent function repeatedly at 0. Sinced

dxtanx = sec2 x and d

dxsec2 x = 2 tanx sec2 x, it is convenient to set

y = tanx and z = sec2 x. We use D in place of d

dxfor convenience. Then

Dy = D tanx = tan0 x = z and D2y = Dz = 2yz.

1. D0y = y.

2. Dy = z.

3. D2y = Dz = 2yz.

4. D3y = D2yz = 2z � z + 2y � 2yz = 2z2 + 4y2z

5. D4y = D(2z2 + 4y2z) = 2 � 2z � 2yz+ 4(2y � z � z + y2 � 2yz)= 16yz2 + 8y3z.

6. D5y = D(16yz2+8y3z) = 16z �z2+16y �2z �2yz+24y2z+8y3 �2yz= 16z3 + 64y2z2 + 24y2z + 16y4z.

7. D6y = D(16z3 + 64y2z2 + 24y2z + 16y4z)

= 48z2 � 2yz + 64(2yz � z2 + y2 � 2z � 2yz) + 24(2yz2 + y2 � 2yz)+16(4y3z � z+ y4 � 2yz)96yz3+128yz3+256y3z2 +48yz2+48y3z

+64y3z2 + 32y5z = 224yz3 + 320y3z2 + 48y3z + 32y5z + 48yz2.

8. D7y = D(224yz3 + 320y3z2 + 48y3z + 32y5z + 48yz2)

= 224(z4 + y � 3z2 � 2yz) + 320(3y2z � z2 + y3 � 2z � 2yz)+48(3y2�z �z+y3 �2yz)+32(5y4�z �z+y5�2yz)+48(z �z2+y �2z �2yz).

Evaluating each of these at 0, noting that tan 0 = 0 and sec 0 = 1, we get:

1. tan0 = 0, 2. tan0 0 = 1,

3. tan00 0 = 0, 4. tan(3) 0 = 2,

5. tan(4) 0 = 0, 6. tan(5) 0 = 16,

7. tan(6) 0 = 0, 8. tan(7) 0 = 272.

The Bernoulli-Euler Triangle.

This third and �nal section provides a straightforward and fast methodfor counting the up-down permutations on [n]. The Bernoulli-Euler (BE)triangle, like Pascal's triangle is a triangular array of numbers each row ofwhich is obtainable from the previous row. The BE triangle:

1

1 0

0 1 1

2 2 1 0

0 2 4 5 5

16 16 14 10 5 0

0 16 32 46 56 61 61

272 272 256 224 178 122 61 0

45

The triangle can be �lled in as follows. In line 1, write 1. Every even lineis �lled in from right to left by writing in each position the sum of all thenumbers of the previous row to the right of the position. Odd lines are �lledin from left to right similarly, summing those numbers in the row above whichlie to the left of the position. For example, the number 46 above is in thesixth row, and it is obtained by adding the two 16's and the 14 which lie toits left in the previous row. We will see below that the entries at the ends ofthe rows are just the numbers we seek.

What do the numbers in the triangle mean? What do the S(n;h) count?Consider the triangle

S(1; 1)

S(2; 1) � S(2; 2)

S(3; 3) �! S(3; 2) �! S(3; 1)

S(4; 1) � S(4; 2) � S(4; 3) � S(4; 4)

S(5; 5) �! S(5; 4) �! S(5; 3) �! S(5; 2) �! S(5; 1)

Here is the answer. If n is odd, S(n; h) counts the number of up-downpermutations of [n] whose last entry is h, and ifn is even, S(n; h) counts thenumber of down-up permutations of [n] whose last entry is h. For example,

(S(5;5); S(5;4); S(5; 3); S(5; 2); S(5; 1)) = (0;2; 4; 5; 5);

which means that

� there are no up-down permutations of [5] with last value 5;

� there are two up-down permutations with last value 4 | they are(1; 3; 2; 5; 4) and (2; 3; 1; 5; 4);

� there are four up-down permutations with last value 3 | they are(1; 4; 2; 5; 3), (2; 4; 1; 5; 3), (1;5; 2; 4; 3) and (2;5; 1; 4; 3);

� there are �ve up-down permutations with last value 2 | they are(1; 4; 3; 5; 2), (3; 4; 1; 5; 2), (1; 5; 3; 4; 2), (3; 5; 1; 4; 2) and (4;5; 1; 3; 2);and

� there are �ve up-down permutations with last value 1 | they are(2; 4; 3; 5; 1), (3; 4; 2; 5; 1), (2; 5; 3; 4; 1), (3; 5; 2; 4; 1) and (4;5; 2; 3; 1).

The next formula is a recursive construction of the triangle. Using thenotation S(n; i) as we did earlier for the i th entry of even numbered rowsand the n� i+1 st entry of odd numbered rows, S(1; 1) = 1, and for n > 1,

S(n; i) =

(0 if i = n ,

S(n; i+ 1) + S(n� 1; n� i) otherwise.

To illustrate why the recursive formulas works, consider the example ob-tained for n = 6 and i = 4:

S(6; 5) + S(5;2) = S(6; 4) .

46

S(6; 5) counts the �ve down-up permutations of [6] which end in 5:(2; 1; 4; 3; 6; 5), (3;1; 4; 2; 6; 5), (3;2; 4; 1; 6; 5), (4; 1; 3; 2; 6; 5), and(4; 2; 3; 1; 6; 5). These give rise to �ve up-down permutations ending in 4

by interchanging the 4 and the 5:

(2; 1; 4; 3; 6; 5) ! (2; 1; 5; 3; 6; 4),(3; 1; 4; 2; 6; 5) ! (3; 1; 5; 2; 6; 4),(3; 2; 4; 1; 6; 5) ! (3; 2; 5; 1; 6; 4),(4; 1; 3; 2; 6; 5) ! (5; 1; 3; 2; 6; 4),(4; 2; 3; 1; 6; 5) ! (5; 2; 3; 1; 6; 4).

Note that S(5; 2) counts the �ve up-down permutations of [5] which endwith 2: (1; 4; 3; 5; 2), (3; 4; 1; 5; 2), (1; 5; 3; 4; 2), (3; 5; 1; 4; 2), and(4; 5; 1; 3; 2). To see how these are transformed into new (other than thosecounted above) permutations counted by S(6; 4), �rst replace (x; y; u; v;w)

with (6� x; 6� y;6 � u; 6� v;6 � w), obtaining a down-up permutationof [5]. Next, replace the 5 in (6 � x; 6 � y;6 � u; 6 � v; 6 � w) with 6,replace the last position 6� w (which must be 6� 2 = 4, because of where(x; y; u; v; w) came from) with 5, and �nally, append the number 4 at theend. Thus, for example,

(1;4; 3; 5; 2)! (5; 2; 3; 1; 4)! (6; 2; 3; 1; 4)!

(6; 2; 3; 1; 5)! (6; 2; 3; 1; 5; 4):

All permutations obtained in this way are di�erent because they end withthe pair 4 5, whereas the ones obtained above all end with 6 4.

References:

1. Combinatorik, E. Netto, 1930.

2. Why Study Mathematics, Vladimir Arnold, Quantum, Sept{Oct, 1994,pp. 25-31.

Harold Reiter

Department of Mathematics

Fretwell Building

University of North Carolina, Charlotte

920 University City Boulevard

Charlotte, NC USA 28223

hbreiter@newmail.uncc.edu

47

PROBLEMS

Problem proposals and solutions should be sent to Bruce Shawyer, De-

partment ofMathematics and Statistics,Memorial University of Newfound-

land, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be ac-

companied by a solution, together with references and other insights which

are likely to be of help to the editor. When a submission is submitted with-

out a solution, the proposer must include su�cient information on why a

solution is likely. An asterisk (?) after a number indicates that a problem

was submitted without a solution.

In particular, original problems are solicited. However, other inter-

esting problems may also be acceptable provided that they are not too well

known, and references are given as to their provenance. Ordinarily, if the

originator of a problem can be located, it should not be submitted without

the originator's permission.

To facilitate their consideration, please send your proposals and so-

lutions on signed and separate standard 812"�11" or A4 sheets of paper.

These may be typewritten or neatly hand-written, and should be mailed to

the Editor-in-Chief, to arrive no later than 1 September 1998. They may also

be sent by email to crux-editors@cms.math.ca. (It would be appreciated if

email proposals and solutions were written in LATEX). Graphics �les should

be in epic format, or encapsulated postscript. Solutions received after the

above date will also be considered if there is su�cient time before the date

of publication. Please note that we do not accept submissions sent by FAX.

2401. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.

In triangle ABC, CD is the altitude from C to AB. E and F are themid-points ofAB andCD respectively. P andQ are points on line segmentsBC and AC respectively, and are such that PQkBA. The projection of Qonto AB is R. PR and EF intersect at S.

Prove that

(a) S is the mid-point of line segment PR,

(b)1

PR2� 1

AB2+

1

CD2.

2402. Proposed by Christopher J. Bradley, Clifton College, Bristol,

UK.

Find all sets of positive integers a, b, c, d, such that

bd > ad+bc and (9ac+bd)(ad+bc) = a2d2+10abcd+b2c2 .

48

2403. Proposed by Christopher J. Bradley, Clifton College, Bristol,

UK.

The positive integers a, b, c have the following properties:

1. a is odd;

2. the greatest common denominator of a, b, c is 1;

3. they satisfy the Diophantine equation:

2

a+

1

b=

1

c.

Prove that abc is a perfect square.

2404?. Proposed by V�aclav Kone �cn �y, Ferris State University, Big

Rapids, Michigan, USA.

Let f(x) =(1� 2x)p + 2(x+ 1)p

2 ((x+ 1)(1� 2x))p+ (x+ 1)2p

, where x 2�0; 1

2

�and

p � 2 is an integer.

The relative maximum of f(x) on�0; 1

2

�has been found by computer

to be at the point xp for several values of p as shown:

p 2 3 4 5 6 7

xp 0:2538 0:1133 0:0633 0:0403 0:0279 0:0205

Explain why xp is \near" 1

p2.

2405. Proposed by G.P. Henderson, Garden Hill, Campbellcroft,

Ontario, (adapted by the editors).

Given two n{sided dice, one with ak sides with k dots (1 � k � n)such that

Pn

k=1 ak = n and the other with bk sides with k dots (1 � k � n)such that

Pn

k=1 bk = n. Both are rolled. Let rk be the probability that thesum of the two faces showing is k.

How should the ak and the bk be chosen to minimize

nXk=1

�rk �

1

n� 1

�2

(a) with n = 6?

(b)? with general n?

2406. Proposed by Paul Yiu, Florida Atlantic University, Boca

Raton, Florida, USA.

Suppose that an integer-sided triangle contains a 120� angle, with thetwo containing arms di�ering by 1. Prove that the length of the longest sideis the sum of two consecutive squares.

49

2407. Proposed by Christopher J. Bradley, Clifton College, Bristol,

UK.

Triangle ABC is given with \BAC = 72�. The perpendicular from B

to CA meets the internal bisector of \BCA at P . The perpendicular fromC to AB meets the internal bisector of \ABC at Q.

If A, P and Q are collinear, determine \ABC and \BCA.

2408. Proposed by Mansur Boase, student, St. Paul's School, Lon-

don, England.

Perpendiculars are dropped from a point P inside an acute-angled tri-angle ABC to the sides BC, CA, AB, meeting them at D, E, F respec-tively.

(a) Prove that the perpendiculars from A to EF , B to FD, C to DE areconcurrent (at K, say).

(b) A point, P , is called \Cevic" with respect to 4ABC if AD, BE andCF are concurrent. Prove that K is Cevic with respect to 4DEF ifand only if P is Cevic with respect to 4ABC.

2409. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.

Triangle ABC with sides a, b, c, has O as the centre of its circumcircle(with radius R) and H as its orthocentre. Suppose that OH intersects CBand CA at P and Q respectively.

(a) Prove that quadrilateral ABPQ is cyclic if and only if a2 + b2 = 6R2.

(b) If quadrilateral ABPQ is cyclic, �nd a formula for the length of PQ interms of a, b and c alone.

2410. Proposed by V�aclav Kone �cn �y, Ferris State University, Big

Rapids, Michigan, USA.For n � 1, de�ne vn = [1; 2; 3; : : : ; n�1; n], where the square bracket

denotes the least common multiple. Let p1 < p2 be twin primes.

Prove or disprove that vp2 = p2 vp1 for p1 > 3.

2411. Proposed by Zun Shan and Edward T.H. Wang, Wilfrid Laurier

University, Waterloo, Ontario.

It is well known and easy to show that the product of four consecutivepositive integers plus one is always a perfect square. It is also easy to showthat the product of any two consecutive positive integers plus one is never aperfect square. Also, note that

2� 3� 4 + 1 = 52 and 4� 5� 6 + 1 = 112 .

(a) Find another natural number n such that n(n+ 1)(n+ 2)+ 1 is aperfect square.

(b)? Are there further examples?

50

2412. Proposed by Darko Veljan, University of Zagreb, Zagreb,

Croatia.

Suppose that A1, A2, A3, A4 are the vertices of a tetrahedron T . Onthe faces opposite A1, A2, A3, construct tetrahedra outside T with apexesA01, A

02, A

03, and volumes V1, V2, V3, respectively.

Let A04 be the point such that

���!A1A

04 =

��!BA4, whereB is the point of in-

tersection of the planes through A0iparallel to the respective bases

(i = 1; 2; 3).

Let V4 be the volume of tetrahedron A1A2A3A04.

Prove that V4 = V1 + V2 + V3.

2413. Proposed by Bill Sands, University of Calgary, Calgary,

Alberta.

A deck of six cards consists of three black cards and three red cards, insome order. The top four cards are picked up, shu�ed randomly, and thenput on the bottom of the deck. This procedure is repeated n times.

Let p(n) be the probability that after n such \shu�es", the deck al-ternates between red and black cards, either colour being the top card. (Sop(n) will, in general, depend on the initial ordering of the deck.)

Find limn!1

p(n) | that is, the \long term" probability that the deck

\tends to be alternating".

Correction

2324. [1998: 109] Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel,

Alcorc �on, Madrid, Spain.

Find the exact value of

1Xn=1

1

un, where un is given by the recurrence

un = n! +

�n� 1

n

�un�1 ,

with the initial condition u1 = 2.

Notice

Proposers are asked to note that we have now published all of theproposals, submitted in 1997, that the Editorial Board has accepted for use.

51

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased to

consider for publication new solutions or new insights on past problems.

2299. [1997: 503] Proposed by Walther Janous, Ursulinengymnas-

ium, Innsbruck, Austria.

Let x, y, z > 0 be real numbers such that x+ y + z = 1. Show that

Ycyclic

�(1� y)(1� z)

x

�(1�y)(1�z)=x� 256

81.

Determine the cases of equality.

Composite solution by Theodore Chronis, Athens, Greece and Kee-Wai

Lau, Hong Kong.The given inequality is clearly equivalent to

A =Xcyclic

�1 +

xy

z

�ln

�1 +

xy

z

�� 4 ln

�4

3

�. (1)

Since the function x ln(x) is convex on (0;1), we have, by Jensen'sInequality, that

A

3� 1

3

�3 +

xy

z+yz

x+zx

y

�ln

�1

3

�3 +

xy

z+yz

x+zx

y

��. (2)

Since (xy)2+(yz)2+(zx)2 � (xy)(yz)+(yz)(zx)+(zx)(xy) = xyz,we get

xy

z+yz

x+zx

y� 1 . (3)

From (2) and (3), we obtain (1) immediately.

It is clear that equality holds if and only if x = y = z = 1

3.

Also solved by NIKOLAOS DERGIADES, Thessaloniki, Greece; FLORIAN HERZIG, stu-dent, Cambridge, UK; RICHARD I. HESS, Rancho Palos Verdes, California, USA; MURRAYS. KLAMKIN, University of Alberta, Edmonton, Alberta; MICHAEL LAMBROU, University ofCrete, Crete, Greece; VEDULA N. MURTY, Visakhapatnam, India; HEINZ-J �URGEN SEIFFERT,Berlin, Germany; and the proposer.

2300. [1997: 503] Proposed by Christopher J. Bradley, Clifton Col-

lege, Bristol, UK.

Suppose thatABC is a triangle with circumradiusR. The circle passingthrough A and touching BC at its mid-point has radius R1. De�ne R2 andR3 similarly. Prove that

R21 + R2

2 + R23 � 27

16R2 .

52

Solution by G. Tsintsifas, Thessaloniki, Greece.

We usemi for the medians and hi for the altitudes. LetM be the mid-point of BC and O1 the centre of the circle through A touching BC at M .From triangle O1MA we have: 1

2m1 = 1

2AM = R1 cos! = R1

h1

m1

, where

! = \O1MA or 1

2m2

1 = R1h1. Similarly, 1

2m2

2 = R2h2;1

2m2

3 = R3h3.Therefore,

3

8(a2 + b2 + c2) =

m21 +m2

2 +m23

2= R1h1 +R2h2 +R3h3 . (1)

From the Cauchy-Schwarz inequality, we have

R1h1 + R2h2 +R3h3 ���R21 + R2

2 + R23

� �h21 + h2

2 + h23

��1=2. (2)

But, it is well known that:

h21 + h2

2 + h23 =

a2b2 + b2c2 + c2a2

4R2. (3)

Using (1), (2), (3) we obtain

9R2

16

�a2 + b2 + c2

�2a2b2 + b2c2 + c2a2

� R21 +R2

2 + R23 . (4)

Obviously we have:

3 ��a2 + b2 + c2

�2�2b2 + b2c2 + c2a2

. (5)

So, from (4), (5), it follows that 27

16R2 � R2

1 + R22 + R2

3.

Also solved by NIELS BEJLEGAARD, Stavanger, Norway; THEODORECHRONIS, Athens, Greece; NIKOLAOS DERGIADES, Thessaloniki, Greece; FLORIAN HERZIG,student, Cambridge, UK; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta; V �ACLAV KONE �CN �Y, FerrisState University, Big Rapids, Michigan, USA; MICHAEL LAMBROU, University of Crete, Crete,Greece; KEE-WAI LAU, Hong Kong; GERRY LEVERSHA, St. Paul's School, London, England;D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer.

2301. [1998: 45] Proposed by Christopher J. Bradley, Clifton Col-

lege, Bristol, UK.

Suppose that ABC is a triangle with sides a, b, c, that P is a pointin the interior of 4ABC, and that AP meets the circle BPC again at A0.De�ne B0 and C0 similarly.

Prove that the perimeter P of the hexagon AB0CA0BC0 satis�es

P � 2�p

ab+pbc+

pca�.

Solution by Florian Herzig, student, Cambridge, UK.

53

Let x = \BPC, y = \CPA, and z = \APB.

Then, asA0CPB is cyclic, the angles in4A0CB are ��x, ��z, ��yand so by the Sine Law:

A0B =a sinz

sinxand A0C =

a siny

sinx.

Thus the perimeter of 4A0CB equals

a(sinx+ siny+ sin z)

sinx.

Analogously we obtain the perimeters of triangles B0AC and C0BA. Sum-ming these yields

(sinx+ siny + sin z)

�a

sinx+

b

siny+

c

sinz

�.

By the Cauchy-Schwarz inequality,

P + a+ b+ c ��p

a+pb+

pc�2

;

or equivalently,

P � 2�p

bc+pca+

pab�.

Equality holds ifa

sin2 x=

b

sin2 y=

c

sin2 z.

Also solved by MANSUR BOASE, student, Cambridge, England; GORAN CONAR, stu-dent, Gymnasium Vara�zdin, Vara�zdin, Croatia; NIKOLAOS DERGIADES, Thessaloniki, Greece;WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, Univer-sity of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul's School, London, England; JOELSCHLOSBERG, student, Bayside, NY, USA; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK,Zaltbommel, the Netherlands; G. TSINTSIFAS, Thessaloniki, Greece; and the proposer.

Most of the submitted solutions are similar to the one given above.

2302. [1998: 45] Proposed by Toshio Seimiya, Kawasaki, Japan.

Suppose that the bisector of angle A of triangle ABC intersects BCat D. Suppose that AB + AD = CD and AC + AD = BC.

Determine the angles B and C.

I. Solution by V�aclav Kone �cn �y, Ferris State University, Big Rapids,

Michigan, USA (somewhat modi�ed by the editor).

54

C

B0

AB

D

E

X

B

C

C

C

B

B

B

The �gure shows the circle with centre A and radius AB intersectingCB at X, CA at E and AB (extended) again at B0. Subtract the �rst ofthe given equalities from the second to get AC � AB = BC � CD, which,because AB = AE, implies that CE = BD (as the �gure shows). Using thesymmetry of AB andAE about the angle bisector AD, we see that BD alsoequals ED, and that B equals the angle at E (\AED). This latter angle isan exterior angle of the isosceles triangle EDC, so that B = \AED = 2C.The exterior angle at A of triangle ABC (namely \B0AE) is therefore 3C,while \B0AX = 4C (since the angle at the centre is twice the angle at thecircumference). This implies that

\XAC = C , (1)

so that triangle XAC is isosceles, with XC equal to the radius AX. The�gure now recalls Archimedes' trisection of \B0AE by compass and markedruler (as in David R. Davis, Modern College Geometry, Addison-Wesley,1958, section 10{8): using CX = AB, we see that the given conditionAB + AD = CD yields XD = AD, so that triangle DXA is isosceles,and (since \DXA is an exterior angle of triangle AXC) \DAX = 2C.This, together with (1) says that half of angle A, namely \DAC, equals 3C,so that A = 6C. Thus 180� = A + B + C = 6C + 2C + C = 9C. Weconclude that A = 120�, B = 40� and C = 20�.

II. Solution by Nikolaos Dergiades, Thessaloniki, Greece.

Let A, B, C be the angles and a, b, c be the sides of 4ABC. Then,from c + AD = CD and b + AD = a, we get a � CD = b � c, so thatBD = b� c and AD = a� b.

If E is the point between A and C with AE = AB, thenEC = AC � AE = b � c = BD = DE, so that 4DEC is isosceles,and B = \DEA = 2C.

[Editor's comment. Note that the argument up to here has repeated the�rst step of solution I, avoiding reference to the �gure. Several other solversderived B = 2C by �rst showing that ca + c2 = b2, noting that theseequivalent conditions are familiar to CRUX with MAYHEM readers, hav-ing appeared in [1976: 74], [1984: 278], [1996: 265], etc.] [It follows that

55

sinA = sin(C+B) = sin(3C), and sin�A

2

�= sin

��

2� (B+C)

2

�= cos

�3C

2

�.]

The Sine Law applied to 4ABD, and then to 4ABC, gives

AD

sinB=

BD

sin(A=2)=) a� b

sinB=

b� c

sin(A=2)

=) sinA� sinB

sinB=

sinB � sinC

sin(A=2)

=) sin(3C)� sin(2C)

sin(2C)=

sin(2C)� sinC

cos(3C=2)

=) 2 sin(C=2)cos(5C=2)

sin(2C)=

2 sin(C=2)cos(3C=2)

cos(3C=2)

=) cos(5C=2) = sin(2C) =) 5C

2+ 2C = 90� .

Thus, C = 20�, B = 40� and A = 120�.Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain;

CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; GORAN CONAR, student, Gym-nasium Vara�zdin, Vara�zdin, Croatia; DAVID DOSTER, Choate Rosemary Hall, Wallingford,Connecticut, USA; IAN JUNE L. GARCES, Ateneo de Manila University, The Philippines andGIOVANNI MAZZARELLO, Ferrovie dello Stato, Florence, Italy; FLORIAN HERZIG, student,Cambridge, UK; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAELLAMBROU, University of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul's School,London, England; ECKARD SPECHT, Otto-von-Guericke-Universit �at, Magdeburg, Germany;D.J. SMEENK, Zaltbommel, the Netherlands; PARAYIOU THEOKLITOS, Limassol, Cyprus;PANOS E. TSAOUSSOGLOU, Athens, Greece; JOHN VLACHAKIS, Athens, Greece; and the pro-poser. There was one incomplete solution.

2303. [1998: 45] Proposed by Toshio Seimiya, Kawasaki, Japan.

Suppose that ABC is a triangle with angles B and C satisfyingC = 90�+ 1

2B, that the exterior bisector of angleA intersects BC atD, and

that the side AB touches the incircle of4ABC at E.

Prove that CD = 2AE.

Solution by Nikolaos Dergiades, Thessaloniki, Greece.

We know that 2AE = b + c � a (see, for example, 290, Roger A.Johnson, Modern Geometry, Houghton-Mi�in, 1929). In the extension ofBA we take AF = b = AC.

Since AD is the external bisector of the angle A of 4ABC, F is thesymmetric point of C with axis of symmetry AD. Then

\AFD = \ACD = 180� \C = 90�\B

2,

so 4BFD is isosceles; that is, BD = BF or

a+CD = b+ c or CD = b+ c� a = 2AE .

56

Also solved byMIGUEL AMENGUALCOVAS, Cala Figuera, Mallorca, Spain (2 solutions);SAMBAETHGE, Nordheim, Texas, USA; FRANCISCOBELLOT ROSADO, I.B. Emilio Ferrari, Val-ladolid, Spain; MANSUR BOASE, student, Cambridge, England; CHRISTOPHER J. BRADLEY,Clifton College, Bristol, UK; GORAN CONAR, student, Gymnasium Vara�zdin, Vara�zdin, Croatia;DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA; FLORIAN HERZIG,student, Cambridge, UK; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; MICHAEL LAMBROU,University of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul's School, London, England;JOEL SCHLOSBERG, student, Bayside, NY, USA; D.J. SMEENK, Zaltbommel, the Nether-lands; ECKARD SPECHT, Otto-von-Guericke Universit �at, Magdeburg, Germany; PARAYIOUTHEOKLITOS, Limassol, Cyprus; JOHN VLACHAKIS, Athens, Greece; PAUL YIU, Florida At-lantic University, Boca Raton, Florida, USA; and the proposer.

2304. [1998: 46] Proposed by Toshio Seimiya, Kawasaki, Japan.

An acute angled triangle ABC is given, and equilateral trianglesABDand ACE are drawn outwardly on the sides AB and AC. Suppose that CDand BE meet AB and AC at F and G respectively, and that CD and BEintersect at P .

Suppose that the area of the quadrilateral AFPG is equal to the areaof the triangle PBC. Determine angle BAC.

Solution by Michael Lambrou, University of Crete, Crete, Greece.

If [AFPG] = [PBC], then [ABG] = [BFC] and so 1

2AGc sinA =

1

2BFa sinB; hence, by the Sine Rule,

AG

BF=

a sinB

c sinA=

b sinA

c sinA=

b

c.

Similarly,AF

CG=c

b, so that

AG

CG=BF

AF.

Writing \AEG = ', \BDF = !, we have from the equilateral trian-gles

sin'

sin(60� ')=

sin\AEG

sin\GEC=AG

CG=BF

AF=

sin\BDF

sin\FDA=

sin!

sin(60� !).

Hence, sin'

p3

2cos! � 1

2sin!

!=

p3

2cos'� 1

2sin'

!sin! and

sin('� !) = 0. This shows that ' = ! (as both are acute).

Recall that P (Fermat or Napoleon point) satis�es \FPG = 120�.Now summing the angles of quadrilateral DAEP , we have

360� = \ADP + \DPE + \PEA+�60� + \BAC + 60�

�= 60� � ! + 120� + '+ 120� + \BAC ,

and hence \BAC = 60�.

57

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, and MAR �IAASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; CHRISTOPHERJ. BRADLEY, Clifton College, Bristol, UK; GORAN CONAR, student, Gymnasium Vara�zdin,Vara�zdin, Croatia; NIKOLAOS DERGIADES, Thessaloniki, Greece; FLORIAN HERZIG, student,Cambridge, UK; GERRY LEVERSHA, St. Paul's School, London, England; D.J. SMEENK, Zalt-bommel, the Netherlands; ECKARD SPECHT, Otto-von-Guericke Universit �at, Magdeburg, Ger-many; PARAYIOU THEOKLITOS, Limassol, Cyprus; and the proposer.

2305. [1998: 46] Proposed by Richard I. Hess, Rancho Palos Verdes,

California, USA.

An integer-sided triangle has angles p� and q�, where p and q arerelatively prime integers. Prove that cos � is rational.

Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Austria

(slightly modi�ed by the editor).

Let Q denote the set of all rationals. By the Cosine Law, we see thatcos(p�) 2 Q and cos(q�) 2 Q.

Furthermore, since cos((p+ q)�) = � cos(� � (p+ q)�), we have aswell that cos((p+ q)�) 2 Q.

From cos(p�) cos(q�) = 1

2(cos((p+ q)�) + cos((p� q)�)), we infer

that cos((p� q)�) 2 Q.Similarly, from cos(p�) cos((p� q)�) = 1

2(cos(2p� q)�+ cos(q�)),

we infer that cos((2p� q)�) 2 Q.Analogously, we have cos((p� 2q)�) 2 Q.A simple induction then yields that cos((lp+kq)�) 2 Q for all integers

l and k. The result now follows, since there exist integers l and k such thatlp + kq = 1.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; THEODORECHRONIS, Athens, Greece; DIANE and RAY DOWLING, University of Manitoba, Winnipeg,Manitoba; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; FLORIAN HERZIG,student, Cambridge, UK; MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta;MICHAEL LAMBROU, University of Crete, Crete, Greece; KEE-WAI LAU, Hong Kong; GERRYLEVERSHA, St. Paul's School, London, England; PAVLOS MARAGOUDAKIS, Pireas, Greece;JOEL SCHLOSBERG, student, Bayside, NY, USA; and the proposer.

Besides Janous, Godin also obtained the stronger result. Klamkin showed thatcos((lp � kq)�) 2 Q for all non-negative integers l and k. Their arguments, which are verysimilar, all used induction. The Dowlings gave a proof based on de Moivre's Theorem and theBinomial Theorem. Bradley gave the following example to show that such triangles do exist:a = 135, b = 144, c = 31. In this case, the angles of the triangle are 2�, 3� and � � 5�,where cos � = 5

6. (Thus, � = cos�1

�56

� � 0:5857 radians, or 33:5573� | Ed.)

2306. [1998: 46, 175] Proposed by Vedula N. Murty, Visakhapat-

nam, India.

58

(a) Give an elementary proof of the inequality:�sin

��x

2

��2>

2x2

1 + x2; (0 < x < 1) . (1)

(b) Hence (or otherwise) show that

tan�x

8<:

<�x(1�x)1�2x ;

�0 < x < 1

2

�,

>�x(1�x)1�2x ;

�1

2< x < 1

�.

(2)

(c) Find the maximum value of f(x) =sin(�x)

x(1� x)on the interval (0;1).

Comments on (a) and (c) by the editor.As pointed out by several solvers (indicated by a dagger y before their

names), part (a) is identical to the �rst half of problem 2296 [1997: 503] withsolution on [1998: 533]. Part (c) follows immediately from the second half

of that problem since f�1

2

�= 4, where f(x) =

sin(�x)

x(1�x) . This, together withthe (corrected) inequality f(x) � 4, shows that max f(x) = 4.

Solution to (b) by yWalther Janous, Ursulinengymnasium, Innsbruck,

Austria.

Consider the function f(x) =�x(1� x)

1� 2x� tan(�x), 0 < x < 1.

Then f(0) = 0. We claim that f 0(x) > 0 for all x 2 (0; 12).

Since f 0(x) = �

�(1� 2x)2 + 2x(1� x)

(1� 2x)2� 1

cos2(�x)

�, it su�ces to

show that

cos2(�x) >(1� 2x)2

(1� 2x)2 + 2x(1� x)(3)

Letting y = 1 � 2x, we have y 2 (0; 1) and x = 1

2(1 � y). Simple

manipulations show that (3) is equivalent to

sin2��y

2

�>

y2

y2 + (1� y)(1+ y)=2or sin2

��y

2

�>

2y2

1 + y2,

which is valid by (a). Hence f(x) > 0 for all x 2 (0; 12).

If 1

2< x < 1, then 0 < 1 � x < 1

2, and hence, by the result just

established, we have f(1� x) > 0; that is,

�x(1� x)

1� 2(1� x)> tan(�(1� x)) , or

�x(1� x)

�(1� 2x)> � tan(�x) ,

from which we obtain that tan(�x) >�x(1� x)

1� 2x.

59

Also solved by GORAN CONAR, student, Gymnasium Vara�zdin, Vara�zdin, Croatia;RICHARD I. HESS, Rancho Palos Verdes, California, USA; yV �ACLAV KONE �CN �Y, Ferris StateUniversity, Big Rapids, Michigan, USA; MICHAEL LAMBROU, University of Crete, Crete,Greece; PHIL McCARTNEY, Northern Kentucky University, Highland Heights, KY, USA;yHEINZ-J�URGEN SEIFFERT, Berlin, Germany; and the proposer. Part (a) only was solved byAISSA GUESMIA, Institut de recherche math �ematique avanc �ee, Universit �e Louis Pasteur etCNRS, Strasbourg, France. There was also one incorrect solution.

Sei�ert also cited the following known inequalities (V.D. Mascioni und K. Sch �utte,Aufgabe 979, Elem. Math. 44, 1989, 20):

8x(1 � x)

�(1 � 2x)< tan(�x) <

�x

1� 4x2, 0 < x < 1

2,

and noted that the right inequality of this re�nes the right inequality in (b). Replacing x byx � 1

2, he derived the following inequalities:

�(1 � x)

(1 � 2x)(3 � 2x)< tan(�x) <

8x(1 � x)

�(1 � 2x), 1

2< x < 1 ,

and remarked that the left inequality of this is sharper than the left inequality in (b). He alsopointed out that this problem and problem 2296 are both simple consequences of problem 519

[1980: 44; 1981: 65] and problem 144 [Matyc J. 14, 1980, 72 and 15, 1981, 156]. All of theseproblems were posed by this proposer.

2307. [1998: 46] Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel,

Alcorc �on, Madrid, Spain.

It is known that every regular 2n{gon can be dissected into�n

2

�rhom-

buses with the same side length.

(a) How many di�erent classes of rhombuses are there?

(b) How many rhombuses are there in each class?

Solution by Michael Lambrou, University of Crete, Crete, Greece.

It is stated on page 10 of Disections, Plane and Fancy by G.N.Frederickson, Cambridge University Press, 1977, that a regular 2n{gon canbe dissected into rhombi with the same side length according to the followingrules:

(a) If n is odd with n = 2m+ 1, then the 2n{gon can be dissected into nsuch rhombi, each of angle j�=n, for each j = 1, 2, : : : , m. In otherwords, we have m = n�1

2classes of rhombi with n rhombi in each

class, making a total of 1

2n(n� 1) =

�n

2

�pieces.

(b) If n is even with n = 2m, then the 2n{gon can be dissected into n suchrhombi, each of angle j�=n, for each j = 1, 2, : : : , m � 1, plus msquares. In other words, we have m = n

2classes of rhombi with a

total of n(m � 1) + m = 2m(m � 1) + m = m(2m � 1) =1

2n(n� 1) =

�n

2

�pieces.

Although the proof is not given in detail in Frederickson, the accompa-nying �gures for the cases n = 2, 3, 4, 5 and 6 make it obvious how to dothe general case, so there is no need to repeat the details here.

Also solved by the proposer.

60

2308. [1998: 46] Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel,

Alcorc �on, Madrid, Spain.

A sequence fvng has initial value v0 = 1 and, for n � 0, satis�es therecurrence relation

vn+1 = 2n+1 �nX

k=0

vk vn�k .

Find a formula for vn in terms of n.

Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Austria.

We use the method of generating functions and show

vn =

�n

bn=2c

�.

where bxc is the greatest integer not exceeding x. Let

f(x) =

1Xn=0

vnxn .

Then using f2(x) to mean [f(x)]2we get

f2(x) =

1Xn=0

nX

k=0

vkvn�k

!xn =

1Xn=0

(�vn+1 + 2n+1)xn .

Hence (via �v0 + 1 = 0) we get

xf2(x) =

1Xn=0

(�vn+1 + 2n+1)xn+1 =

1Xm=1

(�vm + 2m)xm

=

1Xm=0

(�vm + 2m)xm = �f(x) + 1

1� 2x;

that is, 0 = xf2(x) + f(x)� 1

1� 2x,

from which it follows that

f(x) =1

2x

�1�

s1 + 2x

1� 2x

!.

Because of f(0) = 1 we have to use the positive root. Thus

f(x) =1

2x

��1 + (1 + 2x)(1� 4x2)�1=2

�.

It is known that

(1� 4z)�1=2 =

1Xk=0

�2k

k

�zk .

61

Hence

f(x) =1

2x

�1 + (1 + 2x)

1Xk=0

�2k

k

�x2k

!

=

1Xk=0

�2k

k

�x2k +

1Xk=1

1

2

�2k

k

�x2k�1 ;

that is

v2k =

�2k

k

�and v2k�1 =

1

2

�2k

k

�=

�2k � 1

k � 1

�,

showing vn =�

n

bn=2c�, as claimed.

Also solved by MICHEL BATAILLE, Rouen, France; PAUL BRACKEN, CRM, Univer-sit �e de Montr �eal, Montr �eal, Qu �ebec; CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; DAVID DOSTER,Choate Rosemary Hall, Wallingford, Connecticut, USA; FLORIAN HERZIG, student, Cambridge,UK; MICHAEL LAMBROU, University of Crete, Crete, Greece; KEE-WAI LAU, Hong Kong;HEINZ-J�URGEN SEIFFERT, Berlin, Germany; and the proposer. There was one incompletesolution submitted.

Dimminie remarks that if fCng is the sequence of Catalan numbers de�ned by C0 = 1

and

Cn+1 =

nXk=0

CkCn�k

for n � 0, then the solution may also be expressed as

v2n = (n + 1)Cn for n � 0

and

v2n�1 = (2n� 1)Cn�1 for n � 1 .

Janous also observes that by using

(1 � 4z)1=2 = �1Xk=0

�2kk

� zk

2k � 1

we get

f(x) =1

2x

�1 +

(1� 4x2)1=2

1� 2x

!

= � 1

2x

1 +

1Xk=0

�2kk

� x2k

2k � 1�1X`=0

2`x`

!

= � 1

2x

241 + 1X

m=0

0@ X2k+`=m

�2kk

� 2`

2k � 1

1A xm

35

= �1Xm=1

0@ X2k+`=m

�2kk

� 2`�1

2k � 1

1A xm�1

= �1Xn=0

0@ X2k+`=n+1

�2kk

� 2`�1

2k � 1

1A xn .

62

Hence we get the curious (and apparently new) identity [compare coe�cients and replace ` byn + 1� 2k]:

b(n+1)=2cXk=0

�2kk

�2n+1�2k�12k � 1

= �� n

bn=2c�,

which simpli�es to

b(n+1)=2cXk=0

�2kk

� 1

4k(2k � 1)= � 1

2n

� n

bn=2c�.

2310. [1998: 47] Proposed by K.R.S. Sastry, Dodballapur, India.

Let n 2 N. I call a positive integral divisor of n, say d, a unitary divisorif gcd(d; n=d) = 1.

Let �(n) denote the sum of the unitary divisors of n.

Find a characterization of n so that �(n) � 2(mod 4).

Solution by Mansur Boase, student, Cambridge, England.

Suppose n = p�11 p�22 : : : p�rr, where pi are distinct primes and �i > 0

for all i, 1 � i � r. Then

�(n) = (1 + p�11 )(1 + p�22 ) : : : (1 + p�rr) ,

as each term when this is multiplied out is a unitary divisor of n and everyunitary divisor of n occurs exactly once in the expansion.

If�(n) � 2 (mod 4), then it must be even, but not divisible by 4. Forodd pi, the term (1 + p�i

i) is always even, so at most one odd prime can

divide n and n must be of the form 2mpk.

Now �(n) � 2 (mod 4) if and only if (1 + pk) � 2 (mod 4). Ifp � 1 (mod 4), this will be the case for any k; if p � 3 (mod 4), then kmust be even.

Therefore, n = 2mpk with m � 0, p an odd prime, with k an arbi-trary positive integer if p � 1 (mod 4) and k an arbitrary even integer ifp � 3 (mod 4).

Also solved by GORAN CONAR, student, Gymnasium Vara�zdin, Vara�zdin, Croatia;NIKOLAOS DERGIADES, Thessaloniki, Greece; FLORIAN HERZIG, student, Perchtoldsdorf,Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursuli-nengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete, Greece;KEE-WAI LAU, Hong Kong; GERRY LEVERSHA, St. Paul's School, London, England; MICHAELPARMENTER, Memorial University of Newfoundland, St. John's, Newfoundland; ROBERTP. SEALY, Mount Allison University, Sackville, New Brunswick; HEINZ-J�URGEN SEIFFERT,Berlin, Germany; DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA;KENNETH M. WILKE, Topeka, Kansas, USA; and the proposer. There were two incorrect solu-tions submitted.

63

2311. [1998: 47] Proposed by K.R.S. Sastry, Dodballapur, India.

Let �e(n) denote the sum of the even unitary divisors, and�o(n), thesum of the odd unitary divisors, of n. Assume that �e(n)��o(n) = n.

(a) If n is composed of powers of exactly two distinct primes, show that nmust be the product of two consecutive integers, one of which is a Mersenneprime.

(b) Give an example of a natural number n that is composed of powers ofmore than two distinct primes.

Solution by Kenneth M. Wilke, Topeka, Kansas, USA.

For n = 2dpe11 pe2

2 : : : pekk

where d � 1 and each pi denotes an oddprime which divides n, we have

�o(n) =

kYi=1

(peii+ 1) and �e(n) = 2d

kYi=1

(peii+ 1) ,

so

n = �e(n)��o(n) = (2d � 1)

kYi=1

(peii+ 1) . (1)

(a) Here i = 1 so that [writing p for p1 and e for e1]

n = 2dpe = (2d � 1)(pe + 1)

and hence

2d = pe + 1 (2)

where d and e are positive integers. Since p � 3, we must have e � 1 andd � 2. Suppose e is even, say e = 2l. Then p2l � 1 mod 4 regardlessof the choice of the odd prime p and the integer l. Then since 2d � 0 mod 4

for all integers d � 2, we have that

pe + 1 = p2l + 1 � 2 6� 0 � 2d mod 4 ,

which is impossible from (2). Hence e must be odd, say e = 2l + 1. Thenfrom (2) we have

2d = (p+ 1)(p2l � p2l�1 + � � � � p+ 1) .

Since the second factor on the right hand side is odd, equality can be main-tained only if l = 0 and p = 2d � 1, a Mersenne prime.

(b) From (1) and our choice of n,

1 =

2d � 1

2d

!�pe11 + 1

pe11

��pe22 + 1

pe22

�: : :

�pekk

+ 1

pekk

�.

64

We can use the values d = 3, (p1; e1) = (5; 2), (p2; e2) = (7; 2) and(p3; e3) = (13; 1) to �nd

n = 23 � 52 � 72 � 13 = 127400

as a solution for part (b).

Editorial remark. The above solution was also found by solver Hessand by the proposer. The smallest solution to part (b), and the one mostsolvers gave, is n = 180. Other solutions found are

n = 441000 (found only by the proposer)

andn = 2646000 (found by Hess and Leversha).

Are there any others?

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; GORANCONAR, student, Gymnasium Vara�zdin, Vara�zdin, Croatia; NIKOLAOS DERGIADES, Thes-saloniki, Greece; FLORIAN HERZIG, student, Cambridge, UK; RICHARD I. HESS, RanchoPalos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus-tria; MICHAEL LAMBROU, University of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul'sSchool, London, England; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA; JOELSCHLOSBERG, student, Bayside, NY, USA; HEINZ-J�URGEN SEIFFERT, Berlin, Germany;DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA; and the proposer.

In part (a), one solver obtained equation (2) and then concluded rather hastily that e = 1

and so p is a Mersenne prime. Another solver believed (2) to imply that e = 1 but admitted hehad no proof. Both have been given the bene�t of the doubt (in the spirit of the season | thisis being written during the Christmas holidays). Other solvers merely said that this is a knownresult, or proved it themselves (as Wilke did), or gave a reference. For example, Lambroumentioned the article \A note on Mersenne numbers", by Ligh and Neal, on pp. 231{233 ofMath. Magazine 47 (1974), and Manes quoted problem E1221 of the Amer. Math. Monthly,solution in Volume 64 (1957), p. 110. The solution of the Monthly problem also contains somemuch earlier references to this result.

Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell

Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil

Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,

J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

65

Congratulations, Andy Liu!

It is with great pleasure that we congratulate Andy Liu on yet another

award, con�rming the high regard with which he is held in the international

mathematical community. CRUX with MAYHEM is very proud to have had

Andy Liu on its Editorial Board for many years.

The following can be seen on the World Wide Web at

http://www.case.org/awards/canpoy.htm

The Council for Advancement and Support of Education [USA]

and

The Canadian Council for Advancement and Support of Education

1998-1999 CANADIAN PROFESSORS OF THE YEAR

OUTSTANDING UNIVERSITY PROFESSOR

66

Andy Liu joined the faculty of the University of Alberta Mathematical

Sciences department in 1980 and has been making a profound impact on the

institution and its students ever since. His passions include a commitment

to the study of mathematics and to developing innovative techniques that

allow him to share his knowledge with students of all ages. \Students must

not settle into passive learning but must be challenged to participate in the

process," explains Liu.

Liu's success stems from his unique knack for presenting di�cult con-

cepts in a clear and logical manner. Former student William Willette said,

\I have not been inspired to think and achieve by anyone more than Liu. He

is not the type of instructor who just gives answers all the time; he inspires

students to think." By providing practical examples of theoretical concepts,

Liu helps his students understand and learn rather than simply memorize.

\Those of us working in the area of education of gifted and talented chil-

dren have long considered Liu to be a resource par excellence in mathematics

education," said a colleague, Carolyn R. Yewchuk.

In the classroom, Liu uses his lively sense of humour to maintain stu-

dents' interest. His pleasant demeanour creates a comfortable learning envi-

ronment. \In every class, I know the names of all the students by the time of

the midterm test," says Liu. He has even mastered the art of writing upside

down so students can follow his written explanationswhile meeting with him

at his desk.

Liu extends himself beyond the campus in a variety of ways. His on-

and o�-campus lectures as well as the courses he has designed, re ect his

extraordinary talent. Liu is a strong supporter of mathematical competi-

tions as a way to motivate and promote interest in mathematics. He has

participated in mathematical competitions on the local, provincial, national,

and international levels. He has provided training sessions for University of

Alberta undergraduate students, prepared training materials for a number of

national Mathematical Olympiad teams, and conducted training sessions for

the International Mathematical Olympiad (IMO) teams of Australia, Canada,

Hong Kong, Taiwan and the United States. In 1995, he received the IMO

Certi�cate of Appreciation and in 1996 received the David Hilbert Interna-

tional Award for the promotion of mathematics worldwide. Liu explains,

\I feel that the University is an integral part of the community, and its in-

volvement must extend beyond the con�nes of the campus. Also, learning

is a universal endeavour which transcends political boundaries, and the uni-

versity is �rst and foremost an international institution."

67

THE ACADEMY CORNERNo. 23

Bruce Shawyer

All communications about this column should be sent to Bruce

Shawyer, Department of Mathematics and Statistics, Memorial University

of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

Here is the �rst of four articles from the 1998 Canadian Undergraduate Math-

ematics Conference, held at the University of British Columbia in July 1998.

Abstracts � R�esum�es

Canadian Undergraduate Mathematics Conference

1998 | Part 1

Tracer Kinetic Modelling in Dynamic Positron Emission Tomography

Mark Andermann

McGill University

Dynamic positron emission tomography (PET) has often been used in the quan-

titative estimation of physiological parameters in vivo. For example, tracer kinetic

modelling of time-series data has been employed on a voxel by voxel basis using

various methods of non-linear regression analysis. However, due to the low signal-

to-noise ratio and large data sets present in PET studies, such methods are often

unstable and computationally intensive. In this lecture, mathematical theory under-

lying current and past methods which have attempted to increase the precision of

tracer kinetic modelling will be reviewed. Subsequently, a novel technique involving

principal components analysis, split-and-merge image segmentation using hypothe-

sis testing with both heuristic and statistical likelihood tests, as well as some aspects

of spectral analysis will be introduced. Bene�ts and disadvantages of this method

will then be discussed.

You Can Disprove That?

Shabnam Beheshti

McGill University

Prove or disprove the following: If two sets A and B of reals are homeomor-

phic, then they are both dense (respectively meagre). How many times have you

been faced with a question like this on a �nal? If you have experienced the wrath

68

of a professor who enjoys giving \rigorous multiple choice" examinations, then per-

haps the importance of studying and constructing counterexamples has been revealed

already. Every student of mathematics is faced with a choice: either conjuring up

a creative counterexample or waiting for divine intervention (the latter sometimes

takes awhile). Many times, we implicitly assume intuitively sound but analytically

false properties of our system, forcing us to draw incorrect conclusions; the power

of counterexample is thus discovered. Such an error hindered progress towards the

proof of Fermat's Last Theorem. I will attempt to survey the search for a proof to

the theorem of Fermat and present a variation of Kummer's counterexample to the

purported proof due to Lam�e.

Trigonometry, Astronomy, and Computation:

The Historical Quest for an Elusive Constant

Abraham Buckingham

The King's University College

Ancient astronomy gave rise to some of the best historical mathematics, es-

pecially in the realm of trigonometry. Accurate trigonometric tables were critical

to good astronomy, and accurate trigonometric tables relied on an accurate value of

the sine of one degree for the majority of values. Unfortunately an exact value for

this sine is impossible to �nd using ruler and compass methods alone. This prob-

lem was worked on by Claudius Ptolemy (100-175 AD) and Jamsh��d al-K�ash�� (?-1429

AD). Ptolemy's geometric method stood for 1200 years, but al-K�ash�� was not satis-

�ed and constructed an improved geometric estimation for the sine of one degree.

Still not satis�ed, al-K�ash�� went on to develop a �xed point iteration scheme for the

calculation. I will describe and compare these methods within their context.

Invariants and shoelaces

Benoit Charbonneau

Universit �e du Qu�ebec �a Montr �eal

In the Montreal phone directory, there are seven entries for Jacques Labelle.

If X's name is Jacques Labelle, we need more information to identify him. But we

know for sure he is not Nicolas Bourbaki (although this is a subtle question).

One eternal question of mathematics is Are two objects in fact the same object?

Up to what we call isomorphisms, some properties of objects don't change. We

call them invariants. After this talk, you won't look at life the same way!

Invariants et lacets de chaussures

Benoit Charbonneau

Universit �e du Qu�ebec �a Montr �eal

Dans le bottin t �el �ephonique de Montr �eal, il y a sept entr �ees pour Jacques La-

belle. Si le nom de X est Jacques Labelle, on a besoin de plus d'information pour

l'identi�er. On est cependant sur qu'il n'est pas Nicolas Bourbaki (quoique la ques-

tion est subtile).

Une question �eternelle des math �ematiques est Est-ce que deux objets sont

identiques ?

Certaines propri �et �es des objets ne changent pas �a isomorphisme pr �es, on les

appelle des invariants. Suite �a cet expos �e, vous ne verrez plus jamais la vie de la meme

fa�con !

69

Ranking the Participants in a Round-Robin Tournament

Susan Marie Cooper

University of Regina

Various schemes for ranking and comparing the participants in a round-robin

tournament have been proposed. However, none of these are considered entirely

satisfactory. We will consider the Kendall-Wei method, a method of ranking tourna-

ments using iterated strength vectors, which leads to the Perron vector (i.e. vector

of relative strengths) of the corresponding dominance matrix of the tournament. We

will investigate such questions as: \how easily can the iterated strength vectors be

calculated?" and \can we determine the relative ordering of the strengths of the par-

ticipants without calculating the Perron vector?"

What is the (57; 5)-cage?

Jennifer de Kleine

University of Northern British Columbia

An (n; 5)-cage is an n-regular girth-5 graph of smallest order. I will discuss

what this means, and the open question of whether there exists a (57; 5)-cage with

572 + 1 vertices. Time permitting, I will discuss the connection between estimating

the permanent of (0; 1) matrices and a possible approach to using a computer to

search for a 572 + 1 vertex (57; 5)-cage.

Les nombres de Stirling

Caroline Desjardins

Universit �e du Qu�ebec �a Montr �eal

Jacob Stirling, math �ematicien anglais du 18e si �ecle a publi �e un article int �eressant

\Methodus di�erentialis". L'article contient deux triangles de nombres entiers (ap-

parent �es au triangle de Pascal) dont les diverses composantes permettent d'obtenir

des r �esultats int �eressants en analyse, en combinatoire et en probabilit �es. Ce sont

ces nombres que l'on appelle les nombres de Stirling. On peut obtenir ces nombres

en trouvant les matrices de changement de base de deux base de R[[x]] ou bien,

en utilisant les relations de r �ecurrences que ces nombres suivent. Les liens reliant

ces nombres aux nombres de partition en k-classes et de fonction surjectives d'un

ensemble �a n �el �ements vers un ensemble �a k �el �ements, sont quelquesexemples d'ap-

plication des nombres de Stirling.

Applications of Group Theory to Chemistry

Norman Dreger

University of British Columbia

The goal of this paper is to present a number of chemical applications of group

theory. Some key terms will be de�ned and the concepts of symmetry elements and

symmetry groups introduced. The nomenclature of symmetry elements with regards

to chemistry will then be discussed. Finally some applications of symmetry to exper-

imental chemistry will be examined. Only a rudimentary knowledge of group theory

is prerequisite.

70

THE OLYMPIAD CORNERNo. 196

R.E. Woodrow

All communications about this column should be sent to Professor R.E.

Woodrow, Department of Mathematics and Statistics, University of Calgary,

Calgary, Alberta, Canada. T2N 1N4.

We lead o� this issue with the problems of the 19th Austrian-Polish

Mathematics Competitions, written in Poland, June 26{28, 1996. My thanks

go to Ravi Vakil, Canadian Team Leader to the IMO at Mumbai as well as to

regular supportersMarcin E. Kuczma, Warszawa, Poland andWalther Janous,

Ursulinengymnasium, Innsbruck, Austria for supplying copies of the contest

material.

19th AUSTRIAN-POLISH MATHEMATICSCOMPETITION 1996

Problems of the Individual ContextJune 26{27, 1996 (Time: 4.5 hours)

1. Let k � 1 be an integer. Show that there are exactly 3k�1 positive

integers n with the following properties:

(a) The decimal representation of n consists of exactly k digits.

(b) All digits of n are odd.

(c) The number n is divisible by 5.

(d) The numberm = n

5has k odd (decimal) digits.

2. A convex hexagon ABCDEF satis�es the following conditions:

(a) The opposite sides are parallel; that is, ABkDE, BCkEF , CDkFA.

(b) The distances between the opposite sides are equal; that is,

d(AB;DE) = d(BC;EF ) = d(CD;FA), where d(g; h) denotes the

distance between lines g and h.

(c) \FAB and \CDE are right angles.

Show that diagonals BE and CF intersect at an angle of 45�.

3. The polynomials Pn(x) are de�ned recursively by P0(x) = 0,P1(x) = x and

Pn(x) = xPn�1(x) + (1� x)Pn�2(x) for n � 2 .

For every natural number n � 1 �nd all real numbers x satisfying the equa-

tion Pn(x) = 0.

71

4. The real numbers x, y, z, t satisfy the equalities x+ y+ z + t = 0and x2 + y2 + z2 + t2 = 1. Prove that �1 � xy + yz+ zt+ tx � 0.

5. A convex polyhedron P and a sphere S are situated in space in

such a manner that S intercepts on each edge AB of P a segment XY with

AX = XY = Y B = 1

3AB. Prove that there exists a sphere T tangent to

all edges of P .

6. Natural numbers k, n are given such that 1 < k < n. Solve the

system of n equations

x3i� (x2

i+ x2

i+1 + � � �+ x2i+k�1) = x2

i�1 for 1 � i � n

with n real unknowns x1, x2, : : : , xn. Note: x0 = xn, xn+1 = x1,

xn+2 = x2, and so on.

Problems of the Team Contest (Poland)June 28, 1996 (Time: 4 hours)

7. Show that there do not exist non-negative integers k and m such

that k! + 48 = 48(k+ 1)m.

8. Show that there is no polynomial P (x) of degree 998 with real

coe�cients satisfying for all real numbers x the equation

P (x)2� 1 = P (x2 + 1) .

9. We are given a collection of rectangular bricks, no one of which is

a cube. The edge lengths are integers. For every triple of positive integers

(a; b; c), not all equal, there is a su�cient supply of a�b�c bricks. Supposethat the bricks are completely tiling a cubic 10� 10� 10 box.

(a) Assume that at least 100 bricks have been used. Prove that there exist

at least two bricks situated in parallel, in the sense that if AB is an edge of

one of them and A0B0 is an edge of one of the other, and if ABkA0B0, thenAB = A0B0.

(b) Prove the same statement for a number less than 100 (of bricks used).

The smaller number, the better the solution.

Next we move to a country whose contest materials have not been very

often available in CRUX withMAYHEMwith the problems of the 3rd Turkish

Mathematical Olympiad, Second Round, written December 8{9, 1995. My

thanks go to Ravi Vakil, Canadian Team Leader to the IMO at Mumbai for

collecting the problems.

72

3rd TURKISH MATHEMATICAL OLYMPIADSecond Round { First Day

December 8, 1995 (Time: 4.5 hours)

1. Let a1; a2; : : : ; ak and m1;m2; : : : ;mk be integers with 2 � m1

and 2mi � mi+1 for 1 � i � k � 1. Show that there are in�nitely many

integers x which do not satisfy any of the congruences

x � ai (mod m1) , x � a2 (mod m2) , : : : , x � ak (mod mk) .

2. For an acute triangle ABC, k1, k2, k3 are the circles with diam-

eters [BC], [CA], [AB], respectively. If K is the radical centre of these

circles, [AK] \ k1 = fDg, [BK] \ k2 = fEg, [CK] \ k3 = fFg and

Area(ABC) = u, Area(DBC) = x, Area(ECA) = y, and Area(FAB) = z,

show that u2 = x2 + y2 + z2.

3. Let N denote the set of positive integers. Let A be a real number

and fang1n=1 be a sequence of real numbers such that a1 = 1 and

1 <an+1

an� A for all n 2 N .

(a) Show that there is a unique non-decreasing surjective function k : N! N

such that 1 < Ak(n)

an� A for all n 2 N.

(b) If k takes every value at most m times, show that there exists a real

number C > 1 such that Cn � Aan for all n 2 N.

Second Round { Second DayDecember 9, 1995 (Time: 4.5 hours)

4. In a triangle ABC with jABj 6= jACj, the internal and external

bisectors of the angleA intersect the lineBC atD andE, respectively. If the

feet of the perpendiculars from a point F on the circle with diameter [DE] tothe linesBC,CA,AB areK, L,M , respectively, show that jKLj = jKM j.

5. Let t(A) denote the sum of elements of A for a non-empty subset

A of integers, and de�ne t(�) = 0. Find a subset X of the set of positive

integers such that for every integer k there is a unique ordered pair of subsets

(Ak; Bk) of X with Ak \ Bk = � and t(Ak)� t(Bk) = k.

6. Let N denote the set of positive integers. Find all surjective

functions f : N! N satisfying the condition

m j n (==) f(m) j f(n)

for allm, n 2 N.

73

Along with the Turkish Olympiad we have the questions of the Turkish

Team Selection Examination for the 37th IMO, written March 23{24, 1996.

Thanks again go to Ravi Vakil, Canadian Team Leader to the IMO at Mumbai

for forwarding these to me.

TURKISH TEAM SELECTION EXAMINATION FORTHE 37th IMO

First Day | March 23, 1996Time: 4.5 hours

1. LetQ1996

n=1(1 + nx3n) = 1 + a1xk1 + a2x

k2 + � � � + amxkm where

a1, a2, : : : , am are non-zero and k1 < k2 < � � � < km. Find a1996.

2. In a parallelogram ABCD withm(A) < 90�, the circle with diam-

eter [AC] intersects the lines CB and CD at E and F besides C, and the

tangent to this circle at A intersects the line BD at P . Show that the points

P , F , E are collinear.

3. Given real numbers 0 = x1 < x2 < � � � < x2n < x2n+1 = 1 with

xi+1 � xi � h for 1 � i � 2n, show that

1� h

2<

nXi=1

x2i(x2i+1 � x2i�1) �1 + h

2.

Second Day | March 24, 1996Time: 4.5 hours

4. In a convex quadrilateral ABCD, Area(ABC) = Area(ADC) and[AC] \ [BD] = fEg, and the parallels from E to the line segments [AD],[DC], [CB], [BA] intersect [AB], [BC], [CD], [DA] at the points K, L,

M , N , respectively. Compute the ratio

Area(KLMN)

Area(ABCD).

5. Find the maximum number of pairwise disjoint sets of the form

Sa;b = fn2 + an+ b : n 2 Zg with a, b 2 Z.

6. For which ordered pairs of positive real numbers (a; b) is zero the

value of the limit of every sequence fxng satisfying the condition

limn!1

(axn+1 � bxn) = 0 ?

74

To round out the contests for your puzzling pleasure we give the two

papers of the Australian Mathematical Olympiad 1996. My thanks go to

Ravi Vakil, Canadian Team Leader of the IMO at Mumbai, once again, for

providing me with the contest materials.

AUSTRALIANMATHEMATICAL OLYMPIAD 1996Paper 1

February 6, 1996 (Time: 4 hours)

1. Let ABCDE be a convex pentagon such that BC = CD = DE

and each diagonal of the pentagon is parallel to one of its sides. Prove that

all the angles in the pentagon are equal, and that all sides are equal.

2. Let p(x) be a cubic polynomial with roots r1, r2, r3. Suppose that

p�1

2

�+ p

��1

2

�p(0)

= 1000 . Find the value of1

r1r2+

1

r2r3+

1

r3r1.

3. A number of tubes are bundled together into a hexagonal form:

f f f f

f f f f f

f f f f f f

f f f f f f f

f f f f f f

f f f f f

f f f f

A number of tubes in the bundle can be 1, 7, 19, 37 (as shown), 61, 91; : : : .If this sequence is continued, it will be noticed that the total number of tubes

is often a number ending in 69. What is the 69th number in the sequence

which ends in 69?

4. For which positive integers n can we rearrange the sequence

1, 2, : : : , n to a1, a2, : : : , an in such a way that jak � kj = ja1 � 1j 6= 0for k = 2, 3, : : : , n?

Paper 2February 7, 1996 (Time: 4 hours)

5. Let a1, a2, : : : , an be real numbers and s a non-negative real

number such that

(i) a1 � a2 � � � � � an;

(ii) a1 + a2 + � � �+ an = 0;

(iii) ja1j+ ja2j+ � � �+ janj = s.

Prove that

an � a1 �2s

n.

75

6. Let ABCD be a cyclic quadrilateral and let P and Q be points on

the sidesAB andAD respectively such that AP = CD and AQ = BC. Let

M be the point of intersection ofAC andPQ. Show thatM is the mid-point

of PQ.

7. For each positive integer n, let �(n) denote the sum of all positive

integers that divide n. Let k be a positive integer and n1 < n2 < � � � be an

in�nite sequence of positive integers with the property that �(ni)� ni = k

for i = 1, 2, : : : . Prove that ni is a prime for i = 1, 2, : : : .

8. Let f be a function that is de�ned for all integers and takes only the

values 0 and 1. Suppose f has the following properties:

(i) f(n+ 1996) = f(n) for all integers n;

(ii) f(1) + f(2) + � � �+ f(1996) = 45.

Prove that there exists an integer t such that f(n+ t) = 0 for all n for which

f(n) = 1 holds.

Now, an alternate and more general solution to problem 2 of the Dutch

Mathematical Olympiad, Second Round, 1993 than the one given in the Cor-

ner in the October 1998 number [1997: 197], [1998: 330].

2. Given a triangle ABC, \A = 90�. D is the mid-point of BC, F

is the mid-point of AB, E the midpoint of AF andG the mid-point of FB.

AD intersects CE, CF and CG respectively in P , Q and R. Determine the

ratio PQ

QR.

A E F G B

C

P QR

D

q

q

q

q

q

qq

q q q

Alternate Solutionby Geo�rey A. Kandall, Hamden, Connecticut,USA.

We �rst establish the following:

Lemma.PQ

QR=

CP

CE�EF

FG�CG

CR.

Proof.

PQ

QR=

[CPQ]

[CQR]=

[CPQ]

[CEF ]�[CEF ]

[CFG]�[CFG]

[CQR]

=CP � CQCE � CF

�EF

FG�CF � CGCQ �CR

=CP

CE�EF

FG�CG

CR.

76

We now solve the problem, without using the hypothesis that \A = 90�.

By the lemma

PQ

QR=

CP

CE�EF

FG�CG

RC=

CP

CE�CG

RC.

By Menelaus' Theorem we have

CD

DB�BA

AE�EP

PC= 1 , hence

EP

PC=

1

4,

CP

CE=

4

5; (1)

CD

DB�BA

AG�GR

CR= 1 , hence

GR

CR=

3

4,

CG

CR=

7

4. (2)

ConsequentlyPQ

QR=

4

5�7

4=

7

5.

This method can be used with di�erent ratiosCD : DB andAE : EF :FG : GB.

After the February number was �nalized we received a package of so-

lutions from Michael Selby, University of Windsor, Windsor, Ontario. This

included solutions to problems 1 through 4 of the Croatian National Math-

ematics Competition (4th Class) May 13, 1994 for which the problems were

given [1997: 454] and the solutions [1999: 12]. He also sent a solution to

a problem of the Additional Competition for the Olympiad of the Croatian

National Mathematical Competition, given [1997: 454].

1. Find all ordered triples (a; b; c) of real numbers such that for every

three integers x, y, z the following identity holds:

jax+ by+ czj+ jbx+ cy + azj+ jcx + ay + bzj = jxj + jyj+ jzj .

Solution by Michael Selby, University of Windsor, Windsor, Ontario.

Set x = y = z = 1; we obtain ja+ b+ cj = 1 (1)

Set x = 1; y = z = 0 we obtain jaj + jbj+ jcj = 1 (2)

Set x = 1; y = �1, z = 0 we obtain ja� bj+ jb� cj+ jc� aj = 2 (3)

This system is symmetric. Without loss of generality we may assume

a � b � c.

Now (3) becomes 2(a� c) = 2 or a� c = 1. Substituting into (1) and

(2) gives

j1 + b+ 2cj = 1 (4)

77

and

j1 + cj+ jbj+ jcj = 1 . (5)

Squaring (4) and expanding gives

1 + (b+ 2c)2 + 2(b+ 2c) = 1 .

Thus b+ 2c = 0 or b+ 2c = �2.

If b+ 2c = 0, then from (5)

j1 + cj+ 3jcj = 1 .

Since jcj � 1, 1 + c � 0, therefore 1 + c + 3jcj = 1 and c + 3jcj = 0. If

c � 0, we have 4c = 0 and then c = 0. If c � 0, �2c = 0 giving c = 0.Therefore b = �2c = 0, a = 1+ c = 1, in this case.

In case b+ 2c = �2, substitution into (5) yields

j1 + cj+ 2j1 + cj+ jcj = 1 .

Since 1 + c � 0, 3(1 + c) + jcj = 1. If c � 0, 3 + 4c = 1 and c = �1

2. This

is impossible.

If c � 0, 3+3c� c = 1 giving c = �1. Then b = 0 and a = 1+ c = 0.Therefore we have the solution a = 0, b = 0, c = �1, and these are the

solutions for a � b � c.

Hence there are six solutions

(1; 0; 0) , (�1; 0; 0) , (0; 1; 0) , (0;�1; 0) , (0; 0; 1) , (0;0;�1) .

Next we turn to solutions by the readers to problems of the 17th Austrian-

PolishMathematics Competition given in the February 1998number [1998: 4].

17th AUSTRIAN{POLISH MATHEMATICSCOMPETITION

Poland, June 29{July 1, 1994

1. The function f : R! R satis�es for all x 2 R the conditions

f(x+ 19) � f(x) + 19 and f(x+ 94) � f(x) + 94 .

Show that f(x+ 1) = f(x) + 1 for all x 2 R.

78

Solutions by Michel Bataille, Rouen, France; by Pierre Bornsztein,

Courdimanche, France; by Pavlos Maragoudakis, Pireas, Greece; and by

Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give

the solution by Bataille.

Let x be an arbitrary real number. Applying the given conditions to

x� 19 and x� 94 respectively, we obtain

f(x� 19) � f(x)� 19 and f(x� 94) � f(x)� 94 .

Now an easy induction shows that for all n 2 N,

f(x+ 19n) � f(x) + 19n , f(x+ 94n) � f(x) + 94n ,

f(x� 19n) � f(x)� 19n , and f(x� 94n) � f(x)� 94n .

Since 1 = 5� 19� 94 and 1 = 18� 94� 89� 19, we get:

f(x+ 1) = f(x+ 5� 19� 94) � f(x+ 5� 19)� 94

� f(x) + 5� 19� 94

= f(x) + 1 ,

and

f(x+ 1) = f(x+ 18� 94� 89� 19) � f(x+ 18� 94)� 89� 19

� f(x) + 18 + 94� 89� 19

= f(x) + 1 ,

so that f(x+ 1) = f(x) + 1, as required.

Comment: the same result can be obtained from the more general hy-

pothesis: for all x 2 R, f(x + a) � f(x) + a and f(x + b) � f(x) + b

where a and b are positive relatively prime integers. Indeed, the preced-

ing proof adapts easily as we can �nd positive integers m, n, p, q such that

ma� nb = 1 and pb� qa = 1.

2. The sequence fang is de�ned by the formulae

a0 =1

2and an+1 =

2an

1 + a2n

for n � 0 ,

and the sequence fcng is de�ned by the formulae

c0 = 4 and cn+1 = c2n� 2cn + 2 for n � 0 .

Prove that

an =2c0c1 : : : cn�1

cnfor all n � 1 .

Solutions by Michel Bataille, Rouen, France; by Pierre Bornsztein,

Courdimanche, France; by Murray S. Klamkin, University of Alberta, Ed-

monton, Alberta. We give the solution of Klamkin, which gives an indication

of both types of solutions received.

79

Letting xn = cn � 1, we have xn+1 = x2nwhere x0 = 3. Hence,

xn = x2n

0 and cn = 32n

+ 1. Since c1 = 10 and a1 = 4

5it now su�ces to

show that an =2c0c1:::cn�1

cnsatis�es the recurrence an+1 = 2an

1+a2n

for n � 0.

Also since (32n

+1)(32n � 1) = 32

n+1 � 1, it follows (multiplying by 320�1

320�1

)

that2c0c1 : : : cn�1

cn=

32n � 1

32n

+ 1

and by substitution and simpli�cation, this satis�es the recurrence relation

for an.

Comment: We can obtain another representation for an by letting it

equal tanh �n, so that tanh �n+1 = tanh2�n, subject to1

2= tanh �0. It

then follows that an = tanh2n�0 = tanh�2n arctan h1

2

�= tanh

�2n�1 ln 3

�.

4. Let n � 2 be a �xed natural number and let P0 be a �xed vertex

of the regular (n+ 1){gon. The remaining vertices are labelled P1, P2, : : : ,

Pn, in any order. To each side of the (n+1){gon assign a natural number as

follows: if the endpoints of the side are labelled Pi and Pj , then ji� jj is thenumber assigned. Let S be the sum of all the n+ 1 numbers thus assigned.

(Obviously, S depends on the order in which the vertices have been labelled.)

(a) What is the least value of S available (for �xed n)?

(b) How many di�erent labellings yield this minimum value of S?

Solution by Pierre Bornsztein, Courdimanche, France.

Pi2

Pi1

P0

Pn

Pik

(a) Soit P0Pn

l'arc r �eliant P0 �a Pn dans le sens des aiquilles d'une

montre, P0Pn�l'arc r �eliant P0 �a Pn dans le sens contraire.

Notons S� la somme des nombres assign �es sur P0Pn(idem pour S+).

Par d �e�nition,

S� = j0� i1j+ ji1 � i2j+ � � �+ jik�1 � ikj+ jik � nj� j0� i1 + i1 � i2 + � � �+ ik�1 � ik + ik � nj = n

avec egalit �e ssi 0 � i1 � i2 < � � � � ik < n.

80

De meme,

S+ � n

avec egalit �e ssi les sommets sont class �es dans l'ordre croissant de 1 �a n, d'o �u

on en d �eduit S = S� + S+ � 2n.

(b) Pour Pn �x �e il y a i sommets entre P0 et Pn, le long de P0Pn

o �u

i 2 f0, : : : , n�1g. Il y a donc i nombres �a choisir dans f1, : : : ,n�1g, d'o �u�n�1

i

�choix.

Les nombres, une foix choisis, sont alors dispos �es dans l'ordre croissant

de P1 �a Pn : l'ordre est donc impos �e.

De meme sur P0Pn�les nombres restants sont impos �es ainsi que leur

ordre.

Il y a doncPn�1

i=0

�n�1

i

�= 2n�1 choix pour la disposition.

5. Solve the equation

1

2(x+ y)(y+ z)(z+ x) + (x+ y+ z)3 = 1� xyz

in integers.

SolutionsbyMichel Bataille, Rouen, France; by Pierre Bornsztein, Cour-

dimanche, France; and byMurray S. Klamkin, University of Alberta, Edmon-

ton, Alberta. We give the write-up of Bataille, although all three solvers used

the same approach.

Let s = x+ y+ z and

P (X) = (X � x)(X � y)(X � z)

= X3 � sX2 + (xy + yz+ zx)X � xyz .

Then (x+y)(y+ z)(z+x) = P (s) = s(xy+yz+xz)�xyz and the given

equation may be written

s(xy+ yz+ xz)� xyz + 2s3 = 2� 2xyz ,

or 2 + P (�s) = 0.

As P (�s) = �(2x + y + z)(2y + z + x)(2z + x + y), the equation

�nally becomes

(2x+ y + z)(2y+ z + x)(2z + x+ y) = 2 .

Either one of the three factors of the left-hand side is 2 and the other two

are 1, 1 (or �1, �1) or one of the factors is�2 and the other two are 1, �1,(or �1, 1).

The system8<:

2x+ y+ z = 2x+ 2y+ z = 1x+ y+ 2z = 1

is equivalent to x = 1 , y = 0 , z = 0 .

81

The system8<:

2x+ y + z = 2x+ 2y + z = �1x+ y + 2z = �1

is equivalent to x = 2 , y = �1 , z = �1 .

When one of the factors is �2, the two corresponding systems lead to

4(x+ y + z) = �2, which is impossible for integral x, y, z.

Since x, y, z have symmetrical roles, there are six solutions altogether

for the triple (x; y; z):

(1; 0; 0) , (0; 1; 0) , (0; 0; 1) , (2;�1;�1) , (�1; 2;�1) , (�1;�1; 2) .

7. Determine all two-digit (in decimal notation) natural numbers

n = (ab)10 = 10a + b (a � 1) with the property that for every integer

x the di�erence xa � xb is divisible by n.

Solutions by Pierre Bornsztein, Courdimanche, France; and by Edward

T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Wang's

solution.

Clearly, n j xa � xb for all integers x if a = b. We show that besides

11, 22, : : : , 99 there are exactly three more such n's. These are: n = 15, 28,and 48. We assume that a 6= b and start o� by eliminating some impossible

values of n.

(1) If a is even and b is odd, then setting x = �2 leads to n j 2a � 2b andn j 2a+2b. Thusn j 2a+1, which is clearly impossible since the only possible

divisors of 2a+1 are powers of two while n > 1 is odd.

(2) If a is odd and b is even, then setting x = �2 again leads to the same

conclusion that n j 2a+1. Hence n must be a power of two. Since a is odd,

the only possible values are n = 16 and 32. However, 16 6 j 2 � 26 and

32 6 j 23 � 22, showing that there are no solutions in this case either.

(3) If b = 0, then n is even and n j 2a � 1, which is clearly impossible.

Using (1), (2), and (3) we narrow the possible values of n down to the

following set of 32 integers:

f13 , 15 , 17 , 19 , 24 , 26 , 28 , 31 , 35 , 37 , 39 , 42 , 46 , 48 , 51 , 53; 57 ,

59 , 62 , 64 , 68 , 71 , 73 , 75 , 79 , 82 , 84 , 86 , 91 , 93 , 95; 97g .Since n j xa � xb if and only if n j xb � xa we may assume that a > b when

checking whether n satis�es the given property. Note that

23 � 2 = 6 eliminates 13 and 31 ;24 � 22 = 12 eliminates 24 and 42 ;25 � 2 = 30 eliminates 51 (but not 15) ;25 � 23 = 24 eliminates 35 and 53 ;26 � 22 = 60 eliminates 26 and 62 ;26 � 24 = 48 eliminates 46 and 64 ;

82

27 � 2 = 126 eliminates 17 and 71 ;27 � 23 = 120 eliminates 37 and 73 ;27 � 25 = 96 eliminates 57 and 75 ;28 � 22 = 252 eliminates 82 (but not 28) ;28 � 24 = 240 eliminates 84 (but not 48) ;28 � 26 = 192 eliminates 68 and 86 ;29 � 2 = 510 eliminates 19 and 91;29 � 23 = 504 eliminates 39 and 94 ;29 � 25 = 480 eliminates 59 and 95 ;29 � 27 = 384 eliminates 79 and 97 .

Therefore, the only possible values of n are: n = 15, 28 and 48. We now

show that they indeed satisfy the condition that n j xa�xb for all integers x.

(a) For n = 15, we show that x � x5 (mod 15). By Fermat's Little Theorem

(Fthm), we have x3 � x (mod 3) and so x5 � x3 � x (mod 3). Also,

x5 � x (mod 5). Hence x5 � x (mod 15) follows.

(b) For n = 28, we show that x2 � x8 (mod 28). Note that 28 = 22 � 7.By Fthm, we have x7 � x (mod 7) and so x8 � x2 (mod 7). Further, we

claim that x8 � x2 (mod 4). This is obvious if x is even. On the other

hand, if x is odd, then x2 � 1 (mod 4) implies x8 � 1 (mod 4) and so

x8 � x2 (mod 4). Hence x8 � x2 (mod 28) follows.

(c) For n = 48, we show that x4 � x8 (mod 48). Note that 48 = 24 � 3.By Fthm, we have x3 � x (mod 3) and so x4 � x2 (mod 3). Hence

x8 � x4 (mod 3). It remains to show that 16 j x8 � x4. This is clear if

x is even. If x is odd, then x = 2k+ 1 for some integer k and thus

x8 � x4 = x4(x2 � 1)(x2 + 1)

= (2k+ 1)4�4k2 + 4k

� �4k2 + 4k+ 2

�= 8k(k+ 1)

�2k2 + 2k+ 1

�(2k+ 1)

4,

which is divisible by 16 since k(k+ 1) is even.

To summarize, n = 10a + b satis�es n j xa � xb for all integers x if

and only if n = 11, 22, : : : , 99, 15, 28, 48.

Comment: This is one of the most intriguing problems that I have seen

lately. I will be really surprised if there is a much shorter solution!

8. Consider the functional equation f(x; y) = a f(x; z) + b f(y;z)with real constants a, b. For every pair of real numbers a, b give the general

form of functions f : R2 ! R satisfying the given equation for all x, y,

z 2 R.

Solution by Pierre Bornsztein, Courdimanche, France.

Soient a, b 2 R et pour tous x ,y ,z 2 R

f(x; y) = af(x; z) + bf(y;z) . (�)

Alors :

83

Dans le cas o �u x = y = z, f(x; x) = (a + b)f(x;x) donc a + b = 1 ou

f(x; x) = 0. Si a+ b 6= 1, pour tout x 2 R, f(x;x) = 0 et donc pour z = y,

(�) donne

f(x; y) = af(x; y) + bf(y;y) = af(x; y) .

Donc soit a = 1 ou f(x; y) = 0.

Dans le cas ou a = 1

f(x; y) = f(x; z) + bf(y;z) ,

observons qu'avec x = y, f(x;x) = 0 = f(y;z)(1 + b), et donc f � 0 ou

b = �1.

Maintenent si a = 1 et b = �1

f(x; y) = f(x; z)� f(y;z)

ou encore

f(x; z) = f(x; y) + f(y;z)

pour tous x, y, z 2 R.

C'est a dire

f(x; y) = f(x; z) + f(z; y)

pour tous x, y, z 2 R, et donc f(z; y) = �f(y; z). On pose f(x; 0) = g(x),alors f(0; x) = �g(x) et

f(x; y) = f(x; z) + f(z; y)

= f(x; 0) + f(0; y)

= g(x)� g(y) .

Reciproquement, f(x; y) = g(x)� g(y) o �u g est une fonction arbitraire.

Alors f(x; y) = f(x; z) + f(z; y), et f convient.

Dans le cas o �u a+ b = 1, b = 1� a, et (�) s' �ecrit

f(x; y) = af(x; z) + (1� a)f(y;z) , (��)

et alors f(x;x) = f(x; z) et donc pour tous x, y 2 R, f(x; y) = f(x;x).Maintenant (��) donne

f(x;x) = af(x; x) + (1� a)f(y;y) ,

et par cons �equence

(1� a)f(x;x) = (1� a)f(y;y) .

Deux possibilit �es se pr �esentent. Soit a = 1 ou f(x;x) = f(y;y) = f(x; y),et f est constante. Si a = 1, b = 0, alors f(x; y) = f(x; z) pour tous x,

84

y, z 2 R. Donc f(x; y) = f(x; x) ind �ependant de y. On a veri� �e que ces

fonctions conviennent.

En conclusion :

� si (a; b) = (1;�1), f(x; y) = g(x)� g(y) o �u g : R! R est arbitraire ;

� si a+ b 6= 1 et (a; b) 6= (1;�1), f � 0 ;

� si a+ b = 1 et a 6= 1 : f constante ;

� si (a; b) = (1; 0) : f(x; y) = g(x) pour tous x; y 2 R o �u g : R ! R est

arbitraire.

9. On the plane there are given four distinct points A, B, C, D lying

(in this order) on a line g, at distances AB = a, BC = b, CD = c.

(a) Construct, whenever possible, a point P , not on g, such that the angles

\APB, \BPC, \CPD are equal.

(b) Prove that a point P with the property as above exists if and only if the

following inequality holds: (a+ b)(b+ c) < 4ac.

Solution by Michel Bataille, Rouen, France.

(a) If P is a solution, then the lines PB and PC are interior bisectors

in 4APC and 4BPD respectively. Hence we have:PA

PC=

BA

BC

andPB

PD=

CB

CDand P is simultaneously on E1 =

�M :

MA

MC=a

b

�and

E2 =

�M :

MB

MD=b

c

�.

In the general case where a 6= b, denoting byB0 the harmonic conjugate

of B with respect to A and C, E1 is the circle with diameter BB0 and, when

a = b, E1 is the perpendicular bisector of the segment AC. Similar results

hold for E2.

Conversely, we may construct E1 and E2 and, assuming that they are

secant, choose for P one of their two distinct points of intersection symmet-

rical about g. FromPA

PC=BA

BC, we deduce that PB is one of the bisectors

of \APC, more precisely the interior bisector in4APC sinceB is between

A and C. Hence \APB = \BPC. Similarly \BPC = \CPD and �nally:

\APB = \BPC = \CPD.

(b) The above construction provides a point P solution whenever E1 and E2

are secant. We �rst examine the general case where a 6= b and b 6= c: E1 and

E2 are circles with centres I1, I2 and radii r1, r2 respectively. These circles

are secant if and only if:

jr1 � r2j < I1I2 < r1 + r2 (1)

Let us denote by k the real number such thatBI1 =k

bBC (so that jkj = r1).

85

We may compute: I1A = �k + a

bBC and I1C =

b� k

bBC, and from the

Newton's relation, I1B2= I1A � I1C, we obtain easily k =

ab

a� b, so that

r1 =ab

ja� bj. Similarly: r2 =

cb

jc� bj.

We also compute: I1I2 =b2 � ac

(b� a)(b� c)BC so that I1I2 =

bjb2 � acjjb� aj jb� cj

.

The condition (1) may now be successively written:

jcja � bj � ajc� bj j < jb2 � acj < ajc � bj+ cja� bj

a2(c� b)2 + c2(a� b)2 � 2acja� bj jc � bj < (b2 � ac)2

< a2(c� b)2 + c2(a� b)2 + 2acja � bj jc � bj

j(b2 � ac)2 � a2(c� b)2 + c2(a� b)2j < 2acja � bj jc� bj

ja� bj jc� bj jb2 + b(a+ c)� acj < 2acja � bj jc� bj

�2ac < b2 + b(a+ c)� ac < 2ac

�ac < b2 + b(a+ c) < 3ac .

Since b2 + b(a + c) is positive, the latter condition is equivalent to

b2 + b(a+ c) < 3ac or (a+ b)(b+ c) < 4ac.

E1 and E2 are both lines when a = b = c, but in this case they are

strictly parallel so that no point P exists (and the condition

(a+ b)(b+ c) < 4ac is not true either).

Lastly, suppose for instance that E1 is a line and E2 is a circle (that is,

a = b and b 6= c). Since E1 is perpendicular to g at B, E1 and E2 are secant

if and only if I2B < r2. We obtain easily: I2B =b2

jc � bjand the condition

becomes: b < c (and the inequality (a+ b)(b+ c) < 4ac reduces to b < c

as well). The proof of (b) is now complete.

That completes our �le of solutions for problems of the February 1998

number of the Corner. The Olympiad Season is nearly upon us. Sendme your

national and regional Olympiads for use in the Corner. We also welcome your

nice solutions to problems that appear in the Corner.

86

BOOK REVIEWS

ALAN LAW

A Primer of Real Functions, by Ralph Boas Jr.,

published by the Mathematical Association of America, 1996,

ISBN# 0-88385-029-X, softcover, 262+ pages, $32.95.

Reviewed byMurray Klamkin, University of Alberta.

This is the fourth edition of a popular classic Carus monograph which

has been revised, updated and augmented by the author's son Harold P.

Boas. The previous editions covered sets, metric spaces, continuous func-

tions and di�erentiable functions. This edition adds a chapter on measurable

sets and functions, the Lebesgue and Stieltjes integrals, and applications. The

new material is a rewrite of a draft left over by the author at his death. This

book can be likened to a sequence of lectures on a variety of topics selected

from the foundations of analysis and is done in a friendly and lively manner.

Mathematically Speaking: A Dictionary of Quotations, selected and arranged

by Carl C. Gaither and Alma E. Cavazos-Gaither,

published by Institute of Physics Publishing, 1998,

ISBN# 0-7530-0503-7, softcover, 484+xiii pages, $39.00 (US).

Reviewed by Bruce Shawyer, Memorial University of Newfoundland.

The book contains hundreds of quotations from hundreds of authors, as

well as many apocryphal quotations from persons unknown. The quotations

are grouped into 199 sets, ordered by topics, running from ABSTRACTION to

ZERO. Most sections are just a few pages long, except for the topics MATH-

EMATICIAN and MATHEMATICS, which have 29 and 80 pages respectively.

Also included is a complete bibliography of the source material plus

two excellent indices, the SUBJECT BY AUTHOR INDEX and the AUTHOR BY

SUBJECT INDEX.

The quotations vary from the profound to the witty. There are quota-

tions from plays, and quite a few are in poetry, including several mnemonics

for �. Unfortunately, this reviewer's favourite is missing!

How I want a drink 3:1415Alcoholic of course 926After the heavy lectures 5358Involving decimal fractions 979

(Engineers can substitute \quantum mechanics" for \decimal fractions".)

The authors quoted come from all di�erent walks of life, from profes-

sional mathematicians and scientists to historians, journalists, philosophers,

87

poets, rap artists and writers. Some are famous, their names being almost

household words; others are much less known. As might be expected, by far

the majority of the quotations come from mathematicians and scientists.

To give a avour of this book, here is a selection of some of the shorter

quotations:

Some quotations from famous mathematicians

� Richard Guy

Mathematics often owes more to those who ask questions than to

those who answer them.

� Paul Halmos

The only way to learn mathematics is to do mathematics.

� Leopold Kronecker

Number theorists are like lotus-eaters | having once tasted of this

food, they can never give it up.

� George P �olya

Geometry is the art of correct reasoning on incorrect �gures.

Some quotations from others

� Ice-T

I write rhymes with addition and algebra, mental geometry.

� G.K. Chesterton

Poets do not go mad; but chess-players do. Mathematicians go

mad, and cashiers; but creative artists very seldom.

� John Updike

When you look into a mirror rorrim a otni kool uoy nehW

it is not yourself you see ees uoy esruoy ton si ti

but a kind of apish error rorre hsipa fo dnik a tub

posed in fearful symmetry yrtemmys lufraef ni desop

� Mae West

A �gure with curves always o�ers a lot of interesting angles.

� Unknown

Trigonometry is a sine of the times!

The book is a wonderful compendium and a great source of useful wisdom

for teachers of mathematics. It is very readable, and, once one has started

to read it, very di�cult to put down.

88

THE SKOLIAD CORNERNo. 36

R.E. Woodrow

We begin with the problems of the Mini demi-�nale 1996 of the Vingt

et uni �eme Olympiade Math �ematique Belge, organized by the Belgian Math-

ematics Teachers Society. Twenty-�ve of the problems are multiple choice.

The remaining �ve require an integer answer between 0 and 999 (inclusive).

Students are given 90 minutes. My thanks go to Ravi Vakil, Canadian Team

Leader to the InternationalMathematical Olympiad at Mumbai for collecting

the materials.

OLYMPIADE MATH �EMATIQUE BELGEMini demi-�nale 1996Mercredi 6 mars 1996

1. Sans r �eponse pr �eformul �ee | Quel est le nombre premier le plus

proche de 100 ?

2. Laquelle des propositions ci-dessous est la n �egation de: \Chaque

langue europ �eenne est parl �ee par l'un de nos guides au moins." ?

(a) Chacun de nos guides parle toutes les langues europ �eennes.

(b) Chacun de nos guides parle une langue europ �eenne au moins.

(c) Aucun de nos guides ne parle aucune langue europ �eenne.

(d) L'un de nos guides ne parle aucune langue europ �eenne.

(e) L'une des langues europ �eennes n'est parl �ee par aucun de nos guides.

3. Si P d �esigne le p �erim �etre du triangle ABC et kABk, kACk, kBCkles longueures de ses cot �es, laquelle des relations suivantes est correcte ?

(a) kABk = P=3 (b) kABk � P=3 (c) kABk � P=2

(d) kABk+ kACk = 2P=3 (e) kABk+ kACk � 2P=3

4. Un parterre rectangulaire de 8m sur 6m est entour �e ext �erieurement

d'un sentier de 1; 5 m de large. Quelle est l'aire de ce sentier ?

(a) 23; 25 m2 (b) 37;5 m2 (c) 42 m2 (d) 46; 5 m2 (e) 51 m2

89

5. Le graphique ci-dessous donne, pour les quatorze premiers jours de

sa vie, les gains de poids (en grammes, et par rapport au poids �a la naissance)

d'un b �eb �e dont le poids �a la naissance �etait de 3; 250 kg.

�p (g)

�200

0

200

400

t (jours)2 4 6 8 10 12 14

r

r

r rr

r

r

rr

r

rr

rr

r

Quel �etait son poids �a une semaine ?

(a) 0; 100 kg (b) 3; 300 kg (c) 3; 350 kg (d) 3; 650 kg (e) 4; 250 kg

6. SiX, Y et Z sont les sommets d'un triangle, quel est le nombre de

parall �elogrammes admettant X, Y et Z pour sommets ?

(a) 1 (b) 2 (c) 3 (d) 4 (e) 6

7. Que vaut (3x2 � 7x)(2x3 � x2 + x� 2) ?

(a) 6x4 � 17x3 + 10x2 � 13x+ 14.(b) 6x5 � 17x4 + 10x3 � 13x2 + 14x.(c) 6x5 + 17x4 + 10x3 + 13x2 + 14x.(d) 6x6 � 17x5 + 10x4 � 13x3 + 14x2.(e) 6x6 � 3x4 � 14x3 + 10x2 � 7x+ 8.

8. Sans r �eponse pr �eformul �ee | Une bille m �etallique a la propri �et �e de

rebondir �a une hauteur �egale aux huit dixi �emes de sa hauteur initiale; si elle

est lach �ee d'une hauteur de 1; 25 m, quelle sera, en centim �etres, la hauteur

de son troisi �eme rebond ?

9. De combien augmente l'aire totale d'un cube lorsque la longueur de

chacune de ses aretes augmente de 50% ?

(a) 50% (b) 125% (c) 225% (d) 237; 5% (e) 2500%

10. Quatre sacs opaques contiennent :

� Le sac A, une bille blanche et une bille rouge ;� Le sac B, deux billes blanches et deux billes rouges ;

� Le sac C, deux billes blanches, une bille rouge et une bille noire ;

� Le sac D, dix billes blanches et dix billes noires.

De quel sac faut-il tirer une bille au hasard pour avoir le plus de chances que

la bille tir �ee soit blanche ?

(a) Le sac A (b) Le sac B (c) Le sac C (d) Le sac D

(e) Le choix est indi� �erent

90

11. Laquelle des �gures suivantes est le d �eveloppement d'un cube ?

(a) (b) (c) (d) (e)

12. Si la somme de trois nombres naturels est un nombre de deux

chi�res, il est certain que :

(a) chacun des trois nombres est sup �erieur �a 10 ;

(b) deux des nombres, au moins, sont inf �erieurs �a 50 ;

(c) aucun des trois nombres n'est sup �erieur �a 50 ;

(d) les trois nombres sont di� �erents ;

(e) le produit des trois nombres est inf �erieur �a 35 000.

13. Avant son d �epart en vacances, une personne a achet �e 3000 francs

fran�cais pour 18 270 francs belges. Cette personne, en France, a du changer

�a nouveau de l'argent: pour 10000 francs belges, elle a re�cu 1600 francs

fran�cais. Si elle avait achet �e en Belgique, avant son d �epart, tout l'argent

fran�cais dont elle a eu besoin,

(a) elle aurait gagn �e 480 francs belges ;

(b) elle aurait gagn �e 256 francs belges ;

(c) cela serait revenu au meme ;

(d) elle aurait perdu 256 francs belges ;

(e) elle aurait perdu 480 francs belges.

14. Une personne a achet �e des timbres �a 3 F et des timbres �a 5 F

pour un total de 100 F exactement. Parmi les suivants, quel est le nombre

de timbres �a 5 F qu'elle ne peut pas avoir achet �e ?

(a) 5 (b) 8 (c) 9 (d) 11 (e) 17

15. Un cycliste monte une cote �a la vitesse moyenne de 12 km/h, pour

la redescendre ensuite �a la vitesse moyenne de 48 km/h. Si la mont �ee a dur �e

22 min 30 s de plus que la descente, quelle est la longueur de cette cote ?

(a) 6 km (b) 8 km (c) 12 km (d) 13; 5 km (e) 15 km

91

16.

A

B C

D

EF

Voici un hexagone ABCDEF dans

lequel les cot �es oppos �es sont par-

all �eles et de meme longueur. Le tri-

angle ABC est n �ecessairement ap-

pliqu �e sur le triangle DEF par

(a) une translation (b) une sym�etrie orthogonale

(c) une sym�etrie central (d) une rotation de 90�

(e) aucune des transformations pr �ec �edentes

17. Quel est le nombre maximum de points communs �a un cercle et

au bord d'un losange ?

(a) 2 (b) 4 (c) 6 (d) 8 (e) 10

18. Laquelle des a�rmations suivantes est vraie ?

(a) Il existe des carr �es qui ne sont pas des rectangles.

(b) Un carr �e n'est jamais un rectangle.

(c) Tout parall �elogramme est un losange.

(d) Tout losange est un parall �elogramme.

(e) Certains rectangles ne sont pas des parall �elogrammes.

19. Dix nombres sont tous inf �erieurs �a 20 et leurmoyenne (arithm �etique)

vaut 18. Laquelle des a�rmations suivantes est certainement correcte �a pro-

pos de ces nombres ?

(a) L'un d'entre eux, au moins, est �egal �a 18.

(b) Un nombre pair d'entre eux sont �egaux �a 18.

(c) Un nombre impair d'entre eux sont �egaux �a 18.

(d) Ils sont tous sup �erieus �a 16.

(e) Ils sont tous positifs.

20. Sans r �eponse pr �eformul �ee | Les roues avant d'un wagonnet ont

7 cm de circonf �erence et les roues arri �ere ont 9 cm de circonf �erence. Lorsque

les roues avant ont fait 10 tours de plus que les roues arri �ere, de combien de

centim �etres a avanc �e le wagonnet ?

21. La somme de deux nombres premiers est toujours

(a) un nombre pair ; (b) un nombre impair ;

(c) un nombre premier ; (d) strictement sup �erieure �a 3 ;

(e) inf �erieure �a 1000.

22. Le poids moyen des 30 �el �eves d'une classe est de 47 kg; si chacun

de ces �el �eves grossit de 3 kg, de combien augmentera le poids moyen ?

(a) 0; 1 kg (b) 2 kg (c) 3 kg (d) 90 kg (e) Une autre valeur

92

23.

R

S20

10

5

10

5

p

p

p

p

Dans le train d'engrenages

repr �esent �e ci-contre, lorsque la

roue R fait un tour dans le sens des

aiguilles d'une montre, de combien

tourne la roue S ? (Les nombres

indiqu �es donnent le nombre de

dents de chaque engrenage. Deux

cercles concentriques repr �esentent

deux roues solidaires du meme axe.)

(a) D'un tour dans le sens des aiguilles d'une montre.

(b) De 2 tours dans le sens des aiguilles d'une montre.

(c) De 4 tours dans le sens des aiguilles d'une montre.

(d) De 2 tours dans le sens oppos �e �a celui des aiguilles d'une montre.

(e) De 4 tours dans le sens oppos �e �a celui des aiguilles d'une montre.

24. Sans r �eponse pr �eformul �ee | Pour un certain nombre naturel n,

2n+ 3 est un diviseur de 6n+ 43; que vaut n ?

25. Un marchand de jouets a achet �e un lot de 100 oursons en peluche

qui valaient au total 21000 F, mais le grossiste lui a consenti une remise de

10%. Ce marchand souhaite qu'en accordant une remise de 25% sur le prix

de vente a�ch �e, il lui reste encore un b �en �e�ce �egal �a 30% du prix de vente

r �eel. �A combien doit-il a�cher l'ourson ?

(a) 341; 25 F (b) 360 F (c) 400 F (d) 34125 F (e) Une autre r �eponse

26. Soit x et y deux demi-droites contenues dans une meme droits ;

alors,

(a) il existe n �ecessairement une sym�etrie centrale appliquant x sur y ;

(b) il existe n �ecessairement une sym�etrie orthogonale appliquant x sur y ;

(c) il existe n �ecessairement une translation appliquant x sur y ;

(d) il existe n �ecessairement une rotation appliquant x sur y ;

(e) aucune des propositions pr �ec �edentes n'est vraie.

93

27. Un avion vole �a la vitesse de 400 km/h par rapport �a l'air. Pendant

son voyage aller, il a de face un vent de 40 km/h; au retour, il a le meme vent

dans le dos. Quelle est sa vitesse moyenne sur l'ensemble du trajet ?

(a) 390 km/h (b) 396 km/h (c) 398 km/h (d) 400 km/h (e) 410 km/h

Last issue we gave the problems of the Old Mutual Mathematical

Olympiad 1992. Thanks go to John Grant McLoughlin, Faculty of Educa-

tion, Memorial University, St. John's, Newfoundland for the problem set

and solutions. Here are the answers:

1. c 2. b 3. d 4. c

5. a 6. c 7. c 8. d

9. a 10. b 11. e 12. b

13. e 14. a 15. c 16. e

17. a 18. e 19. a 20. e

That completes the Skoliad Corner for this issue. Please send me con-

test materials and suggestions for other features of the Corner.

XXXXXXX X X X X X X X X

Y Y YY Y Y Y Y Y

YYYYYY

X X XX X X XXXXXXXXXYYYYYY Y Y Y Y Y Y Y Y Y

94

MATHEMATICAL MAYHEM

Mathematical Mayhem began in 1988 as a Mathematical Journal for and by

High School and University Students. It continues, with the same emphasis,

as an integral part of Crux Mathematicorum with Mathematical Mayhem.

All material intended for inclusion in this section should be sent to the

Mayhem Editor, Naoki Sato, Department of Mathematics, Yale University,

PO Box 208283 Yale Station, New Haven, CT 06520{8283 USA. The electronic

address is still

mayhem@math.toronto.edu

The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).

The rest of the sta� consists of Adrian Chan (Upper Canada College), Jimmy

Chui (Earl Haig Secondary School), David Savitt (Harvard University) and

Wai Ling Yee (University of Waterloo).

Shreds and Slices

Non-mathematical Problem

InProblemBook for First Year Calculus, by George W. Bluman, Springer-

Verlag, Problem 13.7 asks: Choose a non-mathematician:

(a) John von Neumann. (b) Mick Jagger.

(c) Georg Cantor. (d) Pablo Casals.

(e) Stanley I. Grossman. (f) Ren �e Descartes.

(g) Guy La eur.

The answer at the back of the book? \Possibly (b), (d), (g)."

Awaiting a Combinatorial Proof

Find a combinatorial proof of the following identity:

(n� r)

�n+ r � 1

r

��n

r

�= n

�n+ r � 1

2r

��2r

r

�:

The individualwith the �rst correct solution that is strictly combinatorial will

get a free book prize and the solution published here.

Reference

Tucker, Alan. Applied Combinatorics. JohnWiley & Sons, Inc. Toronto.

1995 pp. 221.

95

Discovering the Human Calculator in You

Richard Hoshinostudent, University of Waterloo

In his Oscar-winning role in the movie Rain Man, DustinHo�man plays

an idiot savant who can perform complex calculations instantly in his head.

Like the \Rain Man", various people have displayed their outstanding capac-

ity for mental arithmetic on TV, and many others have written books teach-

ing these powerful techniques. However, hardly any have ever ventured to

justify the validity of these algorithms, as the mathematics involved is sur-

prisingly elementary. In this article, we detail some of the famous tricks that

the \human calculators" have used over the years, explain why these meth-

ods work, and you will see that, with a little practice, you too can be a human

calculator.

Trick 1. Squaring two-digit numbers ending in 5.

To square any two-digit number that ends in 5, add one to the �rst digit

and multiply that sum by the �rst digit. This will be the �rst two digits of

the answer. The last two digits will always be 25.

For example, 852 = 7225 since 8� (8+1) = 8�9 = 72, and likewise,

252 = 625 since 2 � 3 = 6. We can extend this to larger numbers, for

example, 1952 = 38025, since 19� 20 = 380.

If you are wonderingwhy thismethod works, a little algebrawill quickly

convince you:

(10A+ 5)2 = 100A2 + 100A+ 25 = 100A(A+ 1) + 25 .

Thus, the �rst two digits will be A(A+ 1), and the last two digits will

be 25.

Let us take this idea one step further. Let us multiply pairs of two-digit

numbers whose tens digits are the same, and whose units digits sum to ten.

For example, 37� 33 = 1221, 36� 34 = 1224, and 98� 92 = 9016.

Do you see the pattern? Like in the case above, the �rst two digits of

the answer are determined in the same way. But what about the last two

digits? Do you see how they are obtained? If so, use a little algebra and

convince yourself that it always works.

Copyright c 1999 CanadianMathematical Society

96

Trick 2. Squaring any two-digit number.

Take any two-digit number n. Now we know that

n2 = (n2 � d2) + d2 = (n� d)(n+ d) + d2 ,

so let us try to �nd a value of d so that the product (n� d)(n+ d) can be

easily calculated. Consider the multiple of 10 that is closest to n, and let the

di�erence between the number and this multiple of 10 be d. For example, if

we take n = 87, then the multiple of 10 that is closest to 87 is 90, and since

90� 87 = 3, we have d = 3. Similarly, if n = 94, we have d = 4.

If we perform this calculation for any integer n, then one of n� d or

n+d will be a multiple of 10, and the calculation becomes signi�cantly easier.

The following examples illustrate this technique:

872 = (87 + 3)(87� 3) + 32 = 90� 84 + 9 = 7569 ,

292 = (29 + 1)(29� 1) + 12 = 30� 28 + 1 = 841 ,

962 = (96 + 4)(96� 4) + 42 = 100� 92 + 16 = 9216 .

Use this technique to compute the following: 372, 522, 19992.

Trick 3. The Calendar Trick.

One of the more interesting demonstrations performed by \mathema-

gicians" is the calendar trick. Namely, an audience member calls out her

birthday, or some historical date, and the human calculator is able to tell her

what day of the week that event took place.

The �rst thing to do is to memorize the following table:

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

1 4 4 0 2 5 0 3 6 1 4 6

It appears challenging to remember this, but there is an interesting pat-

tern here. Reading the row of numbers from left to right in threes, we have

144, 025, 036 and 146. Notice that 144, 025 and 036 are perfect squares,

and the last number is just 2 more than the �rst perfect square, 144. This

should make the memorization easier.

Let Y be the last two digits of the year in question. Let D be the day

we are searching for, and letM be the integer that corresponds to the month

in the above table. Thus, if we are searching for July 25th, 1978, Y would

be 78, D would be 25, andM would be 0, since July corresponds to 0.

Compute the value of

Y +

�Y

4

�+D+M ,

and divide that sumby 7. Whatever remainder you get corresponds to the day

of the week you are seeking, namely 0 is Saturday, 1 is Sunday,

2 is Monday, 3 is Tuesday, 4 is Wednesday, 5 is Thursday, and 6 is Friday.

97

A small note to remember. If the year is a leap year, and the month is

January or February, you must subtract 1 from the total. This is due to the

fact that the extra day in a leap year occurs on February 29th, and so if the

day you are searching is before that, then the formula is o� by one day.

Let us look at a historical date in the 20th century. The famous stock

market crash of 1929 occurred on October 29th, so let us use our formula to

determine what day of the week \Black Tuesday" occurred.

We have Y = 29, M = 1, and D = 29. Hence b294c = 7, and our

sum is 29 + 7 + 29 + 1 = 66. Dividing this number by 7, we �nd that the

remainder is 3. We conclude that October 29th, 1929 was indeed a Tuesday.

Unfortunately, this formula only works with dates in the 1900's, be-cause in the Gregorian calendar, not all years that are divisible by 4 are leap

years. For example, 1800 and 1900 are not leap years, but 2000 is. And

thus, we must alter our formula to compensate for this. To calculate dates

in the 1800's, use the same formula, but go forward two days in the week.

To calculate dates in the 2000's, go one day back. If we use our formula, we

�nd that January 1st, 2000 is a Sunday (remember, 2000 is a leap year!). Go

back one day, because it really is a Saturday.

In the past, it was believed that a year had precisely 365:25 days, and

so we compensated for the extra quarter day by adding February 29th to our

calendar once every four years. Unfortunately, a year has 365:2422 days, so

we cannot add an extra day exactly once every four years. It would be nice

however if we could, for then this formula would always hold.

Using this method, determine what day of the week you were born on.

Trick 4. Extracting Cube Roots.

We now detail the method for determining the cube roots of all perfect

cubes under one billion.

First, you must �rst learn the cubes of the integers 0 through 9.

n 0 1 2 3 4 5 6 7 8 9n3 0 1 8 27 64 125 216 343 512 729

Let us �rst �nd the cube root of numbers that are below one million.

Hence, the cube root will be at most 99. Say we want to �nd the cube root of

314; 432. We separate the number into two parts, separated by the comma.

Thus, 314 is the �rst part, and 432 is the second part. The desired cube root

has two digits. We will use the �rst part to get the �rst digit, and we will use

the second part to get the second digit.

Take the �rst part and determine where it lies in the table of cubes. In

our example, 314 lies between 216 and 343. Thus, 216; 000 < 314; 432 <343; 000, which implies that the desired cube root lies between 60 and 70,since 603 = 216; 000 and 703 = 343;000. Hence, it follows that the �rst

digit of our cube root must be 6.

98

Now we determine the second digit. If we look at the table of cubes,

notice that each cube ends in a di�erent digit. So if a certain cube ends in 2,we know that its cube root must end in 8, because 8 is the only digit whose

cube ends in 2. Since 432 ends in 2, the second digit of the cube root must

be 8. Thus, the desired cube root is 68.

Before, you go further, determine the cube roots of the following num-

bers in your head: 157; 464; 185; 193; 778;688; 12; 167.

With a little practice, you will �nd that it is faster to do this exercise in

your head rather than punching it in your calculator!

So that we do not confuse the digits in the following example, let us

introduce some variables. Let n = (100p+10q+r)3; that is n is the perfect

cube, and 100p+10q+r is the cube root of n, where p, q and r are its digits.

To do this trick, once again, you need to memorize a small table:

A 1 2 3 4 5 6 7 8 9 10

B 1 7 9 5 3 8 6 2 4 10

Just keep repeating \one-seven-nine-�ve-three, eight-six-two-four-ten", to

remember the right order in the B row. It might help to notice that the �rst

�ve entries are odd, and the last �ve entries are even.

Now, let us move on to the cubes of three-digit numbers. Let us say

we wanted to �nd the cube root of n = 101; 847; 563. First, let us separatethis large number into three smaller ones, separated at the commas.

As we did before, we look at the �rst part, 101, to determine what the

�rst digit of the cube root is. Since 101 lies between 64 = 43 and 125 = 53,we conclude that the p = 4. The last part is 563, which ends in a 3, and since

73 = 343, we see that r = 7. Hence, we determine the �rst and last digits

the same way as we did before with the smaller numbers.

Now take n, and add and subtract the digits of this number in alternat-

ing fashion starting from the right. Thus, we compute

3� 6 + 5� 7 + 4� 8 + 1� 0 + 1 = �7.Now repeatedly add or subtract 11 to this number until we get a number

between 0 and 10 inclusive (more formally, we say we take this number

modulo 11). Hence, �7 becomes 4. Let this number be A. If you have

ever seen the test for divisibility by 11, you will see that n � A (mod 11).

Now take the number A, and look in the above table to determine

the corresponding number B. We see that A = 4 corresponds to

B = 5; thus we have B = 5. Finally, our second digit q is the value of

p + r � B (mod 11); that is we add or subtract 11 until we get an integer

between 0 and 10 inclusive, and this will be our digit q. Thus, in our case,

p+ r �B = 4+ 7� 5 = 6, and so we have shown that q = 6.

Therefore, p = 4, q = 6 and r = 7; thus the cube root of

n = 101;847; 563 is 467. Checking, we see that this is correct.

99

Let us summarize the algorithm.

(i) Use the �rst part of n to determine the �rst digit of the cube root. Call

this digit p.

(ii) Use the last digit of n to determine the last digit of the cube root. Call

this digit r.

(iii) Take the alternating sum of n to determine A, where A is between 0and 10. That is, �nd the A for which n � A (mod 11).

(iv) Use the table to �nd the number B that corresponds to A.

(v) Determine the sum p+ r � B, and reduce it modulo 11. This is q.

(vi) The number with �rst digit p, second digit q and third digit r is your

desired cube root.

Now try to �nd the cube roots of the following numbers. Note, you can

use a pencil and paper, but you are not allowed to use a calculator!

(a) 17; 173;512, (b) 1; 860; 867, (c) 758; 550; 528,

(d) 84; 604; 519, (e) 170; 953; 875.

Let us now justify why this algorithm works. First of all, take each

value B in the table, and compute B3 (mod 11). You will �nd that we will

get the corresponding value of A in each case. That is how the numbers are

determined. Note that there is a bijection between the elements of the two

rows in the table. If this were not the case, then this algorithm would not

work. We have

n = (100p+ 10q+ r)3

� (p� q + r)3

� A (mod 11) .

And from the table, this implies that (p� q + r) � B (mod 11); thatis q � (p+ r �B) (mod 11).

Therefore, q is uniquely determined, as are p and r from before.

With a little practice, you will become comfortable with all of the meth-

ods described in this article. Maybe you will even be able to come up with

better ways to perform the tricks described. Nevertheless, through practice

and perseverance, you too can be a human calculator.

100

Mayhem Problems

The Mayhem Problems editors are:

Adrian Chan Mayhem High School Problems Editor,

Cyrus Hsia Mayhem Advanced Problems Editor,

David Savitt Mayhem Challenge Board Problems Editor.

Note that all correspondence should be sent to the appropriate editor |

see the relevant section. In this issue, you will �nd only solutions | the

next issue will feature only problems.

We warmly welcome proposals for problems and solutions. With the

new schedule of eight issues per year, we request that solutions from the last

issue be submitted in time for issue 2 of 2000.

High School Solutions

Editor: Adrian Chan, 229 Old Yonge Street, Toronto, Ontario, Canada.

M2P 1R5 <all238@ipoline.com>

H223. In each of the following alphametics, each letter in the addi-

tion represents a unique digit:

1 9 9 7+ O L D

Y E A R

and

1 9 9 8+ O L D

Y E A R

.

For each alphametic, �nd a solution, or prove that a solution does not exist.

Solution. First, we show that the second alphametic has no solution.

If either L or D is greater than 1, then there will be a carry involved in each

step of the addition. Most importantly, from the third column, we obtain

that 1 + 9 + O � E (mod 10), which implies that O � E (mod 10).

Thus, O andE represent the same digit, which we cannot have. Hence,

a solution must satisfy LD = 01; that is L = 0 andD = 1. But this leads to

1 9 9 8+ O 0 1

Y E 9 9,

so A = R = 9, which we cannot have either. Therefore, there is no solution

to this alphametic.

A solution to the �rst alphametic must satisfy L = 0 and D = 1, orL = 0 and D = 2 by the same reasoning as above. Setting L = 0 and

101

D = 1 leads to A = R = 9, but setting L = 0 and D = 1 leads to several

possibilities, for example

1 9 9 7+ 4 0 1

2 3 9 8.

H224. LetABCD be a square. Construct equilateral trianglesAPB,

BQC, CRD, and DSA, where P , Q, R, and S are points outside of the

square.

(a) Prove that PQRS is a square.

(b) Determine the ratio PQ=AB. (See how many ways you can solve this!)

Solution.

A B

CD

P

Q

R

S

60�

60�

(a) In the �gure, \PAS = \PBQ = \QCR = \RDS = 150�, andAP = AS = BP = BQ = CQ = CR = DR = DS, so triangles

PAS, PBQ, QCR, and RDS are congruent. Hence, PQ = QR = RS =SP . Also, \PAS = 150�, so \SPA = \PSA = 15�, and \SPQ =\APB + \BPQ + \BPQ = 15� + 60� + 15� = 90�, and by symmetry,

\PQR = \QRS = \RSP = 90� as well, so PQRS is a square.

(b) By the Cosine Law,

PQ2 = PB2 + BQ2 � 2PB � PQ cos 150�

= 2AB2 � 2AB2 � cos 150� = 2AB2(1 + cos 30�)

=)PQ2

AB2= 2

1 +

p3

2

!= 2 +

p3 =

4 + 2p3

2

=)PQ

AB=

s4 + 2

p3

2=

1 +p3

p2

=

p2 +

p6

2.

102

H225. Consider a row of �ve chairs, numbered 1, 2, 3, 4, and 5. Youare originally sitting on 1. On each move, you must stand up and sit down

on an adjacent chair. Make 19 moves, then take away chairs 1 and 5. Thenmake another 97 moves, with the three remaining chairs. No matter how

the moves are made, you will always end up on chair 3. Why is this the case?

Solution. Say that chairs 1, 3, and 5 are odd chairs and that chairs 2and 4 are even chairs. After 1move, we are on an even chair. After 2moves,

we are on an odd chair. By a simple parity argument, after 19 moves we

must be on an even chair. Thus, we take away chairs 1 and 5, and are left

with two even chairs and one odd chair.

By parity again, after 97 moves, we will be on an odd chair, namely

chair 3. Hence, we will always end up on chair 3.

H226. The smallest multiple of 1998 that consists of only the digits

0 and 9 is 9990.

(a) What is the smallest multiple of 1998 that consists of only the digits 0and 3?

(b) What is the smallest multiple of 1998 that consists of only the digits 0and 1?

Solution. (a) Let N be the smallest multiple of 1998 that only consists

of the digits 0 and 3. Since 1998 = 6� 333, we have that 333 dividesN=6.

By de�nition, N is of the form

N = 3 � 10a1 + 3 � 10a2 + � � �+ 3 � 10an

for some integers a1 > a2 > � � � > an > 0. Then,

N

6= 5 � 10a1�1 + 5 � 10a2�1 + � � �+ 5 � 10an�1 ,

which implies that N=6 is an integer that consists of only the digits 0 and 5.Since 9 divides 333, 9must divideN=6. By a well known divisibility test, the

number of 5's in N=6 must be a multiple of 9. Hence, N=6 � 555 555 555,or N � 3 333 333 330.

Checking, we �nd that 1998 does indeed divide 3 333 333 330, so this

is the multiple of 1998 we seek.

(b) Let N be the smallest multiple of 1998 that consists of only the digits 0and 1. Then N is of the form

N = 10a1 + 10a2 + � � �+ 10an

for some integers a1 > a2 > � � � > an > 0. The last digit of N must

be a 0. Note that 1998 divides N if and only if 999 divides N=2, which

103

in turn occurs if and only if 999 divides N=10, so henceforth we consider

N=10 = 10a1�1 + 10a2�1 + � � �+ 10an�1. Let bi = ai � 1 for all i.

Let a, b, and c be the number of exponents bi which are congruent to

0, 1, and 2 modulo 3 respectively. Then

N

10= 10a1�1 + 10a2�1 + � � �+ 10an�1

= 10b1 + 10b2 + � � �+ 10bn

� a+ 10b+ 100c

� 0 (mod 999) ,

since 103 � 1 (mod 999). Note that a+10b+100c � 0 (mod 999) implies

that 10a + 100b + c � 0 and 100a + b + 10c � 0, which we obtain by

multiplying by 10 and 100 respectively. We consider possible values of a, b,

and c.

Suppose that c � 10. Since a+10b+100c � a+10b+100(c� 10)+1000 � (a+1)+10b+100(c� 10) (mod 999), the triple (a+1; b; c� 10)is also a solution. Similarly, if b � 10 or a � 10, then we can obtain another

solution by using the two congruences derived above. Hence, by using this

reduction, for any given solution, we can obtain a solution where a, b, and c

are all at most 9. There is no danger of being caught in a cycle, since each

application of the reduction decreases the sum a+ b+ c by 9.

Now consider such a reduced solution. Each of a, b, and c is at most 9,but if any is less than 9, then it is apparent that a + 10b+ 100c < 999, sothe congruence a+ 10b+ 100c � 0 (mod 999) cannot hold. Therefore, theonly solution with all of a, b, and c at most 9 is a = b = c = 9. This leadsto the number

N

10= 111:::1| {z }

27 1's

.

We have ruled out the case a + b + c < 27, and all values such that

a+ b+ c � 27 must lead to a number with at least 27 digits. Therefore, the

desired minimal multiple is

N = 111:::1| {z }27 1's

0 .

104

Advanced Solutions

Editor: Donny Cheung, c/o Conrad Grebel College, University of Wa-

terloo, Waterloo, Ontario, Canada. N2L 3G6 <dccheung@uwaterloo.ca>

A209. Is there an in�nite number of squares among the triangular

numbers?

Solution.

The answer is yes. The problem reduces to a special case of Pell's equa-

tion. Pell's equation is the name given to diophantine equations of the form

x2 � dy2 = 1, where d is a positive, non-square integer.

We want to show that there is an in�nite number of integers n such

that the nth triangular number n(n� 1)=2 is a perfect square. Assume that

n(n�1)=2 = a2, or equivalentlyn(n�1) = 2a2. By completing the square,

we obtain (2n� 1)2 � 8a2 = 1, or (2n� 1)2 � 2(2a)2 = 1.

In other words, the problem reduces to showing that there is an in�nite

number of pairs of integer solutions to the equation x2 � 2y2 = 1. Here

x = 2n� 1 and y = 2a. This is a particular case of Pell's equation, which

has in�nitely many solutions. First, note that (x0; y0) = (3; 2) is a speci�c

solution. Then note that for any solution (x; y), (3x+ 4y;2x + 3y) is alsoa solution:

(3x+ 4y)2� 2(2x+ 3y)2 = 9x2 + 24xy + 16y2 � 8x2 � 24xy � 18y2

= x2 � 2y2 = 1 .

Furthermore, if x is odd and y is even, then 3x+4y is odd and 2x+3yis even. In this way, we can generate an in�nite number of solutions to

x2� 2y2 = 1 starting from (3; 2). Thus, setting n = (x+1)=2, we �nd that

there is an in�nite number of integers n such that the nth triangular number

is a perfect square.

A210. Proposed by Naoki Sato.

Let P be a point inside circle C. Find the locus of the centres of all

circles ! which pass through P and are tangent to C.

Solution.

Let the centre of circle C be O and its radiusR. We claim that the locus

is an ellipse with foci at P and O, with semi-major axis length R=2. To see

this, letD be a circle passing through P and tangent to the circle C as shown.Let Q be the centre of this circle.

105

C

D

qq

q

OQ

P

r

r

To show that Q lies on the ellipse with foci P and O, it is su�cient to

show that PQ+QO is a constant. Let the radius of circle D be r. Since the

two circles C and D are tangent, the common tangent point is collinear with

the centres O and Q as shown. Then OQ = R � r, and PQ is the radius of

the circle. Therefore, PQ + QO = r + R � r = R. Thus, the locus is an

ellipse with foci at P and O. Further, the constant is R, so the length of the

ellipse's semi-major axis is R=2.

A211. Do there exist a convex polyhedron and a plane, not passing

through any of its vertices, and intersecting more than 2=3 of all of the edgesof the polyhedron?

(Polish Mathematical Olympiad, �rst round)

Solution.

We �rst prove a lemma.

Lemma. For any polyhedron with E edges and F faces, we have the

inequality 2E � 3F .

Proof. Each face has at least three edges, so there is a total of at least

3F edges, counting each edge twice. Each edge is counted once for the two

faces it attaches. The total number of edges is given by E, so 2E � 3F .

Suppose on the contrary that there are a convex polyhedron and a plane,

not passing through any of its vertices, and intersecting more than 2=3 of all

the edges. LetE, F , andV denote the number of edges, faces, and vertices of

the polyhedron respectively. The plane cuts the polyhedron into a polygonal

region. Suppose the polygon formed in this way has n vertices, and hence

n sides. Each vertex is the intersection of an edge of the polyhedron with

the plane. By assumption, n > 2E=3. Now each edge of the polygon is the

intersection of a face of the polyhedron with the plane. Thus, the number of

faces of the polyhedron is at least n. That is, F � n.

Together with the lemma, we obtain that 2E=3 � F � n > 2E=3,which is a contradiction.

106

Challenge Board Solutions

Editor: David Savitt, Department of Mathematics, Harvard University,

1 Oxford Street, Cambridge, MA, USA 02138 <dsavitt@math.harvard.edu>

This issue, we will be delaying the challenge board solutions until a

later issue. The observant reader will have noticed that the schedule for the

high school and advanced sections have been asynchronous with that of the

challenge board section; this will �x this problem.

Note: Although the editor listed, as the proposer for C83 (CRUX with

MAYHEM, Vol. 25, Issue 1), the person from whom he heard the problem,

the editor is aware that C83 is an old question. In fact, a version of the

problem has appeared in a competition as recently as 1996, when it was used

in a Romanian IMO team selection contest. Thanks to Mohammed Aassila

for bringing this to our attention.

Problem of the Month

Jimmy Chui, student, Earl Haig S.S.

Problem. Let a, b, and c be the lengths of the sides of a triangle. Prove

that pa+ b� c+

pb+ c� a+

pc+ a� b �

pa+

pb+

pc ,

and determine when equality occurs.

(1996 APMO, Problem 5)

Solution. Let s = (a+b+c)=2, the semi-perimeter, and let x = s�a,y = s� b, and z = s� c. Then a = y+ z, b = z+ x, and c = x+ y. Note

that x, y, and z are all positive, since x = (b+c�a)=2, etc., and a, b, and care the lengths of the sides of a triangle. (This is known as the infamous Ravi

Substitution [Ed. at least in Canadian IMO circles].) Hence, the inequality

is equivalent to

p2x+

p2y+

p2z �

pz + x+

px+ y +

py + z .

Recall that the Arithmetic Mean{Quadratic Mean (AM{QM) inequality

states that

a+ b

2�

sa2 + b2

2

107

for all a, b � 0. Then, we have that

p2x+

p2y+

p2z =

p2x+

p2y

2+

p2y+

p2z

2+

p2z +

p2x

2

�r2x+ 2y

2+

r2y+ 2z

2+

r2z + 2x

2(AM{QM)

=pz + x+

px+ y+

py+ z :

Equality holds if and only if x = y = z; that is, if and only if a = b = c.

When dealing with an expression involving the lengths of the sides of a

triangle, implementing the Ravi Substitution will often alter the expression

to a more manageable form.

J.I.R. McKnight Problems Contest 1987

1. (a) Three people, Aretha, Bob, and Chai, throw dice upon the con-

dition that the one who has the lowest result shall give each of

the others the sum of money each of the two winners has already.

Aretha loses �rst, Bob loses second, and Chai loses the third game.

They discovered that each �nishedwith the same amount ofmoney.

Express the amount of money that each one had at the beginning

in terms of the amount that each had at the end of the third game.

(b) Find all integer solutions for x, y, and z:

x(y+ z) = 32 ,

y(x+ z) = 65 ,

z(x+ y) = 77 .

2. (a) Find the sum of the �rst 22 terms of the geometric series having

�rst term i and ratio 1 + i, where i =p�1. Give your answer in

the form a+ bi, where a, b 2 R.(b) A three-dimensional �gure is de�ned by the equation

4jxj+ 3jyj+ 6jzj = 12 .

Identify this �gure and determine its volume.

3. A sequence of integers is de�ned by the following recursion: x1 = 2,x2 = 5, and xk = xk�1 + 2xk�2 for k > 2. Prove that

xn =7 � 2n�1 + (�1)n

3.

108

4. The screen of a drive-in theatre is p units tall and is situated on a hill

q units high. A car is situated a distance x units from the screen such

that the angle � subtended by the screen is a maximum. Show that the

maximum value of � occurs when x =pq(p+ q).

�

-�

6

?

6

?

x

p

q

5. The lengths of the sides of a triangle are 8, 8, and 11. Find the length

of one of the angle trisectors drawn to the longest side.

6. Consider the set of odd numbers f1, 3, 5, : : : , 101g.

(a) How many combinations of two distinct numbers can be formed

from this set?

(b) Determine the sum of the products of the pairs in (a).

7. For any convex quadrilateral ABCD, the diagonalsAC andBD inter-

sect at E. The centroids of trianglesABE, BCE, CDE, andDAE are

P , Q, R, and S respectively.

(a) Prove that PQRS is a parallelogram.

(b) Find the ratio of the area of the parallelogram PQRS to the orig-

inal quadrilateral.

Do question 8 or 9:

8. Seventeen dots are arranged so that no three are collinear. Each pair of

dots is connected by a line segment which may be drawn using one of

three colours. Prove that there are at least three points connected to

each other with the same colour.

9. Prove the following theorem: If the bisectors of a pair of opposite ex-

terior angles of a cyclic quadrilateral are parallel, then the angles at the

other two vertices are right angles.

109

Swedish Mathematics Olympiad 1985

Final Round

4. The polynomial p(x) of degree n has real coe�cients, and p(x) � 0 for

all x. Show that

p(x) + p0(x) + p00(x) + � � �+ p(n)(x) � 0 .

Solution by Hadi Salmasian, Sharif University of Technology, Tehran,

Iran.

Put �(x) = p(x) + p0(x) + � � � + p(n)(x). Since p(n+1)(x) = 0, wehave

�(x) = p(x) + �0(x) . (1)

Now, de�ne (x) = �(x)e�x. Then

ex (x) = p(x) + ex (x) + ex 0(x) ,

which is a result of (1). This gives us the following equation:

0(x) = �p(x)e�x ,

and obviously, for all x 2 R, 0(x) � 0, so (x) is a decreasing function.

On the other hand, the degrees of p0, p00, : : : , are less than n, and the lead-

ing coe�cient in both polynomials p(x) and �(x) are equal. The following

lemma is easy to prove:

Lemma. For a non-zero polynomial p(x), the following two are equiv-

alent:

(i) limx!+1 p(x) = +1.

(ii) The leading coe�cient of p(x) is positive.

Using the lemma, we �nd that �(x) � 0 for su�ciently large values

of x, and the same is true for (x). But (x) is a decreasing function, so

(x) � 0 for all x. The same is also true for �(x).

110

PROBLEMS

Problem proposals and solutions should be sent to Bruce Shawyer, De-

partment ofMathematics and Statistics,Memorial University of Newfound-

land, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be ac-

companied by a solution, together with references and other insights which

are likely to be of help to the editor. When a submission is submitted with-

out a solution, the proposer must include su�cient information on why a

solution is likely. An asterisk (?) after a number indicates that a problem

was submitted without a solution.

In particular, original problems are solicited. However, other inter-

esting problems may also be acceptable provided that they are not too well

known, and references are given as to their provenance. Ordinarily, if the

originator of a problem can be located, it should not be submitted without

the originator's permission.

To facilitate their consideration, please send your proposals and so-

lutions on signed and separate standard 812"�11" or A4 sheets of paper.

These may be typewritten or neatly hand-written, and should be mailed to

the Editor-in-Chief, to arrive no later than 1 October 1999. They may also

be sent by email to crux-editors@cms.math.ca. (It would be appreciated if

email proposals and solutions were written in LATEX). Graphics �les should

be in epic format, or encapsulated postscript. Solutions received after the

above date will also be considered if there is su�cient time before the date

of publication. Please note that we do not accept submissions sent by FAX.

2414. Proposed by Wu Wei Chao, Guang Zhou Normal University,

Guang Zhou City, Guang Dong Province, China, and Edward T.H. Wang, Wil-

frid Laurier University, Waterloo, Ontario.

For 1 < x � e � y or e � x < y, prove that xxyxy

> xyx

yx.

2415. Proposed by Paul Yiu, Florida Atlantic University, Boca Ra-

ton, Florida, USA.

Given a point Z on a line segment AB, �nd a Euclidean construction of

a right-angled triangle ABC whose incircle touches hypotenuse AB at Z.

2416. Proposed by V�aclav Kone �cn �y, Ferris State University, Big

Rapids, Michigan, USA.

Given 4ABC, where C is an obtuse angle, suppose that M is the

mid-point of BC and that the circle with centre A and radius AM meets

BC again at D. Assume also that MD = AB. The circle, �, with centre

M and radiusMB meets AB at E. Let H be the foot of the perpendicular

from A to BC (extended). Suppose that AC and EH intersect at I.

Find the angles \IAH and \AHI as function of \ABC.

[This proposal was inspired by problem 2316.]

111

2417. Proposed by Christopher J. Bradley, Clifton College, Bristol,

UK.

In 4ABC, with AB 6= AC, the internal and external bisectors of

\BAC meet the circumcircle of4ABC again in L andM respectively. The

points L0 and M 0 lie on the extensions of AL and AM respectively, and

satisfy AL = LL0 and AM = MM 0. The circles ALM 0 and AL0M meet

again at P .

Prove that AP k BC.

2418. Proposed by Christopher J. Bradley, Clifton College, Bristol,

UK.

In4ABC, the lengths of the sidesBC, CA, AB are 1998, 2000, 2002respectively.

Prove that there exists exactly one point P (distinct from A and B) on

the minor arc AB of the circumcircle of4ABC such that PA, PB, PC are

all of integral length.

2419. Proposed by K.R.S. Sastry, Dodballapur, India.

Find all solutions to the alphametic:

M I X � E

+ D B A � S

E S U � M

1. The letters before the decimal points represent base ten digits, and ad-

dition is done in that base.

2. The letters after the decimal points represent base six digits, and addi-

tion is done in that base.

3. The same letter stands for the same digit, distinct letters stand for dis-

tinct digits, and initial digits are non-zero.

Readers familiar with cricket will realize that this is a real world problem!

[Ed. Readers not familiar with cricket may be interested to learn that an

`over' consists of six `deliveries'!!]

2420. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.

Suppose that x, y and z are integers. Solve the equation:

x2 + y2 = 2420z2 .

2421. Proposed by Ice B. Risteski, Skopje, Macedonia.

What is the probability that the k numbers in the Las Vegas lottery on

a given payout day do not include two consecutive integers? (The winning

numbers are an unordered random choice of k distinct integers from 1 to n,

where n > k.)

112

2422?. Proposed by Walther Janous, Ursulinengymnasium, Inns-

bruck, Austria.

Let A, B, C be the angles of an arbitrary triangle. Prove or disprove

that1

A+

1

B+

1

C�

9p3

2� (sinA sinB sinC)1=3

.

2423. Proposed by Walther Janous, Ursulinengymnasium, Inns-

bruck, Austria.

Let x1, x2, : : : , xn > 0 be real numbers such that x1+x2+: : :+xn = 1,where n > 2 is a natural number. Prove that

nYk=1

�1 +

1

xk

��

nYk=1

�n� xk

1� xk

�.

Determine the cases of equality.

2424. Proposed by K.R.S. Sastry, Dodballapur, India.

In 4ABC, suppose that I is the incentre and BE is the bisector

of \ABC, with E on AC. Suppose that P is on AB and Q on AC

such that PIQ is parallel to BC. Prove that BE = PQ if and only if

\ABC = 2\ACB.

2425. Proposed by K.R.S. Sastry, Dodballapur, India.

Suppose that D is the foot of the altitude from vertex A of an acute-

angled Heronian triangle ABC (that is, one having integer sides and area).

Suppose that the greatest common divisor of the side lengths is 1. Find the

smallest possible value of the side length BC, given that BD�DC = 6.

NOTE OF THANKS

In the December 1998 issue of CRUX with MAYHEM [1998: 538]

reference was made to whether a copy of the book Exercises de G �eom�etrie by

F. Gabriel-Marie was available anywhere in Canada.

Dr. Kenneth Williams, Department of Mathematics and Statistics, Car-

leton University, has donated a copy to the Canadian Mathematical Society

for the use of the Editors of CRUX with MAYHEM. We are very grateful to

Dr. Williams for his generous gift. Having ready access to this book will be

of considerable assistance to the members of the Editorial Board of CRUX

with MAYHEM.

113

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased to

consider for publication new solutions or new insights on past problems.

2255. [1997: 300; 1998: 378]Proposed by Toshio Seimiya, Kawasaki,

Japan.

Let P be an arbitrary interior point of an equilateral triangle ABC.

Prove that j\PAB � \PACj � j\PBC � \PCBj.

The published solution [1998: 378-379] is incorrect. More precisely,

that solution deals with an inequality that ismuch simpler thanwhat is given;

among other things, our given inequality does not extend to arbitrary isosce-

les triangles. We thank Toshio Seimiya for these remarks. Here is his solu-

tion.

II. Solution by Toshio Seimiya, Kawasaki, Japan.

LetM be the mid-point ofBC. ThenAM is the perpendicular bisector

of BC, and if P lies on it, the given relation holds with both sides zero.

We may therefore assume, without loss of generality, that P is an interior

point of 4ABM . We then have \PAB < \PAC and \PBC > \PCB,

so that j\PAB � \PACj = \PAC � \PAB and j\PBC � \PCBj =\PBC � \PCB. Thus the relation we wish to prove reduces to

\PAC � \PAB > \PBC � \PCB . (1)

Let Q be the re ection of P in the line AM ; then \PAB = \QAC and

\PCB = \QBC. Thus

\PAC � \PAB = \PAC � \QAC = \PAQ ,

and \PBC � \PCB = \PBC � \QBC = \PBQ ,

so that (1) becomes

\PAQ > \PBQ . (2)

Since PQ ? AM and AM ? BC we get PQkBC. Let PQ meet AB and

AC at R and S respectively, and let T be the re ection of B in RS. Then

\PTQ = \PBQ . (3)

Since AM is the perpendicular bisector of both PQ and RS, the circumcen-

tres of 4APQ and 4ARS lie on AM , so that the circumcircle of \ABC

is tangent at A to the circumcircles of both 4ARS and 4APQ, which we

denote by � and �0 respectively. Note that �0 is contained in �. Because

\TRA = \TRQ� \ARQ = \BRQ� \ARQ = 120� � 60� = 60�, while\ASR = 60�, it follows that \TRA = \ASR. Hence RT is tangent to � so

114

that T is a point outside � and, consequently, T is a point outside �0. SinceA and T are on the same side of PQ, we have

\PAQ > \PTQ . (4)

The desired relation (2) is a consequence of (3) and (4).

Summary of Seimiya's further comments.

The above argument extends to triangles for which \B = \C � 60�;that is j\PAB � \PACj � j\PBC � \PCBj also for isosceles triangles

when \A � 60�. On the other hand, when \B = \C > 60� one can �nd

positions for P where the given inequality fails to hold.

2309. [1998: 47] Proposed by Christopher J. Bradley, Clifton Col-

lege, Bristol, UK.

Suppose thatABC is a triangle and that P is a point of the circumcircle,

distinct from A, B and C. Denote by SA the circle with centre A and radius

AP . De�ne SB and SC similarly. Suppose that SA and SB intersect at P

and PC. De�ne PB and PA similarly.

Prove that PA, PB and PC are collinear.

Solution by 12 of the 17 solvers; the notation of our featured solution

is by Florian Herzig, student, Cambridge, UK.

Since SA and SB are symmetric with respect to their line of centres

AB, their intersections P and PC are also symmetric with respect to this

line. Let QC be the mid-point of PPC , etc. By the above conclusion, QC is

the foot of the perpendicular from P onto AB. The points QA, QB and QC

are collinear since they determine the Simson [or Wallace] line (of P with

respect to 4ABC). A dilatation with centre P and factor 2 takes QA to PA,

etc. Hence PA, PB and PC are collinear as well [in a line parallel to the

Simson line].

Also solved by MICHEL BATAILLE, Rouen, France; NIELS BEJLEGAARD, Stavanger,Norway; FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; MANSURBOASE, student, Cambridge, England; GORAN CONAR, student, Gymnasium Vara�zdin,Vara�zdin, Croatia; NIKOLAOS DERGIADES, Thessaloniki, Greece; WALTHER JANOUS, Ursu-linengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete, Greece;GERRY LEVERSHA, St. Paul's School, London, England; JOEL SCHLOSBERG, student, Bay-side, NY, USA; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel, the Nether-lands; PARAYIOU THEKLITOS, Limassol, Cyprus; G. TSINTSIFAS, Thessaloniki, Greece; JOHNVLACHAKIS, Athens, Greece; and the proposer.

Schlosberg found the problem in a slightly di�erent form on page 43 of RossHonsberger's Episodes in 19th and 20th Century Euclidean Geometry. Bellot informs us thatthe line PAPBPC is called the Steiner line of P with respect to4ABC; its properties are dis-cussed in Y. and R. Sortais, La g �eom �etrie du triangle, Exercises r �esolus, Hermann, Paris, 1987.Both he and Seimiya report that the line passes through the orthocentre of4ABC.

115

2312. [1998: 47] Proposed by K.R.S. Sastry, Dodballapur, India.

The rth n{gonal number is given by P (n; r) = (n� 2)r2

2� (n� 4)r

2,

where n � 3, r = 1, 2, : : : .

Prove that, in the interval [P (n; r); P (n; r+ 1)], there is an (n� 1){gonal number.

Solution by Florian Herzig, student, Cambridge, UK.

Assume that for some n � 3 and r, s � 1,

P (n� 1; s) < P (n; r) < P (n; r + 1) < P (n� 1; s+ 1) . (1)

Therefore P (n� 1; s+ 1)� P (n� 1; s) > P (n; r+ 1)� P (n; r).

Now

P (n; r + 1)� P (n; r) = (n� 2)

�(r+ 1)2 � r2

2

�� (n� 4)

1

2

= r(n� 2) + 1 .

Similarly, P (n� 1; s+ 1)� P (n� 1; s) = s(n� 3) + 1.

Hence, from the last inequality above, s(n�3)+1 > r(n�2)+1, and thus

n � 4 and s >r(n� 2)

n� 3.

Using the fact that P (n � 1; s), as a real function in s, is strictly in-

creasing for all s � 1 (veri�ed by di�erentiation), we have

P (n� 1; s) > P

�n� 1;

r(n� 2)

n� 3

�=

(n� 2)2

(n� 3)

r2

2�

(n� 2)(n� 5)

(n� 3)

r

2

� P (n; r);

since(n� 2)2

n� 3� n� 2 and

(n� 2)(n� 5)

n� 3� n� 4 .This contradicts (1)

and hence it follows that in [P (n; r); P (n; r + 1)] there is an (n� 1){gonalnumber.

Also solved by MANSUR BOASE, student, Cambridge, England;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; RICHARD I. HESS, RanchoPalos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;MICHAEL LAMBROU, University of Crete, Crete, Greece; and the proposer.

It is worth noting that all seven solutions involved di�erent methods. Boase showed

that P�n� 1;

lqn�2

n�3

m�always lies in the interval [P (n; r); P (n; r+1)]. Janous and others

noted that the involvement of (n� 1){gonal numbers implies that n� 1 � 3; that is, n � 4.Referring to the geometry involved, the proposer noted that the problem can be reformulated:Prove that there are just one or just two (n� 1){gonal numbers in [P (n; r); P (n; r + 1)].

116

2313. [1998: 47] Proposed by Heinz-J �urgen Sei�ert, Berlin, Ger-

many.

Let N be a non-negative integer and let a and b be complex numbers

with a, b 62 f0, �1, �2, : : : , �(n� 1)g. Find a closed form expression for

nXk=0

(�1)k

(a)k (b)n�k,

where (a)k denotes the Pochhammer symbol, de�ned by (a)0 = 1,(a)k = a(a+ 1) : : : (a+ k � 1), k 2 N.

Solution by Michael Lambrou, University of Crete, Crete, Greece.

If S(a; b; n) denotes the sought sum, we show by induction on n that

S(a; b;n) =(a� 1)n+1 + (�1)n(b� 1)n+1

(a+ b+ n� 2)(a)n(b)n.

Indeed, for n = 1 it is true since

S(a; b; 1) =1

b�

1

a=

(a� b)(a+ b� 1)

(a+ b� 1)ab=

(a� 1)a� (b� 1)b

(a+ b� 1)ab.

The induction step uses the facts

c(c+ 1)t = (c)t+1 and (c+ t)(c)t = (c)t+1

for all c and t, and the observation

S(a; b; n+ 1)

=

n+1Xk=0

(�1)k

(a)k(b)n+1�k

=1

b

nXk=0

(�1)k

(a)k(b+ 1)n�k+

(�1)n+1

(a)n+1

=1

bS(a; b+ 1; n) +

(�1)n+1

(a)n+1

=1

b

�(a� 1)n+1 + (�1)n(b)n+1

(a+ b+ n� 1)(a)n(b+ 1)n

�+

(�1)n+1

(a)n+1

=(a� 1)n+2 + (�1)n(b)n+1(a+ n) + (�1)n+1(a+ b+ n� 1)(b)n+1

(a+ b+ n� 1)(a)n+1(b)n+1

=(a� 1)n+2 + (�1)n+1(b� 1)n+2

(a+ b+ n� 1)(a)n+1(b)n+1

,

as required.

Also solved by KEE-WAI LAU,Hong Kong; and the proposer. One other reader submitteda solution which was not in closed form.

Lau notes that a+b+n�2 6= 0 is required for this problem, and that if a+b+n�2 = 0

then the given expression likely does not have a closed form.

117

The proposer gives a similar expression fornXk=0

(�1)k(a)k(b)n�k, and lists some com-

binatorial identities as special cases of these two results. For instance, from the formula forS(1=2; 1=2; 2n) and the observation that (1=2)k = 4�k(2k)!=k! he obtains

2nXk=0

(�1)k�4n

2k

��2n

k

� =1

1 � 2n;

and as a special case of his formula for

nXk=0

(�1)k(a)k(b)n�k he gets

2nXk=0

(�1)k�2n

k

� =2n + 1

n+ 1.

2314. [1998: 107] Proposed by Toshio Seimiya, Kawasaki, Japan.

Given triangle ABC with AB < AC. The bisectors of angles B and C

meet AC and AB at D and E respectively, and DE intersects BC at F .

Suppose that \DFC = 1

2(\DBC � \ECB). Determine angle A.

Solution by Florian Herzig, student, Cambridge, UK.

First we show that DI = IE where I is the incentre. We have

1

2\DIC =

1

2(\DBC + \ECB) = \DFC + \ECB

= \ECF + \EFC = \DEC .

Since \DIC = \DEC + \EDB, it follows that DI = EI. Now by the

Sine Law:AI

sin\ADI=

ID

sin A

2

=IE

sin A

2

=AI

sin\AEI

and hence either \ADI = \AEI or \ADI+\AEI = 180�. From the �rst

of these it easily follows that the angles B and C are equal, a contradiction.

From the second it follows that the quadrilateral AEID is cyclic. Hence

180� = A+ \EID = A+

�90� +

A

2

�,

and hence A = 60�.

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain;MANSUR BOASE, student, Cambridge, England; CHRISTOPHER J. BRADLEY, Clifton College,Bristol, UK; GORAN CONAR, student, Gymnasium Vara�zdin, Vara�zdin, Croatia; NIKOLAOSDERGIADES, Thessaloniki, Greece; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Con-necticut, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEE-WAI LAU,Hong Kong; GERRY LEVERSHA, St. Paul's School, London, England; D.J. SMEENK, Zaltbom-mel, the Netherlands; PARAYIOU THEOKLITOS, Limassol, Cyprus; and the proposer. Therewere two incorrect solutions.

118

2315. [1998: 107] Proposed by V�aclav Kone �cn �y, Ferris State Uni-

versity, Big Rapids, Michigan, USA.

Prove or disprove that F (n) =

sn

�1�

1

n

�n�1, where F (n) is the

maximum value of

f(x1; x2; : : : ; xn) = sinx1 cosx2 : : : cosxn + cosx1 sinx2 : : : cosxn

+ : : :+ cosx1 cosx2 : : : sinxn ,

xk 2 [0; �=2], k = 1, 2, : : : , n, and n > 1 is a natural number.

Solution by Heinz-J �urgen Sei�ert, Berlin, Germany.

Let n � 2 and let

C(n) =n� 1

n�1pnn�2

.

Then [1998: 119] if a1, : : : , an � 0,

nXk=1

pak �

nYk=1

pak + C(n) .

Replacing ak by C(n)bk and noting that C(n)(n�1)=2 = F (n) give

nXk=1

pbk � F (n)

nYk=1

pbk + 1 , (1)

valid for all b1, : : : , bn � 0.

If x1, : : : , xn 2 [0; �=2), let bk = tan2 xk. Sincepbk + 1 = secxk,

we obtain, from (1), after multiplying byQn

j=1 cosxj ,

nXk=1

sinxk

nYj=1;j 6=k

cosxj � F (n) .

Clearly, the latter inequality remains valid if xk = �=2 for some k. Since

there is equality if xk = arcsin(1=pn) for all k, it follows that F (n), in fact,

is the maximum value of the considered expression.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; THEODORECHRONIS, student, Aristotle University of Thessaloniki, Greece; NIKOLAOSDERGIADES, Thes-saloniki, Greece; RICHARD I. HESS, RanchoPalos Verdes, California, USA; WALTHER JANOUS,Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete,Greece; and PANOS E. TSAOUSSOGLOU, Athens, Greece.

The problem was inspired by problem 2214. Janous's solution also made use of 2214.

119

2316. [1998: 108] Proposed by Toshio Seimiya, Kawasaki, Japan.

Given triangle ABC with angles B and C satisfying C = 90� + 1

2B.

Suppose that M is the mid-point of BC, and that the circle with centre A

and radius AM meets BC again at D. Prove that MD = AB.

Solution by David Doster, Choate Rosemary Hall, Wallingford, Con-

necticut, USA.

Let E be the other point where the circle centred at A with radius AC

meets line BC. Then \AEC = \ACE, so that \AED = \ACM: Also,

AD = AM and \ADE = \AMC. Therefore, 4ADE �= 4AMC. Hence,

DE = CM . But CM = MB; therefore, DE =MB, so that MD = BE.

Note that \AEC = \ACE = 90� � 1

2\B. Hence,

\BAE = 180� ��\B + 90� �

1

2\B

�= 90� �

1

2\B .

Thus, AB = BE. Therefore, MD = AB.

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; FRANCISCOBELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; MANSURBOASE, student, Cambridge,England; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; GORAN CONAR, stu-dent, Gymnasium Vara�zdin, Vara�zdin, Croatia; NIKOLAOS DERGIADES, Thessaloniki, Greece;FLORIAN HERZIG, student, Cambridge, UK; WALTHER JANOUS, Ursulinengymnasium, Inns-bruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA;MICHAEL LAMBROU, University of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul's School,London, England; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; JOEL SCHLOSBERG,student, Robert Louis Stevenson School, New York, NY, USA; RON SHEPLER, Ferris State Uni-versity, Big Rapids, Michigan, USA; D.J. SMEENK, Zaltbommel, the Netherlands; PARAYIOUTHEOKLITOS, Limassol, Cyprus; PANOS E. TSAOUSSOGLOU, Athens, Greece; PAUL YIU,Florida Atlantic University, Boca Raton, Florida, USA; and the proposer.

2317. [1998: 108]Proposedby Richard I. Hess, Rancho PalosVerdes,California, USA.

� 2�

4� 3�

b a

c

de

�

The quadrilateral shown at the left

has integer elements a through e.

The angles as shown are integer

multiples of the smallest.

(a) What is the smallest possible

value of c?

(b) What is the smallest possible

value of c if �must be obtuse?

Solution by Michael Lambrou, University of Crete, Crete, Greece.

We show that the answer to (a) is c = 5 � 17 � 29 � 1049 = 2585785 and

to (b) it is c = 19 � 41 � 59 � 139 � 50539 = 322872394081.

120

(a) From the Law of Sines on the upper triangle we have

a

sin�=

b

sin 2�=

c

sin 3�

and as sin2� = 2sin � cos � and sin3� = (4cos2 � � 1) sin �, this reducesto

a =b

2 cos �=

c

4 cos2 � � 1.

Eliminating cos � we �nd cos � = b=2a and so

a2 + ac� b2 = 0 . (1)

As a quadratic in a this last must have discriminant a perfect square, say x2.

That is,

c2 + (2b)2 = x2 . (2)

Hence there exist s, t, m 2 N with (s; t) = 1 and

c = (s2 � t2)m , 2b = 2stm , x = (s2 + t2)m . (3)

(Note that if c is odd this is certainly so, but when c is even we may have the

dual relations:

2b = (s2 � t2)m , c = 2stm , x = (s2 + t2)m .

However, this amounts to the same thing because then, by (2), x is even; so

eitherm is even or s2+ t2 is even. In the �rst case setm = 2M , s1 = s+ t,t1 = s� t and we then have c = (s21 � t21)M , b = s1t1M , x = (s21 + t21)Mwhich is the same as (3). If, instead,m is odd, so s2+ t2 is even, we see thats, t are of the same parity; set s1 = (s+ t)=2, t1 = (s � t)=2, M = 2m,

and we recapture (3), since s1, t1,M 2 N).

From (1) (keeping the positive sign only) we have

a =�c+

px2

2= t2m , so cos � =

b

2a=

s

2t.

From the lower triangle we have 0 < 3� + 4� < 180�; that is

1 > cos � > cos �

7� 0:90096887 > 0:9 .

Using cos � = s=2t we have

1:8t < 2t cos �

7< s < 2t . (4)

In particular t cannot be 1, 2, 3, 4, or 5, as there is no integer s in the open

interval�2t cos �

7; 2t�. The least allowable t is t = 6 (whence s = 11). For

the record the next few allowable pairs are (t; s) = (7;13), (8;15), (9; 17).

121

From the Law of Sines on the lower triangle we have

d

sin 3�=

e

sin4�=

c

sin7�.

Hence d sin 4� = e sin 3� or

d(8 cos3 � � 4 cos �) sin � = e(4 cos2 � � 1) sin � .

Using cos � = s=2t we get

ds(s2 � 2t2) = et(s2 � t2) . (5)

Recall that (s; t) = 1. We show that we also have

(s; s2 � t2) = (t; s2 � 2t2) = (s2 � 2t2; s2 � t2) = 1 .

Indeed, if p is a prime and pjs, pjs2� t2, then pjt2, so pjt. Thus pj(s; t) = 1,showing that (s; s2 � t2) = 1. Similarly (t; s2 � 2t2) = 1, and if pjs2 � 2t2,pjs2 � t2, then pjs2 � t2 � (s2 � 2t2) = t2, so pjt, etc., as before.

These conditions of relative primality applied to (5) show that there

exists a constant � 2 N such that

d = �t(s2 � t2) and e = �s(s2 � 2t2) . (6)

Now

c =d sin 7�

sin3�=

d(64 cos6 � � 80 cos4 � + 24 cos2 � � 1) sin �

(4 cos2 � � 1) sin �

=�(s6 � 5s4t2 + 6s2t4 � t6)

t3by (6) .

From (3) we have

mt3(s2 � t2) = �(s6 � 5s4t2 + 6s2t4 � t6) . (7)

Observe that

(t; s6 � 5s4t2 + 6s2t4 � t6) = 1 = (s2 � t2; s6 � 5s4t2 + 6s2t4 � t6) .

Indeed the �rst, as above, is clear. For the second, note that

s6 � 5s4t2 + 6s2t4 � t6 = (s2 � t2)(s4 � 4s2t2 + 2t4) + t6 ,

so any prime dividing both terms must divide s2� t2 and t6 (and so t), which

is impossible, as above. We conclude from (7) that there is a � 2 N such that

m = (s6 � 5s4t2 + 6s2t4 � t6)� , � = t3(s2 � t2)� ,

122

and hence

c = (s2 � t2)m = (s2 � t2)(s6 � 5s4t2 + 6s2t4 � t6)�

= (s2 � t2)(s3 + s2t� 2st2 � t3)(s3 � s2t� 2st2 + t3)� .

For the least c, we clearly need � = 1 (any solution for (s; t; � ) is larger thanfor (s; t; 1)). We shall show that the least c comes from t = 6, s = 11 (and

� = 1).

Denote by c(t; s) the value of c at the pair (t; s) (and � = 1). For

6 � t � 12 the relatively prime pairs (t; s) satisfying (4) are (6; 11), (7; 13),(8; 15), (9; 17), (10; 19), (11;20), (11; 21), and (12; 23). They give

c(6; 11) = 2585785 (the least), c(7; 13) � 1:7 � 107, c(8; 15) � 7:4� 107,c(9; 17) � 2:4 � 108, c(10; 19) � 6:6 � 108, c(11; 20) � 1:6 � 108,c(11; 21) � 1:6 � 109, and c(12; 23) � 3:6 � 109. We now show that

for t � 13 we still get larger values than c(6; 11) by eliminating c as follows.

By (4) we have for the factors of c:

s2 � t2 � (1:8t)2 � t2 = 2:24t2 ;

s3 + s2t� 2st2 � t3 � (1:8t)3 + (1:8t)2 � 2(2t)t2 � t3

= 4:072t3 .

The other factor must be positive (since c, and thus the product of the re-

maining two factors, is positive). As this factor is an integer, it must be at

least 1. Hence

c � (2:24t2)(4:072t3)1 > 9t5 .

Hence if t � 13 we have c > 9 � 135 � 3:3� 106 > c(6; 11).

To conclude part (a) the least value of c is c(6; 11) = 2585785.

(b) If we insist that � is obtuse, we have equivalently 3� + 4� < 90�;that is, � < �

14. So instead of (4) we have the sharper requirements:

s

2t= cos � > cos �

14� 0:9749279 ;

that is,

(1:949)t < 2t cos �

14< s < 2t . (8)

It follows that t � 20, as for 1 � t � 19 there is no s 2 N in the open

interval (1:949t; 2t). The least t is t = 20 with corresponding s = 39 (and

the next few pairs are (21; 41), (22; 43), (23; 45), : : : ). We show that the

least value of c is c(20; 39) = 19 � 41 � 59 � 139 � 50539 � 3:2 � 1011.Certainly, c(20; 39) < c(21; 41) as the expression for c gives c(21; 41) =1240 � 58799 � 6719 � 4:89� 1011. Moreover, for larger t we use estimates

123

arising from (8), namely

s2 � t2 > (1:949t)2 � t2 > 2:798t2 ;

s3 + s2t� 2st2 � t3 > (1:949t)3 + (1:949t)2t� 2(2t)t2 � t3

> 6:202t3 ;

s3 � s2t� 2st2 + t3 > (1:949t)3 � (2t)2t� 2(2t)t2 + t3

> 0:4034t3 .

Multiplying together we �nd c > 7t8. Thus for t � 22 we have

c > 7 � 228 � 3:8� 1011 > c(20; 39). This concludes the proof.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; KENNETH

M. WILKE, Topeka, Kansas, USA; and the proposer.

2318. [1998: 108] Proposed by V�aclav Kone �cn �y, Ferris State Uni-

versity, Big Rapids, Michigan, USA.

Suppose that ABC is a triangle with circumcentre O and circumra-

dius R.

Consider the bisector (`) of any side (say AC), and let P (the \pedal

point") be any point on ` inside the circumcircle.

Let K, L,M denote the feet of the perpendiculars from P to the lines

AB, BC, CA respectively.

Show that [KLM ] (the area of the pedal triangleKLM ) is a decreasing

function of � = OP , � 2 (0; R).

Combination of the solutions by Francisco Bellot Rosado, I.B. Emilio

Ferrari, Valladolid, Spain; Michael Lambrou, University of Crete, Crete,

Greece; Gerry Leversha, St. Paul's School, London, England; and Toshio

Seimiya, Kawasaki, Japan.

Either by referring to problem 2236, or R.A. Johnson'sAdvanced Euclid-ean Geometry, Dover, 1960, theorem 198, page 139, it is known that

[KLM ] =

�R2 � OP 2

4R2

�[ABC] .

Thus, it follows that [KLM ] is a decreasing function of �.

Also solved (in full) by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; GORANCONAR, student, Gymnasium Vara�zdin, Vara�zdin, Croatia; NIKOLAOS DERGIADES, Thessa-loniki, Greece; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; D.J. SMEENK,Zaltbommel, the Netherlands; PARAYIOU THEOKLITOS, Limassol, Cyprus; and the proposer.

Bellot Rosado and Leversha note that there is no need for P to lie on a perpendicularbisector of a side.

Dergiades and Janous observe that if d = R, then [KLM ] = 0, and soK, L,M lie onthe Simson line.

124

Dergiades recalls the result as being known as Steggals' Theorem. He observes that if[ABC] = E and if E0, E1, E2 and E3 are the areas of the triangle [KLM ] when P is theincentre, and the three excentres respectively, then we can deduce, from Euler's Theorem, that

d2 = R(R� 2r) d21 = R(R� 2r1) d2

2 = R(R� 2r2) d23 = R(R� 2r3) .

Steggal's Theorem now gives

E0

E=

r

2R,

E1

E=

r1

2R,

E2

E=

r2

2R,

E3

E=

r3

2R,

and since r1 + r2 + r3 = r + 4R (Bobillier's Theorem), we have

E1 + E2 +E3 = E0 + 2E .

2319. [1998: 108] Proposed by Florian Herzig, student, Cambridge,

UK.

Suppose that UV is a diameter of a semicircle, and that P , Q are two

points on the semicircle with UP < UQ. The tangents to the semicircle at P

andQmeet at R. Suppose that S is the point of intersection of UP and V Q.

Prove that RS is perpendicular to UV .

Editorial note. Some solvers sent in more than one correct solution.

They are indicated by a y after their names.

I. Solution by Diane and Roy Dowling, University of Manitoba,

Winnipeg, Manitoba.

The problem may be generalized slightly as follows: UV is a diameter

of a circle; P and Q are distinct points on the circle; PQ is not a diameter;

P 6= U andQ 6= V ; the tangents to the circle at P andQmeet at R; the lines

UP and V Q meet at S; prove that RS is perpendicular to UV . (The given

conditions ensure that the points of intersection R and S exist, are unique

and distinct from each other.)

Choose coordinate axes and scale so that the origin O is the centre of

the circle, V = (1;0) and consequently U = (�1; 0). Let P = (a; b); thena2 + b2 = 1. Let Q = (c; d); then c2 + d2 = 1. Let R = (x1; y1) and

S = (x2; y2).

The equation of PR is ax + by = 1 and the equation of QR is

cx + dy = 1. Since PR and QR intersect at exactly one point, the de-

terminant of the coe�cients of this system is non-zero. Solving the system

for x we get the x{coordinate of R:

x1 =d� b

ad� bc.

The equation of UP is bx � (a + 1)y = �b and the equation of V Q

is dx � (c � 1)y = d. Since UP and V Q intersect at exactly one point,

namely S, the determinant of the coe�cients of this system is also non-zero.

Solving the system for x we get the x{coordinate of S:

x2 =ad+ bc+ d� b

ad� bc+ d+ b.

125

Therefore

x1 � x2 =d� b

ad � bc�ad+ bc+ d� b

ad� bc+ d+ b=

d2(1� a2)� b2(1� c2)

(ad� bc)(ad� bc+ d+ b)

=d2b2 � b2d2

(ad� bc)(ad� bc+ d+ b)= 0 .

It follows that RS is perpendicular to UV .

II. Solution by Toshio Seimiya y, Kawasaki, Japan.Let T be the intersection of UQ and V P , and letM be the mid-point

of ST . Since\UPV = \UQV = 90�, T is the orthocentre of triangle SUV ,

so that ST ? UV . Hence we have \PTS = \PUV . Since \SPT = 90�

andM is the mid-point of ST we getMP =MT . Thus

\MPV = \MPT = \MTP = \PTS = \PUV .

Hence PM is tangent to the semicircle. SimilarlyQM is tangent to the semi-

circle. Therefore M coincides with R. As ST ? UV , we have RS ? UV .

III. Solution by Keivan Mallahi y, student, Sharif University of Tech-

nology, Tehran, Iran.

Let T be the intersection of UQ and PV . Consider the hexagon

QQUPPV . Note that all of its vertices lie on the semicircle. Applying Pas-

cal's Hexagon Theorem we see that the following points are collinear:

QQ \ PP = R , QU \ PV = T , UP \ V Q = S

(where QQ and PP are intended to be the tangents at Q and P , respec-

tively). Thus it is su�cient to show that the line ST is perpendicular to UV .

To this end note that \UQV = \V PU = 90�, so the lines UQ and PV

are the altitudes of the triangle SUV . Since the altitudes of a triangle are

concurrent, ST must be the third altitude, which completes the proof.

IV. Solution by Michael Lambrou y, University of Crete, Crete, Greece.Consider the nine-point (Feuerbach) circle of triangleUV S. This passes

through P and Q (as P and Q are the feet of the perpendiculars from V

and U , respectively, because \UPV = \UQV = 90�, being angles on the

semicircle). If PV , QU meet at T , the orthocentre, then ST is also an al-

titude meeting UV at D, say. Note that the nine-point circle also passes

through D, through the mid-pointM of ST , and through the mid-point O

of UV (so O is the centre of the semicircle on UV ). Moreover, MO is a

diameter of the nine-point circle (as \MDO = 90�). Thus we also have

PM ? PO. But OP is a radius of the semicircle on UV , so PM , being per-

pendicular to PO at its endpoint P , is a tangent to the semicircle. Similarly

QM is a tangent. In other words the pointM is where these two tangents

meet. ThusM and R coincide and clearly SR (being the same as SM ) is an

altitude of SUV . This conclude the proof that SR ? UV .

126

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; MICHELBATAILLE, Rouen, France; FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid,Spain; MANSUR BOASE, student, Cambridge, England; CHRISTOPHER J. BRADLEY, CliftonCollege, Bristol, UK; GORAN CONAR, student, Gymnasium Vara�zdin, Vara�zdin, Croatia;NIKOLAOS DERGIADES, Thessaloniki, Greece; DAVID DOSTER, Choate Rosemary Hall, Wall-ingford, Connecticut, USA; STERGIOU HARAFAPOS y, Greece; RICHARD I. HESS, RanchoPalos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus-tria; MASOUD KAMGARPOUR, Carson Graham Secondary School, North Vancouver, BritishColumbia; KATHLEEN E. LEWIS, SUNY Oswego, Oswego, NY, USA; JOS �E H. NIETO, Univer-sidad del Zulia, Maracaibo, Venezuela; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Aus-tria; D.J. SMEENK, Zaltbommel, the Netherlands; PARAYIOU THEOKLITOS, Limassol, Cyprus;GEORGE TSAPAKIDIS, Agrinio, Greece; PAUL YIU, Florida Atlantic University, Boca Raton,Florida, USA; and the proposer y. There was one incorrect solution submitted.

Although four methods of solving are given above, this was not exhaustive. Bataille usedinversion and Theoklitos used radical axes in their proofs.

2320. [1998: 108] Proposed by D.J. Smeenk, Zaltbommel, the Neth-

erlands.

Two circles on the same side of the line ` are tangent to it at D. The

tangents to the smaller circle from a variable point A on the larger circle

intersect ` at B and C. If b and c are the radii of the incircles of triangles

ABD and ACD, prove that b+ c is independent of the choice of A.

Solution by Florian Herzig, student, Cambridge, UK.

Let m be the parallel to ` which is also tangent to the smaller circle.

We will show that

b + c is constant if A is a point outside the region between `

andm [so that D is between B and C], and b : c is constant if Ais between these lines.

Denote the larger circle by C1(M ;R) and the smaller circle by C2(N ; r).Let S 2 AB and T 2 AC be the points of contact with C2. Also let IBand IC be the incentres of triangles ABD and ACD. U and V are the feet

of the perpendiculars from IB and IC to `. We will now calculate the ratio

AS : AD which is constant in both cases: let P be the second intersection of

C2 and AD. Then AP � AD = AS2. Also by similarity PD : AD = r : R.Hence

AS2 = AP � AD = AD2

�1�

r

R

�,

and thus the ratio AS : AD is constant for all positions of A.

If A is not between ` andm, then

DU =BD + AD� AB

2=

AD � AS

2=

CD + AD �AC

2= DV

(because BD = BS, CD = CT and AS = AT ). Therefore the two

incircles touch line AD at the same point, call it Y . It also follows that

127

IBIC ? AD. Moreover DN is the middle parallel between IBU and ICV ,

whence 2DX = b + c, where X = DNTIBIC. If L is the mid-point of

AD, then4DLM � 4DY X so that DX : DM = DY : DL.

Using 2DX = b + c, DM = R, and 2DY = 2DU = AD � AS it

follows that

b+ c =2R(AD � AS)

AD= 2R

�1�

r1�

r

R

�= 2(R�

pR2 � Rr)

is constant.

Otherwise, if A is between lines ` and m assume, without loss of

generality, that DB > DC. Then 2DV = AD � AS as before, but

DU =AD +BD � AB

2=

AD + AS

2.

Therefore b : c = DU : DV =AD + AS

AD � AS=R+

pR2 � Rr

R�pR2 � Rr

is constant.

Finally, note that all the calculations of the �rst case remain valid

(mutatis mutandis) if instead of the incircle in the second case we take the

excircle opposite B [so that b is the radius of the excircle of 4ABD that

touches the side AD].

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; GORAN CONAR, student, Gymna-sium Vara�zdin, Vara�zdin, Croatia; NIKOLAOS DERGIADES, Thessaloniki, Greece; WALTHERJANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete,Crete, Greece; GERRY LEVERSHA, St. Paul's School, London, England; TOSHIO SEIMIYA, Ka-wasaki, Japan; PARAYIOU THEOKLITOS, Limassol, Cyprus; HOE TECK WEE, Singapore; andthe proposer.

Bellot found the problem in the Japanese Encyclopedia of Geometry, volume 4, page 261,which provided the reference Journal deMath �ematiques �El �ementaires 34 (1909), problem 7000.Several readers noted that the statement of our problem was not entirely correct, but Seimiyawas the only solver besides Herzig to provide a correct alternative using an excircle.

2323. [1998: 109] Proposed by K.R.S. Sastry, Dodballapur, India.

Determine a positive constant c so that the Diophantine equation

uv2 � v2 � uv � u = c

has exactly four solutions in positive integers u and v.

Solution by Digby Smith, Mount Royal College, Calgary, Alberta.

Since u = �1� c < 0 when v = 1, we have v � 2. Then

v2� v � 1 = (v � 2)2 + 3(v� 2) + 1 > 0

128

and

u =v2 + c

v2 � v � 1> 0 (1)

When c = 61, one veri�es easily that (1) admits no positive integer

solutions for u if v = 5, 6, 7, 8, while v = 2, 3, 4, 9 would yield four

solutions in u and v: (u; v) = (65; 2), (14; 3), (7; 4) and (2; 9). Finally, if

v > 9, then (v � 9)(v+ 7) > 0 would imply that 2(v2� v� 1) > v2 + 61,

making u = v2+61

v2�v�1< 2. However, if u = 1, then v = �62 < 0. Hence (1)

admits no positive integer solutions for u if v > 9. Therefore (1) has exactly

four solutions as listed above.

Also solved by SAM BAETHGE, Nordheim, Texas, USA; MANSUR BOASE, student,St. Paul's School, London, England; CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; FLORIANHERZIG, student, Cambridge, UK; RICHARD I. HESS, Rancho Palos Verdes, California, USA;WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, Uni-versity of Crete, Crete, Greece; KEE-WAI LAU, Hong Kong; GERRY LEVERSHA, St. Paul'sSchool, London, England; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA; PANOS E.TSAOUSSOGLOU, Athens, Greece; and the proposer. There was also one incorrect solution.

While most solutions gave only one value of c, Manes gave two. Baethge and Chroniseach gave three values of c, while Hess gave the following twelve values and claimed that theyare the twelve \lowest" values of c: 51, 61, 156, 321, 336, 402, 431, 486, 526, 611, 761 and771. In terms of the \frequency" of the value of c given, the top three are c = 61, 336 (eachgiven six times) and c = 51 ( given �ve times). The largest value of c given was c = 26461,obtained by Janous. Both Hess and Herzig considered similar problems in which one seeksthe value of c for which the given equation has exactly n solutions for n = 4, 5, 6, : : : .Herzig used a computer to �nd that, for 4 � n � 10, the corresponding minimum values of care 51, 1381, 3966, 33776, 51816 and 14686766, respectively, while Hess list the ten lowestvalues of c for each n such that 4 � n � 8. His lowest values agree with Herzig's.

For those curious, the editor has obtained (by computer) the next few lowest values of cfor n = 4: 776, 816, 1066, 1071, 1146, 1153, 1172, 1201, 1271 and 1360 as those less than1381, the lowest value with �ve solutions.

Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell

Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil

Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,

J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

129

THE ACADEMY CORNERNo. 24

Bruce Shawyer

All communications about this column should be sent to Bruce

Shawyer, Department of Mathematics and Statistics, Memorial University

of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

Here is the second of four articles from the 1998 Canadian UndergraduateMathematics Conference, held at the University of British Columbia in July1998.

Abstracts � R�esum�es

Canadian Undergraduate Mathematics Conference

1998 | Part 2

Topological Censorship:

A Theorem in Global General Relativity

Joel K. Erickson

University of British Columbia

Since the 1960's, modern di�erential geometry and algebraic topology have

been used to study the global properties of space-times. In particular, these tools

have been used to study the questions \what is the topology of our universe?" and

\why don't we observe any non-trivial topology?" Neither the Einstein equations

nor physical considerations provide grounds for restricting the range of possible spa-

tial topologies; there is no reason to assume that spatial slices of our universe have

one of the trivial topologies S3 or R3. Friedman, Schleich, and Witt have proved

that if a number of physically reasonable conditions are satis�ed by space-time, then

non-trivial topological structures cannot be probed; that is, a topological censorship

principle holds. Following a brief introduction to causal structure, the theorem and

its proof will be described. Current e�orts to generalize the theorem to cosmological

models will then be discussed.

Modern Answers to an Old Number Theory Question

Alexandru E. Gh�tz�a

McGill University

What integers can be expressed as a sum of two cubes of rational numbers? This

simply stated problem, more than 350 years old, has been investigated by some of the

130

big names in mathematics: Fermat, Lagrange, Euler, Legendre, Dirichlet. Nonethe-

less, the question has only been solved in a few particular cases. We shall present

some of these from two completely di�erent points of view: that of classical num-

ber theory, illustrated by a 19th century theorem of Sylvester, and that of modern

number theory, represented by the Heegner point construction which has been used

recently to attack the problem.

Les partages d'entiers

Philippe Girard

Universit �e du Qu�ebec �a Montr �eal

Le probl �eme g �en �eral de la th �eorie additive des nombres consiste �a d �eterminer

le nombre de fa�cons d'exprimer un entier naturel en une somme de ces derniers. Le

partages d'entiers (en anglais \integer partitions") s'int �eressent �a ces d �ecompositions

dans la mesure o �u l'ordre des termes est sans importance. Le premier math �ematicien

�a se pencher sur la question et qui apporta des r �esultats int �eressant est Euler dans

les ann �ees 1740. Dans cet expos �e, une l �eg �ere introduction �a la th �eorie des partages

d'entiers sera pr �esent �ee �a l'aide de quelques d �e�nitions, d'exemples, de th �eor �emes

propos �es par Euler et de d �emonstrations.

Variations sur quelques g �en �eralisations en analyse

Alexandre Girouard

Universit �e de Montr �eal

L'analyse c'est vue, au cours du dernier si �ecle, completement transform�ee.

Nous pr �esentons ici quelques une des g �en �eralisations lui ayant donn �ee sa puissance

actuelle ! Le mat �eriel couvert ici est tr �es facilement disponible dans la litt �erature.

Pour cette raison, ce document n'est presque rien de plus qu'une liste de r �ef �erence.

Finite-Di�erence Approximations for Partial Di�erential Equations

Jacinthe Granger-Pich �e

Universit �e de Montr �eal

Partial di�erential equations are used to represent a wide range of phenom-

ena. In particular, they are well used to describe cloud formation. Often, analytical

solutions of such equations cannot be found, so numerical techniques are needed. In

this paper, we will present di�erent �nite-di�erence approximations applied to the

advection and the heat equations, as well as their application to cloud formation.

Cli�ord Algebras and Spinors

Marco Gualtieri

McGill University

Cli�ord algebras are ideally suited for the study of rotations (for any signature

metric). In fact, Cli�ord called his invention \Geometric Algebra."

I will describe the Real and Complex Cli�ord algebras, and the marvellous Bott

8-periodicity. Then I will give a general description of a spinor, and how it is used

in physics to represent \spin - 1/2 particles" like the electron. I may �nally discuss

131

the mysterious coincidence called \Triality" which can be used to understand the

octonions, the exceptional Lie groups, and other exotic animals.

For maximumenjoyment, you should know some linear algebra, (especially the

Sylvester Inertia theorem for real symmetric bilinear forms).

Witten's Formulas for Symplectic Volumes of Moduli Spaces

Patrick Hayden

McGill University

This paper develops the geometric and algebraic tools necessary to understand

Witten's calculation of the symplectic volumes of moduli spaces of at connections

over Riemann surfaces and then goes on to supply a detailed account of his argument.

We summarize the basic constructions and theorems from di�erential geometry that

are required for our investigation of moduli spaces before de�ning the Reidemeister

torsion and proving some useful results for evaluating the torsion of complexes de�ned

over surfaces. Next, we introduce the moduli space of at connections on a Riemann

surface, giving both the geometric construction and the equivalent representation-

theoretic one. Our account ends with a proof of the equivalence of the symplectic

volume and Reidemeister torsion and a calculation for the symplectic volume of the

moduli space of at connections over compact oriented surfaces of genus g � 2.

Reduced Decompositions of Permutations

Sylvie H�ebert

Universit �e du Qu�ebec �a Montr �eal

After having recalled that the symmetric group is generated by adjacent trans-

positions, we will de�ne reduced expressions (or decompositions) of permutations.

The length of such a decomposition is the number of inversions of the permutation.

We will show how to pass from a reduced decomposition to another, and how one

can draw such a decomposition by a con�guration of lines in the plane. Finally we

will deduce the Coxeter presentation of the symmetric group.

A New Characterization of Topology?

Patrick Ingram

Simon Fraser University

In most classical texts on topology we are given several equivalent methods of

representing topological spaces : by open sets, by closed sets, by an interior function,

and by a closure function. We are also given methods of representing a topological

space in terms of another topological space and a function between them that we

wish to be continuous. This author, however, was not able to �nd a characterization

from sequence convergence and/or clustering, and as such this paper explores these

possibilities. This paper contains elementary de�nitions to make it possible for a

student with a minimal pure math background to read it.

132

THE OLYMPIAD CORNERNo. 197

R.E. Woodrow

All communications about this column should be sent to Professor R.E.

Woodrow, Department of Mathematics and Statistics, University of Calgary,

Calgary, Alberta, Canada. T2N 1N4.

We begin this number with the 47th Polish Mathematical Olympiad1995{96 and the problems of the Final Round, written in March 1996. Mythanks go to Marcin E. Kuczma, Warszawa, Poland and to Ravi Vakil, Cana-dian Team Leader to the IMO at Mumbai, for collecting the materials for theCorner.

47th POLISH MATHEMATICAL OLYMPIAD

Problems of the Final Round (March 29{30, 1996)

First Day (Time: 5 hours)

1. Find all pairs (n; r), with n a positive integer, r a real number, forwhich the polynomial (x+ 1)n � r is divisible by 2x2 + 2x+ 1.

2. Given is a triangleABC and a point P inside it satisfying the condi-tions: \PBC = \PCA < \PAB. Line BP cuts the circumcircle of ABCat B and E. The circumcircle of triangle APE meets line CE at D and F .Show that the points A, P , E, F are consecutive vertices of a quadrilateral.Also show that the ratio of the area of quadrilateral APEF to the area oftriangle ABP does not depend on the choice of P .

3. Let n � 2 be a �xed natural number and let a1, a2, : : : , an bepositive numbers whose sum equals 1.

(a) Prove the inequality

2Xi<j

xixj �n� 2

n� 1+

nXi=1

aix2

i

1� ai

for any positive numbers x1; x2; : : : ; xn summing to 1.

(b) Determine all n{tuples of positive numbers x1; x2; : : : ; xn summing to 1for which equality holds.

133

Second Day (Time: 5 hours)

4. Let ABCD be a tetrahedron with

\BAC = \ACD and \ABD = \BDC .

Show that edges AB and CD have equal lengths.

5. For a natural number k � 1 let p(k) denote the least prime numberwhich is not a divisor of k. If p(k) > 2, de�ne q(k) to be the product of allprimes less than p(k), and if p(k) = 2, set q(k) = 1. Consider the sequence

x0 = 1 , xn+1 =xnp(xn)

q(xn)for n = 0, 1, 2, : : : .

Determine all natural numbers n such that xn = 111111.

6. From the set of all permutations f of the set f1, 2, : : : , ng thatsatisfy the condition

f(i) � i� 1 for i = 1, 2, : : : , n ,

choose one, with an equal probability of each choice. Let pn be the proba-bility that the permutation (chosen) satis�es the condition

f(i) � i+ 1 for i = 1, 2, : : : , n .

Find all natural numbers n with pn > 1=3.

Continuing our Northern European theme, we give the problems of the10th Nordic Mathematical Contest of April 1996. My thanks go to Ravi Vakil,Canadian Team Leader to the IMO at Mumbai, for collecting the questions.

10th NORDIC MATHEMATICAL CONTESTApril 11, 1996 (Time: 4 hours)

1. Prove the existence of a positive integer divisible by 1996, the sumof whose decimal digits is 1996.

2. Determine all real x such that xn + x�n is an integer for anyinteger n.

134

3. A circle has the altitude from A in a triangle ABC as a diameter,and intersects AB and AC in the points D and E, respectively, di�erentfrom A. Prove that the circumcentre of triangle ABC lies on the altitudefrom A in triangle ADE, or it produced.

4. A real-valued function f is de�ned for positive integers, and a pos-itive integer a satis�es

f(a) = f(1995) , f(a+ 1) = f(1996) , f(a+ 2) = f(1997) ,

f(n+ a) =f(n)� 1

f(n)+ 1for any positive integer n .

(a) Prove that f(n+ 4a) = f(n) for any positive integer n.

(b) Determine the smallest possible value of a.

As a �nal contest set for your enjoyment we give the problems of theDutch Mathematical Olympiad of September 1995. Again my thanks go toRavi Vakil, Canadian Team Leader to the IMO at Mumbai.

DUTCH MATHEMATICAL OLYMPIADSeptember 15, 1995

Second Round

1. A kangaroo jumps from lattice-point to lattice-point in the (x; y){plane. She can make only two kinds of jumps:

Jump A: 1 to the right (in the positive x{direction) and 3 up (in thepositive y{direction).

Jump B: 2 to the left and 4 down.

(a) The start position of the kangaroo is the origin (0; 0). Show that thekangaroo can jump to the point (19;95), and determine the number of jumpsshe needs to reach that point.

(b) Take the start position to be the point (1; 0). Show that it is impossiblefor her to reach the point (19; 95).

(c) The start position of the kangaroo is once more the origin (0; 0). Whichpoints (m;n) with m, n � 0 can she reach, and which points can she notreach?

2. On a segment AB a point P is chosen. On AP and PB isoscelesand right-angled triangles AQP and PRB are constructed with Q and R atthe same side of AB. M is the midpoint of QR. Determine the set of allpointsM for all points P on the segment AB.

135

r r r

r

r

r

A P B

M

R

Q

3. 101 marbles are numbered from 1 to 101. The marbles are dividedover two baskets A and B. The marble numbered 40 is in basket A. Thismarble is removed from basket A and put in basket B. The average of allthe numbers on the marbles in A increases by 1

4. The average of all the

numbers of the marbles in B increases by 1

4too. How many marbles were

there originally in basket A?

4. A number of spheres, all with radius 1, are being placed in the formof a square pyramid. First, there is a layer in the form of a square with n�nspheres. On top of that layer comes the next layer with (n� 1) � (n� 1)

spheres, and so on. The top layer consists of only one sphere. Determinethe height of the pyramid.

5. We consider arrays (a1; a2; : : : ; a13) containing 13 integers. An ar-ray is called \tame" when for each i 2 f1, 2, : : : , 13g the following conditionholds: if you leave ai out, the remaining twelve integers can be divided intotwo groups in such a way that the sum of the numbers in one group is equalto the sum of the numbers in the other group. A \tame" array is called \turbotame" if you can always divide the remaining twelve numbers in two groupsof six numbers having the same sum.

(a) Give an example of an array of 13 integers (not all equal!) that is \tame".Show that your array is \tame".

(b) Prove that in a \tame" array all numbers are even or all numbers are odd.

(c) Prove that in a \turbo tame" array all numbers are equal.

An astute reader kept track of the problems proposed to the jury butnot used at the International Mathematical Olympiad at Mumbai in 1996given on [1997: 450], for which we did not publish solutions. Thanks goto Mohammed Aassila, now of the Centre de Recherches Math �ematiques,Montr �eal, Qu �ebec, for �lling the gaps.

6. Let n be an even positive integer. Prove that there exists a positiveinteger k such that

k = f(x)(x+ 1)n + g(x)(xn+ 1)

for some polynomials f(x), g(x) having integer coe�cients. If k0 denotesthe least such k, determine k0 as a function of n.

136

SolutionbyMohammed Aassila, Centre de RecherchesMath �ematiques,

Montr �eal, Qu �ebec.

The statement of the problem is equivalent to: prove that for any posi-tive integer n, there exist polynomials f(x), g(x) having integer coe�cientssuch that

f(x)(x+ 1)2n + g(x)(x2n+ 1) = 2 ,

or equivalently to �nd f(x) and g(x) such that

f(x)x2n + g(x)((x� 1)2n + 1) = 2 ,

or equivalently, to �nd g(x) such that g(x)((x� 1)2n + 1)� 2 is divisibleby x2n.

But, this is not di�cult to prove. If

g(x) = a0 + a1x+ � � �+ a2n�1x2n�1 ,

then we have only to choose ai, 0 � i � 2n� 1 as follows:

a0 = 1 and 2ak ��2n

1

�ak�1 +

�2n

2

�ak�2 + � � �+ (�1)k

�2n

k

�= 0

for 1 � k � 2n� 1.

8. Let the sequence a(n), n = 1, 2, 3, : : : , be generated as follows:a(1) = 0, and for n > 1,

a(n) = a(bn=2c) + (�1)n(n+1)=2 .

(Here btc = the greatest integer � t.)

(a) Determine the maximum and minimum value of a(n) over n � 1996, and�nd all n � 1996 for which these extreme values are attained.

(b) How many terms a(n), n � 1996, are equal to 0?

SolutionbyMohammed Aassila, Centre de RecherchesMath �ematiques,

Montr �eal, Qu �ebec.A simple induction shows that a(n) = b(n)� c(n) where b(n) (resp.

c(n)) denote the total number of couplings 00, 11 (resp. 01, 10) in the binaryrepresentation of n.

(To see this, it su�ces to observe that the binary representation of�n2

�is obtained from that of n by deleting the last digit.)

(a) The maximum value of a(n) is the largest n � 1996 for whichc(n) = 0. It is attained for 1111111111 or 1023 where b(n) = 9 and hencea(n) = 9. The minimum is the largest n � 1996 for which b(n) = 0. It isattained for 10101010101 or 1365 where a(n) = �10.

(b) a(n) = 0 if and only if, in the binary representation of n, the num-ber of the same two consecutive digits is equal to the number of di�erent

137

two consecutive digits. Noting that the �rst digit has to be 1 and that suchrepresentation of n can be formed in

�m

m=2

�ways for evenm, we deduce that

there are

�0

0

�+

�2

1

�+

�4

2

�+

�6

3

�+

�8

4

�+

�10

5

�= 351

positive integers n < 211 = 2048 with a(n) = 0. Finally, since 2002, 2004,2006, 2010, 2026 are such that a(n) = 0 but exceed 1996, there are only346 numbers � 1996 with a(n) = 0.

12. Let the sides of two rectangles be fa; bg and fc; dg respectively,with a < c � d < b and ab < cd. Prove that the �rst rectangle can be placedwithin the second one if and only if

(b2 � a2)2 � (bc� ad)2 + (bd� ac)2 .

SolutionbyMohammed Aassila, Centre de RecherchesMath �ematiques,

Montr �eal, Qu �ebec.

F

E H

G

C

D

A

B

c

d

��

a

b

Let ABCD and EFGH

denote respectively the �rstand second rectangle.

ABCD can be placedwithin EFGH if and onlyif(a cos � + b sin� � c ,

b cos � + a sin� � d .

This means that the point of intersection of a cos � + b sin � = c and

b cos � + a sin� = d, which is�bd�acb2�a2 ;

bc�adb2�a2

�, lies outside or on the unit

circle, or equivalently

(b2 � a2)2 � (bd� ac)2 + (bc� ad)2 .

Editorial note: The condition, a < c � d < b, does not appear to havebeen used. However, it is needed to ensure that the intersection point is inthe �rst quadrant.

14. Let ABCD be a convex quadrilateral, and let RA, RB , RC , RD

denote the circumradii of the triangles DAB, ABC, BCD, CDA respec-tively. Prove that RA+RC > RB+RD if and only if \A+\C > \B+\D.

138

SolutionbyMohammed Aassila, Centre de RecherchesMath �ematiques,

Montr �eal, Qu �ebec.

Let the diagonals meet in E. We may suppose that \AED > 90�.Then all of \EDA, \DAE, \EBC, \BCE are acute. Label them as �, �,�0, �0, respectively. Then � + � = �0 + �0.

D

C

B

A

E

�

�

�0

�0

Now, if A, B, C and D were concyclic we would have � = �0 and � = �0.

On the other hand, if \A + \C > 180�, then � < �0 and � > �0.This is easily seen by drawing a circumcircle of 4ABC and examining thepossible locations of B on the extension of DE.

The purpose of isolating acute angles was for a convenient applicationof the Sine Law. We have

RA =AB

2 sin �; RB =

AB

2 sin �0; RC =

DC

2 sin�0; RD =

DC

2 sin�.

Thus if \A + \C = 180� then �0 = � and � = �0 and thereforeRA +RC = RB + RD.

More generally,

\A+ \C > \B + \D (==) \A+ \C > 180�

===) �0 < � , � < �0

(==) RC > RD , RA > RB

===) RA + RC > RB + RD .

The fact that we have equivalence follows by symmetry.

15. On the plane are given a point O and a polygonF (not necessarilyconvex). Let P denote the perimeter of F , D the sum of the distances fromO to the vertices of F , and H the sum of the distances from O to the linescontaining the sides of F. Prove that D2 �H2 � P 2=4.

SolutionbyMohammed Aassila, Centre de RecherchesMath �ematiques,

Montr �eal, Qu �ebec.

We shall prove that D2 �H2 � P 2=4 for any n{gon. We use \induc-tion" on n.

Ifn = 2, the inequalityD2�H2 � P 2=4 is equivalent to (a+b)2�4h2 � c2.

139

A BD c

ah

b

O

To prove this, consider a line (L) passing throughO and parallel toAB,and let B1 be the image of B under re ection in (L). ThenOA+ OB = OA+ OB1 � AB1; that is a+ b �

p4h2 + c2 , as required.

Now, let the n{gon F be P1P2 : : : Pn and let di = OPi, pi = PiPi+1,hi = distance from O to PiPi+1. Then by the \induction hypothesis" wehave for each i

di + di+1 �q4h2i + p2i .

Summing these inequalities over i = 1, 2, : : : , n we obtain

2D �Xi

q4h2i + p2i ,

or equivalently 4D2 ��P

i

p4h2i + p2i

�2.

Now, 4H2 + P 2 = 4(P

i hi)2 + (

Pi pi)

2, so if we prove that

Xi

q4h2i + p2i �

vuut4

Xi

hi

!2

+

Xi

pi

!2

,

then we are done. But the above inequality follows from the triangle in-equality applied to the sum of the vectors (2hi; pi).

17. A �nite sequence of integers a0, a1, : : : , an is called quadratic if

for each i in the set f1, 2, : : : , ng we have the equality jai � ai�1j = i2.

(a) Prove that for any two integers b and c, there exist a natural number nand a quadratic sequence with a0 = b and an = c.

(b) Find the smallest natural number n for which there exists a quadraticsequence with a0 = 0 and an = 1996.

SolutionbyMohammed Aassila, Centre de RecherchesMath �ematiques,

Montr �eal, Qu �ebec.This problem is well known | it is problem 250 from the book of Sier-

pinski!!

Next, a bit more tidying up. Last issue we gave solutions by the readersto problems of the 17th Austrian-Polish Mathematics Competition[1998: 4{6; 1999: 77{85]. Next we give a solution to problem 6.

140

6. Let n > 1 be an odd positive integer. Assume that the integers x1,x2, : : : , xn � 0 satisfy the system of equations

(x2 � x1)2 + 2(x2 + x1) + 1 = n2 ,

(x3 � x2)2 + 2(x3 + x2) + 1 = n2 ,

...............................(x1 � xn)

2 + 2(x1 + xn) + 1 = n2 .

Show that either x1 = xn, or there exists j with 1 � j � n� 1, such thatxj = xj+1.

Solutionadapted from one by Pierre Bornsztein, Courdimanche, France.We show, by induction onm, that, if non-negative integers x1, x2, : : : ,

xm � 0, with m > 1 being an odd integer, satisfy

(x2 � x1)2 + 2(x2 + x1) + 1 = n2 ,

(x3 � x2)2 + 2(x3 + x1) + 1 = n2 ,

................................(x1 � xm)2 + 2(x1 + xm) + 1 = n2 ,

then either x1 = xm, or there is j with 1 � j � m � 1, such thatxj = xj+1. We adopt the convention that subscripts are read modulo m,so that xm+1 = x1, etc..

Now notice that for each j, xj�1 andxj+1 are solutions (possibly equal)to the quadratic equation

X2 +X(2� 2xj) + (xj + 1)2 � n2 = 0 (Ej) ,

for which the discriminant�j = 4(n2� 4xj) � 0 for there to be a real root.Moreover as n is odd, �j � 1.

Also because (Ej) has integral coe�cients and at least one integral root,both roots are integers, and they are distinct as �j � 1. Denote the rootsby �j and �j with �j < �j .

Also we have �j + �j = 2xj � 2, and that fxj�1; xj+1g � f�j; �jg.We claim that either �j = xj � 2 and �j = xj or �j < xj < �j. (Indeedif �j ; �j � xj then �j + �j > 2xj � 2, and if �j ; �j < xj then �j + �j �2xj � 3.)

Now let j be such that xj = maxfx1; : : : ; xng. Now xj�1 and xj+1

are roots of Ej. Unless xj�1 = xj or xj = xj+1 we must have

xj�1 = xj+1

or xj�1 < xj < xj+1

or xj+1 < xj < xj�1 .

The last two cases contradict the choice of j, so xj�1 = xj+1. Ifm = 3 weare done since xj�1 and xj+1 are cyclically adjacent. Otherwise m > 3 andby removing xj�1; xj we obtain a solution with m� 2 values, to which theinduction hypothesis applies.

141

Now we turn our attention to solutions from the readers to �ve of theproblems of the Iranian National Mathematical Olympiad, February 6, 1994,Second Round [1998: 6{7].

IRANIAN NATIONALMATHEMATICAL OLYMPIADFebruary 6, 1994Second Round

1. Suppose that p is a prime number and is greater than 3. Prove that7p � 6p � 1 is divisible by 43.

Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;

by Michel Bataille, Rouen, France; by Pierre Bornsztein, Courdimanche,

France; by Yeo Keng Hee, Hwa Chong Junior College, Singapore; by Murray

S. Klamkin, University of Alberta, Edmonton, Alberta; by Pavlos

Maragoudakis, Pireas, Greece; by Bob Prielipp, University of Wisconsin{

Oshkosh, Wisconsin, USA; and by Edward T.H. Wang, Wilfrid Laurier Uni-

versity, Waterloo, Ontario. We give Klamkin's generalization.

More generally we show that [(x + 1)n � xn � 1]=[x2 + x + 1] isan integral polynomial for all natural odd n not divisible by 3. The givenproblem corresponds to the special case x = 6.

By the Remainder Theorem (x + 1)n � xn � 1 will be divisible byx2 + x+ 1 if it vanishes for the zeros of the latter, which are ! and !2, thetwo complex cube roots of unity. Here,

(!+ 1)n � !n � 1 = (�1)n!2n� !n � 1 .

As known, !2m+!m+1 = 0 if and only ifm is not divisible by 3. The samealso applies to the other zero !2. Also we must have (�1)n to be 1, so thatn is even.

Comment. Similarly, [(x + 1)n + xn + 1] is divisible by [x2 + x + 1]

for all natural even n not divisible by 3. As a special case, 7n + 6n + 1 isdivisible by 43 for any of these values of n.

3. Let n and r be natural numbers. Find the smallest natural numberm satisfying this condition: For each partition of the set f1, 2, : : : , mg intor subsets A1, A2, : : : , Ar, there exist two numbers a and b in some Ai

(1 � i � r) such that 1 < ab� 1 + 1

n.

Solution by Yeo Keng Hee, Hwa Chong Junior College, Singapore.

I claim the answer is (nr + r). Suppose m < nr + r. Then foreach i � m, let i be put into the subset Aj , (j = 1, 2, : : : , r) such thati � j (mod r). Then for any 2 numbers a, b in the same subset witha > b, we have b � (a� r). Thus

a=b = (1 + (a� b)=b) � (1 + r=b) > (1 + r=nr) = (1 + 1=n) .

142

Therefore the condition in the question cannot hold. For m = (nr + r),consider the (r+1) numbers nr; (nr+1); : : : ; (nr+ r). By the PigeonholePrinciple, among them there exists a > b such that a and b are in the samesubset. Then

a=b = (1 + (a� b)=b) � (1 + r=b) � (1 + r=nr) = (1 + 1=n) ,

and 1 < a=b and we are done.

4. G is a graph with n vertices A1, A2, : : : , An such that for eachpair of non-adjacent vertices Ai and Aj there exists another vertex Ak thatis adjacent to both Ai and Aj.

(a) Find the minimum number of edges of such a graph.

(b) If n = 6 and A1, A2, A3, A4, A5, A6 form a cycle of length 6, �ndout the number of edges that must be added to this cycle such that the abovecondition holds.

Solution by Yeo Keng Hee, Hwa Chong Junior College, Singapore.

(a) G must be a connected graph. Thus it must have at least (n � 1)

edges. If the (n� 1) edges are the edges that join Ai, (i = 1, 2, : : : , n� 1)

to An, the condition is indeed satis�ed. Thus the minimum number of edgesis (n� 1).

(b) Without loss of generality, let Ai be joined to Ai�1 and Ai+1,(i = 1, 2, : : : , 6), A0 = A6, A7 = A1. Then consider the three pairs ofvertices (A1, A4), (A2, A5), (A3, A6). In each pair of vertices, the degree ofat least one vertex needs to be increased by 1, because any path joining thetwo vertices has length at least 3, so one vertex must be joined to the othervertex or a vertex adjacent to that vertex.

Thus the sum of degrees of all the vertices must increase by at least 3;that is, at least two edges must be added. Adding the edgesA1A4 andA2A6

satis�es the conditions.

5. Show that if D1 and D2 are two skew lines, then there are in�-nitely many straight lines such that their points have equal distance from D1

and D2.

SolutionsbyMichel Bataille, Rouen, France; and byMurray S. Klamkin,

University of Alberta, Edmonton, Alberta. We give Bataille's solution.

Let H1 on D1, H2 on D2 be such that: H1H2 ? D1 and H1H2 ? D2

and let O be the midpoint ofH1H2. We will work in the following system ofrectangular axes: we take O as origin, the z{axis alongH1H2 and, as x{axisand y{axis, we take the bisectors of the angle formed at O by the parallelsto D1 and D2. Then there exist non-zero real numbers a andm such that:

D1 is the intersection of the planes (P1) : z = a and (Q1) : y = mx;

D2 is the intersection of the planes (P2) : z = �a and (Q2) : y = �mx.

143

Now, letA(�;0; 0) be a point lying on the x{axis and�A be the line throughA directed by ~u(0;1; t). We show that it is possible to choose t (dependingon �) such that each point M(�; k; kt), (k 2 R) of �A has equal distancefrom D1 and D2.

As the planes (P1) and (Q1) are perpendicular, we have:

[d(M;D1)]2 = [d(M;P1)]

2 + [d(M;Q1)]2 = (kt� a)2 +

(k�m�)2

1 +m2.

Similarly: [d(M;D2)]2 = (kt+ a)2 +

(k+m�)2

1 +m2.

Hence, d(M;D1) = d(M;D2) is equivalent to 4k

�at+

m�

1 +m2

�= 0

so that, by choosing t = � m�

a (1 +m2), we have d(M;D1) = d(M;D2) for

allM on �A.

Since �A 6= �B whenever A 6= B on the x{axis (�A and �B are instrictly parallel planes), the family of lines�A, when � takes all real values,answers the question.

6. f(x) and g(x) are polynomials with real coe�cients such that for

in�nitely many rational values x, f(x)

g(x)is rational. Prove that f(x)

g(x)can be

written as the ratio of two polynomials with rational coe�cients.

Solution by Murray S. Klamkin, University of Alberta, Edmonton, Al-

berta.With little change, a solution of this problem is given in [1]. Without

loss of generality we can assume that f and g are relatively prime polynomialsand let r denote the sum of the degrees of f and g. Also letR(x) = f(x)=g(x), assuming that the degree of f is � than the degreeof g. If not, we consider R�1(x). For r = 0, the result is obvious. Now leta be one of the rational numbers such that g(a) 6= 0 and R(a) is rational.Now de�ne f1(x) by

f1(x)=g(x) = fR(x)� f(a)=g(a)g=(x� a) .

Then f1(x)=g(x) is also rational for all the rational values that f(x)=g(x) isrational except x = a. Since

f1(x) = [g(a)f(x)� f(a)g(x)]=g(a)(x� a) ,

the degree of f1(x) is less than that of f(x). Thus the sum of the degreesof f1(x) and g(x) is less than that of f(x) and g(x). The desired result nowfollows by mathematical induction.Reference: [1] G. Polya, G. Szego, Problems and Theorems in Analysis,Springer-Verlag, II, NY, 1976, pp. 130, 321, Problem 92.

144

Next we look at readers' solutions to two problems of the Japan Math-ematical Olympiad, Final Round, February 1994 given in [1998: 68{69].

1. For a positive integer n, let an be the nearest positive integer topn, and let bn = n+ an. Dropping all bn (n = 1, 2, : : : ) from the set of

all positive integers N , we get a sequence of positive integers in ascendingorder fcng. Represent cn by n.

Solution by Pierre Bornsztein, Courdimanche, France.

On a

a1 = 1; b1 = 2

a2 = 1; b2 = 3

a3 = 2; b3 = 5

etc..

La suite fang est croissante, donc fbng est strictement croissante.

Soit n 2 N�. On a n <

pn2 + n < n + 1

2, donc an2+n = n, et alors

bn2+n = (n + 1)2 � 1. En plus n + 1

2<pn2 + n+ 1 < n + 1, donc

an2+n+1 = n+ 1, et alors bn2+n+1 = (n+1)2 + 1. On en d �eduit que fbngne contient aucun carr �e.

Pour n 2 N�, la cardinalit �e de l'ensemble fb1; : : : ; bn2+ng = n2 + n, et lacardinalit �e de fp 2 N� : p � (n+1)2�1 et p n'est pas un carr �eg = n2+n.

De plus,

fb1 , : : : ,bn2+ng � fp 2 N� : p � (n+ 1)2 � 1 , p non carr �eg

d'o �u

fb1 , : : : ; bn2+ng = fp 2 N� : p � (n+ 1)2 � 1 , p non carr �eg ,

et comme n est arbitraire

fb1 ,b2 , : : : g = N�nf1 ,4 ,9 ,16 , : : : g .

On en obtient pour n � 1, cn = n2 .

4. We consider a triangle ABC such that \MAC = 15� where M isthe midpoint of BC. Determine the possible maximum value of \B.

Solutions by Pierre Bornsztein, Courdimanche, France; and by D.J.

Smeenk, Zaltbommel, the Netherlands. We give the solution of Smeenk.

Let �(O;R) be the circumcircle of 4AMC, with centre O andradius R. Denote \B = �. Now � has its maximum when AB touches� at A. We denote \BAM = \ACM = x.

145

O

C

A

B M

� x

�

x15�

Since AM is a median in4ABC we have:

sinx : sin 15� = sin� : sinx ,

2 sin2 x = 2sin� sin15� ,

1� cos 2x = cos(� � 15�)� cos(� + 15�) .

Since � + 15� + 2x = 180�, we have 2x = 165� � �. Thus1 + cos(� + 15�) = cos(� � 15�)� cos(� + 15�), or

2 cos(� + 15�)� cos(� � 15�) + 1 = 0 . (1)

It is easy to verify that this is satis�ed by � = 105�.

Write � = 105� + y. From (1) we obtain

2 cos(120� + y)� cos(90� + y) + 1 = 0 ,2 cos 120� cos y � 2 sin120� siny + siny + 1 = 0 ,

1� cos y = siny�p

3� 1�,

2 sin2�1

2

�y = 2 sin

�1

2y

�cos

�1

2y

�(p3� 1) .

Therefore, sin�1

2y�= 0 ===) y = 0 or

tan

�1

2y

�=

p3� 1

===) y = 72:4 : : :�

===) � = 177:4 : : :� .

Since x > 15�, this does not hold. So � has a maximum of 105�.

That completes the Corner for this issue. Do not forget to sendme Olympiadcontests and your nice solutions to the problems we have given.

146

BOOK REVIEWS

ALAN LAW

Interdisciplinary Lively ApplicationProjects (ILAPs), edited byDavid C. Amey,Published by The Mathematical Association of America, USA, 1997,ISBN# 0-88385-706-5, softcover, 222+ pages.Reviewed by T. W. Leung, Hong Kong Polytechnic University.

This book is a collection of eight modelling projects (interdisciplinarylively application projects) developed by academics from a consortium oftwelve schools led by the United States Military Academy. It is not a bookon modelling theories and techniques [1], nor a collection of many modellingcases [3]. Rather, every project is a step-by-step guide to how a situationarises that requires mathematical modelling, about complications involvedwhen trying to set up a model, and means to solve the model. In spirit, theprojects are akin to the UMAP modules.

Mathematics backgrounds employed in these projects are calculus, lin-ear algebra and basic statistics. For instance, in the project \Getting Fitwith Mathematics", simple regression analysis is used to compare severalvariables, whereas in \Parachute Panic", the basic ingredient is a �rst orderordinary di�erential equation, while in \Flying with Di�erential Equation",it is in essence a demonstration of oscillation and resonance. Making useof vector calculus, Stokes's theorem and the like, the project \ContaminantTransport" is mathematically the most sophisticated.

Nevertheless because these projects were produced as \interdiscipli-nary" e�orts, new elements were added. We learn about oxygen consump-tion, ATP, VO2 and others in \Getting Fit with Mathematics", and it is in-teresting to observe so many variables playing a part. In the \Decked Out",only simple algebraic manipulations are used to solve the model, but wealso learn about Bill of Materials (BOM). In \Parachute Panic", besides thefree fall, we see also \severity index" and the interplay of di�erent forces. Anovelty of these projects is that they were produced by cooperative e�ortsfrom di�erent disciplines, which makes the projects more realistic, insteadof merely exercises in mathematics (for comparison, see the review [4]).

The authors take a \no nonsense" approach, analyze the situation, setup a model using simplemathematics, solve the model, freely using computerpackages if necessary. The hard part of each project is the formulation ofthe model, and the problem of veri�cations of a model is often unanswered.They demonstrated using mathematics as a \tool", and a tool should be sim-ple to use (user-friendly), and serves few special functions, I believe. Timeand again I was told by my colleagues from engineering departments thatwe should teach mathematics as a tool, and that it may help them to solveproblems. This is a point often neglected by mathematicians, and our sense

147

of importance in mathematics may di�er from the engineers or the scientists.Concerning using computer packages, it seems the authors concede that stu-dents may not understand the theories involved, and it is OK. A similar viewwas echoed in [2].

Another point seemingly to be advocated by the authors is that studentsmay learn the necessary mathematics during the modelling process. Thisis in line with the recent movement of curriculum reform (see p.205, alsofor example [5]). In these days the students' backgrounds are so weak andvaried, perhaps it is better that students take basic courses in calculus andlinear algebra �rst, before entering into this kind of endeavour.

In short it is nice to try these projects in a modelling course, askingstudents to look for variations and veri�cations. Therefore this book servesas a good reference for investigation.

References

1. D.E. Edwards and M. Hamson, Guide to Mathematical Modelling,Macmillan Education Limited, London, 1989.

2. M.S. Jones, Teaching Mathematical Modelling, Int. J. Math. Edn. inSci. & Tech., Vol. 28, # 4, July/August, 1997, pp. 553{560.

3. M.S. Klamkin (Editor), Mathematical Modelling, Classroom Notes in

Applied Mathematics, SIAM, 1987.

4. D. Lawson, Review of Industrial Mathematics: A Course in Solving

Real World Problems, Bull. IMA, Vol. 31, # 7/8, July/August, 1995,pp. 124{125.

5. D. O'Shea and H. Pollatsek, Do we need prerequisites? Notices of theAMS, Vol. 44, # 5, 1997, pp. 564{570.

148

Euler's Triangle Theorem

G.C. Shephard

1. In 1780, Leonhard Euler proved the following remarkable result[4, p. 96]:Theorem: Let [A;B;C] be an arbitrary triangle and O any point of the planewhich does not lie on a side of the triangle. Let AO, BO, CO meet BC, CAand AB in the points D, E, F respectively. Then

jjAOjjjjODjj +

jjBOjjjjOEjj +

jjCOjjjjOF jj =

jjAOjjjjODjj �

jjBOjjjjOEjj �

jjCOjjjjOF jj + 2 . (1)

B D C

A

E

FO

Figure 1

To begin with we shall assume, like Euler, that O lies in the interior of the

triangle, see Figure 1. For the present, the notationjjXY jjjjY Zjj will simply in-

dicate the ratio of the lengths of the indicated segments. We describe thistheorem as remarkable not only because of its early date, but also becauseit connects the value of a cyclic sum (on the left side of (1)) with the valueof a cyclic product (on the right side of (1)). There exist a great number ofresults concerning cyclic products for triangles and n{gons, the best knownof which are the theorems of Ceva andMenelaus. Manymore such theoremsare given in [6] and [7]. On the other hand there are few papers on cyclicsums, such as [5] and [8]. Apart from [1], Euler's paper [4] is, so far as weare aware, the only one which related these two concepts.

Euler's proof of his theorem is by algebra and trigonometry; he calcu-lates the ratios of the lengths of the line segments in (1) using trigonometricalformulae involving the sines of the angles between the lines at O. The prooftakes two and a half pages, and whilst we do not wish to criticize the workof one of the most illustrious of mathematicians, it is worth considering theshortcomings of his proof. Apart from its length and complexity, the proofgives no insight as to why relation (1) is true, and therefore does nothing to

Copyright c 1999 Canadian Mathematical Society

149

suggest the existence of other relations of a similar nature. The followingapproach seems to overcome these criticisms; it depends on what has beencalled the \area principle" in [6] and the \area method" in [2].

B C

A

rq

p

O

Figure 2

2. Let p = jjCOBjj; q = jjAOCjj; r = jjBOAjj be the areas of the indicatedtriangles (see Figure 2). Then

jjBDjjjjDCjj =

jjBOAjjjjAOCjj =

r

q.

The �rst equality follows since the triangles [B;O; A] and [A;O;C] have thesame base [A; O] and their heights are proportional to jjBDjj : jjDCjj. Bycyclic permutation of the letters,

jjBDjjjjDCjj =

r

q,jjCEjjjjEAjj =

p

r,jjAF jjjjFBjj =

q

p. (2)

AlsojjADjjjjODjj =

jjABCjjjjOBCjj =

p+ q + r

p. The �rst equality follows since the

triangles [A;B;C] and [O;B;C] have the same base [B; C] and their heightsare proportional to jjADjj : jjODjj. Thus we obtain three more relations

jjADjjjjODjj =

p+ q+ r

p,jjBEjjjjOEjj =

p+ q + r

q,jjCF jjjjOF jj =

p+ q + r

r. (3)

Finally,

jjAOjjjjODjj =

jjADjj � jjODjjjjODjj =

jjADjjjjODjj � 1 =

p+ q+ r

p� 1 =

q+ r

p

leading in a similar fashion to the three relations

jjAOjjjjODjj =

q + r

p,jjBOjjjjOEjj =

r + p

q,jjCOjjjjOF jj =

p+ q

r. (4)

150

Relations (4) can also be proved directly (without using (3)) by the area prin-ciple. For our proof of (1) we only need relations (4):

jjAOjjjjODjj �

jjBOjjjjOEjj �

jjCOjjjjOF jj =

q + r

p� r + p

q� p+ q

r

=q2r + r2q+ r2p+ p2r + p2q+ q2p+ 2pqr

pqr

=q + r

p+r + p

q+p+ q

r+ 2

=jjAOjjjjODjj +

jjBOjjjjOEjj +

jjCOjjjjOF jj + 2 ,

as required.

3. We can exploit (2), (3) and (4) to yield many more relations between theratios of lengths in the triangle. The �rst two of these are well known.

1 =r

q� qp� pr

=jjBDjjjjDCjj �

jjAF jjjjFBjj �

jjCEjjjjEAjj ,

which is Ceva's Theorem for the triangle. Also, from (3),

jjODjjjjADjj =

p

p+ q + r,jjOEjjjjBEjj =

q

p+ q + r,jjOF jjjjCF jj =

r

p+ q + r,

and hence

jjODjjjjADjj +

jjOEjjjjBEjj +

jjOF jjjjCF jj =

p+ q+ r

p+ q+ r= 1 , (5)

which is mentioned in [5] and [8]. Euler also gives a proof of this result[4, p. 102].

Almost as easy is

jjAOjjjjODjj =

q + r

p=

q

p+r

p=

jjAF jjjjFBjj +

jjAEjjjjECjj ,

which is not trivial, as one �nds if one tries to prove it directly by calculat-ing the lengths of the line segments by trigonometry and not using the areaprinciple.

Also we observe:

jjAOjjjjODjj �

jjCEjjjjEAjj =

q + r

p� pr

= 1+q

r= 1+

jjCDjjjjDBjj .

Another example is:

jjAOjjjjODjj

� jjBOjjjjOEjj

=(q+ r)(r+ p)

pq= 1 +

r(p+ q+ r)

pq

= 1 +jjADjjjjODjj �

jjBDjjjjDCjj = 1+

jjBEjjjjOEjj �

jjAEjjjjECjj .

151

In fact, every algebraic identity in the variables p, q, r which is homogeneousof degree zero, leads, in at least one way, to an identity between ratios oflengths in the triangle.

4. However, this is not all. We supposed that O was in the interior of thetriangle, but this restriction is not necessary; we need only assume that Odoes not lie on a side (or side extended) of the triangle. Euler does notdiscuss this case, but the result is true since the above proof still holds withappropriate interpretation.

Firstly, we must use signed or directed lengths; this is the meaningof the doubled modulus lines. We assign a positive direction to every line in

the plane and then jjXY jj is positive if�!XY lies in the assigned direction and

is negative otherwise. Thus jjY Xjj = �jjXY jj and if X, Y , Z are distinct

collinear points thenjjXY jjjjY Zjj is positive if and only ifY lies betweenX andZ.

Secondly, we must use signed areas, which means that the area jjXY Zjj of atriangle [X; Y; Z] is taken to be positive if the vertices are named in a coun-terclockwise and negative if the vertices are named in a clockwise direction.All of the equations in Sections 2 and 3 have been written in such a way that,with these sign conventions, they hold regardless of the relative positionsof the points. In particular they remain true whether the point O is in theinterior of the triangle or not.

5. The primary purpose of Euler's paper was not to prove the \remarkabletheorem" discussed above, but to solve the following problem:

Given positive numbers x, y, z, is it possible to construct a triangle[A;B;C] so that, with the notation of Figure 1,

x =jjAOjjjjODjj , y =

jjBOjjjjOEjj , z =

jjCOjjjjOF jj ?

The theorem shows that this is impossible unless xyz = x+y+ z+2,

but the question remains as to whether this condition is su�cient. Eulerproves that it is so, but again his proof takes two and a half pages of algebraicand trigonometrical calculation.

Our approach, using the area principle, also gives an easy solution tothis problem. If a triangle satisfying the given condition exists, then by (4),

x =q + r

p, y =

r + p

q, z =

p+ q

r,

from which we obtain,

p =4

x+ 1, q =

4y + 1

, r =4

z + 1,

where4 = p+q+r is the total area of the triangle, and this may be chosenarbitrarily.

152

B C

A

��

�

a

c

b

AAU

O

� = 2�3

Figure 3

Now consider the triangle shown in Figure 3. Here the three line segments

[AO]; [BO]; [CO] are equally inclined at angles2�

3to each other, and the

lengths of the segments are denoted by a, b, c respectively. Then, by thewell-known formula for the area of a triangle,

p = bc sin2�

3=

p3bc

2, q =

p3ca

2, r =

p3ab

2,

and so we can solve for a, b, c in terms of the areas of the triangles p, q, robtained above. Clearly,

a2 =2qrp3p

, b2 =2rpp3q

, c2 =2pqp3r

,

from which the (positive) values of a, b, c can be determined. With thesevalues the triangle in Figure 3 has the required properties.

The reader may ask why we were justi�ed in taking the lines OA, OBand OC as equally inclined to each other. The reason is that the theorem,our proof, and the problem are a�ne-invariant. This means that, in partic-ular, if any triangle can be found which solves the problem, then the resultof applying a non-singular a�ne transformation (that is, a change of scalepossibly followed by a shear) yields another triangle which is also a solutionto the problem. It is well known that an a�ne transformation can be foundwhich will transform any three intersecting lines into three such lines whichare equally inclined (at angle 2�

3) to each other.

6. The study of early mathematical papers is fraught with di�culty. Some-times the notation is strange and the language is often obscure. In somecases (as with Euler's paper) it is well worth the e�ort to overcome thesedi�culties.

The study of such papers also gives an insight into the knowledge andthought processes of early authors. Clearly Euler, in 1780, did not know ofthe area principle, though this was certainly known to A.L. Crelle in 1816since he uses it in his book [3]. Euler knew about a�ne transformations,for he mentions them in his Introd. in Analysin In�nitorum II of 1748. It is

153

therefore surprising that he did not use them to simplify his solution of theproblem stated in Section 5. Perhaps he was not aware that it is possible,using a�ne transformations, to convert any three concurrent lines into threesuch lines which are equally inclined to each other.

These trivial observations do not, of course, detract from the fact thatEuler's contributions to very many branches of mathematics, are both pro-found and extensive.

Acknowledgements: My thanks are due to Miss Lucy Parker for her help inthe translation of Euler's paper and to the referees for their helpful commentsand suggestions.

References

[1] C.J. Bradley, Solution to Problem 1997, Crux Math. 21 (1995), 283-285.

[2] S.-C. Chou, X.-S. Gao, J.-Z. Zhang,Machine Proofs in Geometry, World-Scienti�c,

Singapore-New Jersey-London-Hong Kong, 1994.

[3] A.L. Crelle, �Uber einige Eigenschaften des ebenen geradlinigen Dreiecks r �uck-

sichtlish dreier durch die Winkel-Spitzen gezogenen geraden Linien, Berlin 1816.

[4] L. Euler, Geometrica et sphaerica quaedam,M�emoires de l'Acad �emie des Sciences

de St. Petersbourg 5 (1812), 96-114 (Received 1 May 1780) = L. Euleri, OperaOmnis, Series 1, Volume 26 (1953), 344-358.

[5] Branko Gr �unbaum, Cyclic ratio sums and products, Crux Math. 24 (1998), 20-25.

[6] Branko Gr �unbaum, and G.C. Shephard, Ceva, Menelaus and the area principle,

Math. Magazine 68 (1995), 254-268.

[7] Branko Gr �unbaum and G.C. Shephard, Some new transversality properties, Geom.

Dedicata 71 (1998), 179-208.

[8] G.C. Shephard, Cyclic sums for polygons, Math. Magazine (to appear).

G.C. ShephardSchool of Mathematics

University of East AngliaNorwich NR4 7TJ

England, U.K.g.shephard@uea.ac.uk

154

THE SKOLIAD CORNERNo. 37

R.E. Woodrow

We begin this number with the problems of Part I of the Alberta HighSchool Prize Exam, written November 1998. My thanks go to the organiz-ing committee, chaired by Ted Lewis, University of Alberta, Edmonton, forsupplying the contest and its problems.

THE ALBERTA HIGH SCHOOLMATHEMATICS COMPETITION

Part I | November 1998

1. A restaurant usually sells its bottles of wine for 100% more than itpays for them. Recently it managed to buy some bottles of its most popularwine for half of what it usually pays for them, but still charged its customerswhat it would normally charge. For these bottles of wine, the selling pricewas what percent more than the purchase price?

(a) 50 (b) 200 (c) 300 (d) 400 (e) not enough information

2. How many integer solutions n are there to the inequality34n � n2 + 289?

(a) 0 (b) 1 (c) 2 (d) 3 (e)more than 3

3. A university evaluates �ve magazines. Last year, the rankings wereMacLuck with a rating of 150, followed by MacLock with 120, MacLick with100, MacLeck with 90, and MacLack with 80. This year, the ratings of thesemagazines are down 50%, 40%, 20%, 10% and 5% respectively. How does theranking change for MacLeck?

(a) up 3 places (b) up 2 places (c) up 1 place (d) unchanged (e) down 1 place

4. Parallel lines are drawn on a rectangular piece of paper. The paperis then cut along each of the lines, forming n identical rectangular strips. Ifthe strips have the same length to width ratio as the original, what is thisratio?

(a)pn : 1 (b) n : 1 (c) n :

p5+1

2(d) n : 2 (e) n2 : 1

5. \The smallest integer which is at least a% of 20 is 10." For howmany integers a is this statement true?

(a) 1 (b) 2 (c) 3 (d) 4 (e) 5

155

6. Let S = 1 + 2 + 3 + � � � + 10n. How many factors of 2 appear inthe prime factorization of S?

(a) 0 (b) 1 (c) n� 1 (d) n (e) n+ 1

7. When 1 + x+ x2 + x3 + x4 + x5 is factored as far as possible intopolynomials with integral coe�cients, what is the number of such factors,not counting trivial factors consisting of the constant polynomial 1?

(a) 1 (b) 2 (c) 3 (d) 4 (e) 5

8. In triangle ABC, AB = AC. The perpendicular bisector of ABpasses through the midpoint of BC. If the length of AC is 10

p2 cm, what

is the area of ABC in cm2 ?

(a) 25 (b) 25p2 (c) 50 (d) 50

p2 (e) none of these

9. If f(x) = xx, what is f(f(x)) equal to?

(a) x2x (b) x(x2) (c) x(x

x) (d) x(x

(x+1)) (e) x(x

(xx))

10. A certain TV station has alogo which is a rotating cube in whichone face has anA on it and the other �vefaces are blank. Originally the A-face isat the front of the cube as shown on theright. Then you perform the followingsequence of three moves over and over:rotate the cube 90� around the verti-cal axis v, so that the front face movesto the left; then rotate the cube 90�

around a horizontal axis h, so that thenew front face moves down; then rotatethe cube 90� around the vertical axisagain, so that the new front face movesto the left. Suppose you perform thissequence of three moves a total of 1998times. What will the front face look likewhen you have �nished?

(a) (b) (c) (d) (e)

11. How many triples (x; y; z) of real numbers satisfy the simultane-ous equations x+ y = 2 and xy � z2 = 1?

(a) 1 (b) 2 (c) 3 (d) 4 (e) in�nitely many

156

12. In the diagram, ABC is an equilateral triangle of side length 3

and PA is parallel to BS. If PQ = QR = RS, what is the length of BR?

A

B C

P

Q

R

S

(a)p6 (b)

p6 (c) 3

p3

2(d)p7 (e) none of these

13. Let a, b, c and d be the roots of x4�8x3�21x2+148x�160 = 0.

What is the value of1

abc+

1

abd+

1

acd+

1

bcd?

(a) � 4

37(b) � 1

20(c) 1

20(d) 4

37(e) none of these

14. Wei writes down, in order of size, all positive integers b with theproperty that b and 2b end in the same digit when they are written in base 10.What is the 1998th number in Wei's list?

(a) 19974 (b) 19976 (c) 19994 (d) 19996 (e) none of these

15. Suppose that x = 31998. How many integers are there betweenpx2 + 2x+ 4 and

p4x2 + 2x+ 1?

(a) 31998 � 2 (b) 31998� 1 (c) 31998 (d) 31998 + 1 (e) 31998 + 2

16. The lengths of all three sides of a right triangle are positive inte-

gers. The area of the triangle is 120. What is the length of the hypotenuse?

(a) 13 (b) 17 (c) 26 (d) 34 (e) none of these

Last number we gave the problems of the Mini demi-�nale 1996 of theOlympiade Mathematique Belge. Somehow the last three problems wereoverlooked.

28. Lors d'un championnat de poursuite, deux cyclistes partent enmeme temps de deux points diam �etralement oppos �es d'un v �elodrome de250mde tours. Le vainqueur a rattrap �e son rival apr �es avoir parcouru 8 tours.Quel est le rapport de la vitesse moyenne du vainqueur �a celle du perdant ?

(a) 9

8(b) 8

7(c) 17

16(d) 16

15(e) 7

8

157

29. Il �etait une fois : : : deux fontaines et une citerne. La premi �erefontaine mettait un jour entier �a remplir la citerne, la seconde quatre. Encombien de temps la citerne peut-elle etre remplie par les deux fontainescoulant ensemble?

(a) 1 h 15 min (b) 8 h (c) 18 h (d) 19 h 12 min (e) 30 h

30. Laquelle des propositions suivantes est vraie dans tout carr �e ?

(a) La longueur d'un cot �e est �egale �a la moiti �e de la longueur d'une diagonale.(b) La longueur d'un cot �e est �egale �a la racine carr �ee du p �erim �etre.(c) La longueur d'un cot �e est sup �erieure aux deux tiers de la longueur d'unediagonale.(d) La longueur d'un cot �e est inf �erieure aux deux tiers de la longueur d'unediagonale.(e) La longueur d'un cot �e est �egale �a la longueur d'une diagonale multipli �eepar

p2.

And now the answers.

1. 101 2. e 3. c 4. e 5. c

6. c 7. b 8. 64 9. b 10. e

11. e 12. b 13. b 14. c 15. a

16. c 17. d 18. d 19. e 20. 315

21. d 22. c 23. d 24. 7 25. b

26. e 27. b 28. d 29. d 30. c

That completes the Skoliad Corner for this issue. Sendme your suitablecontest materials, as well as suggestions for features for this Corner.

158

MATHEMATICAL MAYHEM

Mathematical Mayhem began in 1988 as a Mathematical Journal for and by

High School and University Students. It continues, with the same emphasis,as an integral part of Crux Mathematicorum with Mathematical Mayhem.

All material intended for inclusion in this section (except for the prob-

lems section, which have their own editors) should be sent to theMayhemEd-

itor, Naoki Sato, Department ofMathematics, Yale University, POBox 208283

Yale Station, New Haven, CT 06520{8283 USA. The electronic address is still

mayhem@math.toronto.edu

The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).The rest of the sta� consists of Adrian Chan (Upper Canada College), JimmyChui (Earl Haig Secondary School), David Savitt (Harvard University) andWai Ling Yee (University of Waterloo).

Mayhem Problems

The Mayhem Problems editors are:

Adrian Chan Mayhem High School Problems Editor,Donny Cheung Mayhem Advanced Problems Editor,David Savitt Mayhem Challenge Board Problems Editor.

Note that all correspondence should be sent to the appropriate editor |see the relevant section. In this issue, you will �nd only problems | thenext issue will feature only solutions.

We warmly welcome proposals for problems and solutions. With thenew schedule of eight issues per year, we request that solutions from thisissue be submitted in time for issue 4 of 2000.

High School Problems

Editor: Adrian Chan, 229 Old Yonge Street, Toronto, Ontario, Canada.M2P 1R5 <all238@ipoline.com>

H253. Find all real solutions to the equationp3x2 � 18x+ 52 +

p2x2 � 12x+ 162 =

p�x2 + 6x+ 160 .

159

H254. Proposed by Alexandre Trichtchenko, 1st year, Carleton Uni-

versity.Let p and q be relatively prime positive integers, and n a multiple of

pq. Find all ordered pairs (a; b) of non-negative integers that satisfy thediophantine equation n = ap+ bq.

H255. We have a set of tiles which contains an in�nite number ofregular n{gons, for each n = 3, 4, : : : . Which subsets of tiles can be chosen,so that they �t around a common vertex? For example, we can choose foursquares, or four triangles and a hexagon.

H256. Let A = 2apb1pc2, where p1 and p2 are primes, possibly equal

to each other and to 2, and a, b, and c are positive integers. It is knownthat p1 � p2 (mod 4), b � c (mod 2), and that 2a, pb

1, pc

2are three consecu-

tive terms of an arithmetic sequence, not necessarily in that order. Find allpossible values for A.

Advanced Problems

Editor: Donny Cheung, c/o Conrad Grebel College, University of Wa-terloo, Waterloo, Ontario, Canada. N2L 3G6 <dccheung@uwaterloo.ca>

A229. In tetrahedron SABC, the medians of the faces SAB, SBC,and SCA, taken from the vertex S, make equal angles with the edges thatthey lead to. Prove that jSAj = jSBj = jSCj.

(Polish Olympiad)

A230. Proposed by Naoki Sato.For non-negative integersn and k, let Pn;k(x) denote the rational func-

tion(xn � 1)(xn � x) � � � (xn � xk�1)

(xk � 1)(xk � x) � � � (xk � xk�1).

Show that Pn;k(x) is actually a polynomial for all n, k.

A231. Proposed by Mohammed Aassila, Centre de Recherches

Math �ematiques, Montr �eal, Qu �ebec.For the sides of a triangle a, b, and c, prove that

13

27� (a+ b+ c)(a2 + b2 + c2) + 4abc

(a+ b+ c)3� 1

2.

A232. Five distinct points A, B, C, D, and E lie on a line (in thisorder) and jABj = jBCj = jCDj = jDEj. The point F lies outside theline. Let G be the circumcentre of triangle ADF and H the circumcentre oftriangle BEF . Show that the lines GH and FC are perpendicular.

(1997 Baltic Way)

160

Challenge Board Problems

Editor: David Savitt, Department of Mathematics, Harvard University,1 Oxford Street, Cambridge, MA, USA 02138 <dsavitt@math.harvard.edu>

C85. Proposed by Christopher Long, graduate student, Rutgers Uni-

versity. (From a set of course notes on analytic number theory.)

Let C = (ci;j) be anm� n matrix with complex entries, and let D bea real number. Show that the following statements are equivalent:

(i) For any complex numbers aj , j = 1, : : : , n,

mXi=1

������nXj=1

ajci;j

������2

� D

nXj=1

jaj j2 .

(ii) For any complex numbers bi, i = 1, : : : , m,

nXj=1

�����mXi=1

bici;j

�����2

� D

mXi=1

jbij2 .

C86. Let Kn denote the complete graph on n vertices; that is, thegraph on n vertices with all possible edges present. Show that Kn can bedecomposed into n�1 disjoint paths of length 1, 2, : : : , n�1. For example,for n = 4, the graphK4, with vertices A, B, C, andD, decomposes into thepaths fACg, fBD;DAg, and fAB;BC;CDg.

Can we require that the paths in the decomposition be simple? (We saythat a path is simple if it passes through each vertex at most once; that is, ifno vertex is the end-point of more than two edges along the path.)

161

Problem of the Month

Jimmy Chui, student, Earl Haig S.S.

Problem. Let ABC be an isosceles triangle with AB = AC. Supposethat the angle bisector of \B meets AC at D and that BC = BD + AD.Determine \A.

(1996 CMO, Problem 4)

Solution.

A

B CE

D

x

x

180�4x

2x

2x

Construct E on BC such that DE = EC. Let x = \ABD = \DBC.Then, \DCE = \ACB = 2x. Hence, \CDE = 2x, and further\ADE = 180� � 2x. Thus, quadrilateral ABED is cyclic.

Drawing AE, we see that \DAE = \DBE = x, and \DEA =

\DBA = x. Therefore, triangle ADE is isosceles, and so AD = DE.

Now, BC = BD + AD = BD + DE = BD + EC. Also,BC = BE + EC. So, BD + EC = BE +EC. Thus, BD = BE.

Now, \BED = 180� � \DEC = 4x, and \BDE = 180� � 5x, so4x = 180� � 5x.

Solving for x, we obtain x = 20�.

Thus, \A = 180� � 4x = 100�.

162

Self-Centred Triangles

Cyrus Hsia

student, University of Toronto

There are four famous triangular centres that all students of classicalgeometry shouldbe familiar with. Readers already familiar with these shouldgo on to the next section to see how the information on one will be useful toanother. Otherwise, the centres are as follows:

Central De�nitions

The following de�nitions all pertain to an arbitrary triangle in Euclideanspace.

Centroid

The centroid, denoted by G, is the point of intersection of the mediansof a triangle. A median is a line from a vertex of a triangle to the mid-pointof the opposite side.

G

Circumcentre

The circumcentre, denoted byO, is the point of intersection of the rightbisectors of the three sides. A right bisector of a line segment is a line per-pendicular to it and divides it into two equal segments. It should be notedthat O is also the center of the circle that passes through the vertices of thetriangle. This circle is called the circumcircle and hence O is referred to asthe circumcentre. The con�rmation that these two de�nitions are equivalentis left to the reader.

Copyright c 1999 CanadianMathematical Society

163

O

Incentre

The incentre, denoted by I, is the point of intersection of the threeangle bisectors of the triangle. An angle bisector is a line which divides agiven angle into two equal angles. It should be noted that in this case, I isthe centre of the circle inside the triangle tangent to all three sides. Thesede�nitions are equivalent and we leave the work to the reader.

I

�

�

� �

Orthocentre

The orthocentre, denoted byH, is the point of intersection of the threealtitudes of the triangle. An altitude is a line through a vertex that is per-pendicular to the side opposite this vertex.

H

164

The Central Problem

Although it was taken for granted in the de�nitions, it is not trivial toshow that the points above exist. After all, why do three line segments haveto intersect at a common point? We say the lines concur if they do. In fact,draw three lines from the vertices of a triangle to their opposite sides. Theselines are called cevians. It is easy to see that these cevians need not concur.

There are many ways to show that the points described above exist andto show that certain cevians always concur for any triangle (See [1] and [2]).Here we will consider something of purely mathematical interest. There isa certain similarity between the de�nitions of the four centres and so it maybe possible that the existence of one centre is enough to prove the existenceof another. For example, if it has been shown that the centroid exists, thenis it possible to show that the incentre exists from this fact? We present twosuch scenarios here.

Circumcentre implies Orthocentre

Given triangle ABC, let hA, hB, hC be the altitudes from verticesA, B, C respectively. We wish to show that these three altitudes (cevians)concur. At vertex A, draw a line perpendicular to hA and similarly draw linesfor the other two sides as shown.

����

����

�����

�����o o

���

��� A

B C

D

EF

hA

Let the lines intersect at points D, E, F opposite A, B, C respectively asshown. Now FE is perpendicular to hA and BC is also perpendicular to hAso FE and BC are parallel. Likewise, FD and AC are parallel and ED andAB are parallel.

ThusFA = BC, since FACB is a parallelogram, andAE = BC, sinceAECB is also a parallelogram. Thus FA = AE, so A is the mid-point ofFE and hA is perpendicular to FE. In other words, hA is the perpendicularbisector of FE. Likewise, hB is the perpendicular bisector of FD and hC isthat of DE.

Now we see that if the circumcentre of a triangle exists, in this case hA,hB , and hC of triangle DEF concur, means that the altitudes of triangleABC concur. Thus the orthocentre exists by de�nition.

165

Incentre implies Orthocentre

Given a triangle ABC, let the altitudes from vertices A, B, C, be hA,hB , hC respectively. Let the feet of the altitudes be D, E, F as shown.

A

B CD

E

F

�

�

��

Let �, �, be the angles of triangle ABC at vertices A, B, C re-spectively. Now \FEB = \FCB since FECB is a cyclic quadrilateral.\FEB = 90� � �. Also, \BED = \BAD = 90� � �. In other words,hB = BE is the angle bisector of \FED. Likewise, hA and hC are anglebisectors of vertices D and F respectively in triangle FED.

As the reader has suspected, if the incentre exists, the three angle bi-sectors hA, hB , hC concur in triangleDEF , and since these are the altitudesof triangle ABC, then the orthocentre exists.

Epilogue

Here, we have shown that the existence of the incentre or the circum-centre implies the existence of the orthocentre. These and the other impli-cations are listed in the table below. We welcome the reader to explore theother possibilities and to send us feedback on any results they �nd. A fur-ther exploration, for the brave at heart, would be to investigate other famouscentres of triangles as well.

implies centroid circumcentre incentre orthocentre

centroid clear

circumcentre clear proved

incentre clear proved

orthocentre clear

Exercises

1. Provide a proof that the existence of the orthocentre implies the exis-tence of the incentre and the circumcentre by a similar method to theones given.

2. Are there other proofs to the implications above?

166

3. For exploration, are other well-known centres of triangles related asnicely to the orthocentre, circumcentre and incentre? Try the following:

4. (a) The nine-point circle of a triangle is de�ned as the circle throughthe three mid-points of the three sides, the three feet of the alti-tudes, and the three mid-points of the line segments joining thevertices to the orthocentre. The centre of the nine-point circle isdenoted by the letter N . By de�nition, the existence ofN dependson the existence of the orthocentre. How about the other centres?

(b) The Fermat point, F , of a triangle is the point in a triangle thatminimizes the sum of the lengths from this point to each of thethree vertices. Is it possible to determine the existence of thispoint by the existence of the other centres?

Acknowledgements

The proof of the �rst result was modi�ed from the lecture notes inMAT325, by Professor J.W. Lorimer, University of Toronto, Toronto, On-tario, and was the inspiration for this article.

References

1. Grossman, J.P. Ye Olde Geometry Shoppe - Part I, Mathematical

Mayhem, Volume 6, Issue 2. November/December 1993. pp. 13-17.

2. Grossman, J.P. Ye Olde Geometry Shoppe - Part II, Mathematical

Mayhem, Volume 6, Issue 3. January/February 1994. pp. 7-12.

167

J.I.R. McKnight Problems Contest 1988

TIME: 212HOURS

1. Calculators are permitted.

2. Complete solutions are necessary to all problems for full marks.

3. Do all questions in Part A and Part B.

PART A

(6 questions � 5 marks = 30 marks)

1. Find all ordered pairs of real numbers (x; y) such that log(x2

y3) = 1 and

log(x2y3) = 7.

2. (a) Determine the di�erence between the number of zeroes at the endof each of the numbers (103)! and (103)100.

(b) Determine n, if n! ends in 17 zeroes.

3. A particle's position at time t seconds, from a pointP , is given inmetres(s is distance) by

s = t3 � 3t2 + 4 , t � 0 .

Find the time(s) when the particle is speeding up.

4. Determine all integer solutions to the following system of equations:

a+ b+ c = 0 ,

ab+ ac+ bc = �19 ,abc = �30 .

5. In triangle ABC, the angle at B is obtuse and AB > BC. An angle

bisector of an exterior angle at Ameets CB atD, and an angle bisectorof an exterior angle at B meets AC at E. If AD = AB = BE, �nd\BAC.

6. Find the measure of the acute angle � for which it is true that

�16

81

�sin2 �+

�16

81

�cos2 �=

26

27.

PART B

(5 questions � 10 marks = 50 marks)

1. Find the sum of the �rst 100 terms of the following double arithmeticseries:

1� 4 + 5� 7 + 9� 10 + 13� 13 + 17� 16 + � � � .

168

2. A

P

CB

D

Points A, B, C, and P are lo-cated on a circle as shown. Provethat if triangle ABC is equilat-eral andAP intersects BC atD,then

1

PD=

1

PB+

1

PC.

3. Determine all triangles ABC which satisfy the condition

tan(A�B) + tan(B �C) + tan(C � A) = 0 .

4. Find all integers a such that (n� a)(n � 10) + 1 can be written as aproduct (n+ b)(n+ c), where b and c are integers.

5. A transformation is given by

P (x; y)! P 0(y+ 246;�x� 430) .

Giving reasons, describe this transformation in simplest possible terms.

Swedish Mathematics Olympiad

1990 Qualifying Round

1. A boy has got two-thirds of the way over a railroad bridge, when hecatches sight of a train coming towards him. He can just get o� thebridge, and so escape the train, by running as fast as he can either to-wards the train or away from it. The train is approaching at a speed of60 km/h. How fast can the boy run?

2. Which six-digit numbers of the form abcabc are divisible by 33?

3. A large cube, 6 � 6 � 6, is constructed of 216 unit cubes. These arenumbered from 1 to 216 as shown in the �gure on the next page. The�rst layer is numbered from 1 to 36, the �rst row in that layer from 1

to 6, the second from 7 to 12, and so on, always from left to right. Thenext layer is similarly numbered from 37 to 72, and so on. Choose 36unit cubes so that no two of the chosen cubes come from the same row,column, or rank (parallel to one of the edges of the cube). Let S denotethe sum of the numbers assigned to the 36 cubes. What are the possiblevalues of S?

169

1 2 3 4 5 6

37 38 39 40 41 42

73 74 75 76 77 78

109 110 111 112 113 114

145 146 147 148 149 150

181 182 183 184 185 186

1 2 3 4 5 67 8 9 10 11 12

13 14 15 16 17 1819 20 21 22 23 24

25 26 27 28 29 3031 32 33 34 35 36

186

150

114

78

42

6

192

156

120

84

48

12

198

162

128

90

54

18

204

168

132

96

60

24

210

174

138

102

66

30

216

180

144

108

72

36

4. A is one of the angles in a given triangle. The corresponding side haslength a, and the other two sides have lengths b and c. Show that

sinA � a

2pbc:

5. Find all real solutions of the system of equations

x1jx1j = x2jx2j+ (x1 � a)jx1 � ajx2jx2j = x3jx3j+ (x2 � a)jx2 � aj

� � �xnjxnj = x1jx1j+ (xn � a)jxn � aj

where a is a given positive integer.

6. Four houses are situated at the corners of a rectangle ABCD with sides3000 metres and 500 metres. The owners of the houses intend to sinka common well and run water mains from the well to each house. Theyhave access to 1020 metres of pipe and a device that can join togethertwo or more pieces of pipe. Where should the well be placed so thatthe length of pipe is su�cient?

170

1990 Final Round

1. The positive divisors of n = 1900! are d1, d2, : : : , dk. Show that

d1pn+

d2pn+ � � �+ dkp

n=

pn

d1+

pn

d2+ � � �+

pn

dk.

2. The points A1, A2, : : : , A2n lie, in this order, on a straight line, andjAkAk+1j = k for k = 1, 2, : : : , 2n� 1. The point P is situated on

the line so that the sum

2nXk=1

jPAkj is as small as possible.

Find this sum.

3. The numbers a and b are such that sinx+sina � b cosx for all x. Finda and b.

4. A quadrilateral ABCD is inscribed in a circle. The angle bisectors of Aand B meet at a point E. Draw a line through E parallel to CD whichmeets AD in L and BC inM . Show that jLAj + jMBj = jLM j.

5. Find all (not necessarily strict) monotonic, positive functions f whichare de�ned on the positive reals, and which satisfy

f(xy) � f�f(y)

x

�= 1

for all x, y > 0.

6. Find all positive integers x and y such that y � 500 and

117

158>

x

y>

97

131.

171

PROBLEMSProblem proposals and solutions should be sent to Bruce Shawyer, Department

of Mathematics and Statistics, Memorial University of Newfoundland, St. John's,Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution,

together with references and other insights which are likely to be of help to the edi-

tor. When a submission is submitted without a solution, the proposer must includesu�cient information on why a solution is likely. An asterisk (?) after a number

indicates that a problem was submitted without a solution.

In particular, original problems are solicited. However, other interesting prob-lems may also be acceptable provided that they are not too well known, and refer-

ences are given as to their provenance. Ordinarily, if the originator of a problem can

be located, it should not be submitted without the originator's permission.

To facilitate their consideration, please send your proposals and solutions

on signed and separate standard 812"�11" or A4 sheets of paper. These may be

typewritten or neatly hand-written, and should be mailed to the Editor-in-Chief,to arrive no later than 1 November 1999. They may also be sent by email to

crux-editors@cms.math.ca. (It would be appreciated if email proposals and solu-

tions were written in LATEX). Graphics �les should be in epic format, or encapsulatedpostscript. Solutions received after the above date will also be considered if there

is su�cient time before the date of publication. Please note that we do not accept

submissions sent by FAX.

2388 [1998, 503] Correction. Proposed by Daniel Kupper, B �ullin-

gen, Belgium.

Suppose that n � 1 2 N is given and that, for each integerk 2 f0, 1, : : : , n� 1g, the numbers ak, bk, zk 2 C are given, with the z2kdistinct. Suppose that the polynomials

An(z) = zn +

n�1Xk=0

ak zk and Bn(z) = zn +

n�1Xk=0

bk zk

satisfy An(zj) = Bn(z2

j ) = 0 for all j 2 f0, 1, : : : , n� 1g.Find an expression for b0, b1, : : : , bn�1 in terms of a0, a1, : : : , an�1.

2389 [1998, 503] Correction. Proposed by Nikolaos Dergiades,

Thessaloniki, Greece.

Suppose that f is continuous on Rn and satis�es the condition thatwhen any two of its variables are replaced by their arithmetic mean, thevalue of the function increases; for example:

f(a1; a2; a3; : : : ; an) � f

�a1 + a3

2; a2;

a1 + a3

2; a4; : : : ; an

�.

Let m =a1 + a2 + : : :+ an

n. Prove that

f(a1; a2; a3; : : : ; an) � f (m;m;m; : : : ;m) :

172

2426. Proposed by Mohammed Aassila, Strasbourg, France.

(a) Show that there are two polynomials, p(x) and q(x), both having threeinteger roots and such that p(x)� q(x) is a non-zero constant.

(b)? Do there exist two polynomials, p(x) and q(x), both having n > 3

integer roots and such that p(x)� q(x) is a non-zero constant?

2427. Proposed by Toshio Seimiya, Kawasaki, Japan.Suppose that G is the centroid of triangle ABC, and that

\GAB + \GBC + \GCA = 90� :

Characterize triangle ABC.

2428. Proposed by Toshio Seimiya, Kawasaki, Japan.Given triangle ABC with \BAC = 90�. The incircle of triangle ABC

touches AB and AC at D and E respectively. Let M be the mid-pointof BC, and let P and Q be the incentres of triangles ABM and ACM

respectively. Prove that

1. PD k QE ;

2. PD2 +QE2 = PQ2 .

2429. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.Suppose that D, E and F are points on the sideAB (or its production)

of triangle ABC. Suppose further that CD is a median, that CE is thebisector of \ACB, and that CF is its external bisector.

The circumcircle, �, of triangle EFC intersects CD again at P .Suppose that �A and �B are the circumcircles of triangles CPA and CPBrespectively.

Show that �A and �B are tangent to AB at A and B respectively.

2430. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.Points A and B lie outside circle �. Find a point C on � with the

following property:

AC andBC intersect � again atD andE respectively, withDE k AB.

2431. Proposed by Jill Taylor, student, Mount Allison University,

Sackville, New Brunswick.Let n 2 N. Prove that there exist triangles with integer area, integer

side lengths, one side n and perimeter 4n, where n is not necessarily prime.

For a given n, are such triangles uniquely determined?

[Compare problem 2331.]

173

2432. Proposed by K.R.S. Sastry, Bangalore, India.In 4ABC, we use the standard notation: O is the circumcentre, H

is the orthocentre. Let M be the mid-point of BC, OH = m, OM = n

(m; n 2 N), and suppose that OH kBC.

How many sides of 4ABC can have integer lengths?

2433. Proposed by K.R.S. Sastry, Bangalore, India.In 4ABC, let e denote the length of the segment of the Euler line

between the orthocentre and the circumcentre.

Prove or disprove that 4ABC is right angled if and only if e equalsone half of the length of one of the sides of 4ABC.

2434. Proposed by K.R.S. Sastry, Bangalore, India.In 4ABC, let \ABC = 60�. Point P is on the line segment AC

such that \CBP = \BAC. Point Q is on the line segment BP such thatBQ = BC.

Prove that Q lies on the altitude through A of 4ABC if and only if\BAC = 40�.

2435. Proposed by V�aclav Kone �cn �y, Ferris State University, Big

Rapids, Michigan, USA.Show that, for x > 0, the following functions are increasing:

f(x) =

�1 + 1

x

�x(1 + x)

1x

and g(x) =

�1 +

1

x

�x� (1 + x)

1x :

2436. Proposed by V�aclav Kone �cn �y, Ferris State University, Big

Rapids, Michigan, USA.Find all real solutions of

2 cosh(xy) + 2y ��(2 cosh(x))

y+ 2

�= 0 :

2437. Proposed by Paul Yiu, Florida Atlantic University, Boca Ra-

ton, Florida, USA.Let P be a point in the plane of triangle ABC. If the mid-points of the

line segments AP , BP , CP all lie on the nine-point circle of triangle ABC,prove that P must be the orthocentre of triangle ABC.

2438. Proposed by Peter Hurthig, Columbia College, Burnaby, BC.Show how to tile an equilateral triangle with congruent pentagons.

Re ections are allowed. (Compare problem 1988.)

174

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased to

consider for publication new solutions or new insights on past problems.

2321. [1998: 109] Proposed by David Doster, Choate Rosemary

Hall, Wallingford, Connecticut, USA.Suppose that n � 2. Prove that

nXk=2

�n2

k

�=

n2Xk=n+1

�n2

k

�.

Here, as usual, bxc means the greatest integer less than or equal to x.

Solution by Florian Herzig, student, Cambridge, UK.

We show more generally that

nXk=2

�m

k

�=

n2Xk=n+1

�m

k

�(1)

for all n2 � m < (n + 1)2 and n � 2. The proof is by induction on m.For m = 4 this is easily veri�ed. For the induction step, �rst assume thatn2 < m < (n + 1)2 for some n and that (1) holds for m � 1. Then theleft-hand side increases (from its value for m� 1) by the number of divisorsd of m with 2 � d � n. If n2 < m < (n + 1)2, and d divides m, where2 � d � n, then m

dalso dividesm, and

n <n2 + 1

n� n2 + 1

d� m

d� n2 + 2n

d� n2 + 2n

2� n2 ;

that is,

n+ 1 � m

d� n2 .

Conversely, if n+ 1 � md� n2, where d dividesm, then

1 <m

n2� d � m

n+ 1< n + 1 ;

that is,2 � d � n .

Consequently, there is a one-to-one correspondence between the divisors ofm in the interval [2; n] and those in [n+ 1; n2]. Therefore both sides of (1)increase by the same amount asm increases by 1.

On the other hand, ifm = n2 > 4 for some n and (1) is true form�1,then the left-hand side increases by the number N of divisors d of n2 with

2 � d < n and byjn2

n

k= n; that is, byN+n. The right-hand side increases

175

by the number of divisors d of n2 with n � d � (n� 1)2, which is N + 1,and by

nXk=(n�1)2+1

�n2

k

���n2

n

�= (2n � 1) � n = n� 1 .

Therefore both sides increase by the same value also in this case, and theresult follows by induction.

Also solved by ZAVOSH AMIR-KHOSRAVI, student, North Toronto Collegiate Institute,

Toronto; MICHEL BATAILLE, Rouen, France; MANSUR BOASE, student, Cambridge, England;

CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; THEODORE CHRONIS, student, Aris-

totle University of Thessaloniki, Greece; WALTHER JANOUS, Ursulinengymnasium, Innsbruck,

Austria; MICHAEL LAMBROU, University of Crete, Crete, Greece; KEE-WAI LAU, Hong Kong;

GERRY LEVERSHA, St. Paul's School, London, England; PARAGIOU THEOKLITOS, Limassol,

Cyprus; KENNETH M. WILKE, Topeka, Kansas, USA; and the proposer.

2322. [1998: 109] Proposed by K.R.S. Sastry, Dodballapur, India.

Suppose that the ellipse E has equationx2

a2+y2

b2= 1. Suppose that

� is any circle concentric with E. Suppose that A is a point on E and B is apoint of � such that AB is tangent to both E and �.

Find the maximum length of AB.

Solution by Nikolaos Dergiades, Thessaloniki, Greece.Let the common tangent touch the ellipse at A = (x1; y1) and the

circle (of radius R) at B = (x2; y2). Assume, without loss of generality, thatb < R < a and (since it is not perpendicular to an axis of the ellipse) thatAB has the equation

y = px+ q . (1)

(x1; y1) satis�es (1) and also the equation

x2

a2+y2

b2= 1 . (2)

Plugging (1) into (2) and setting the discriminant equal to zero, we �nd

x1 = �pa2

q, q2 = b2 + p2a2 . (3)

Similarly, (x2; y2) satis�es (1) and also the equation

x2 + y2 = R2 ,

so that

x2 = �pR2

q, q2 = R2(1 + p2) . (4)

176

From (3) and (4) we �nd that

p2 =R2 � b2

a2 � R2.

Since y2 � y1 = p(x2 � x1) and x2 � x1 =p

q

�a2 � R2

�, we have

AB2 = (x2 � x1)2 + (y2 � y1)

2

= (1 + p2)(x2 � x1)2

=(R2 � b2)(a2� R2)

R2

= a2 + b2 �R2 � a2b2

R2

= (a� b)2 ��R� ab

R

�2� (a� b)2 ,

with equality if and only if R =pab, in which case the maximum value of

AB is a� b.

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain;

THEODORE CHRONIS, Athens, Greece; RICHARD I. HESS, Rancho Palos Verdes, California,

USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU,

University of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul's School, London, England;

KATHLEEN E. LEWIS, SUNYOswego, Oswego, NY, USA; D.J. SMEENK, Zaltbommel, the Neth-

erlands; NICOLAS TH �ERIAULT, �etudiant, Universit �e Laval, Qu �ebec; and the proposer.

2325?. [1998: 109] Proposed by Walther Janous, Ursulinengymnas-

ium, Innsbruck, Austria.Suppose that q is a prime and n is a positive integer. Suppose that

fakg (0 � k � n) is given by

nXk=0

akxk =

1

qn

nXk=0

�qn

qk

�(qx� 1)k .

Prove that each ak is an integer.

Solution by G.P. Henderson, Garden Hill, Campbellcroft, Ontario.We are to prove that

F =1

qn

nXk=0

�qn

qk

�(qx� 1)k

is an integer polynomial.

Let the (complex) q-th roots of the number qx � 1 be y1, y2, : : : , yqso that yqr = qx� 1 for r = 1, 2, : : : , q.

177

Lemma. If P (y) is an integer polynomial, then

1

q

qXr=1

P (yr)

is an integer polynomial in x.

Proof. Suppose

P (y) =

bXm=0

cmym .

Then1

q

qXr=1

P (yr) =1

q

bXm=0

cm

qXr=1

ymr .

Ifm is a multiple of q, saym = qk, the inner sum is

qXr=1

yqkr =

qXr=1

(qx� 1)k = q(qx� 1)k ,

and if m is not a multiple of q, the inner sum is zero. [Editorial comment:This follows because, with m and q relatively prime, ym

1, : : : , ymq are the

(distinct) q-th roots of (qx� 1)m and thus sum to zero.] Therefore

1

q

qXr=1

P (yr) =

bb=qcXk=0

cqk(qx� 1)k , (1)

which proves the lemma. 2

Set b = qn and cm =�qn

m

�. Then

P (yr) =

qnXm=0

�qn

m

�ymr = (1 + yr)

qn

for r = 1, : : : , q. Thus, dividing (1) by qn:

1

qn+1

qXr=1

(1 + yr)qn =

1

qn

nXk=0

�qn

qk

�(qx� 1)k = F . (2)

We have

(1 + yr)q = 1 +

q�1Xm=1

�q

m

�ymr + yqr = q[x+Q(yr)] ,

where Q(y) is the integer polynomial

Q(y) =

q�1Xm=1

1

q

�q

m

�ym .

178

Using this in (2),

F =1

q

qXr=1

[x +Q(yr)]n =

nXs=0

�n

s

�xn�s

1

q

qXr=1

[Q(yr)]s . (3)

Applying the lemma to the integer polynomials [Q(y)]s, we see that F is aninteger polynomial.

Note. If q = 2, we can obtain an explicit expression for F . The rootsy1 and y2 are �

p2x� 1, and Q(y) = y. From (3),

F =1

2

h(x+

p2x� 1)n + (x�

p2x� 1)n

i

= xn +

�n

2

�xn�2(2x� 1) +

�n

4

�xn�4(2x� 1)2

+ � � �+�n

2r

�xn�2r(2x� 1)r ,

where r = bn=2c.No other solutions were received.

2326?. [1998: 175, 301] Proposed by Walther Janous, Ursulinen-

gymnasium, Innsbruck, Austria.

Prove or disprove that if A, B and C are the angles of a triangle, then

2

�<

Xcyclic

�1� sin A

2

� �1 + 2 sin A

2

�� � A

� 9

2�.

Solution by Michael Lambrou, University of Crete, Crete, Greece.

SetA0 = (��A)=2 and similarly forB andC so thatA0+B0+C0 = �

and A0, B0, C0 are the angles of an acute-angled triangle. Conversely, everyacute-angled triangle arises in this way if we set A = � � 2A0, etc. So thesubstitution A = � � 2A0 transforms the inequality to be proved into

4

�<X cosA0 � cos 2A0

A0 � 9

�(1)

for all acute-angled triangles. For notational convenience drop the primesand write A, B, C in place of A0, B0, C0 once again. We may further assume

�=2 > A � B � C > 0. For 0 < x � �=2 set f(x) =cosx� cos 2x

x,

and extend this, by continuity, to x = 0 [Editorial note: by de�ning f(0) = 0

| it is easy to show with calculus that limx!0 f(x) = 0].

For x 2 [0; �=2], we have f 00(x) = g(x)=x3, where

g(x) = (cosx� cos 2x)00x2 � 2(cosx� cos 2x)0x+ 2(cosx� cos 2x) ,

179

so that

g0(x) = (cosx� cos 2x)000x2 = x2 sinx(1� 16 cosx) .

Hence g0(x) = 0 if and only if x = 0 or x = arccos(1=16) � 86�, andg decreases in [0; arccos(1=16)] and increases in [arccos(1=16); �=2]

and so has absolute minimum at x = arccos(1=16). But g(0) = 0,g(�=2) = ��2 + � + 2 < 0 so g(x) � 0 for all x in [0; �=2]. It followsthat f 00(x) � 0 for all x in [0; �=2] and so f is concave down. By Jensen'sinequality we have

f(A) + f(B) + f(C) � 3f

�A+ B + C

3

�= 3f

��

3

�=

9

�,

showing the right hand side of (1).

The concavity of f together with f(�=4) = 2p2=� > 2=� = f(�=2)

shows that the least value of f in the restricted interval [�=4; �=2] occurs atx = �=2. We also have f(0) = 0 < f(�=2) so f(x) � 0 for all x 2 [0; �=2]

with strict inequality f(x)> 0 if x 6= 0.

Finally, since �=2 � A � B � C, we have that B � �=4, so thatf(A), f(B)� f(�=2) = 2=�. Hence

f(A) + f(B) + f(C) > f(A) + f(B) � f

��

2

�+ f

��

2

�=

4

�,

which proves the left hand side of (1).

Editorial note. Both bounds in (1) are best possible: the lower bound4=� is attained by the degenerate �=2, �=2, 0 triangle, and the upper bound9=� by the equilateral triangle.

Also solved by HAYO AHLBURG, Benidorm, Spain; NIKOLAOS DERGIADES, Thessa-

loniki, Greece; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; and

HEINZ-J �URGEN SEIFFERT, Berlin, Germany.

Kone�cn �y, Sei�ert and the proposer rewrote the sum in the problem as

X sin(A=2) + cosA

� �A;

which may be a more attractive form, though not perhaps quite as attractive as Lambrou's

equivalent form (which was also found by Ahlburg).

The problem arose from Janous's solution to his CRUX with MAYHEM problem 2190.

180

2327. [1998: 175] Proposed by Christopher J. Bradley, Clifton College,

Bristol, UK.

The sequence fang is de�ned by a1 = 1, a2 = 2, a3 = 3, and

an+1 = an � an�1 +a2n

an�2; n � 3 :

Prove that each an 2 N, and that no an is divisible by 4.

Composite solutions by Edward J. Barbeau, University of Toronto,

Toronto, Ontario and Florian Herzig, student, Cambridge, UK.

Note �rst that fang is increasing and so has no zero terms.[Ed.: A simple induction su�ces.]

Rearranging, we havean+1 + an�1

an + an�2=

an

an�2.

Setting n = 3, 4, : : : , m, and taking the product, we get, for allm � 2,

am+1 + am�1

a3 + a1=

amam�1

a2a1.

Since a1 = 1, a2 = 2, a3 = 3, we get am+1 + am�1 = 2amam�1, oram+1 = am�1 (2am � 1); whence an 2 N for all n, by induction. Since2am � 1 is odd, another induction shows that am+1 is not divisible by 4,since am�1 is not.

Also solved by ZAVOSH AMIR-KHOSRAVI, student, North-Toronto Collegiate Insti-

tute, Toronto, Ontario; MICHEL BATAILLE, Rouen, France; JAMES T. BRUENING, South-

east Missouri State University, Cape Girardeau, MO, USA; NIKOLAOS DERGIADES, Thessa-

loniki, Greece; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; DAVID

DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA; ANTHONY FONG, student,

Eric Hamber Secondary School, Vancouver, BC; RICHARD I. HESS, Rancho Palos Verdes,

California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL

LAMBROU, University of Crete, Crete, Greece; KEE-WAI LAU, Hong Kong; MICHAEL

PARMENTER, Memorial University of Newfoundland, St. John's, Newfoundland; HADI

SALMASIAN, student, Sharif University of Technology, Tehran, Iran; JOEL SCHLOSBERG, stu-

dent, Bayside, NY, USA; NICOLAS TH �ERIAULT, student, Universit �e Laval, Montr �eal, Qu �ebec;

TODD THOMPSON, student, University of Arizona, Tucson, AZ, USA; and the proposer. There

was also a partially incorrect solution submitted.

The recurrence relation shown in the solution above was also obtained by nine other

solvers. However, only Herzig, Lambrou and the proposer actually derived the relation, while

the others all used induction. Strictly speaking, a complete proof should include a statement

and proof for the fact that an 6= 0 for all n. However, only a few solvers pointed this out

explicitly.

Janous considered more generally the sequence de�ned by

an+1 = kan � ka

n�1 +a2n

an�2

, (n � 3) ,

where k, a1, a2, a3 2 N, and showed that ifka1 + a3

a1a22 N and

ka1 + a3

a1a2> k,

then an 2 N for all n.

181

2328?. [1998: 176] Proposed by Walther Janous, Ursulinengymnas-

ium, Innsbruck, Austria.

It is known from Wilson's Theorem that the sequence fyn : n � 0g,with yn =

n! + 1

n+ 1, contains in�nitely many integers; namely, yn 2 N if and

only if n+ 1 is prime.

(a) Determine all integer members of the sequences fyn(a) : n � 0g, withyn =

n! + a

n+ a, in the cases a = 2, 3, 4.

(b) Determine all integer members of the sequences fyn(a) : n � 0g, withyn =

n! + a

n+ a, in the cases a � 5.

Solution by Nikolaos Dergiades, Thessaloniki, Greece.

For all a 2 N, we have that yn(a) =n! + a

n+ ais an integer when n = 1

and n = 2. If n = a > 2, then yn(a) =(a� 1)! + 1

2is not an integer.

Case 1: n+ a is not prime.

Suppose n > a. There are two possibilities:

(i) n + a = pq where gcd(p; q) = 1. Then 2n > n + a � 2q whichimplies q � n�1. Similarly p � n�1. This means pj(n�1)! and qj(n�1)!,which further impliespqjn!. Thusn!+a � a (modn+a), whence the numbern! + a

n+ ais not an integer.

(ii) n + a = pk, where k > 1 and p is prime. Arguing as above we

get pk�1j(n� 1)!. If the numbern! + a

n+ ais an integer, then pk�1j(n! + a).

[Ed. note: we actually have pkj(n!+a), but that is not needed in the proof.]From pk�1jn! we conclude that pk�1ja; from this and pk�1j(n + a) we getpk�1jn and �nally

pk�1j(n� 1)! and pk�1jn =) pk+k�2jn! =) n+ a = pkjn!

or n! + a � a (mod n+ a). So the numbern! + a

n+ ais not an integer.

Therefore, in case 1 if the numbern! + a

n+ ais an integer we must have n < a.

Case 2: n+ a is prime.

From Wilson's Theorem we have (n+ a � 1)! + 1 � 0 (mod n+ a).

182

If n! + a � 0 (mod n+ a), then

n!(n+ 1) � � � (n+ a� 1) + a(n+ 1) � � � (n+ a� 1) � 0 (mod n+ a) ,

�1 + a(n+ 1) � � � (n+ a� 1) � 0 (mod n+ a) ,

�1 + a(�a+ 1)(�a+ 2) � � � (�2)(�1) � 0 (mod n+ a) ,

�1� (�1)aa! � 0 (mod n+ a) .

In this case the numbern! + a

n+ ais an integer if n + a is a prime divisor of

s = �1� (�1)aa!. So there are a �nite number of terms of yn(a) such thatn! + a

n+ ais an integer, since when a 6= 1 we have s 6= 0.

Examples:

Let a = 2. Then s = �3 with 3 the only prime divisor, so n+ 2 = 3 gives

n = 1. Thus the numbern! + 2

n+ 2is an integer only when n = 1, 2.

Let a = 3. Then s = 5. So n+3 = 5 gives n = 2 and the numbern! + 3

n+ 3is

an integer only when n = 1, 2.

Let a = 4. Then s = �25 with 5 the only prime divisor, so n+ 4 = 5 gives

n = 1. We also need to check n < a: if n = 3, thenn! + 4

n+ 4=

10

7, which is

not an integer. Thus the numbern! + 4

n+ 4is an integer only when n = 1, 2.

Let a = 5. Then s = 119 with prime divisors 7 and 17, so n+ 5 = 7 or 17implies n = 2 or 12, respectively. We also need to check n < a: if n = 3, 4,

thenn! + 5

n+ 5=

11

8,29

9, respectively, neither of which is an integer. So the

numbern! + 5

n+ 5is an integer only when n = 1, 2, 12.

As a last example, if a = 22, then the numbern! + 22

n+ 22is an integer only

when n = 1; 2, 12, 499, 93 799 610 095 769 625.

Also solved by MANSUR BOASE, student, Cambridge, England; CHRISTOPHER

J. BRADLEY, Clifton College, Bristol, UK; THEODORE CHRONIS, student, Aristotle Univer-

sity of Thessaloniki, Greece; RICHARD I. HESS, Rancho Palos Verdes, California, USA; and the

proposer.

Hess was the only solver who speci�cally showed that yn(a) is never an integer when

n = 0 and a � 2. He also gave a complete list of the values of n which yield integers for

yn(a) for values of 6 � a � 17 (the list below gives only the values of n > 2):

183

a = 6 : 4 , 97

a = 7 : 5 032

a = 8 : 6 , 53 , 653

a = 9 : 2 990

a = 10 : 329 881

a = 11 : 6 , 12 , 7 842

a = 12 : 10 , 2 834 317

a = 13 : 1 720 , 3 593 190

a = 14 : 9 , 12 , 3 790 360 473

a = 15 : 6 , 16 , 38 , 1 510 244

a = 16 : 4 , 45 , 121 , 123 , 1 059 495

a = 17 : 56 , 256 443 711 660

Janous observes that there are quite a few new questions coming from this problem,

such as:

What is ` = lim supa!1

i(a), where i(a) is the number of integer members of the set�n! + a

n + a: n = 1; 2; 3 : : :

�?

If n(k) = #fa � 2 : i(a) = kg, what can be said about n(k)? Especially, in the event

of ` =1, is it true that, for all k � 2, we always have n(k) � 1? If not, for what values of k

is n(k) = 0?

2330. [1998: 176] Proposed by Florian Herzig, student, Perchtolds-

dorf, Austria.Prove that

e = 3� 1!

1 � 3 +2!

3 � 11 �3!

11 � 53 +4!

53 � 309 �5!

309 � 2119 + : : : ,

where11 = 3 � 3 + 2 � 1 ,53 = 4 � 11 + 3 � 3 ,309 = 5 � 53 + 4 � 11 ,2119 = 6 � 309 + 5 � 53 ,

.

.

.

Solution by Con Amore Problem Group, Royal Danish School of Edu-

cational Studies, Copenhagen, Denmark.We have to prove that

e = 3� 1!

a1a2+

2!

a2a3� 3!

a3a4+ � � �+ (�1)n � n!

anan+1

+ � � � , (1)

where a1, a2, : : : , are such that a1 = 1, a2 = 3, and for n � 3:

an = nan�1 + (n� 1)an�2 . (2)

184

Along with a1, a2, : : : , we consider d1, d2, : : : , de�ned by

dn = n!

�1

0!� 1

1!+

1

2!+ � � �+ (�1)n

n!

�(3)

for n = 1, 2, : : : . These are well-known as the derangement (or displace-ment) numbers, so called because dn is the number of permutations of(1, 2, : : : , n) where all the elements are displaced, but this need not concernus here. We just note for later use that

d3 = 2 and d4 = 9 (4)

and

dn+1

(n+ 1)!� dn

n!=

(�1)n+1

(n+ 1)!,

or (n+ 1)dn � dn+1 = (�1)n (5)

for n = 1, 2, : : : , whence (n+ 2)dn+1 � dn+2 = dn+1 � (n+ 1)dn, or

dn+2 = (n+ 1)dn+1 + (n+ 1)dn (6)

for n = 1, 2, : : : . Now de�ne for n = 1, 2, : : : :

�n =dn+2

n+ 1. (7)

Then by (4) and (6), �1 = d3=2 = 1, and �2 = d4=3 = 3, and for n � 3 wehave

(n+ 1)�n = (n+ 1)n�n�1 + (n+ 1)(n� 1)�n�2 ,

or �n = n�n�1 + (n � 1)�n�2. Comparison with (2) shows that forn = 1, 2, : : : , we have �n = an. Thus by (7):

an =dn+2

n+ 1(8)

for n = 1, 2, : : : . Now (8) and (5) imply

(�1)n � n!

anan+1

= (�1)n+2 � n!(n+ 1)(n+ 2)

dn+2dn+3

=�(n+ 3)dn+2 � dn+3

�� (n+ 2)!

dn+2dn+3

,

and so, for n = 1, 2, : : : :

(�1)n � n!

anan+1

=(n+ 3)!

dn+3

� (n+ 2)!

dn+2

. (9)

185

Using (9) and (3), and noting that 3 = 3!=d3, we get

3� 1!

a1a2+

2!

a2a3� 3!

a3a4+ � � �+ (�1)n � n!

anan+1

=3!

d3+

�4!

d4� 3!

d3

�+

�5!

d5� 4!

d4

�+ � � �+

�(n+ 3)!

dn+3

� (n+ 2)!

dn+2

�

=(n+ 3)!

dn+3

=

1

0!� 1

1!+

1

2!+ � � �+ (�1)n+3

(n+ 3)!

!�1

! (e�1)�1 = e as n!1 ,

which proves (1).

Also solved by WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL

LAMBROU, University of Crete, Crete, Greece; KEE-WAI LAU, Hong Kong; HEINZ-J �URGEN

SEIFFERT, Berlin, Germany; and the proposer.

Herzig used continued fractions to solve the problem.

2331. [1998: 176] Proposed by Paul Yiu, Florida Atlantic University,

Boca Raton, Florida, USA.

Let p be an odd prime. Show that there is at most one non-degenerateinteger triangle with perimeter 4p and integer area. Characterize those primesfor which such triangles exist.

Composite solutionby Jill S. Taylor, student,MountAllisonUniversity,

Sackville, New Brunswick; and the proposer.

Consider such a triangle of sides a, b, c, and semi-perimeter s, witharea

ps(s� a)(s� b)(s� c). Then 1 � a, b, c � 2p� 1.

Since s = 2p and the area of the triangle is an integer, one of (s� a),(s � b), (s � c) must be divisible by p. Without loss of generality, assumethat pj(s� a). Then s� a = p since 1 � s� a � 2p� 1.

Hence (s � b) + (s � c) = a = p and (s � b)(s� c) must be twicea square. Clearly, (s � b) and (s � c) are relatively prime. [Ed.: If d 2 Nis such that dj(s � b) and dj(s � c), then djp, and thus d = 1 or d = p.However, if d = p, then pj(2p� b), (2p� c) would imply that b = c = p,and so, 2s = a+ b+ c = 3p, which is a contradiction.]

It follows that s� b = x2 and x� c = 2y2 for relatively prime integersx and y. Hence p = x2 + 2y2.

Conversely, if p = x2 + 2y2, then p, p+ x2 and 2p� x2 would be thesides of a triangle with the described properties, since

2s = p+ (p+ x2) + (2p� x2) = 4p and

s(s� p)(s� (p+ x2))(s� (2p� x2)) = 2p2(p� x2)x2

= 2p2(2x2y2) = (2pxy)2 .

186

By well-known results in elementary number theory (see [3]), therepresentation p = x2 + 2y2 is possible and unique if and only ifp � 1, 3 (mod 8). Hence such a triangle exists (and is unique) if and only ifp � 1, 3 (mod 8).

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; FLORIAN

HERZIG, student, Cambridge, UK; RICHARD I. HESS, Rancho Palos Verdes, California, USA;

WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAVKONE �CN �Y, Ferris State

University, Big Rapids, Michigan, USA; MICHAEL LAMBROU, University of Crete, Crete,

Greece; GERRY LEVERSHA, St. Paul's School, London, England; JOEL SCHLOSBERG, student,

Bayside, NY, USA; HEINZ-J �URGENSEIFFERT, Berlin, Germany; KENNETHM.WILKE, Topeka,

Kansas, USA.

The �rst four triangles with the described proprties are (3; 4; 5; 6), (11; 13; 20; 66),

(17; 25; 26; 204), and (19; 20; 37; 114), (where the components denote the side lengths and

the areas of the triangles, respectively). These were given by Hess, Leversha and the proposer.

Hess and Leversha actually listed the �rst six and eleven such triangles, respectively.

The result cited in the solution above can be found in many books on number theory.

Besides [3], Kone �cn �y quoted [1] and two others. Both Sei�ert and Wilke quoted [2].

References:

[1] Ethan D. Bolker, Elementary Number Theory, W.A. Benjamin Inc., New York, 1970,

pp. 113{116.

[2] T. Nagell, Introduction to Number Theory, 2nd Ed., Chelsea, 1964, pp. 188{191.

[3] W. Sierpi �nski, A Selection of Problems in the Theory of Numbers, Pergamon Press

Ltd., Oxford, England, 1964, p. 72.

2332. [1998: 177] Proposed by D.J. Smeenk, Zaltbommel, the Neth-

erlands.

Suppose x and y are integers. Solve the equation

x2y2 � 7x2y+ 12x2 � 21xy � 4y2 + 63x+ 70y � 174 = 0 .

Solution by Michel Bataille, Rouen, France.

The equation reduces to the equivalent form

(x� 2)(y� 3)((x+ 2)(y� 4)� 21) = 0 :

The �rst two factors provide two families of solutions: (2; a) and (b;3), wherea and b are integers. Any other solution is such that (x+2)(y�4) = 21; thatis, (x+2; y� 4) is one of the pairs (1; 21), (21;1), (3; 7), (7; 3), (�1;�21),(�21;�1), (�3;�7), and (�7;�3). Examining each case separately, weobtain seven new solutions:

(�1;25) , (19;5) , (1;11) , (5;7) , (�3;�17) , (�5;�3) , and (�9; 1) .

Also solved by HAYO AHLBURG, Benidorm, Spain; MIGUEL AMENGUAL COVAS,

Cala Figuera, Mallorca, Spain; ZAVOSH AMIR-KHOSRAVI, student, North Toronto Colle-

giate Institute, Toronto; SAM BAETHGE, Nordheim, Texas, USA; EDWARD J. BARBEAU, Uni-

versity of Toronto, Toronto, Ontario; CHRISTOPHER J. BRADLEY, Clifton College, Bristol,

UK; THEODORE CHRONIS, Athens, Greece; GORAN CONAR, student, Gymnasium Vara�zdin,

187

Vara�zdin, Croatia; NIKOLAOS DERGIADES, Thessaloniki, Greece; CHARLES R. DIMINNIE,

Angelo State University, San Angelo, TX, USA; DAVID DOSTER, Choate Rosemary Hall,

Wallingford, Connecticut, USA; ANTHONY FONG, student, Eric Hamber Secondary School,

Vancouver; IAN JUNE L. GARCES, Ateneo de Manila University, The Philippines and GIO-

VANNI MAZZARELLO, Ferrovie dello Stato, Florence, Italy; DOUGLASS L. GRANT, Univer-

sity College of Cape Breton, Sydney, Nova Scotia; YEO KENG HEE, Hwa Chong Junior Col-

lege, Singapore; FLORIAN HERZIG, student, Cambridge, UK; RICHARD I. HESS, Rancho

Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus-

tria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; MICHAEL

LAMBROU, University of Crete, Crete, Greece; KEE-WAI LAU, Hong Kong; JESSIE LEI, stu-

dent, Vincent Massey Secondary School, Windsor, Ontario; GERRY LEVERSHA, St. Paul's

School, London, England; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA;

CHRISTO SARGOTIS, Thessaloniki, Greece; HEINZ-J �URGEN SEIFFERT, Berlin, Germany; BOB

PRIELIPP, University of Wisconsin{Oshkosh, Wisconsin, USA; DIGBY SMITH, Mount Royal

College, Calgary, Alberta; PANOS E. TSAOUSSOGLOU, Athens, Greece; ERIC UMEGA�

RD,

V�astera�

s, Sweden; KENNETH M. WILKE, Topeka, Kansas, USA; and the proposer. There was

one incomplete solution submitted.

Most of the submitted solutions are similar to the one given above.

2333. [1998: 177] Proposed by D.J. Smeenk, Zaltbommel, the Neth-

erlands.

You are given that D and E are points on the sides AC and AB re-spectively of4ABC. Also,DE is not parallel to CB. Suppose F andG arepoints of BC and ED respectively such that

BF : FC = EG : GD = BE : CD .

Show that GF is parallel to the angle bisector of \BAC.

I. Solution by Michael Lambrou, University of Crete, Crete, Greece.

With respect to position vectors originating from A, write�!AB = b;�!

AC = c, so that for some p, q 2 (0; 1) (see remark) we have e :=�!AE = pb,

d :=�!AD = qc. If the given equal ratios are denoted by � > 0 (see remark),

we have

f :=�!AF =

�c + b

�+ 1and g :=

�!AG =

�d + e

�+ 1=

�qc + pb

�+ 1.

Moreover, the condition BE : CD = � becomes����!BE��� = �

����!CD���, and so

(1� p)jbj = �(1� q)jcj. Thus,

�!GF = f � g =

�(1� q)

�+ 1c +

1� p

�+ 1b

=�(1� q)jcj�+ 1

c

jcj +(1� p)jbj�+ 1

b

jbj =(1� p)jbj�+ 1

�c

jcj +b

jbj

�,

which is parallel to

�c

jcj +b

jbj

�. But this last is a sum of unit vectors along

AB, AC and so is along the bisector of A, completing the proof.

188

Remarks. 1. The condition \DE not parallel to CB" has not been used. Inthe parallel case it turns out that GF coincides with the bisector.

[Editor's additional remark: Most solvers made a similar remark, with dif-ferent interpretations according to whether or not one's de�nition of parallellines permits the lines to coincide.]

2. We have taken here D, E, F in the interior of the correspondingsides of 4ABC. Thus we have 0 < p, q < 1 and � > 0. However, this isnot necessary and the proof can be adapted to the more general case. Forexample, if � > 0 then p, q must be both greater than 1 or both less than 1.

II. Solution by Nikolaos Dergiades, Thessaloniki, Greece.

Let K, L be chosen so that the quadrangles EBKG and GLCD areparallelograms, and therefore KG is equal and parallel to BE while LGis equal and parallel to CD. Consequently, \BAC = \KGL, and sinceBKjjLC and

BK

LC=

EG

GD=

BF

FC

[the last equality by assumption], we deduce that 4BKF is similar to4CLF , so that KL passes through F . Therefore,

KF

FL=

BF

FC=

BE

CD=

KG

LG

[the middle equality by assumption], which implies that in4KGL, we havethat GF is the bisector of \KGL and (sinceKGjjBA and LGjjCA) parallelto the bisector of \BAC.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; FLORIAN

HERZIG, student, Cambridge, UK; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus-

tria; GERRYLEVERSHA, St. Paul's School, London, England; HEINZ-J �URGENSEIFFERT, Berlin,

Germany; TOSHIO SEIMIYA, Kawasaki, Japan; and the proposer.

2334. [1998: 177] Proposed by Toshio Seimiya, Kawasaki, Japan.

Suppose that ABC is a triangle with incentre I, and that BI, CI meetAC, AB atD, E respectively. Suppose that P is the intersection ofAI withDE. Suppose that PD = PI.

Find angle ACB.

Solution by Gerry Leversha, St. Paul's School, London, England.

We are given that PD = PI; hence \AID = \EDI. But

\AID = \ABI + \BAI =A+ B

2

Therefore \EDI =A+ B

2.

189

We need to calculate some angles.

\DEI = 180� � \EDI � \EID

= 180� � A+B

2��180� � B +C

2

�=

C � A

2,

\DEA = 180� � \EAD � \ADE

= 180� � A� (180� � \EDI � \BDC)

= �A+A+ B

2+

�180� � C � B

2

�

= 180� � C � A

2.

Now let ED = x and use the Sine Rule on triangles AED and CED:

x

AD=

sinA

sin�C + A

2

� andx

DC=

sin C

2

sin�C�A2

� .

Now we divide and recall thatDC

AD=a

cby the Angle Bisector Theorem.

ThereforesinA

sinC=

a

c=

sinA sin�C�A2

�sin C

2sin

�C + A

2

� .Thus,

sinC

2sin

�C +

A

2

�= sinC sin

�C �A

2

�

= 2 sinC

2cos

C

2sin

�C � A

2

�,

and sin

�C +

A

2

�= 2 cos

C

2sin

�C �A

2

�= sin

�C � A

2

�� sin

A

2.

Hence sinA

2= sin

�C � A

2

�� sin

�C +

A

2

�= �2 cosC sin

A

2,

and cosC = �1

2, so that C = 120�.

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain;

CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; NIKOLAOS DERGIADES, Thessa-

loniki, Greece; FLORIAN HERZIG, student, Cambridge, UK; RICHARD I. HESS, Rancho Palos

Verdes, California, USA; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan,

USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; KEE-WAI LAU, Hong Kong;

D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer. There was one incorrect solu-

tion.

190

2335. [1998: 177] Proposed by Toshio Seimiya, Kawasaki, Japan.

Triangle ABC has circumcircle �. A circle �0 is internally tangent to �

at P , and touches sidesAB, AC at D, E respectively. Let X, Y be the feetof the perpendiculars from P to BC, DE respectively.

Prove that PX = PY sin A

2.

Solution by Florian Herzig, student, Cambridge, UK.

Rename the circles � and �0 (respectively) as �1 and �2 to avoid laterconfusion. Invert the con�guration in any circle with centre P , and denotethe image of a point or circle X by X0. Then �0

1and �0

2are parallel lines,

and the latter touches the circumcircles of PC0E0A0 and PA0D0B0 (becausethe corresponding lines touch �2). Therefore B

0C0 = 2D0E0 and also

\E0PD0 = \E0PA0+\A0PD0 =1

2(\C0PA0+\A0PB0) =

1

2\C0PB0 .

Moreover, \C0PB0 = \BPC = 180o � A. Since X is the foot of theperpendicular to BC from P (the centre of inversion), the image of BC isthe circle B0C0P having PX0 as diameter.

[Editor's comment. Herzig provided a simple justi�cation of the claim, butproofs are readily found in any text with a section on inversions, such asCoxeter and Greitzer's Geometry Revisited.]

Similarly, PY 0 is a diameter of circle E0PD0. Hence PX0 is twice the cir-cumradius of circle B0PC0, and PY 0 of E0PD0. By the Sine Law, therefore,

PX0 =B0C0

sinA, and PY 0 =

D0E0

sin\E0PD0 =D0E0

sin(90o � A2).

Finally,

PX

PY=

PY 0

PX0 =D0E0 � sinAB0C0 � cos A

2

= sinA

2,

as we wanted to show.

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain;

NIKOLAOS DERGIADES, Thessaloniki, Greece; WALTHER JANOUS, Ursulinengymnasium,

Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete, Greece; D.J. SMEENK,

Zaltbommel, the Netherlands; and the proposer.

The con�guration investigated here has appeared before in CRUX | in Leon Banko�'s

\Mixtilinear Adventure" [1983: 2-7] and in S. Shirali's \On a Generalized Ptolemy Theorem"

[1996: 49-53]. See also Nathan Altshiller Court, College Geometry, p. 239 theorem 537. We

thank Bellot and Seimiya for these references. Bellot also found the con�guration in a 1991

Bulgarian journal and in the Cambridge Mathematical Tripos of 1929. It is remarkable how he

is so often able to �nd relevant references. This editor wonders if the method can be applied to

�nding my glasses (which occasionally get misplaced.)

191

2336. [1998: 177] Proposed by Toshio Seimiya, Kawasaki, Japan.

The bisector of angleA of a triangleABC meetsBC atD. Let � and �0

be the circumcircles of triangles ABD and ACD respectively, and let P , Qbe the intersections of AD with the common tangents to �, �0 respectively.

Prove that PQ2 = AB � AC.

Solution by Francisco Bellot Rosado, I.B. Emilio Ferrari, Valladolid,

Spain.

We �rst will prove that, with the notation of the problem, calling t thelength of the segment of common tangent to both circles (AD the commonchord),

PQ2 = AD2 + t2 . (1)

It is easy to see that P is the midpoint of the common tangentsegment (for example, calculating the power of P with respect to bothcircles in two forms: if L, N are the points of tangency, thenPL2 = PA � PD = PN2 ===) PL = PN .)

Then calling x = AP = QD, we have

PQ = AD+ 2x ===) PQ2 = AD2 + 4x(AD+ x) ,

and as �t

2

�2= x(AD+ x) ,

the formula (1) follows.

As AD is the internal bisector of angle A, we have the well knownformulae

AD =2bc

b+ ccos

A

2, (2)

AD2 = cb

�1� a2

(b+ c)2

�. (3)

Therefore, we must prove that, if AD is the bisector of the angle A,

AD2 + t2 = cb or, using (3), t =apbc

b+ c. (4)

To this end, we �rst �nd the radius R1 of the circle ABD with aid ofthe Sine Law:

BD

sin A

2

= 2R1 (==) R1 =ab

2(b+ c) sin A

2

, (5)

192

and analogously the radius of the circle ADC:

R2 =ab

2(b+ c) sin A2

. (6)

If O1, O2 are the centres of these circles, we have

t2 = O1O2

2 � (R1 � R2)2 , (7)

and, callingM = O1O2 \ AD, we obtain:

O1M2

= R2

1� AD2

4; O2M

2

= R2

2� AD2

4,

from which, using (2), (5) and the Sine Law, we �nd

O1M2

=c2(a2 � b2 sin2A)

4(b+ c)2 sin2 A2

=c2a2 cos2B

4(b+ c)2 sin2 A2

;

that is, O1M =ca cosB

2(b+ c) sin A2

, and analogously, O2M =ba cosC

2(b+ c) sin A2

,

from which results

O1O2 =a2

2(b+ c) sin A2

.

Going back to (7), we obtain

t2 =a2[a2 � (c� b)2]

4(b+ c)2 sin2 A2

=a2(a+ c� b)(a� c+ b)

4(b+ c)2 sin2 A2

=a2bc

(b+ c)2,

which is precisely (4), and the problem is solved.

Also solved by NIKOLAOS DERGIADES, Thessaloniki, Greece; WALTHER JANOUS,Ursu-

linengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete, Greece;

D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer. There was one incorrect solu-

tion.

Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell

Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil

Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,

J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

193

THE ACADEMY CORNERNo. 25

Bruce Shawyer

All communications about this column should be sent to BruceShawyer, Department of Mathematics and Statistics, Memorial Universityof Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7.

Abstracts � R�esum�es

Canadian Undergraduate Mathematics Conference

1998 | Part 3

The Brachistochrone Problem

Nils Johnson

The University of British Columbia

The brachistochrone problem is to �nd the curve between two points down

which a bead will slide in the shortest amount of time, neglecting friction and as-

suming conservation of energy. To solve the problem, an integral is derived that

computes the amount of time it would take a bead to slide down a given curve y(x).

This integral is minimized over all possible curves and yields the di�erential equation

y(1 + (y0)2) = k2 as a constraint for the minimizing function y(x). Solving this

di�erential equation shows that a cycloid (the path traced out by a point on the rim

of a rolling wheel) is the solution to the brachistochrone problem. First proposed

in 1696 by Johann Bernoulli, this problem is credited with having led to the devel-

opment of the calculus of variations. The solution presented assumes knowledge of

one-dimensional calculus and elementary di�erential equations.

The Theory of Error-Correcting Codes

Dennis Hill

University of Ottawa

Coding theory is concerned with the transfer of data. There are two issues of

fundamental importance. First, the data must be transferred accurately. But equally

important is that the transfer be done in an e�cient manner. It is the interplay of

these two issues which is the core of the theory of error-correcting codes.

Typically, the data is represented as a string of zeros and ones. Then a code

consists of a set of such strings, each of the same length. The most fruitful approach

to the subject is to consider the set f0; 1g as a two-element �eld. We will then only

194

consider codes which are vector spaces over this �eld. Such codes will be called linear

codes.

Since there is no way to always ensure that the message has been transferred

accurately, it is important to add redundant structure so that, except for extreme

cases, the receiver will realize that there has been an error. This is called error-

detection. It would be even better if the receiver were able to not only detect that an

error has occurred, but to determine the location of the incorrect digit or digits, and

thus determine the original message. This is the notion of error correction.

This theory uses in a fundamental way ideas from linear algebra over �nite

�elds, such abstract algebra as principal ideal domains, and metric space theory. One

of the reasons this is such an exciting subject is that it combines principles of abstract

algebra with very concrete and important applications. Among the many applications

are the minimization of noise from compact disc recordings, transmission of data from

satellites, and transmission of �nancial data.

I will introduce the basic concepts of coding theory, focusing on linear codes. I

will discuss several more specialized topics, such as the most important methods of

error correction. I will also discuss a number of speci�c examples. I will conclude

with a brief discussion of convolutional codes, one of the more recent advances in the

�eld.

Fixed Points and Diagonalization: An Abstraction of G�odel's Theorem

Todd A. Kemp

University of Calgary

G�odel's Incompleteness Theorem is one of themost provocative results inMath-

ematical Logic. It may be summarized \In any `su�ciently powerful' consistent the-

ory, there must always be theorems (true sentences) which are unprovable."

In this paper, I address the question of what this heuristic phrase `su�ciently

powerful' means. I will appeal to abstract conditions put forward by Smullyan, and

show that any system meeting them also satis�es G�odel's Theorem. Further, af-

ter outlining some of the basic terminology and major results in the area, I will

demonstrate that Robinson's Arithmetic|awell-known G�odel system|indeed meets

Smullyan's conditions, hence lending some credibility to the claim that these condi-

tions are not only su�cient but necessary.

Smullyan also goes on to show that his abstract form of G�odel's Theorem is

actually a special case of a general Fixed Point Theorem|one which is also the prim-

itive framework behind a major result in Recursive Function Theory, and Smullyan's

favourite puzzle|the Mockingburd Puzzle. I will discuss this Fixed Point Theorem,

and how it relates to G�odel's theorem.

Misuse of Statistics

Theodoro Koulis

McGill University

Statistics play a big role in the sciences. Statistics provide scientists with tools

that help them verify the signi�cance of their data. However, since these tools are

widely distributed in popular software packages such as Microsoft Excel, they are

195

often misused. Using some test statistics, I will show how some basic procedures can

be misdirected and how conclusions can be misinterpreted. These will be:

1. The sign test and hypothesis testing:

Foundations of hypothesis testing (the likelihood ratio principle)

2. Goodness of �t test statistics:

Good and bad measures: Pearson's �2 and Kolmogorov-Smirnov tests

3. Correlation and rank correlation (if time permits)

An Introduction to Random Walks from P �olya to Self-Avoidance

Michael Kozdron

University of British Columbia

This paper provides an introduction to random walks. We begin with some ba-

sic de�nitions and culminate with the classical theorem of P �olya that a simple random

walk in Zd, d � 3 is transient and recurrent otherwise. We then discuss the more

contemporary topic of self-avoiding random walks and survey some currently open

problems. This paper assumes only a minimal background in probability including

the notion of a random variable and an expectation.

Of Graphs and Co� Grounds: Decompiling Java

Patrick Lam

McGill University

Java programmers write applications and applets in plain English-like text, and

then apply a java compiler to the text to obtain class �les. Class �les, which are

typically transmitted across the Web, are a low-level representation of the original

text; they are not human-readable. Consider a compiler as a function from text to

class �les. My goal is to compute the inverse function: given the compiled class �le,

I wish to �nd the best approximation to the original text possible. This is called

decompilation.

Given a class �le, one can build an unstructured graph representation of the

program. The main goal of my work is to develop graph algorithms to convert these

unstructured graphs into structured graphs corresponding to high-level program con-

structs, and I will discuss these in detail. I shall also mention some results concerning

possibility and impossibility which show that decompilation is always possible if the

program may be lengthened.

196

THE OLYMPIAD CORNERNo. 198

R.E. Woodrow

All communications about this column should be sent to Professor R.E.Woodrow, Department of Mathematics and Statistics, University of Calgary,Calgary, Alberta, Canada. T2N 1N4.

We begin this number with the problems of the XXXIX Republic Com-

petition of Mathematics in Macedonia and the problems of the third Mace-

donian Mathematical Olympiad. My thanks go to Ravi Vakil who collected

them for me while he was Canadian Team Leader to the IMO at Mumbai.

XXXIX REPUBLIC COMPETITION OFMATHEMATICS IN MACEDONIA

Class I

1. The sum of three integers a, b and c is 0. Prove that 2a4+2b4+2c4

is the square of an integer.

2. Prove that if

aa10

= aa21

= � � � = aa19961995

= aa01996

; a1 2 R� ,

then

a0 = a1 = � � � = a1996 .

3. Let ha, hb and hc be the altitudes of the triangle with edges a, b

and c, and r be the radius of the inscribed circle in the triangle. Prove that

the triangle is equilateral if and only if ha + hb + hc = 9r.

4. Prove that each square can be cut into n (n � 6) squares.

Class II

1. Prove that for positive real numbers a and b

2 �pa+ 3 � 3

pb � 5 � 5

pab .

2. The point M is the mid-point on the side BC = a of a triangle

ABC. Let r1, r2, r3 be the radii of the inscribed circles in the triangles

ABC, ABM and ACM respectively. Prove that

1

r1+

1

r2� 2

�1

4+

2

a

�.

197

3. Let A = fz1, z2, : : : , z1996g be a set of complex numbers and for

each i 2 f1, 2, : : : , 1996g suppose fz1z1, z1z2, : : : , z1z1996g = A.

(a) Prove that jzij = 1 for each i.

(b) Prove that z 2 A implies z 2 A.

4. Find the biggest value of the di�erence x�y if 2(x2+y2) = x+ y.

Class III

1. Solve the equation x1996�1996x1995+ � � �+1 = 0 (the coe�cients

in front of x, : : : , x1994 are unknown), if it is known that its roots are positive

real numbers.

2. LetAH, BK andCL be the altitudes of an arbitrary triangleABC.

Prove that

AK � BL � CH = AL � BH �CK = HK �KL � LH .

3. An initial triple of numbers 2,p2, 1p

2is given. A new triple may be

obtained from an old one as follows: two numbers a and b of the triple are

changed to(a+b)p

2and

(a�b)p2

and the third number is unchanged. Is it possible

after a �nite number of such steps to obtain the triple 1,p2, 1 +

p2 ?

4. A �nite number of points in the plane are given such that not all

of them are collinear. A real number is assigned to each point. The sum of

the numbers for each line containing at least two of the given points is zero.

Prove that all numbers are zeros.

Class IV

1. Let a1, a2, : : : , an be real numbers which satisfy:

There exists a real number M such that jaij �M for each i 2 f1, : : : , ng.

Prove that a1 + 2a2 + � � �+ nan �Mn2

4.

2. Two circles with radii R and r touch from inside. Find the side of

an equilateral triangle having one vertex at the common point of the circles

and the other two vertices lying on the two circles.

3. The same problem as problem 3 given for Class III.

4. The same problem as problem 4 given for Class III.

198

PROBLEMS ON THE THIRD MACEDONIANMATHEMATICAL OLYMPIAD

1. Let ABCD be a parallelogram which is not a rectangle and E be

a point in its plane, such that AE ? AB and BC ? EC. Prove that

\DAE = \CEB. [Ed. We know this is incorrect | can any reader supply

the correct version?]

2. Let P be the set of all polygons in the plane and letM : P ! R be

a mapping which satis�es:

(i) M(P ) � 0 for each polygon P ;

(ii) M(P ) = x2 if P is an equilateral triangle of side x;

(iii) If P is a polygon separated into two polygons S and T , then

M(P ) = M(S) +M(T ); and

(iv) If P and T are congruent polygons, then M(P ) = M(T ).

FindM(P ) if P is a rectangle with edges x and y.

3. Prove that if �, � and are angles of a triangle, then

1

sin�+

1

sin��

8

3 + 2 cos .

4. A polygon is called \good" if the following conditions are satis�ed:

(i) all angles belong to (0; �) [ (�; 2�);

(ii) two non-neighbouring sides do not have any common point; and

(iii) for any three sides, at least two are parallel and equal.

Find all non-negative integers n such that there exists a \good" polygon with

n sides.

5. Find the biggest number n such that there exist n straight lines in

space, R3, which pass through one point and the angle between each two

lines is the same. (The angle between two intersecting straight lines is de-

�ned to be the smaller one of the two angles between these two lines.)

Next we give the problems of the Ninth Irish Mathematical Olympiad,

written Saturday, May 4, 1996. My thanks again go to Ravi Vakil for col-

lecting the problems and sending them to me while he was Canadian Team

Leader to the IMO at Mumbai.

199

NINTH IRISH MATHEMATICAL OLYMPIADFirst Paper | May 4, 1996

Time: 3 hours

1. For each positive integer n, let f(n) denote the greatest common

divisor of n!+1 and (n+1)! (where ! denotes \factorial"). Find, with proof,

a formula for f(n) for each n.

2. For each positive integer n, let S(n) denote the sum of the digits

of n (when n is written in base 10). Prove that for every positive integer n,

S(2n) � 2S(n) � 10S(2n) .

Prove also that there exists a positive integer n with S(n) = 1996S(3n),

3. Let K be the set of all real numbers x with 0 � x � 1. Let f

be a function from K to the set of all real numbers R with the following

properties:

(i) f(1) = 1;

(ii) f(x) � 0 for all x 2 K;

(iii) if x, y and x+ y are all inK, then f(x+ y) � f(x) + f(y).

Prove that f(x) � 2x for all x 2 K.

4. Let F be the mid-point of the sideBC of the triangleABC. Isosce-

les right-angled triangles ABD and ACE are constructed externally on the

sides AB and AC with the right angles at D and E, respectively.

Prove that DEF is a right-angled isosceles triangle.

5. Show, with proof, how to dissect a square into at most �ve pieces

in such a way that the pieces can be reassembled to form three squares no

two of which have the same area.

Second Paper | May 4, 1996Time: 3 hours

6. The Fibonacci sequence F0, F1, F2, : : : is de�ned as follows: F0 = 0,

F1 = 1 and for all n � 0, Fn+2 = Fn+Fn+1. (So F2 = 1, F3 = 2, F4 = 3,

F5 = 5, F6 = 8, : : : .) Prove that

(i) The statement \Fn+k � Fn is divisible by 10 for all positive inte-

gers n" is true if k = 60 but it is not true for any positive integer k < 60.

(ii) The statement \Fn+t � Fn is divisible by 100 for all positive inte-

gers n" is true if t = 300 but it is not true for any positive integer t < 300.

7. Prove that the inequality 21=2 � 41=4 � 81=8 � � � (2n)1=2n < 4 holds

for all positive integers n.

200

8. Let p be a prime number and a and n positive integers. Prove that

if 2p + 3p = an, then n = 1.

9. Let ABC be an acute-angled triangle and let D, E, F be the feet of

the perpendiculars from A, B, C onto the sides BC, CA, AB, respectively.

Let P , Q, R be the feet of the perpendiculars from A, B, C onto the lines

EF , FD, DE respectively. Prove that the lines AP , BQ, CR (extended)

are concurrent.

10. We are given a rectangular chessboard of size 5� 9 ( so there are

�ve rows of squares, each row containing nine squares). The following game

is played: Initially, a number of discs are randomly placed on some of the

squares, no square being allowed to contain more than one disc. A complete

move consists of moving every disc from the square containing it to another

square subject to the following rules:

(i) each disc may be moved one square up or down, or left or right, of the

square it occupies, to an adjoining square;

(ii) if a particular disc is moved up or down as part of a complete move,

then it must be moved left or right in the next complete move;

(iii) if a particular disc is moved left or right as part of a complete move,

then it must be moved up or down in the next complete move;

(iv) at the end of each complete move, no square can contain two or more

discs.

The game stops if it becomes impossible to perform a complete move. Prove

that if initially 33 discs are placed on the board then the game must eventu-

ally stop. Prove also that it is possible to place 32 discs on the board in such

a way that the game could go on forever.

Now we turn to solutions by our readers to problems of the 30th

SpanishMathematical Olympiad, Final Round, November 26{27, 1993 [1998:

69{70].

30th SPANISH MATHEMATICAL OLYMPIADFirst Round | November 26{27, 1993

1. Show that, for all n 2 N, the fractions

n� 1

n,

n

2n+ 1,

2n+ 1

2n2 + 2n,

are irreducible.

201

Solution by Pierre Bornsztein, Courdimanche, France.

Soit n 2 N�.

On a, pour n � 2, (n;n � 1) = 1. En plus(n;2n + 1) = 1 et doncn

2n+1est irr �eductible. En�n (2n + 1; 2n) = 1 et (2n + 1; n + 1) = 1, car

si p divise 2n+ 1 et n + 1 alors p divise (2n+ 1) � (n+ 1) = n, donc p

divise (n;n + 1) = 1. D'o �u (2n + 1; 2n(n + 1)) = 1 et donc 2n+1

2n2+2nest

irr �eductible.

3. Solve the following system of equations:

x � jxj + y � jyj = 1; bxc + byc = 1 ,

in which jtj and btc represent the absolute value and the integer part of the

real number t.

Solutions by Pierre Bornsztein, Courdimanche, France; and EdwardT.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Wang'ssolution.

We show that there are only two solutions: (x; y) = (1; 0) and (0;1).

Clearly, x and y cannot both be negative. Hence by symmetry, there are two

cases to be considered:

(i) If x � 0 and y � 0, then from x2 + y2 = 1 we get 0 � x, y � 1. If x < 1

and y < 1, then bxc + byc = 0, a contradiction. Hence x = 1 or y = 1.

Then from x2 + y2 = 1 we obtain the two solutions (1;0) and (0; 1).

(ii) If x � 0 and y < 0 then x2� y2 = 1, and from bxc � x, byc � y we get

x+ y � 1. Since x� y > x+ y � 1 we have x2�y2 = (x�y)(x+ y) > 1,

a contradiction. Therefore, there are no solutions in this case.

4. Let AD be the internal bisector of the triangle ABC (D 2 BC), E

the point symmetric to D with respect to the mid-point of BC, and F the

point of BC such that \BAF = \EAC. Show that BFFC

= c3

b3.

Solutions by Pierre Bornsztein, Courdimanche, France; by ToshioSeimiya, Kawasaki, Japan; and byD.J. Smeenk, Zaltbommel, theNetherlands.We give Seimiya's solution.

202

A

B CD M EF

T

S

Let T be the point on AF such that BT kAC and let S be the point on

AE such that CS k AB. Then \ABT + \BAC = 180� and

\ACS + \BAC = 180�.

Thus we have \ABT = \ACS. Since \BAT = \BAF = \CAE =

\CAS we have 4ABT � 4ACS so that

BT

CS=

AB

AC=

c

b. (1)

As BT k AC we get

BF

FC=

BT

AC=

BT

b. (2)

As AB kCS we have

BE

EC=

AB

CS=

c

CS. (3)

Let M be the mid-point of BC.

Since E is the point symmetric to D with respect to M we have

EC = BD and BE = DC ,

so thatBE

EC=

DC

BD.

Since AD is the bisector of \BAC, we getDC

BD=

AC

AB=

b

c. Thus

we have

BE

EC=

b

c. (4)

203

From (3) and (4) we have

c

CS=

b

c; so that CS =

c2

b.

Hence, from (1), we get BT =c3

b2; whence from (2),

BF

FC=

c3

b3.

5. Find all the natural numbers n such that the number

n(n+ 1)(n+ 2)(n+ 3)

has exactly three prime divisors.

Solutions by Pierre Bornsztein, Courdimanche, France; and by EdwardT.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Wang'ssolution.

The only such n's are n = 2, 3 and 6.

Let P (n) = n(n+ 1)(n+ 2)(n+ 3). Then P (1) = 23 � 3 and thus

n = 1 is not a solution. Hence we assume n � 2. Note �rst that for all

k 2 N, (k; k+1) = (2k� 1; 2k+1) = 1 and hence if n is odd, then each of

n, n+1 and n+2must be a distinct prime power. We are led to two cases:

(a) If n is odd, then since n+1 is even we must have n = pa, n+1 = 2b and

n+ 2 = qc where a, b, c, p, q 2 N with p and q being distinct odd primes.

Note that n+3 = 2b+2 = 2(2b�1+1) where b � 2. Since the only possible

prime divisors of n+ 3 are 2, p or q, we have either

(i) 2b�1 + 1 = p� for some � 2 N

or

(ii) 2b�1 + 1 = q� for some � 2 N .

In case (i) we have 2p� = n+ 3 = pa + 3. Clearly � � a and thus p� j pa.Hence p� j 3 which implies p = 3, � = 1. Thus b = 2 and n = 3. Indeed,

n = 3 is a solution since P (3) = 23 � 32 � 5.

In case (ii) we have 2q� = n+ 3 = qc + 1, which is clearly impossible

since q 6 j 1.(b) If n is even, then by the same argument, we have n+1 = pa, n+2 = 2b

and n + 3 = qc where a, b, c, p, q 2 N with p and q being distinct odd

primes. Note that n = 2b � 2 = 2(2b�1 � 1) where b � 2. If b = 2, then

n = 2, which is indeed a solution since P (2) = 23 � 3� 5. If b > 2, then

we must have either

(iii) 2b�1 � 1 = p� for some � 2 N

or

(iv) 2b�1 � 1 = q� for some � 2 N .

204

In case (iii) we have 2p� = n = pa � 1 which is clearly impossible since

p 6 j 1.In case (iv) we have 2q� = n = qc � 3. Clearly � � c and thus q� j 3

which implies q = 3, � = 1. Thus b = 3 and n = 6, which is indeed a

solution since P (6) = 24 � 33 � 7.

To summarize, n(n+1)(n+2)(n+3)has exactly three prime divisors

if and only if n = 2, 3 or 6.

6. An ellipse is drawn taking as major axis the biggest of the sides of

a given rectangle, such that the ellipse passes through the intersection point

of the diagonals of the rectangle.

Show that, if a point of the ellipse, external to the rectangle, is joined

to the extreme points of the opposite side, then three segments in geometric

progression are determined on the major axis.

Solutions by Pierre Bornsztein, Courdimanche, France; and by D.J.Smeenk, Zaltbommel, the Netherlands. We give Smeenk's solution.

AB

C D

PQR

O

y

xD1C1

An equation of the ellipse E is b2x2 + a2y2� a2b2 = 0. The rectangle

ABCD has sides AB = CD = 2a, AD = BC = 2b. A point P of E is

P (a cos'; b sin'), with sin' > 0. Let the line through P parallel to AB

meet BC andAD inR andQ, respectively. Let PD andPC meet the major

axis at D1, C1 respectively.

From 4DPQ and4DD1A, we have

D1A : PQ = 2b : (2b+b sin') , D1A : a(1�cos') = 2 : (2+sin') ;

===) D1A =2a(1� cos')

2 + sin'. (1)

In the same way, from 4PCD and 4PC1D1, we have

C1D1 =2a sin'

2 + sin', (2)

205

and from 4CPR and4CC1B, we have

BC1 =2a(1 + cos')

2 + sin'. (3)

From (1), (2) and (3) D1A : C1D1 : BC1 = (1� cos') : sin' : (1+ cos').

And as sin2 ' = 1� cos2' the three segments are in geometric progression

indeed.

7. Let a 2 R be given. Find the real numbers x1, : : : , xn which satisfy

the system of equations

x21+ ax1 +

�a�12

�2= x2 ,

x22+ ax2 +

�a�12

�2= x3 ,

..........................

x2n�1 + axn�1 +�a�12

�2= xn ,

x2n + axn +�a�12

�2= x1 .

Solutions by Pierre Bornsztein, Courdimanche, France; by PavlosMaragoudakis, Pireas, Greece; and by PanosE. Tsaoussoglou,Athens, Greece.We give thewrite-up ofMaragoudakis, although themethodswere the same.

The �rst equation becomes

x21+(a�1)x1+

�a� 1

2

�2= x2�x1 or

�x1 +

a� 1

2

�2= x2�x1 .

Similarly �x1 +

a�12

�2= x2 � x1 ,�

x2 +a�12

�2= x3 � x2 ,

................�xn�1 + a�1

2

�2= xn � xn�1 ,�

xn + a�12

�2= x1 � xn .

Adding, we get:

�x1 +

a�12

�2+�x2 +

a�12

�2+ � � � +

�xn�1 + a�1

2

�2+�xn + a�1

2

�2= 0 .

So x1 = x2 = � � � = xn�1 = xn = 1�a2

.

We next give solutions to two of the problems from Peru's Selection

Test for the XII IberoAmerican Olympiad [1998: 131].

206

1. Given an integer a0 > 2, the sequence a0, a1, a2, : : : is de�ned as

follows:ak+1 = ak(1 + ak); if ak is an odd number

ak+1 = ak2; if ak is an even number.

Prove that there is a non-negative integer p such that ap > ap+1 > ap+2.

Solution by Pierre Bornsztein, Courdimanche, France.On montre facilement par r �ecurrence sur k que pour tout k 2 N, ak 2

N.

Alors pour tout k � 0, ak+1 6= ak si ak est impair alors ak+1 = ak +

a2k > ak. Cependant si ak est pair alors ak+1 < ak. Donc ap > ap+1 >

ap+2 si et seulement si ap � 0 (mod 4).

Lemme : Soit n 2 N, n � 2. S'il existe k � 0 tel que ak = 2 + 2nq o �u q

impair, q � 1, alors il existe p � k tel que ap > ap+1 > ap+2.

Preuve du Lemme : Par r �ecurrence sur n

Pour n = 2; si ak = 2+ 4q o �u q impair, q � 1 .

Alors

ak+1 = 1 + 2q impair, d'o �u

ak+2 = (1 + 2q)(2 + 2q) = 2(1 + 2q)(1+ q)

avec 1 + q pair, donc ak+2 � 0 (mod 4), et il su�t de choisir p = k + 2.

Soit n � 2 �x �e. Supposons la propri �et �e vraie pour ce n. Soit ak tel que

ak = 2+ 2n+1q, o �u q impair, q � 1. Alors

ak+1 = 1 + 2nq; impair et

ak+2 = (1 + 2nq)(2 + 2nq) = 2 + 2n(q+ 2q + 2nq2) ,

et comme q+2q+2nq2 est impair, d'apr �es l'hypoth �ese de r �ecurrence il existe

p � k+ 2 � k tel que ap � 0 (mod 4), d'o �u le r �esultat pour n+ 1.

Ce qui ach �eve la recurrence et prouve le Lemme.

On distingue quatre cas.

1. a0 � 0 (mod 4) : il su�t de choisir p = 0.

2. a0 � 3 (mod 4) : alors a0 est impair, d'o �u a1 = a0(1 + a0) et

1 + a0 � 0 (mod 4). Donc a1 � 0 (mod 4), et il su�t de choisir p = 1.

3. a0 � 2 (mod 4) : alors il existe n � 2, il existe q, impair, q � 1 (car

a0 > 2) tel que a0 = 2+ 2nq. Le Lemme permet de conclure.

4. a0 � 1 (mod 4) : alors a0 = 1 + 4k, k � 1, car a0 > 2. D'o �u, a0 impair

et

a1 = (1 + 4k)(2 + 4k)

= 2 + 4(3k+ 4k2)

207

donc a1 � 2 (mod 4) et on est ramen �e au 3 �eme cas.

Donc, dans tous les cas, il existe p � 0 tel que ap � 0 (mod 4), c. �a.d. tel

que ap > ap+1 > ap+2.

2. A positive integer is called \almost-triangular" if the number is

itself triangular or is the sum of di�erent triangular numbers. How many

almost-triangular numbers are there in the set f1, 2, 3, : : : , 1997g?Note: The triangular numbers are a1, a2, a3, : : : , ak, : : : , where

a1 = 1, and ak = k + ak�1, for all k � 2.

Solution by Pierre Bornsztein, Courdimanche, France.

On pose Ti = i �eme nombre triangulaire =i(i+1)

2, i � 1. On a T1 = 1,

T2 = 3, T3 = 6, T4 = 10, T5 = 15, T6 = 21, T7 = 28, T8 = 36, T9 = 45,

T10 = 55, T11 = 66.

On v �eri�e facilement que si n � 33 alors n est presque triangulaire

sauf pour n 2 I = f2, 5, 8, 12, 23, 33g.Supposons n � 34.

On v �eri�e que : si 34 � n � 66 alors n convient et que n =PTi avec

Ti < 66.

On en d �eduit que si n = 66 + a = T11 + a, o �u a 2 f1, : : : , 66g et

a 62 I, alors n convient (on est sur que l'on peut d �ecomposer a sans utiliser

T11). On v �eri�e que si n = 66 + a avec a 2 I, alors n convient puisque :

68 = 55 + 10 + 3 78 = 36 + 28 + 10 + 3 + 1

71 = 55 + 10 + 6 89 = 55 + 28 + 6

74 = 55 + 10 + 6 + 3 99 = 55 + 28 + 10 .

Par cons �equent si n � 132 alors n convient et n =PTi avec Ti � 66. Or

T13 = 91 et donc tout n = 91 + a convient pour a 2 f34, : : : , 132g. Doncsi n � 223, alors n convient et n =

PTi avec Ti � 91.

Or T19 = 190. On en d �eduit que tout n = 190 + a, o �u a 2 f34, : : : ,223g convient. Donc si n � 413 alors n convient et n =

PTi avec Ti � 190.

Or T25 = 325. Donc tout n = 325+ a o �u a 2 f34, : : : , 413g convient.Donc si n � 738 alors n convient et n =

PTi avec Ti � 325.

Or T37 = 703. Donc tout n = 703+ a o �u a 2 f34, : : : , 738g convient,donc si n � 1441 alors n convient et n =

PTi avec Ti � 703. Or

T52 = 1378.

Donc tout n = 1378 + a o �u a 2 f34, : : : , 1441g convient, donc si

n � 2819 alors n convient.

Finalement : les seuls n 2 f1; : : : ; 1997g qui ne sont pas presque triangu-

laires sont 2, 5, 8, 12, 23, 33. Il y a donc 1991 nombres presque triangulaires

dans f1, : : : , 1997g.

208

Remarque : \Tout nombre n � 34 est presque triangulaire". Ce r �esultat

serait du �a R. Graham et P. Erd }os, \L'enseignement math �ematiques" 1980.

We now turn to solutions from the readers to problems for the Third

Grade of the 38thMathematics Competition of the Republic of Slovenia [1998:

132].

1. Let n be a natural number. Prove: if 2n+1 and 3n+ 1 are perfect

squares, then n is divisible by 40.

SolutionsbyMiguel Amengual Covas, Cala Figuera,Mallorca, Spain; byPierre Bornsztein, Courdimanche, France; by Masoud Kamgarpour, CarsonGraham Secondary School, North Vancouver, BC; by Pavlos Maragoudakis,Pireas, Greece; by Panos E. Tsaoussoglou, Athens, Greece; and by EdwardT.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give the solu-tion by Amengual Covas.

Since 40 = 23 � 5, it is su�cient to prove that n is divisible by 8 and 5.

Set

2n+ 1 = x2 (1)

and

3n+ 1 = y2 (2)

where x, y are natural numbers.

We note that the number x2 is odd, and thus also the number x is odd;

consequently x = 2a+ 1, where a is a natural number.

Equation (1) implies the equality 2n + 1 = (2a + 1)2, whence

n = 2a2 + 2a.

The number n, as the sum of two even numbers, is even. It follows

from equation (2) that the number y2 is odd, and thus also the number y is

odd; consequently y = 2b+ 1, where b is a natural number.

1� We subtract (1) from (2) and �nd that

n = y2 � x2 = (2b+ 1)2� (2a+ 1)2 = 4(b+ a+ 1)(b� a) .

Since both of b+a, b�a are either even or odd, one of the numbers b+a+1,

b� a is even, whence the number n is divisible by 8.

2� We can eliminate n between (1) and (2) to get

3x2 � 2y2 = 1 .

Since the square of an odd number ends in 1, 5, or 9, each of the num-

bers x2 and y2 ends in 1, 5, or 9. Therefore the number 3x2 ends in 3, 5, or

7 and 2y2 ends in 2, 0 or 8.

209

Since 3x2 � 2y2 = 1, 3x2 must have ended in 3 and 2y2 must have

ended in 2, whence both of the numbers x2 and y2 end in 1.

Hence n = y2 � x2 ends in 0, and consequently n is divisible by 5.

Comment: A related problem appears in Arthur Engel's Problem-Solving Strategies, Springer-Verlag 1998, page 131. If 2n + 1 and 3n + 1

are squares, then 5n+ 3 is not a prime.

2. Show that cos(sinx) > sin(cosx) holds for every real number x.

SolutionsbyMohammed Aassila, Univ �ersit �e LouisPasteur, Strasbourg,France; by Pierre Bornsztein, Courdimanche, France; by PavlosMaragoudakis, Pireas, Greece; and by Edward T.H. Wang, Wilfrid LaurierUniversity, Waterloo, Ontario. We give the two solutions of Maragoudakis.

First Solution. ��2< �1 � sinx � 1 < �

2===) cos(sinx) > 0 for

x 2 R.If cosx � 0, then cosx 2 (��

2; 0], so sin(cosx) � 0 < cos(sinx).

If cosx > 0, then cosx 2 (0; �2)]. It is known that siny < y for

y 2 (0; �2). So

sin(cosx) < cosx . (1)

Also cos y � 1� y2

2for y 2 (��

2; �2). Since sinx 2 (��

2; �2) we have

cos(sinx) � 1�sin2 x

2=

1 + cos2 x

2. (2)

But, using (1) and (2),1 + cos2 x

2� cosx ===) cos(sinx) > sin(cosx) .

Second Solution.

cos(sinx)� sin(cosx)

= cos(sinx)� cos

��

2� cosx

�

= 2 sin

�sinx� cosx+ �

2

2

�sin

� �2� sinx� cosx

2

�

= 2 sin

p2

2sin

�x�

�

4

�+�

4

! sin

�

4�p2

2sin

�x+

�

4

�!!.

It is easy to prove that

0 <�

4�p2

2�

p2

2sin

�x�

�

4

�+�

4,

�

4�p2

2sin

�x+

�

4

��

�

4+

p2

2<�

2,

210

so that

sin

p2

2sin

�x�

�

4

�+�

4

!, sin

�

4�p2

2sin

�x+

�

4

�!> 0 .

3. The polynomial p(x) = x3+ax2+bx+c has only real roots. Show

that the polynomial q(x) = x3�bx2+acx�c2 has at least one non-negativeroot.

Solutions by Pierre Bornsztein, Courdimanche, France; by PavlosMaragoudakis, Pireas, Greece; and by Edward T.H. Wang, Wilfrid LaurierUniversity, Waterloo, Ontario. We give Bornsztein's solution.

Les hypoth �eses sont inutiles car :

limx!+1

q(x) = +1 et q(0) = �c2 � 0 .

Donc, d'apr �es le th �eor �eme des valeurs intermediaires (q est continue

sur R+), il existe � 2 R+ tel que q(�) = 0.

4. Let the point D on the hypotenuse AC of the right triangle ABC

be such that jABj = jCDj. Prove that the bisector of the angle \A, the

median through B and the altitude through D of the triangle ABD have a

common point.

Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;by Pierre Bornsztein, Courdimanche, France; by Masoud Kamgarpour,Carson Graham Secondary School, North Vancouver, BC; and by PavlosMaragoudakis, Pireas, Greece. We give the solution of Kamgarpour.

A

B

CDM

K

O

H

LetM be the foot of the median from B,K the point of intersection of

the angle bisector withBD andH the foot of the altitude fromD in4ABD.

We use Ceva's Theorem to prove that AM , DH and AK have a

common point.

MA

MD= 1 (because BM is a median)

KD

KB=

AD

AB(Bisector Property)

211

4AHD � 4ABC ===)HB

HA=

DC

DA=

AB

AD.

Thus

MA

MD�KD

KB�HB

HA=

AD

AB�AB

AD= 1 .

That completes the Corner this issue. Sendme your Olympiad contests,

your nice solutions, and generalizations.

Challenge AnswerIn the February 1999 issue [1999: 32], we issued the challenge:

What is the 10th term in the following sequence, and why?

n xn

0 0

1 1

16

�p30 +

p10 �

p6 �

p2 � 2

�p3� 1

�q5 +

p5

�

2 1

8

�q30� 6

p5 �

p5 � 1

�

3 1

8

�p10 +

p2� 2

q5 �

p5

�

4 1

8

�q10 + 2

p5 �

p15 +

p3

�

5 1

4

�p6 �

p2�

6 1

4

�p5 � 1

�

7 1

16

�2�p

3 + 1�q

5�p5�

p30 +

p10 �

p6 +

p2

�

8 1

8

�p15 +

p3�

q10 � 2

p5

�

9 1

8

�2

q5 +

p5�

p10 +

p2

�

10 ?

The answer, sent in by Luyun Zhong-Qiao, Columbia International College,

Hamilton, Ontario, is 1

2. He notes that Tn = sin(3n�).

[Ed.: note there was a typo in T7.]

212

BOOK REVIEWS

ALAN LAW

Elementary Mathematical Models by Dan Kalman,

published the The Mathematical Association of America, 1997

ISBN# 0-88385-707-3, soft cover, 340+ pages.

Reviewed by Richard Charron, Memorial University of Newfoundland,St. John's, Newfoundland.

Subtitled Order Aplenty and a Glimpse of Chaos, this book is intended

for post-secondary students who are pursuing a course of study requiring no

more mathematics than college algebra. The idea is to embed the rudiments

of college algebra into a course whose focus is not algebra itself but rather

the use of algebra in model problems common to the sciences, economics and

business.

The book starts with basic notions of sequences, recursion, and dif-

ference equations to then introduce arithmetic growth models, quadratic

growth, geometric growth and ultimately logistic growth models. The mod-

els are all presented in a particular context, going from problems dealingwith

pollution data, consumption of non-renewable resources, on to population

growth. As the models are presented, a set of natural questions arises. It is

in answering these questions that the usual algebraic techniques are inter-

twined, covering the usual range of questions from determining the equation

of a line, computing slope and intercepts, to quadratic functions and their

roots, polynomials, rational functions and their graphs, exponential and log-

arithmic functions. The book does end with a problem giving a glimpse into

the interesting dynamics of the logistic equation.

As the focus of this book is not algebra but rather algebra to assist in

answering modeling questions, the instructor/student who prefers a course

emphasizing de�nitions, theorems and techniques will not enjoy this book.

Students looking for a litany of worked-out examples will not be entirely sat-

is�ed either. Those who teach/learn algebra but prefer to do so in a broader

scienti�c or social context will �nd the text to be quite interesting. The au-

thor does not shy away from the fact that life, data and models are never

in agreement and makes a conscientious e�ort to convey to the reader the

importance of understanding the limitations of one's models. On the minus

side, the text is a bit verbose. The author admits his guilt in this respect in

the introductory remarks indicating he preferred to err in this fashion.

Overall it stands as a recommended text for the subject.

213

Principles of Mathematical Problem Solving by Martin J. Erickson and Joe

Flowers, published by Prentice-Hall, Inc., 1999.

ISBN # 0-13-096445-X, hardcover, 252+ pages.

Reviewed by Christopher Small, University of Waterloo, Waterloo, On-tario.

Teaching mathematical problem solving is a bit like trying to teach

someone to �nd a light switch in a dark room in the middle of the night.

Unlike the tidy presentation of theorem, corollary and lemma that makes up

the standard course, training people to be good problem solvers is a very

inexact science. Nevertheless, like �nding the light switch in a dark room,

mathematical problem solving has many rules of thumb to get you through

the task.

Principles of Mathematical Problem Solving by Martin J. Erickson and

Joe Flowers contains many of the standard rules of thumb, and is pitched at

a level for the undergraduate university student possibly preparing for the

William Lowell Putnam Competition. While the book requires some math-

ematical expertise beyond the high school level, the advanced high school

student will �nd much that is useful as well. In many respects, it is similar

in academic level and coverage of topics to Loren C. Larson's Problem Solv-ing Through Problems. However the overall level of presentation is easier

and more suitable for the novice who has little experience with mathematics

competitions.

The book explains many standard methods with careful attention to

the fundamental arguments. Many examples are well developed, and the

problems at the end are judiciously chosen, although many are fairly well

known.

It is good to note that the problems have been selected in part because

of their elegance. The importance of this cannot be overestimated in pro-

viding students with an eye for mathematical beauty and an appreciation for

the subject.

214

THE SKOLIAD CORNERNo. 38

R.E. Woodrow

In this issue we give an example of a rather di�erent contest, the New-

foundland and Labrador Teachers' Association Senior Mathematics League,

for 1998{1999. The League started in 1987 as a contest in the St. John's area

and grew to a province-wide event, with schools competing in local leagues

at several sites. The same contest occurs simultaneously at the sites, and the

top schools in each district coming together for the provincial �nals. The em-

phasis is on cooperative problem solving. A school team consists of four stu-

dents who work together on problems. Individual work is rewarded, but to

foster collaboration there are bonus marks for correct work as a team. A typ-

ical competition consists of ten questions and a relay. Unlike most contests,

the problems are presented separately, answers collected, and solutions dis-

cussed before going on to the next problem. Individual student solutions

may earn 1 mark each, while a correct team solution gains 5 marks. Incor-

rect answers score 0. The relay has four points, with the answer to each part

being an input to the next. One point is awarded if only part 1 is correct, two

for parts 1 and 2, three for parts 1, 2, and 3 and �ve marks for all four parts

of the relay. Bonus points are awarded for each minute remaining in the �rst

ten minutes allotted for the relay. (If an incorrect answer is submitted early,

the team is told the answer is wrong but not why, and they may go back to

rework it.) A tie-breaker may be required. This is based on speed, but to de-

ter silly guesses, an incorrect answer means the team cannot answer for one

minute. My thanks go to John Grant McLoughlin, Memorial University of

Newfoundland, for forwarding me the contest and background information.

NLTA SENIOR MATH LEAGUEGAME 1 | 1998{99

1. Find a two-digit number that equals twice the product of its digits.

2. The degree measures of the interior angles of a triangle are A, B, C

where A � B � C. If A, B, and C are multiples of 15, how many possible

values of (A;B;C) exist?

3. Place an operation (+, �, �, �) in each square so that the expres-

sion using 1, 2, 3, : : : , 9 equals 100.

1 2 2 2 3 2 4 2 5 2 6 2 7 2 8 2 9 = 100

You may also freely place brackets before/after any digits in the expression.

Note that the squares must be �lled in with operational symbols only.

215

4. A, B and C are points on a line that is parallel to another line

containing pointsD andE, as shown. Point F does not lie on either of these

lines.r r r

r r

r

A B C

F

D E

How many distinct triangles can be formed such that all three of their

vertices are chosen from A, B, C, D, E, and F ?

5. Michael, Jane and Bert enjoyed a picnic lunch. The three of them

were to contribute an equal amount of money toward the cost of the food.

Michael spent twice as much money as Jane did buying food for lunch. Bert

did not spend any money on food. Instead, Bert brought $6 which exactly

covered his share. How much (in dollars) of Bert's contribution should be

given to Michael?

6. Two semicircles of radius 3 are inscribed in a semicircle of radius 6.

A circle of radius R is tangent to all three semicircles, as shown. Find R.

7. If 5A = 3 and 9B = 125, �nd the value of AB.

8. The legs of a right-angled triangle are 10 and 24 cm respectively.

Let A = the length (cm) of the hypotenuse,

B = the perimeter (cm) of the triangle,

C = the area (cm2) of the triangle.

Determine the lowest common multiple of A, B, and C.

9. A lattice point is a point (x; y) such that both x and y are integers.

For example, (2;�1) is a lattice point, whereas, (3; 12) and (�1

3; 23) are not.

How many lattice points lie inside the circle de�ned by x2 + y2 = 20?

(Do NOT count lattice points that lie on the circumference of the circle.)

10. The quadratic equation x2 + bx + c = 0 has roots, r1 and r2,

that have a sum which equals 3 times their product. Suppose that (r1 + 5)

and (r2 + 5) are the roots of another quadratic equation x2 + ex + f = 0.

Given that the ratio of e : f = 1 : 23, determine the values of b and c in the

original quadratic equation.

216

RELAY

R1. Operations � and � are de�ned as follows:

A �B =AB +BA

A+ Band A � B =

AB � BA

A� B.

Simplify N = (3 � 2) � (3 � 2). Write the value of N in Box #1 of the relay

answer sheet.

R2. A square has a perimeter of P cm and an area of Q cm2? Given

that 3NP = 2Q, determine the value of P . Write the value of P in Box #2

of the relay answer sheet.

R3. List all two-digit numbers that have digits whose product is P .

Call the sum of these two-digit numbers S. Write the value of S in Box #3

of the relay answer sheet.

R4. How many integers between 6 and 24 share no common factors

with S that are greater than 1? Write the number in Box #4 of the relay

answer sheet.

TIE-BREAKER

Find the maximum value of

f(x) = 14�px2 � 6x+ 25 .

Last number we gave the problems of Part I of the Alberta High School

Mathematics Competition written in November 1998. Next we give the of-

�cial solutions. My thanks go to the organizing committee, chaired by Ted

Lewis, University of Alberta, Edmonton, for supplying the contest problems

and the o�cial solutions.

THE ALBERTA HIGH SCHOOLMATHEMATICS COMPETITION

Part I | November 1998

1. A restaurant usually sells its bottles of wine for 100% more than it

pays for them. Recently it managed to buy some bottles of its most popular

wine for half of what it usually pays for them, but still charged its customers

what it would normally charge. For these bottles of wine, the selling price

was what percent more than the purchase price?

Solution. The answer is: (c) 300. Let the old buying price be 2x. Then

the new buying price is x and the selling price is 4x. The latter exceeds the

former by 3x.

217

2. How many integer solutions n are there to the inequality

34n � n2 + 289?

Solution. The answer is: (b) 1. The only solution is n = 17 since the

inequality can be rewritten as 0 � (n� 17)2.

3. A university evaluates �ve magazines. Last year, the rankings were

MacLuck with a rating of 150, followed by MacLock with 120, MacLick with

100, MacLeck with 90, and MacLack with 80. This year, the ratings of these

magazines are down 50%, 40%, 20%, 10% and 5% respectively. How does the

ranking change for MacLeck?

Solution. The answer is: (a) up 3 places. The new ratings are 75

for MacLuck, 72 for MacLock, 80 for MacLick, 81 for MacLeck and 76 for

MacLack. This problem illustrates that quality and ranking are two di�erentthings.

4. Parallel lines are drawn on a rectangular piece of paper. The paper

is then cut along each of the lines, forming n identical rectangular strips. If

the strips have the same length to width ratio as the original, what is this

ratio?

Solution. The answer is: (a)pn : 1. Suppose the length to width ratio

for the original rectangle is a : b with a > b. Then the ratio for the strips is

b : an= a : b. Hence a

b=pn.

5. \The smallest integer which is at least a% of 20 is 10." For how

many integers a is this statement true?

Solution. The answer is: (e) 5. The largest possible value of a is 50 and

the smallest is 46.

6. Let S = 1 + 2 + 3 + � � � + 10n. How many factors of 2 appear in

the prime factorization of S?

Solution. The answer is: (c) n � 1. We have S = 1

210n(10n + 1) =

2n�15n(10n + 1).

7. When 1 + x+ x2 + x3 + x4 + x5 is factored as far as possible into

polynomials with integral coe�cients, what is the number of such factors,

not counting trivial factors consisting of the constant polynomial 1?

Solution. The answer is: (c) 3. We have (1+x+x2)+(x3+x4+x5) =

(1+x+x2)(1+x3) = (1+x+x2)(1+x)(1�x+x2). The two quadratic

factors are irreducible over polynomials with integral coe�cients since the

only possible linear factors are 1 + x and 1� x, but neither divides them.

8. In triangle ABC, AB = AC. The perpendicular bisector of AB

passes through the midpoint of BC. If the length of AC is 10p2 cm, what

is the area of ABC in cm2 ?

Solution. The answer is: (e) none of these. The line joining the mid-

points of AB and BC is parallel to AC. Hence \CAB = 90� and the area

218

of triangle ABC is 1

2AB � AC = 100.

9. If f(x) = xx, what is f(f(x)) equal to?

Solution. The answer is: (d) x(x(x+1)

). We have

f(f(x)) = (xx)(xx) = xxx

x

.

10. A certain TV station has a

logo which is a rotating cube in which

one face has anA on it and the other �ve

faces are blank. Originally the A-face is

at the front of the cube as shown on the

right. Then you perform the following

sequence of three moves over and over:

rotate the cube 90� around the verti-

cal axis v, so that the front face moves

to the left; then rotate the cube 90�

around a horizontal axis h, so that the

new front face moves down; then rotate

the cube 90� around the vertical axis

again, so that the new front face moves

to the left. Suppose you perform this

sequence of three moves a total of 1998

times. What will the front face look like

when you have �nished?

Solution. The answer is: (a) . When the sequence is performed,

the A-face �rst goes to the left, pointing up, then stays at the left but pointing

to the front, and �nally goes to the back, pointing to the left. When the

sequence is performed again, the A-face �rst goes to the right, pointing to

the back, then stays at right, pointing up, and �nally returns to the front,

pointing up.

11. How many triples (x; y; z) of real numbers satisfy the simultane-

ous equations x+ y = 2 and xy � z2 = 1?

Solution. The answer is: (a) 1. We have 1 + z2 = x(2 � x) which is

equivalent to (x� 1)2 + z2 = 0. Hence x = y = 1 and z = 0.

219

12. In the diagram, ABC is an equilateral triangle of side length 3

and PA is parallel to BS. If PQ = QR = RS, what is the length of BR?

A

B C

P

Q

R

S

Solution. The answer is: (d)p7. Triangles PRA and SRC are similar.

Since PR = 2RS and AC = 3, we have CR = 1. Let the foot of perpen-

dicular from R to BC be T . Since \ACB = 60�, we have RT = 1

2

p3 and

CT = 1

2, so that BT = 5

2. By Pythagoras' Theorem, BR2 = RT 2 + BT 2.

13. Let a, b, c and d be the roots of x4�8x3�21x2+148x�160 = 0.

What is the value of 1

abc+ 1

abd+ 1

acd+ 1

bcd?

Solution. The answer is: (b) � 1

20. We have a + b + c + d = 8 while

abcd = �160. The desired expression is equal to a+b+c+dabcd

.

14. Wei writes down, in order of size, all positive integers b with the

property that b and 2b end in the same digit when they are written in base

10. What is the 1998th number in Wei's list?

Solution. The answer is: (b) 19976. Since 2b is even, so is b. When b is

even, 2b ends alternately in 4 and 6. For 2 � b � 20, the only matches are

b = 14 and b = 16. Since everything repeats in a cycle of 20, Wei's list is 14,

16, 34, 36, : : : . The (2n � 1)-st number is 10(2n � 1) + 4 and the 2n-th

number is 10(2n� 1) + 6.

15. Suppose x = 31998. How many integers are betweenpx2 + 2x+ 4 and

p4x2 + 2x+ 1?

Solution. The answer is: (b) 31998 � 1. Note that

x+ 1 <px2 + 2x+ 4 < x+ 2 ,

while

2x <p4x2 + 2x+ 1 < 2x+ 1 .

Hence the number of integers between the two radicals is x� 1.

16. The lengths of all three sides of a right triangle are positive inte-

gers. The area of the triangle is 120. What is the length of the hypotenuse?

Solution. The answer is: (c) 26. Suppose that the three side lengths arex � y � z. Then x2 + y2 = z2 since we have a right triangle. Hence1

2xy = 120 and xy = 243 � 5.

220

Of the numbers 32+802, 52+482, 152+162, 62+402, 102+242 and

122 + 202, we only have an integral value for z when x = 10 and y = 24,

namely, z2 = 22(52 + 122) = (2 � 13)2.

In the February 1999 number of the Corner we gave the solutions to

the problems of the OldMutualMathematical Olympiad 1991, Final Paper 1.

One of our readers proposes an alternate \variable free" solution to one of

them.

2. [1998: 477; 1999: 28] What is the value ofq17� 12

p2 +

q17 + 12

p2

in its simplest form?

Alternate solutionby Luyun ZhongQiho,Mathematics Teacher, Hamil-ton, Ontario.q

17� 12p2 +

q17 + 12

p2

=

r9� 2(3)

�2p2�+ 8 +

r9 + 2(3)

�2p2�+ 8

=

r�3� 2

p2�2

+

r�3 + 2

p2�2

=�3� 2

p2�+�3 + 2

p2�

since 3 > 2p2

= 3 + 3

= 6 .

That completes the Skoliad Corner for this month. Send me suitable

contest materials, novel solutions, and suggestions for future directions.

221

MATHEMATICAL MAYHEM

Mathematical Mayhem began in 1988 as a Mathematical Journal for and by

High School and University Students. It continues, with the same emphasis,

as an integral part of Crux Mathematicorum with Mathematical Mayhem.

All material intended for inclusion in this section should be sent to the

Mayhem Editor, Naoki Sato, Department of Mathematics, Yale University,

PO Box 208283 Yale Station, New Haven, CT 06520{8283 USA. The electronic

address is still

mayhem@math.toronto.edu

The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).

The rest of the sta� consists of Adrian Chan (Upper Canada College), Donny

Cheung (University of Waterloo), Jimmy Chui (Earl Haig Secondary School),

David Savitt (Harvard University) and Wai Ling Yee (University of Waterloo).

Shreds and Slices

A Combinatorial Proof

In Issue 2, this volume, we issued the following challenge to our audi-

ence.

Problem. Find a combinatorial proof of the following identity:

(n� r)

�n+ r � 1

r

��n

r

�= n

�n+ r � 1

2r

��2r

r

�.

Dave Arthur, of Toronto, Ontario, was the �rst, and so far only person

to give a combinatorial proof. His prize is a copy of \Riddles of the Sphinx",

by Martin Gardner; we print his solution here. The proposer still seeks a

combinatorial proof that proves the identity in one step, and so an additional

prize will be o�ered for such a solution.

Solution by Dave Arthur.

Lemma 1. Let us consider the number of ways of choosing two distinct

sets, A and B, each with r elements from a set of n+ r � 1 elements.

There are�n+r�1

2r

�ways to choose the elements that will belong to their

union, and there are�2r

r

�ways to choose the elements of these that will

belong to A. Therefore, the number of ways is�n+ r � 1

2r

��2r

r

�.

222

However, we also note there are�n+r�1

r

�ways to choose A, and

�n�1r

�ways to choose B from the remaining elements. It follows that�

n+ r � 1

2r

��2r

r

�=

�n+ r � 1

r

��n� 1

r

�.

Lemma 2. Let us consider the number of ways of choosing two distinct

sets, C and D, such that C has 1 element and D has r elements, from a set

of n elements.

There are n ways to choose C �rst, and�n�1r

�ways to choose D from

the remaining elements, so the number of ways is n�n�1r

�. Also, there are�

n

r

�ways to choose D �rst, and n� r ways to choose C from the remaining

elements, so it follows that

(n� r)

�n

r

�= n

�n� 1

r

�.

Therefore, by lemmas 1 and 2,

n

�n+ r � 1

2r

��2r

r

�= n

�n+ r � 1

r

��n� 1

r

�= (n�r)

�n

r

��n+ r � 1

r

�.

Erratum

On page 169, Issue 3 of this Volume, in problem 6 of the Qualifying

Round of the 1990 Swedish Mathematical Olympiad, the dimensions of rec-

tangle ABCD are described as 3000 metres by 500 metres. This is a typo:

the dimensions should be 300 metres by 500 metres. Thanks to Jim Totten

for the correction.

Mayhem Problems

The Mayhem Problems editors are:

Adrian Chan Mayhem High School Problems Editor,Donny Cheung Mayhem Advanced Problems Editor,David Savitt Mayhem Challenge Board Problems Editor.

Note that all correspondence should be sent to the appropriate editor |

see the relevant section. In this issue, you will �nd only solutions | the

next issue will feature only problems.

We warmly welcome proposals for problems and solutions. With the

new schedule of eight issues per year, we request that solutions from the last

issue be submitted in time for issue 4 of 2000.

223

High School Solutions

Editor: Adrian Chan, 229 Old Yonge Street, Toronto, Ontario, Canada.

M2P 1R5 <all238@sprint.com>

H228. Verify that the following three inequalities hold for positive

reals x, y, and z:

(i) x(x � y)(x� z) + y(y � x)(y � z) + z(z � x)(z � y) � 0. (This is

known as Schur's Inequality.)

(ii) x4 + y4 + z4 + xyz(x+ y + z) � 2(x2y2 + y2z2 + z2x2).

(iii) 9xyz + 1 � 4(xy+ yz + zx), where x+ y + z = 1.

(Can you derive an ingenious method that allows you to solve the problem

without having to prove all three inequalities directly?)

Additional solution by Murray S. Klamkin, University of Alberta, Ed-monton, Alberta.

(i) As indicated, this is the special case n = 1 of Schur's Inequality:

xn(x� y)(x� z) + yn(y� z)(y� x) + zn(z � x)(z� y) � 0

for n real. For a simple proof, let x � y � z without loss of generality. Then

for n � 0,

xn(x� y)(x� z) � yn(y� z)(x� y) and zn(x� z)(y� z) � 0 .

For n � 0,

zn(x� z)(y� z) � yn(y� z)(x� y) and xn(x� y)(x� z) � 0 .

There is equality if and only if x = y = z. It also follows that if n is an

even integer, then x, y and z need not be positive.

(ii) Since as known

2(y2z2 + z2x2 + x2y2)� (x4 + y4 + z4)

= (x+ y + z)(y+ z � x)(z + x� y)(x+ y � z)

(related to Heron's formula for the area of a triangle), the inequality reduces

to

xyz � (y+ z � x)(z + x� y)(x+ y� z) . (1)

In terms of the elementary symmetric functions T1 = x + y + z,

T2 = yz + zx+ xy, and T3 = xyz, (1) becomes T 3

1+ 9T3 � 4T1T2, which

is the same as (i).

224

(iii) In homogeneous form, the inequality is equivalent to

9xyz + (x+ y+ z)3 � 4(x+ y+ z)(yz+ zx+ xy) ,

which is the same as (1).

For a generalization of (1) to

(ux+ vy+ wz)(vx+ wy+ uz)(wx+ uy + vz)

� (y+ z � x)(z + x� y)(x+ y � z) ,

where u+ v +w = 1 and 0 � u; v; w � 1, see [1].

Reference

1. Klamkin, M.S., Inequalities for a triangle associated with n given tri-angles, Publ. Electrotechn. Fak. Ser. Mat. Fiz. Univ. Beograd, No.

330 (1970), pp. 3{7.

H237. The letters of the word MATHEMATICAL are arranged at ran-

dom. What is the probability that the resulting arrangement contains no

adjacent A's?

Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,Ontario.

First, the total number of arrangements is

12!

3!2!2!.

Now, to count the number of arrangements with no adjacent A's, we

�rst arrange the other 9 letters, for a total of

9!

2!2!

ways. For each such arrangement, we choose any 3 of the 10 \spaces" that

are between consecutive letters, including those on the ends. This can be

done in a total of�10

3

�ways.

Now, we insert the 3 A's into the 3 spaces that we have chosen. This

corresponds to a unique arrangement with no adjacent A's. The total number

is then9!

2!2!��10

3

�.

The required probability is then equal to

9!

2!2!��10

3

�12!

3!2!2!

=6

11.

225

H238. Johnny is dazed and confused. Starting atA(0; 0) in the Carte-sian grid, he moves 1 unit to the right, then r units up, r2 units left, r3 units

down, r4 units right, r5 units up, and continues the same pattern inde�-

nitely. If r is a positive number less than 1, he will be approaching a point

B(x; y). Show that the length of the line segment AB is greater than 7

10.

Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,Ontario.

Let Pn = (xn; yn) denote the point where Johnny is at after n moves,

n = 0, 1, 2, : : : . So, P0 = A = (0; 0), P1 = (1;0), P2 = (1; r),

P3 = (1�r2; r), etc. A simple pattern reveals itself, such that for allm � 2,

x2m�1 = x2m = 1� r2 + r4 � � � �+ (�1)m�1r2m�2 ,y2m = y2m+1 = r � r3 + r5 � � � �+ (�1)m�1r2m�1 .

Hence,

x =1

1� (�r2)=

1

1 + r2,

y =r

1� (�r2)=

r

1 + r2.

So,

AB2 = x2 + y2 =1

1 + r2.

And since r < 1,

AB =1

p1 + r2

>1p2

=5p2

10>

7

10.

H239. Find all pairs of integers (x; y) which satisfy the equation

y2(x2 + 1) + x2(y2 + 16) = 448.

Solution. We have

y2(x2 + 1) + x2(y2 + 16) = 448

=) 2x2y2 + 16x2 + y2� 448 = 0

=) 2x2(y2 + 8) + y2 + 8 = (2x2 + 1)(y2+ 8)

= 456 = 23 � 3� 19 .

Since x and y are integers, both 2x2+1 and y2+8 are positive integers.

Since 2x2 + 1 is odd, it must equal one of the odd factors of 456, namely 1,

3, 19, and 57. Checking each of these cases, we �nd that x = 0, �1, and �3are the only solutions.

If x = 0, then y2 + 8 = 456, for which there is no solution.

If x = �1, then y2 + 8 = 152, from which we obtain y = �12.

226

If x = �3, then y2 + 8 = 24, from which we obtain y = �4.Therefore, there are eight solutions, namely (�1;�12) and (�3;�4).Also solved by Edward T.H. Wang, Wilfrid Laurier University, Water-

loo, Ontario.

H240. Proposed by Alexandre Trichtchenko, Brook�eld High School,Ottawa, ON.

A Pythagorean triple (a; b; c) is a triple of integers satisfying the equa-

tion a2 + b2 = c2. We say that such a triple is primitive if gcd(a; b; c) = 1.

Let p be an odd integer with exactly n prime divisors. Show that there exist

exactly 2n�1 primitive Pythagorean triples where p is the �rst element of

the triple. For example, if p = 15, then (15;8; 17) and (15;112; 113) are the

primitive Pythagorean triples with �rst element 15.

Solution. We may write p uniquely in the form

p = pe11pe22� � � penn ;

where p1, p2, : : : , pn are distinct odd primes, and e1, e2, : : : , en are positive

integers. We seek all primitive Pythagorean triples (p; b; c). For such a triple,

p2+b2 = c2, or c2�b2 = (c+b)(c�b) = p2. Suppose pi divides both c+b

and c� b for some i. Then pi divides (c+ b)� (c� b) = 2b, and since pi is

odd, it follows that pi divides b. Similarly, pi divides (c+ b)+ (c� b) = 2c,

so pi divides c. Then (p; b; c) fails to be a primitive Pythagorean triple, since

pi divides all three numbers, so for each i, pi divides at most one of c + b

and c� b.

This implies that in the factorization p2 = (c+ b)(c� b), for each i, all

the factors of pi must reside in c+ b or c� b. In other words, for each i, we

can make one of two choices of where to place all the factors of pi, for a total

of 2n factorizations. However, half of them must be discarded, since c + b

must be the greater number, and c� b the lesser. (We cannot have c+ b and

c � b equal, since they have di�erent prime factors.) Each of the other half,

however, does lead to a unique solution: If p2 = xy, where x > y and x and

y are relatively prime, then b = (x� y)=2 and c = (x+ y)=2. Hence, there

are 2n�1 such Pythagorean triples.

Also solved by Edward T.H. Wang, Wilfrid Laurier University, Water-loo, Ontario.

Advanced Solutions

Editor: Donny Cheung, c/o Conrad Grebel College, University of Wa-

terloo, Waterloo, Ontario, Canada. N2L 3G6 <dccheung@uwaterloo.ca>

A212. Let A and B be real n�nmatrices such that A2+B2 = AB.

Prove that if AB � BA is an invertible matrix, then n is divisible by 3.

227

(International Competition in Mathematics for University Students)

Solution. Let S = A + !B, where ! is a primitive cube root of unity.

Then we have that

SS = (A+ !B)(A+ !B)

= (A+ !B)(A+ !B) = A2 + !BA+ !AB + B2

= AB + !BA+ !AB = !(BA� AB) ,

since ! + 1 = �!. Also, det(SS) = detS � detS is a real number and

det!(BA�AB) = !n det(BA�AB) 6= 0, so !n must be a real number.

Hence, n is divisible by 3.

A213. Show that the number of positive integer solutions to the

equation a + b + c + d = 98, where 0 < a < b < c < d, is equal to the

number of positive integer solutions to the equation p+2q+3r+4s = 98,

where 0 < p, q, r, s.

I. Solutionby Edward T.H. Wang, Wilfrid Laurier University, Waterloo,Ontario.

Suppose that a, b, c, and d are positive integers such that a < b < c <

d and satisfy a+b+c+d = 98. Then letting p = d�c, q = c�b, r = b�a,and s = a, we see that p, q, r, and s are positive integers satisfying

p+2q+3r+4s = (d�c)+2(c�b)+3(b�a)+4a= a+b+c+d = 98 .

Conversely, suppose that p, q, r, and s are positive integers satisfying

p+ 2q + 3r + 4s = 98. Then letting a = s, b = r + s, c = q + r + s, and

d = p+ q + r + s, we see that a, b, c and d are positive integers satisfying

a < b < c < d, and

a+ b+ c+ d = p+ 2q+ 3r + 4s = 98 .

This establishes a bijection between the solutions (a; b; c; d) and (p; q; r; s).

II. Solution. First count the number of ways of distributing 98 identical

marbles into 4 distinct boxes, with the �rst box containing the least number

of marbles, the second box containing the second least number, and so on.

This value is equal to the number of solutions to the �rst equation with the

appropriate conditions. Another way of counting this is to distribute the

same number h to each of the 4 boxes. Then distribute g to each of the

second, third and fourth boxes. This ensures that the �rst box has the least.

Distribute f to each of boxes three and four. This ensures that the second

box is the second least and so on. The number of ways to do this is equal to

the number of solutions to the second equation.

A214. Show that any rational number can be written as the sum of

a �nite number of distinct unit fractions. A unit fraction is of the form 1=n,

where n is an integer.

228

Solution.

We solve the problem by step-climbing through several cases.

Case I: Positive rationals in [0; 1].

Let r be a positive rational number in [0; 1]. Then we show that r can

be written as a sum of distinct unit fractions by providing an algorithm that

explicitly �nds the unit fractions, namely the greedy algorithm.

Let r = a=b, in reduced terms. Let n = db=ae. In other words, n is

the integer b=a rounded up, so b=a � n < b=a+ 1. Then 1=n is the largest

unit fraction which is less than or equal to a=b, and when we subtract 1=n

from a=b, we obtain r0 =a

b�

1

n=

an� b

bn.

Since b=a � n < b=a + 1, b � an < b+ a, or 0 � an � b < a. So,

the numerator in r0 is less than the numerator in r. If r0 is reducible, thenwe reduce, and the numerator decreases still. We now subtract the largest

unit fraction from r0, and so on. Since the numerator decreases by at least 1

each step, the algorithm must stop at some point, in fact after at most a� 1

steps. Then r is the sum of the unit fractions produced by the algorithm.

For example, for r = 6=7, we have that

6

7�

1

2=

5

14,

5

14�

1

3=

1

42,

so that6

7=

1

2+

1

3+

1

42.

Case II: Positive rationals greater than 1.

Let r be a positive rational greater than 1. Since the harmonic seriesP1i=1

1=i diverges, there exists a positive integer k such that

kXi=1

1

i� r <

k+1Xi=1

1

i.

If r =Pk

i=11=i, then we have an expression of r as a sum of distinct unit

fractions, so assume that r >Pk

i=11=i. Consider the rational

s = r �kXi=1

1

i, so that 0 < s <

1

k + 1.

Then s is a positive rational in [0; 1], so by Case I, s is the sum of distinct

unit fractions

s =

mXj=1

1

nj.

229

Since s < 1=(k + 1), each nj is at least k + 1 (otherwise nj � k

=) 1=nj � 1=k =) s � 1=k, contradiction), so that

r =

kXi=1

1

i+

mXj=1

1

nj

is an expression of r as a sum of distinct unit fractions.

Case III: Negative rationals and 0.

If r < 0, then �r, by Cases I and II, can be written as the sum of

distinct unit fractions. Negate each term to get such a sum for r. Finally, 0

can be written as 0 =1

1�

1

2�

1

3�

1

6.

A215. For a �xed integer n � 2, determine the maximum value of

k1+� � �+kn, where k1, : : : , kn are positive integers with k31+� � �+k3n � 7n.

(Polish Mathematical Olympiad)

Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,Ontario.

Let T =Pn

i=1k3i and S =

Pn

i=1ki. We claim that

S � n+

�6n

7

�,

where bxc denotes, as usual, the greatest integer less than or equal to x.

This maximum is attained if and only if b6n=7c of the ki equal 2 and the rest

equal 1.

First note that if ki � 2 for all i, then T � 8n, a contradiction. Hence

ki = 1 for at least one i, say k1 = 1. If kj � 3 for some j > 1, then we

consider T 0 obtained from T as follows: Replace k1 and kj by k01= 2 and

k0j = kj � 1 respectively, and leave all the other ki unchanged. Then clearly

the value of S is unchanged. On the other hand, T 0 � T , which means that

(kj�1)3+8 � k3j +1. Rewriting this, we obtain (kj+1)(kj�2) � 0. This

is true as long as kj � 3. Hence T 0 � 7n, and to obtain the maximum value

of S, we may assume, without loss of generality, that ki = 1 or 2 for all i.

Suppose among all the ki, that there are m 2's and n � m 1's. Then

T = 8m + (n � m) � 7n, which implies that m � b6n=7c. Clearly, the

maximum value of S is attained whenm = b6n=7c, in which case we obtain

S = 2b6n=7c+ n� b6n=7c = n+ b6n=7c.

A216. Given a continuous function f : R ! R satisfying the condi-

tions:

f(1000) = 999 ,

f(x) � f(f(x)) = 1 for all x 2 R .

Determine f(500).

(Polish Mathematical Olympiad)

230

Solution. By the second condition, f(1000)f(f(1000)) = 1, so we

have that 999f(999) = 1 or f(999) = 1=999.

Since f is a continuous function, by the Intermediate Value Theorem,

there exists an a 2 [999; 1000] such that f(a) = 500.

Then f(a)f(f(a)) = 1, giving 500f(500) = 1, so f(500) = 1=500.

In fact, f(x) = 1=x for all x 2 [1=999;999]. To complete the

function, for any x outside this range, set f(x) to any value, within the

interval [1=999;999]. Then for any x, 1=999 � f(x) � 999, and so

f(f(x)) = 1=f(x).

Note: Because f(1000) = 999 6= 1=1000, f(x) can never equal 1000.

Challenge Board Solutions

Editor: David Savitt, Department of Mathematics, Harvard University,

1 Oxford Street, Cambridge, MA, USA 02138 <dsavitt@math.harvard.edu>

C77. Let Fi denote the ith Fibonacci number, with F0 = 1 and

F1 = 1. (Then F2 = 2, F3 = 3, F4 = 5, etc.)

(a) Prove that each positive integer is uniquely expressible in the form

Fa1+� � �+Fak , where the subscripts form a strictly increasing sequence

of positive integers no pair of which are consecutive.

(b) Let � = 1

2(1 +

p5), and for any positive integer n, let f(n) equal the

integer nearest to n� . Prove that ifn = Fa1+� � �+Fak is the expressionfor n from part (a) and if a2 6= 3, then f(n) = Fa1+1 + � � �+ Fak+1.

(c) Keeping the notation from part (b), if a2 = 3 (so that a1 = 1), it is

not always true that the formula f(n) = Fa1+1 + � � � + Fak+1 holds.

For example, if n = 4 = F3 + F1 = 1 + 3, then the closest integer to

n� = 6:47 : : : is 6, not F2+ F4 = 2+5 = 7. Fortunately, in the cases

where the formula fails, we can correct the problem by setting a1 = 0

instead of a1 = 1: For example, 4 = F0 + F3 = 1 + 3 as well, and

indeed 6 = F1+ F4 = 1+5. Determine for which sequences of ai this

correction is necessary.

Solution.

(a) Supposewe know that every positive integer less thanFn is uniquely

expressible in the desired form. (The base case n = 1 is vacuous.) If N

satis�es Fn � N < Fn+1, then by our inductive assumption, N � Fn is

expressible in the desired form, say N � Fn = Fa1 + � � � + Fak. Since

231

N � Fn < Fn+1 � Fn = Fn�1, we must have that ak � n� 2, and so the

expression

N = Fa1 + � � �+ Fak + Fn

is of the desired form. It remains to show that this is the unique expressionfor N as such a sum; observe, by induction, that it su�ces to show that any

such sum for N must include Fn.

Given an expression

N = Fa1 + : : :+ Fak

of the necessary form, if we assume that ak < n, then we know that ak is at

most n� 1, so ak�1 is at most n� 3, and so on. Thus, N is at most

F1 + F3 + F5 + � � �+ Fn�1

if n is even and

F2 + F4 + � � �+ Fn�1

if n is odd. However, it is a straightforward induction argument to prove

that in fact for any positive integer m,

F1 + F3 + F5 + � � �+ F2m�1 = F2m � 1

and

F2 + F4 + � � �+ F2m = F2m+1 � 1 ,

so we obtain a contradiction.

(b) Since f(n) is the unique integer N such that jN � n� j < 1=2, we

need only show that

jFa1+1 + � � �+ Fak+1 � � (Fa1 + � � �+ Fak)j <1

2.

Recall that Fi = � i+1��i+1p5

, where � = 1

2(1�

p5). Hence, we want to

prove that �����kXi=1

�(�ai+2 � �ai+2)� � (�ai+1 � �ai+1)

������ <p5

2.

Using the fact that �� = �1, this becomes the same as showing that�����kXi=0

(�ai+2 + �ai)

����� <p5

2,

and since �2 + 1 = ��p5, we are further reduced to showing that�����

kXi=0

�ai+1

����� < 1

2.

232

Let P be the sum of the positive terms in the sumP�ai+1, and let N

be the sum of the negative terms.

Then P is at most

1Xi=1

�2i =�2

1� �2< 0:62,

and N is at least

1Xi=1

�2i+1 =�3

1� �2> �0:39 . In particular, the only

way that jP +Nj � 1=2 is if P �1

2+ jN j .

If a2 6= 3, then either �2 or �4 is not in the sum forP. Since �2 > 0:38

and �4 > 0:14, it follows that P < 1=2, so P < 1=2 + jN j and the desired

formula holds.

(c) Keeping our notation from (b), if P + N > 1=2, then a1 = 1 and

a2 = 3. Setting a01= 0 and a0i = ai for i > 1, we �nd that

�����kXi=0

�a0

i+1

����� =��(P � �2) + (N + �)

�� = jP +N � 1j <1

2,

so the suggested correction does indeed work when necessary.

We would like to decide precisely when this correction is necessary; that

is, when

S =

kXi=0

�a0

i+1 > �1

2.

Let us retain the assumption that a1 = 1 (so that the de�nition of the a0imakes sense) but drop the assumption that a2 = 3. This will turn out to be

more convenient, in the end.

Observing, by direct calculation, that

1Xi=0

�3i+1 = �1

2, let j be the

smallest (non-negative) integer i such that a0i+26= 3(i+ 1).

(If a0i+2= 3(i+ 1) for all i � k, then put j = k� 1.) Then

S +1

2=

1

2+ � + � � �+ �3j+1 +

kXi=j+2

�a0

i+1 =1

2�3j+3 +

kXi=j+2

�a0

i+1 .

We will attempt to show (in what will amount to a clumsy veri�cation) that

S + 1=2 has the same sign as 1

2�3j+3, and so the correction is necessary

precisely when j is odd.

Remembering the de�nition of j and the fact that the ai are non-

consecutive, using a0j+1= 3j, we break into three cases: a0j+2

= 3j + 2,

a0j+2= 3j + 4, or a0j+2

� 3j + 5.

233

To begin with, in the �rst case,������kX

i=j+3

�a0

i+1

������ <���3j+5

�� 1Xi=0

�2i =���3j+4

��<

����32 �3j+3

���� =

����12 �3j+3 + �a0

j+2+1

���� ,as desired. Similarly, in the second case,������

kXi=j+3

�a0

i+1

������ <���3j+7

�� 1Xi=0

�2i =���3j+6

��<

����12 �3j+3 + �3j+5

���� =

����12 �3j+3 + �a0

j+2+1

���� .Finally, in the third case,������

kXi=j+2

�a0

i+1

������ <���3j+6

�� 1Xi=0

�2i =���3j+5

�� < ����12�3j+3

���� ,and we are done.

Thus, we have shown: If a1 = 1, then if j is the smallest non-negativeinteger i such that ai+2 6= 3(i + 1), then the correction is necessary if andonly if j is odd.

Remark: In the solution of part (b), we never fully used the fact that

we were using the representation from part (a). In particular, the proof of

(b) actually showed that if n = Fa1 + � � � + Fak and the ai are distinct

positive integers, then as long as 1 and 3 are not both among the ai, we can

conclude that f(n) = Fa1+1+ � � �+Fak+1. Unfortunately, as should not be

a surprise, not every integer can be expressed in this way, n = 4 being the

smallest example.

Problem of the Month

Jimmy Chui, student, Earl Haig S.S.

Problem. A rectangular wine rack, PQRS, holds �ve rows of identical

bottles (Figure 1). The bottom row contains enough room for three bottles

(A, B, and C) but not enough room for a fourth bottle. The second row,

consisting of just two bottles (D and E), holds B in place somewhere be-

tween A and C, and pushesA and C to the sides of the rack. The third row,

234

consisting of three bottles (F , G, and H), lies on top of those two bottles,

and F and H rest against the sides of the rack. The fourth layer holds two

bottles (I and J) and the �fth layer contains three bottles (K, L and M ).

Prove that the �fth row is perfectly horizontal regardless of how A, B and C

are positioned.

P

Q R

S

p p p

p

p

p

p

p

p

p

p p p

A B C

DE

FG

H

IJ

K L M

P

Q R

S

p p p

p

p

p

p

p

p

p

p p p

A B C

DE

FG

H

IJ

K L M

Figure 1. Figure 2.

Solution. All the bottles are identical, and so F and K are the same

distance away from the wall (Figure 2). Since FK is vertical, we only need

to show that \FKL = 90�.

The distance between the centres of touching bottles is constant; it is

the diameter of a bottle. Therefore, IF , IK, and IL are all equal, and hence

I is the circumcentre of triangleFKL. In order for \FKL to be a right angle,

then from properties of right angle triangles, the circumcentre I is also the

mid-point of FL. Hence, we will show that I is the mid-point of FL.

Note that the four quadrilaterals GDFI, GILJ , GJHE, GEBD are

all rhombi (they have side length of a bottle diameter). So�!FI =

�!BE and�!

IL =��!EH. Furthermore, since EB, EC, and EH are all equal, E is the

circumcentre of triangle BCH. However, we know that BCH is a right

triangle. Hence, E is the mid-point of BH. Thus, I is the mid-point of FL,

and it follows that \FKL = 90�.

Similarly, \HML is a right angle. Therefore, the top row is perfectly

horizontal, QED.

235

Four Ways to Count

Jimmy Chuistudent, Earl Haig Secondary School

Problem. Evaluate�n

1

�+ 2

�n

2

�+ 3

�n

3

�+ � � �+ n

�n

n

�,

where n is a positive integer.

Solution 1. Let the given sum be equal to S. Now,

S = 0

�n

0

�+ 1

�n

1

�+ 2

�n

2

�+ 3

�n

3

�+ � � �+ n

�n

n

�

= 0

�n

n

�+ 1

�n

n� 1

�+ 2

�n

n� 2

�+ 3

�n

n� 3

�+ � � �+ n

�n

0

�

= n

�n

0

�+ (n� 1)

�n

1

�+ (n� 2)

�n

2

�+ (n� 3)

�n

3

�+ � � �+ 0

�n

n

�.

Adding the �rst and third equations and dividing by 2, we obtain

S =n

2

��n

0

�+

�n

1

�+

�n

2

�+

�n

3

�+ � � �+

�n

n

��

=n

2� 2n

= n2n�1 .

Solution 2. We claim that�n

1

�+ 2

�n

2

�+ 3

�n

3

�+ � � �+ n

�n

n

�= n2n�1 .

We will prove this by mathematical induction. The base case n = 1 is trivial.

Now we assume, for some n = k, that

�k

1

�+ 2

�k

2

�+ 3

�k

3

�+ � � �+ k

�k

k

�= k2k�1 .

Copyright c 1999 CanadianMathematical Society

236

Then,�k+ 1

1

�+ 2

�k + 1

2

�+ 3

�k+ 1

3

�+ � � �+ k

�k+ 1

k

�+ (k+ 1)

�k+ 1

k+ 1

�

= 1

��k

0

�+

�k

1

��+ 2

��k

1

�+

�k

2

��+ 3

��k

2

�+

�k

3

��

+ � � �+ k

��k

k� 1

�+

�k

k

��+ (k+ 1)

�k

k

�

= 1

�k

0

�+ 3

�k

1

�+ 5

�k

2

�+ 7

�k

3

�+ � � �+ (2k+ 1)

�k

k

�

=

�k

0

�+

�k

1

�+

�k

2

�+

�k

3

�+ � � �+

�k

k

�

+ 2

�0

�k

0

�+ 1

�k

1

�+ 2

�k

2

�+ 3

�k

3

�+ � � �+ k

�k

k

��= 2k + 2(k2k�1)

= (k+ 1)2k .

Hence, the claim is true for n = k + 1, and by the principle of mathe-

matical induction, for all positive integers n.

Solution 3. Let

f(x) =

nXi=0

�n

i

�xi = 1 +

nXi=1

�n

i

�xi .

Then

f 0(x) =

nXi=1

i

�n

i

�xi�1 ,

so that

f 0(1) =

nXi=1

i

�n

i

�.

But

f(x) = (1 + x)n

by the Binomial Theorem. Then

f 0(x) = n(1 + x)n�1 ,

so that

f 0(1) = n2n�1 .

Hence,nXi=1

i

�n

i

�= n2n�1 .

237

Solution 4. Consider a set of n people. We wish to count the number

of di�erent teams that can be formed, with the condition that there is one

and only one leader.

One way is to �nd the total number of team members �rst, and then

select a leader from those chosen. Suppose you choose i members. (Note

that 1 � i � n.) There are�n

i

�such subsets. In each of these subsets,

there are i possible leaders. Hence, the total number of teams that can be

formed with i members is i�n

i

�. Therefore, the total number of teams with

any number of members is merely the sum

nXi=1

i

�n

i

�.

Another way is to choose the leader �rst, and the rest of the members

afterwards. The leader can be chosen in n ways. The members can be chosen

out of the other n� 1 people in any way, and there are 2n�1 ways of doingso. Hence the total number of teams is n2n�1.

Thus,nXi=1

i

�n

i

�= n2n�1 .

Comment. This question appears in the strangest of places, and it is

pleasant to see four very di�erent, yet equally elegant, solutions.

238

PROBLEMS

Problem proposals and solutions should be sent to Bruce Shawyer, Department

of Mathematics and Statistics, Memorial University of Newfoundland, St. John's,

Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution,

together with references and other insights which are likely to be of help to the edi-

tor. When a submission is submitted without a solution, the proposer must include

su�cient information on why a solution is likely. An asterisk (?) after a number

indicates that a problem was submitted without a solution.

In particular, original problems are solicited. However, other interesting prob-

lems may also be acceptable provided that they are not too well known, and refer-

ences are given as to their provenance. Ordinarily, if the originator of a problem can

be located, it should not be submitted without the originator's permission.

To facilitate their consideration, please send your proposals and solutions

on signed and separate standard 812"�11" or A4 sheets of paper. These may be

typewritten or neatly hand-written, and should be mailed to the Editor-in-Chief,

to arrive no later than 1 December 1999. They may also be sent by email to

crux-editors@cms.math.ca. (It would be appreciated if email proposals and solu-

tions were written in LATEX). Graphics �les should be in epic format, or encapsulated

postscript. Solutions received after the above date will also be considered if there

is su�cient time before the date of publication. Please note that we do not accept

submissions sent by FAX.

G.P. Henderson, Garden Hill, Campbellcroft, Ontario asks us to point

out that problem 2405 is not the problem that he submitted. It is a modi-

�cation made by the editors. It should be regarded as an unsolved problem

proposed by the editors.

2439. Proposed by Toshio Seimiya, Kawasaki, Japan.

Suppose that ABCD is a square with side a. Let P and Q be points

on sides BC and CD respectively, such that \PAQ = 45�. Let E and

F be the intersections of PQ with AB and AD respectively. Prove that

AE + AF � 2p2a.

2440. Proposed by Toshio Seimiya, Kawasaki, Japan.

Given: triangleABC with \BAC = 90�. The incircle of triangleABCtouches BC at D. Let E and F be the feet of the perpendiculars from D to

AB and AC respectively. Let H be the foot of the perpendicular from A to

BC.

Prove that the area of the rectangle AEDF is equal toAH2

2.

239

2441. Proposed by Paul Yiu, Florida Atlantic University, Boca Ra-ton, Florida, USA.

Suppose that D, E, F are the mid-points of the sides BC, CA, AB

of4ABC. The incircle of4AEF touches EF at X, the incircle of4BFDtouches FD at Y , and the incircle of 4CDE touches DE at Z.

Show that DX, EY , FZ are collinear. What is the intersection point?

2442. Proposed by Michael Lambrou, University of Crete, Crete,Greece.

Let fang11 , fxng11 , fyng11 , : : : , fzng11 , be a �nite number of given

sequences of non-negative numbers, where all an > 0. Suppose thatPan

is divergent and all the other in�nite series,Pxn,

Pyn, : : : ,

Pzn, are

convergent. Let An =Pn

k=1ak.

(a) Show that, for every � > 0, there is an n 2 N such that, simultaneously,

0 �Anxn

an< � , 0 �

Anyn

an< � , : : : , 0 �

Anzn

an< � .

(b) From part (a), it is clear that if limn!1

Anxn

anexists, it must have value zero.

Construct an example of sequences as above such that the stated limit does

not exist.

2443. Proposed by Michael Lambrou, University of Crete, Crete,Greece.

Without the use of any calculating device, �nd an explicit example of an

integer, M , such that sin(M) > sin(33)(� 0:99991). (Of course, M and 33

are in radians.)

2444. Proposed by Michael Lambrou, University of Crete, Crete,Greece.

Determine limn!1

0@ ln(n!)

n�

1

n

nXk=1

ln(k)

0@ nXj=k

1

j

1A1A.

2445. Proposed by Michael Lambrou, University of Crete, Crete,Greece.

Let A, B be a partition of the set C = fq 2 Q : 0 < q < 1g (so that

A, B are disjoint sets whose union is C).

Show that there exist sequences fang, fbng of elements of A and B

respectively such that (an � bn)! 0 as n!1.

240

2446. Proposed by Catherine Shevlin, Wallsend upon Tyne, Eng-land.

A sequence of integers, fang with a1 > 0, is de�ned by

an+1 =

8>><>>:

an2

if n � 0 (mod 4),

3an + 1 if n � 1 (mod 4),

2an � 1 if n � 2 (mod 4),an+1

4if n � 3 (mod 4).

Prove that there is an integer m such that am = 1.

(Compare OQ.117 in OCTOGON, vol 5, No. 2, p. 108.)

2447. Proposed by Gerry Leversha, St. Paul's School, London, Eng-land.

Two circles intersect at P and Q. A variable line through P meets the

circles again at A and B. Find the locus of the orthocentre of triangle ABQ.

2448. Proposed by Gerry Leversha, St. Paul's School, London, Eng-land.

Suppose that S is a circle, centre O, and P is a point outside S. The

tangents from P to S meet the circle at A andB. Through any point Q on S,

the line perpendicular to PQ intersects OA at T and OB at U . Prove that

OT � OU = OP 2.

2449. Proposed by Gerry Leversha, St. Paul's School, London, Eng-land.

Two circles intersect atD andE. They are tangent to the sidesAB and

AC of4ABC at B and C respectively.. IfD is the mid-point of BC, prove

that DA�DE = DC2.

2450. Proposed by Gerry Leversha, St. Paul's School, London, Eng-land.

Find the exact value of

1Xk=0

(2k)!

(k!)3

1Xk=0

1

(k!)2

.

241

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.

2329?. [1998: 176, 301] Proposed by Walther Janous, Ursulinen-gymnasium, Innsbruck, Austria.

Suppose that p and t > 0 are real numbers. De�ne

�p(t) := tp + t�p + 2p and �p(t) :=�t+ t�1

�p+ 2 :

(a) Show that �p(t) � �p(t) for p � 2.

(b) Determine the sets of p: A, B and C, such that

1. �p(t) � �p(t),

2. �p(t) = �p(t),

3. �p(t) � �p(t).

Solution by Michael Lambrou, University of Crete, Crete, Greece.

(a) This part of the problem is included in (b), so we pass on to the latter.

(b) We show that

A := fpj�p(t) � �p(t) for all t > 0g = [0; 1] [ [2;1) ,

B := fpj�p(t) = �p(t) for all t > 0g = f0, 1, 2g ,

and

C := fpj�p(t) � �p(t) for all t > 0g = (�1; 0] [ [1; 2] .

The case for B will be settled once we observe B = A \ C, so we

determine A and C.

We shall make repeated use of the inequalities: for any x > 0,

xq � 1 � q(x� 1) if q � 0 or q � 1 , (1)

xq � 1 � q(x� 1) if 0 � q � 1 (2)

(see for example Hardy, Littlewood and P �olya, Inequalities, Theorem 42,

page 40).

Set fp(t) = �p(t)� �p(t). First suppose p � 2. For 0 < t < 1, and

by using (1) twice with q = p � 1 � 1 [and with �rst x = t2 + 1 and then

242

x = t2], we have (since 1� t2 � 0)

dfp(t)

dt= p(t+ t�1)p�1

�1�

1

t2

�� ptp�1 + pt�p�1

=�ptp+1

�(1� t2)(t2 + 1)p�1 + t2(t2)p�1 � 1

��

�ptp+1

h(1� t2)

�1 + (p� 1)t2

�+ t2

�1 + (p� 1)(t2� 1)

�� 1

i= 0 . (3)

Thus fp(t) is decreasing in (0;1]. Note that fp(1=t) = fp(t) so fp(t) is in-

creasing in [1;1), so t = 1 gives an absolute minimum. Thus

fp(t) � fp(1) = 0 for all t > 0, showing that [2;1) � A.

If now 1 � p � 2 then q = p � 1 2 [0; 1], so we use (2), showing

that inequality (3) is reversed and hence that [1; 2] � C. If 0 � p � 1 then

q = p� 1 � 0 so we use (1) again to show [0; 1] � A. Finally, if p � 0 then

q = p� 1 � 0 and �p � 0, so by use of (1) we get dfp(t)=dt � 0 instead of

(3), so (�1; 0] � C, completing the proof.

Remark. It is easy to see that for 0 < p < 1 and 1 < p < 2 the

limit limt!1 fp(t) exists and equals 2� 2p. Thus by the monotonicity of fpshown above we have the best possible inequalities

2� 2p < �p(t)� �p(t) � 0 for 1 < p < 2 ,

0 � �p(t)� �p(t) < 2� 2p for 0 < p < 1 .

For the rest of p it is easy to see that limt!1 fp(t) = �1 so no correspond-

ing inequality exists.

Also solved by NIKOLAOS DERGIADES, Thessaloniki, Greece; V �ACLAV KONE �CN �Y, Fer-ris State University, Big Rapids, Michigan, USA; KEE-WAI LAU, Hong Kong, China; and theproposer. One other reader sent in a counterexample to the incorrect version of this problempublished on [1998: 176].

2337. [1998: 177] Proposed by Iliya Bluskov, Simon Fraser Univer-sity, Burnaby, BC.

Let F (1) =

�n2 + 2n+ 2

n2 + n+ 1

�, and, for each i > 1, let

F (i) =

�n2 + 2n+ i+ 1

n2 + n+ iF (i� 1)

�.

Find F (n).

Solution by Michael Lambrou, University of Crete, Crete, Greece.

If the given n is n = 1, then F (n) = F (1) =

�12 + 2 + 2

12 + 1 + 1

�=

�5

3

�= 2.

Let us then do the more interesting case when the given n is � 2.

243

We show that for each i with 1 � i � n we have F (i) = i+ 1, so that

in particular F (n) = n+ 1. We use induction on i.

For i = 1 we have

F (i) = F (1) =

�n2 + 2n+ 2

n2 + n+ 1

�=

�1 +

n+ 1

n2 + n+ 1

�.

Thus

1 < F (1) =

�1 +

n+ 1

n2 + n+ 1

���1 +

n+ 1

n2 + n

�=

�1 +

1

n

�= 2 ,

from which we see that F (1) = 2, as required.

If we assume validity for i = m� 1 (that is, F (m� 1) = m,

where 2 � m � n) then

F (m) =

�n2 + 2n+m+ 1

n2 + n+m�m

�=

�m+

(n+ 1)m

n2 + n+m

�

��m+

(n+ 1)m

n2 + n

�=

�m+

m

n

�� dm+ 1e = m+ 1 ,

but as(n+ 1)m

n2 + n+m� 0, we havem <

�m+

(n+ 1)m

n2 + n+m

�.

Thus we have m < F (m) � m + 1, showing that F (m) = m + 1,

completing the induction step and proving the claim.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; DAVIDDOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA; FLORIAN HERZIG, student,Cambridge, UK; RICHARD I. HESS, Rancho Palos Verdes, California, USA; JOEL SCHLOSBERG,student, Bayside, NY, USA; HEINZ-J �URGEN SEIFFERT, Berlin, Germany; and the proposer.

2338. [1998: 234] Proposed by Toshio Seimiya, Kawasaki, Japan.Suppose ABCD is a convex cyclic quadrilateral, and P is the intersec-

tion of the diagonals AC and BD. Let I1, I2, I3 and I4 be the incentres of

triangles PAB, PBC, PCD and PDA respectively. Suppose that I1, I2,

I3 and I4 are concyclic.

Prove that ABCD has an incircle.

Solution by Peter Y. Woo, Biola University, La Miranda, California,USA.

ABCD does not have to be cyclic. More precisely,

When a convex quadrilateral is subdivided into 4 triangles by itstwo diagonals, then the incentres of the 4 triangles are concyclicif and only if the quadrilateral has an incircle.

244

Notation. Let P be the point where the diagonals intersect and let the tri-

angles be T1, T2, T3, T4 (labelled counterclockwise as in the �gure), with

the respective incentres I1, I2, I3, I4. Denote the 8 angles formed by the

diagonals with the four sides by 2�, 2�0, 2�, 2�0, 2 , 2 0, 2�, 2�0 (counter-clockwise with 2�, 2�0 in T1, etc.).

q

q

qq

I1

I2

I3

I4

P s1

s2

s3

s4

2�

C

2 0

2�0

D

2�

2�0

A2�

2

B

2�0

Step 1. In the usual notation (used only here in step 1) for4ABC with

sides a, b, c, incentre I, and semiperimeter s = a+b+c

2, AI satis�es

AI2 = bc tanB

2tan

C

2.

The proof follows from familiar formulas. In E.W. Hobson's Treatise on Planeand Advanced Trigonometry, for example, in section 123 the author shows

that

AI =r

sin A2

, tanB

2=

r

s� b, tan

C

2=

r

s� c,

and sin2A

2=

(s� b)(s� c)

bc,

which, when combined, proves the claim.

Step 2. I1, I2, I3, I4 are concyclic if and only if

tan� tan�0 tan tan 0 = tan� tan�0 tan � tan �0 .

Proof. Since PI1 and PI3 bisect vertically opposite angles, as do PI2and PI4, the Intersecting Chords Theorem says that I1, I2, I3, I4 are con-

cyclic if and only if PI1 �PI3 = PI2 �PI4. The desired equality then follows

from step 1.

Step 3. A quadrilateral has an incircle if and only if the sum of one pairof opposite sides equals the sum of the other. This is a standard result of

elementary geometry. See, for example, Nathan Altshiller Court's CollegeGeometry, p. 135.

Step 4. Let ABCD be any quadrilateral (that is, any four points, no

three collinear), I, I0 be incentres of4ABC and4ADC, and IN , I0N 0 be

245

perpendiculars to the diagonal AC from I and I0. Then CN � CN 0 if andonly if AD + BC � AB + CD, with equality for both or for neither.

Proof. CB�AB = CN�AN [because CN = s�c andAN = s�ain the notation of step 1] and AD � CD = AN 0 � CN 0. Add these two

equalities [noting that AN = AC � CN and AN 0 = AC � CN 0].

Step 5. AD + BC � AB + CD if and only if

tan\BAI tan\DCI0 � tan\BCI tan\DAI0 ,

with equality for both or for neither.

Proof. This follows from step 4 since we have tan\BAI = INAC�CN ,

tan\DCI0 = I0N0

CN0, tan\BCI = IN

CN, and tan\DAI0 = I0N0

AC�CN0.

[Note that the incentres here generally do not coincide with those of the

main result. The key observation is that IA (for example) bisects the angle

between a diagonal and side of the quadrilateral, while the angles�, �0, etc.in step 2 each are equal to half the angle between a diagonal and side.]

Proof of the main result. If ABCD has an incircle then

AD + BC = AB + CD (step 3), so that tan� tan = tan�0 tan �0 andtan�0 tan 0 = tan� tan � (step 5). By step 2, I1, I2, I3, I4 are there-

fore concyclic. On the other hand, if ABCD does not circumscribe some

circle, then let si be the side of Ti opposite P (for i = 1, 2, 3, 4). Without

loss of generality, assume s2 + s4 > s1 + s3. Then by step 5,

tan� tan > tan�0 tan �0 and tan�0 tan 0 > tan� tan �, so that by step

2, I1, I2, I3, I4 are not concyclic.

Also solved by NIELS BEJLEGAARD, Stavanger, Norway; FRANCISCOBELLOT ROSADO,I.B. Emilio Ferrari, Valladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; NIKOLAOS DERGIADES, Thessaloniki, Greece; WALTHER JANOUS, Ursulinengymnasium,Innsbruck, Austria; VJECKOSLAV KOVA�C, student, Univ. Zagreb, Croatia; D.J. SMEENK, Zalt-bommel, the Netherlands; JEREMY YOUNG, student, Nottingham, England; and the proposer.

All solvers except Woo and Janous proved the theorem as stated (with ABCD cyclic).Janous remembers having seen the stronger version before, but he did not recall the reference.Can any reader provide a reference?

2340. [1998: 235] Proposed by Walther Janous, Ursulinengymnas-ium, Innsbruck, Austria.

Let � > 0 be a real number and a, b, c be the sides of a triangle. Prove

that Ycyclic

s+ �a

s� a� (2�+ 3)3:

[As usual s denotes the semiperimeter.]

246

I. Solution by Jeremy Young, student, Nottingham High School, Not-tingham, England.

By the common substitution that a, b, c are the sides of a triangle if

and only if a = y + z, b = z + x, c = x+ y for x, y, z > 0, the inequality

is equivalent to

1

(2�+ 3)3

Ycyclic

�x+ (�+ 1)y+ (�+ 1)z

�� xyz . (1)

Now by the weighted AM{GM inequality,�1

2�+ 3

�x+

��+ 1

2�+ 3

�y+

��+ 1

2�+ 3

�z � x

12�+3y

�+12�+3z

�+12�+3 ,

��+ 1

2�+ 3

�x+

�1

2�+ 3

�y+

��+ 1

2�+ 3

�z � x

�+12�+3y

12�+3z

�+12�+3 ,

and ��+ 1

2�+ 3

�x+

��+ 1

2�+ 3

�y+

�1

2�+ 3

�z � x

�+12�+3y

�+12�+3 z

12�+3 .

Multiplying these three lines together gives (1), the required result.

II. Solution by Michael Lambrou, University of Crete, Crete, Greece.Set

f(�) = s(s+ �a)(s+ �b)(s+ �c)� (2�+ 3)3�2

where � � 0 and � is the area of the triangle. Then

f 0(�) = sXcyclic

a(s+ �b)(s+ �c)� 6(2�+ 3)2�2

= s�s2(a+ b+ c) + 2�s(ab+ bc+ ca) + 3�2abc

�� 6(2�+ 3)2�2.

Using a+ b+ c = 2s and

s2 � 3�p3 , ab+ bc+ ca � 4�

p3 , 9abc � 8s�

p3

(see for example Bottema et al, Geometric Inequalities, items 4.2, 4.5, 4.13)

we have

f 0(�) �2

3s2�

p3(9 + 12�+ 4�2)� 6(2�+ 3)2�2

� 6�2(2�+ 3)2� 6(2�+ 3)2�2 = 0 .

Thus, f is increasing in [0;1), and so

f(�) � f(0) = s4 � 27�2 � 0 .

247

Hence by Heron's formula,

f(�) = s(s+�a)(s+�b)(s+�c)� (2�+3)3s(s�a)(s� b)(s�c) � 0 ,

which reduces to the given inequality.

Also solved by ED BARBEAU, University of Toronto, Toronto, Ontario; CHRISTOPHERJ. BRADLEY, Clifton College, Bristol, UK; THEODORE CHRONIS, Athens, Greece; NIKOLAOSDERGIADES, Thessaloniki, Greece; FLORIAN HERZIG, student, Cambridge, UK; RICHARDI. HESS, Rancho Palos Verdes, California, USA; V �ACLAV KONE �CN �Y, Ferris State University,Big Rapids, Michigan, USA; VJEKOSLAV KOVA �C, student, University of Zagreb, Croatia; KEE-WAI LAU, Hong Kong, China; JESSIE LEI, student, Vincent Massey Secondary School, Windsor,Ontario; VEDULA N. MURTY, Dover, PA, USA; HEINZ-J �URGEN SEIFFERT, Berlin, Germany;PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer.

Herzig's solution is the same as Young's. Herzig also points out that the inequality istrue for all � � �1, and that equality holds if and only if � = �1 or x = y = z; that is,a = b = c.

2341. [1998: 235] Proposed by Walther Janous, Ursulinengymnas-ium, Innsbruck, Austria.

Let a, b, c be the sides of a triangle. For real � > 0, put

s(�) :=

������Xcyclic

"�a

b

����b

a

��#������and let �(�) be the supremum of s(�) over all triangles.

1. Show that �(�) is �nite if � 2 (0;1] and �(�) is in�nite for � > 1.

2.? What is the exact value of �(�) for � 2 (0; 1)?

I. Solution to part 1 by Florian Herzig, student, Cambridge, UK.Note that

s(�) =

�����(c� � b�)(c� � a�)(b� � a�)

(abc)�

����� .First, I show that, for positive reals a and b, the inequality ax + bx �

(a + b)x holds for all 0 � x � 1. Without loss of generality, assume that

a + b = 1. Then 0 < a, b < 1 and hence ax � a and bx � b. Therefore

ax + bx � a+ b = (a+ b)x as claimed.

Suppose a � b � c without loss of generality. By the above and the

triangle inequality, ax + bx � (a+ b)x > cx. Hence ax > cx � bx � 0 and

bx > cx � ax � 0. Trivially also cx > bx � ax � 0. Thus for all 0 < � � 1,

s(�) �a�b�c�

(abc)�= 1 ,

248

and so �(�) � 1.

If � > 1 then consider the triangle with sides 1 + 1

n, n+ 1

n, n+ 1 for

a positive integer n. Then in the expression for s(�) [that is,

s(�) =

������n+ 1

n2 + 1

��+

�n2 + 1

n2 + n

��+n� �

�n2 + 1

n+ 1

����n2 + n

n2 + 1

����1

n

�������| Ed.] two terms tend to 0 and two tend to 1 as n ! 1. Consider the

remaining two terms:

n� ��n2 + 1

n+ 1

��� n� �

�n�

1

2

��. (1)

where the inequality holds for all n � 3. Since

n� =

�n�

1

2+

1

2

����n�

1

2

��+�

2

�n�

1

2

���1

[by the Binomial Theorem], the left-hand side of (1) and hence s(�) tends

to in�nity as n!1. It follows that �(�) is in�nite.

II.Partial solutionto part 2 by NikolaosDergiades, Thessaloniki, Greece(with editorial comments).

All readers who attempted part 2 agreed that there will be no explicit

formula for�(�) for arbitrary 0 < � < 1. They all gave � = 1=2 as a special

case (and so did the proposer, in fact), and further agreed that in this case

�(1=2) � 0:0740033.

Dergiades, however, managed to �nd the exact value of �(1=2). First

he puts a � b � c without loss of generality, in which case (as in Solution I)

s(�) =(c� � b�)(c� � a�)(b�� a�)

(abc)�

=

�a

b

��+

�b

c

��+

�c

a

����b

a

����c

b

����a

c

��.

Considering a, b and � as constant and calling this function F (c), he gets

F 0(c) =��c2� � (ab)�

� �b� � a�

�c(abc)�

� 0 ,

so F is increasing. Thus to maximize s he puts c = a+ b, b = x and without

loss of generality a = 1, which for � = 1=2 means that s can be rewritten as

the function

f(x) =1px+

px

p1 + x

+p1 + x�

px�

p1 + xpx

�1

p1 + x

,

249

which now must be maximized over x � 1. Setting the derivative of this to

zero results in a sixth degree polynomial equation

x6 � 4x5 � 32x4 � 58x3 � 32x2 � 4x+ 1 = 0 ,

whose root � � 8:57318922 can be substituted into f(x) to obtain the above

maximum value of �(1=2). This is where the other solvers stopped. How-

ever, Dergiades then transforms the equation for f(x) by putting

x = tan2 y and sin2y = z ; that is, z =2px

1 + x,

getting the function

h(z) =

p1� z � (2 + z � 2

p1 + z)

z.

Solving h0(z) = 0 gives a cubic equation

z3 � 3z2 + 8z � 4 = 0

which can be solved. He gets the real root

r = 1��1 +

2

9

p114

�1=3+

5

3

�1 +

2

9

p114

��1=3,

so the exact value of �(1=2) is h(r) (which, by my calculator, is indeed the

previously found 0:0740033).

Part 1 also solved by V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan,USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; HEINZ-J �URGEN SEIFFERT,Berlin, Germany; and the proposer. The approximate value of �(1=2) was found by Kone�cn�y,Lambrou and the proposer.

2342. [1998: 235] Proposed by D.J. Smeenk, Zaltbommel, the Neth-erlands.

Given A and B are �xed points of circle �. The point C moves on �,

on one side of AB. D and E are points outside 4ABC such that 4ACDand 4BCE are both equilateral.

(a) Show that CD and CE each pass through a �xed point of � when C

moves on �.

(b) Determine the locus of the mid-point of DE.

250

Solutionby Jeremy Young, student, NottinghamHigh School, Notting-ham, England.

A B

D0

D00

C0

C00

X(120���)

?

�

60�

�

Let X be the point subtending

angles 120� and 60� with A. By

\angles in the same segment",

CD always passes through X.

Two possible positions of C are

shown: C0 and C00. Similarly,

two corresponding positions ofD

are shown: D0 and D00.

A B

C

(�a; 0) (a; 0)

�

�

�+�

Similarly, de�ne Y relative to B.

Introduce a Cartesian coordinate

system with A, B as (�a; 0),(a; 0) respectively. Let C be a

point with positive y{coordinate

such that \ACB = �.

By the Sine Rule, we haveAC = 2asin(�+ �)

sin�andBC = 2a

sin�

sin�.

Therefore D has position vector��a0

�+ 2a

sin(�+ �)

sin�

�cos(� + 60�)sin(� + 60�)

�.

Similarly, E has position vector�a

0

�+ 2a

sin�

sin�

�cos(�+ � � 60�)sin(�+ � � 60�)

�.

Hence, the mid-point has position vector

a

sin�

�sin(�+ �) cos(� + 60�) + sin� cos(�+ � � 60�)sin(�+ �) sin(� + 60�) + sin� sin(�+ � � 60�)

�

=a

2 sin�

�sin(�+ 2�)

2 cos(�� 60�)� cos(�+ 2�)

�

=a

2 sin�

�sin(�+ 2�)

� cos(a+ 2�)

�+

a

sin�

�0

cos(�� 60�)

�,

where 0 � � � 18� � �. Thus the required locus is an arc of a circle,

radius R=2 (where R = a=2 sin� is the circumradius of4ABC) and centre

(0; 2R cos(�� 60�)), which is exterior to the equilateral triangle with AB

as base.

Also solved by NIELS BEJLEGAARD, Stavanger, Norway; FRANCISCOBELLOT ROSADO,I.B. Emilio Ferrari, Valladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol,

251

UK; NIKOLAOS DERGIADES, Thessaloniki, Greece; WALTHER JANOUS, Ursulinengymnasium,Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA;VJEKOSLAV KOVA �C, student, University of Zagreb, Croatia; MICHAEL LAMBROU, Univer-sity of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul's School, London, England; TOSHIOSEIMIYA, Kawasaki, Japan; and the proposer.

Most solvers used pure geometric methods. Bellot Rosado and L �opez Chamorro solvedthe problem entirely with the use of complex numbers.

2343. [1998: 235] Proposed by Doru Popescu Anastasiu, Liceul\Radu Greceanu", Slatina, Olt, Romania.

For positive numbers sequences fxngn�1, fyngn�1, fzngn�1 with con-ditions: for n � 1, we have

(n+ 1)x2n + (n2 + 1)y2n+ (n2 + n)z2n

= 2pn�nxnyn +

pnxnzn + ynzn

�, (1)

and for n � 2, we have

xn +pnyn � nzn = xn�1 + yn�1 �

pn� 1zn�1 . (2)

Find limn!1

xn, limn!1

yn and limn!1

zn.

Solution by Christo Saragiotis, Thessaloniki, Greece.

Equation (1) can be re-written as

�pnxn � nyn

�2+ (xn � nzn)

2+�yn �

pnzn

�2= 0 .

Thus, yn =xnpn, zn =

xn

nand yn =

pnzn for all n � 1. Substituting these

into equation (2) yields xn = xn�1 for all n � 2.

Therefore, limn!1xn = x1 and lim

n!1yn = limn!1 zn = 0.

Also solved by THEODORE CHRONIS, Athens, Greece; OSCAR CIAURRI, Logro ~no,Spain; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS,Ursulinengymnasium, Innsbruck, Austria; VJEKOSLAV KOVA �C, student, University of Zagreb,Croatia; MICHAEL LAMBROU, University of Crete, Crete, Greece; LAURENT LESSARD,student, Le Coll �ege fran�cais, Toronto, Ontario; GERRY LEVERSHA, St. Paul's School, London,England; and the proposer. There were two partially incorrect solutions.

The solutions of almost all the solvers were virtually identical to the one given above.

2344. [1998: 235] Proposed by Murali Vajapeyam, student, Camp-ina Grande, Brazil and Florian Herzig, student, Perchtoldsdorf, Austria.

Find all positive integers N that are quadratic residues modulo all

primes greater than N .

252

Solution by Mansur Boase, student, Cambridge, England.Clearly any square is a quadratic residue modulo any prime. Suppose

N is not a square. Then we can write N as m2p1p2 � � � pr where pi 6= pjfor any pair (i; j) with i 6= j, and where r is some positive integer. Without

loss of generality let us assume p1 < p2 < � � � < pr. We shall show that N

cannot be a perfect square modulo all primes congruent to 1 (mod 4).

Let q be a prime congruent to 1 (mod 4). Then, introducing Legendre sym-

bols, we have�N

q

�=

�m2p1p2 � � � pr

q

�=

�m2

q

��p1

q

��p2

q

�� � ��pr

q

�

=

�p1

q

��p2

q

�� � ��pr

q

�

=

�q

p1

��q

p2

�� � ��q

pr

�.

The latter equality follows from the Law of Quadratic Reciprocity since

q � 1 (mod 4).

Suppose p1 = 2. Then if q � 1 + 4p2p3 � � � pr (mod 8p2p3 � � � pr),we have q � 1 (mod pi) for 2 � i � r, whence�

q

pi

�=

�1

pi

�= 1 for 2 � i � r:

Also, since all the p2, p3, : : : , pr are odd, q � 5 (mod 8), and therefore,�2

q

�= (�1)(q2�1)=8 = �1. Thus

�Nq

�= �1, which would be a contradic-

tion. It thus remains to show that a prime greater than N exists satisfying

q � 1 + 4p2p3 � � � pr (mod 8p2p3 � � � pr). But by Dirichlet's Theorem there

are in�nitely many such primes. Thus we cannot have p1 = 2, and all the

primes are odd.

There are (pi� 1)=2 quadratic residues modulo the prime pi, (pi � 3)

and this is a positive integer. Suppose n is a quadratic non-residue

modulo p1. Then by the Chinese Remainder Theorem, there exists a solution

modulo 4p1p2 � � � pr to the set of congruences:

x � 1 (mod 4) , x � n (mod p1) , x � 1 (mod pi) for 2 � i � r .

as (2; pi) = 1 for all i, 1 � i � r, and (pi; pj) = 1 for all i 6= j. Suppose

the prime q satis�es q � x (mod 4p1p2 � � � pr) and hence q satis�es the set

of congruences above. Then�N

q

�=

�q

p1

��q

p2

�� � ��q

pr

�

=

�n

p1

��1

p2

�� � ��1

pr

�= �1 ,

253

which would be a contradiction. Thus we must prove that there can be

no primes greater than N congruent to x (mod 4p1p2 � � � pr), as all such

numbers satisfy the requirement of being congruent to 1 (mod 4). But by

Dirichlet's Theorem, as (x; 4p1p2 � � � pr) = 1 there must be in�nitely many

primes in this arithmetic progression. [n 6= 0, so the set of congruences

above shows that x is relatively prime to each of 4, p1, p2, : : : , pr, and

hence also to the product.]

Thus N can satisfy the conditions of the problem only if N is a perfect

square.

Also solved by WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEE-WAILAU, Hong Kong, China; and the proposer. There was one incorrect solution submitted.

Lau comments that Theorem 3 of Chapter 5 of Ireland and Rosen's \A Classical Intro-duction to Modern Number Theory" (Springer-Verlag, 1982) proves that if N is a non-squareinteger, then there are in�nitely many primes p for whichN is a quadratic nonresidue. Combin-ing this with the observation that the squares are quadratic residues modulo all primes greaterthan them solves the problem.

2345. [1998: 236] Proposed by Vedula N. Murty, Visakhapatnam,India.

Suppose that x > 1.

(a) Show that ln(x) >3(x2 � 1)

x2 + 4x+ 1.

(b) Show thata� b

ln(a)� ln(b)<

1

3

�2pab+

a+ b

2

�,

where a > 0, b > 0 and a 6= b.

Solution by Kee-Wai Lau, Hong Kong, China.

(a) For x � 1, let f(x) = ln(x)�3(x2 � 1)

x2 + 4x+ 1. Then straightforward

computations show that f 0(x) =(x� 1)4

x (x2 + 4x+ 1)2, and so f 0(x) > 0 for

all x > 1. Since f(1) = 0, we have f(x) > 0 for x > 1, and the inequality

follows.

(b) Clearly, we can assume that a > b. Then the desired inequality

follows readily by substituting x =pa=b into the inequality in (a).

Also solved by THEODORE CHRONIS, Athens, Greece; OSCAR CIAURRI; Logro ~no,Spain; NIKOLAOS DERGIADES, Thessaloniki, Greece; RICHARD B. EDEN, student, Ateneode Manila University, Quezon City, Philippines; FLORIAN HERZIG, student, Cambridge, UK;RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengym-nasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michi-gan, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; LAURENT LESSARD,student, Le Coll �ege Fran�cais, Toronto, Ontario; GERRY LEVERSHA, St. Paul's School, Lon-don, England; PHILIP McCARTNEY, Northern Kentucky University, Highland Heights, Ken-tucky, USA; BOB PRIELIPP, University of Wisconsin{Oshkosh, Wisconsin, USA; HEINZ-J �URGEN SEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal College, Calgary, Alberta;

254

PANOS E. TSAOUSSOGLOU, Athens, Greece; JEREMY YOUNG, student, Nottingham HighSchool, Nottingham, England; and the proposer. There was one incorrect solution.

Solutions virtually identical to the one given above were submitted by about half of thesolvers.

Herzig remarked that the right hand side of the inequality in (a) is a Pad �e approximationto ln x at x = 1. Janous commented that this inequality is \quite sharp" in positive neigh-bourhoods of x = 1. Lambrou remarked that part (b) is a repetition of problem 2206 [1997:

46]. He and Kone �cn �y both pointed out that the inequality in (a) is reversed for 0 < x < 1.(Ed: This is obvious from the solution given above.) Leversha considered some generalizations

of (b) by setting x =�a

b

�1=n

in the inequality of (a); for example, putting n = 4 would give

a� b

ln a� ln b<

1

12

�pa+

pb+ 4

4pab

��pa +

pb

�,

which is sharper than the inequality in (b). Although a few solvers commented that the in-equality in (b) is known as P �olya's Inequality, Sie�ert (who was the proposer of problem 2206

mentioned above) stated that it should really be called the P �olya{Szeg�o Inequality, since it �rstappeared in a joint paper by them in 1951.

2347. [1998: 236] Proposed by �Sefket Arslanagi�c, University of Sara-jevo, Sarajevo, Bosnia and Herzegovina.

Prove that the equation x2+y2 = z1998 has in�nitely many solutions

in positive integers, x, y and z.

Solution by Richard I. Hess, Rancho Palos Verdes, California, USA. Letk be any positive integer. Multiply both sides of the equality 32 + 42 = 52

by 51996k1998. The result is

(3 � 5998k999)2 + (4 � 5998k999)2 = (5k)1998 .

Hence (3 � 5998k999 , 4 � 5998k999 , 5k) is an in�nite family of solutions to the

given equation.

Also solved by CHARLES ASHBACHER, Cedar Rapids, Iowa, USA; NIELSBEJLEGAARD, Stavanger, Norway; M. BENITO and E. FERN �ANDEZ, Logro ~no, Spain (twosolutions); MANSUR BOASE, student, Cambridge, England; CHRISTOPHER J. BRADLEY,Clifton College, Bristol, UK; THEODORE CHRONIS, Athens, Greece; GORAN CONAR, stu-dent, Gymnasium Vara�zdin, Vara�zdin, Croatia; NIKOLAOS DERGIADES, Thessaloniki, Greece;CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; KEITH EKBLAW,Walla Walla, Washington, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;GEOFFREY A. KANDALL, Hamden, Connecticut, USA; V �ACLAV KONE �CN �Y, Ferris StateUniversity, Big Rapids, Michigan, USA (two solutions); MICHAEL LAMBROU, Universityof Crete, Crete, Greece; KEE-WAI LAU, Hong Kong, China; GERRY LEVERSHA, St. Paul'sSchool, London, England; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria (threesolutions); BOB PRIELIPP, University of Wisconsin{Oshkosh, Wisconsin, USA; CHRISTOSSARAGIOTIS, Thessaloniki, Greece; DIGBY SMITH, Mount Royal College, Calgary, Alberta;PANOS E. TSAOUSSOGLOU, Athens, Greece; STAN WAGON, Macalester College, St. Paul,Minnesota, USA (two solutions); JEREMY YOUNG, student, Nottingham High School,Nottingham, England; and the proposer.

255

2349. [1998: 236] Proposed by V�aclav Kone �cn �y, Ferris State Uni-versity, Big Rapids, Michigan, USA.

Suppose that 4ABC has acute angles such that A < B < C. Prove

that

sin2B sin�A

2

�sin�A+ B

2

�> sin2A sin

�B

2

�sin�B + A

2

�.

Solution by Florian Herzig, student, Cambridge, UK.LetAD,BE be the angle bisectors in the triangle. By the angle bisector

property, EC = aba+c

. Hence in triangle BEC,

sin B2

sin(A+ B2)

=EC

BC=

b

a+ c.

A similar result holds in triangle ADC. Since also sinB : sinA = b : a, the

given inequality is equivalent to b2

a2> b+c

a� ba+c

, or b(a + c) > a(b + c) ;

that is, b > a, and this is, of course, true.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; THEODORECHRONIS, Athens, Greece; GORAN CONAR, student, Gymnasium Vara�zdin, Vara�zdin, Croa-tia; NIKOLAOS DERGIADES, Thessaloniki, Greece; RICHARD I. HESS, Rancho Palos Verdes,California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAELLAMBROU, University of Crete, Crete, Greece; KEE-WAI LAU, Hong Kong, China; GERRYLEVERSHA, St. Paul's School, London, England; HEINZ-J �URGEN SEIFFERT, Berlin, Germany;TOSHIO SEIMIYA, Kawasaki, Japan; PANOS E. TSAOUSSOGLOU, Athens, Greece; PAUL YIU,Florida Atlantic University, Boca Raton, Florida, USA; JEREMY YOUNG, student, NottinghamHigh School, Nottingham, England; and the proposer.

Most of the solutions were along the same lines as the featured one, which was chosenfor its brevity.

2350. [1998: 236] Proposed by Christopher J. Bradley, Clifton Col-lege, Bristol, UK.

Suppose that the centroid of 4ABC is G, and that M and N are the

mid-points of AC and AB respectively. Suppose that circles ANC and

AMB meet at (A and) P , and that circle AMN meets AP again at T .

1. Determine AT : AP .

2. Prove that \BAG = \CAT .

Identical independent solutionsby FlorianHerzig, student, Cambridge,UK and by Vjekoslav Kova�c, student, U. of Zagreb, Croatia.

Invert in a circle with centre A, and denote the image of a point X by

X0. Then C0 is the mid-point of AM 0 since AC : AM = AM 0 : AC0.Similarly B0 is the mid-point of AN 0. P 0 is the intersection of lines M 0B0

and N 0C0, so that it is the centroid of 4AM 0N 0. T 0 is then the point of

256

intersection of line AP 0 and line M 0N 0 (that is, the mid-point of segment

M 0N 0). By the property of medians,

AT : AP = AP 0 : AT 0 = 2 : 3 .

Note that 4ABC � 4ANM � 4AM 0N 0 and hence

\CAT = \C0AT 0 = \M 0AP 0 = \BAG

as we wanted to show.

Also solved by NIELS BEJLEGAARD, Stavanger, Norway; FRANCISCOBELLOT ROSADO,I.B. Emilio Ferrari, Valladolid, Spain; NIKOLAOS DERGIADES, Thessaloniki, Greece; WALTHERJANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete,Crete, Greece; TOSHIO SEIMIYA, Kawasaki, Japan; and the proposer.

Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell

Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical Mayhem

Founding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil

Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,

J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

257

THE ACADEMY CORNERNo. 26

Bruce ShawyerAll communications about this column should be sent to Bruce

Shawyer, Department of Mathematics and Statistics, Memorial University

of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

Abstracts � R�esum�es

Canadian Undergraduate Mathematics Conference

1998 | Part 4

The Dirichlet-to-Neumann Map

Scott MacLachlan

University of British Columbia

There are two classic problems in potential theory - the Dirichlet and Neumannproblems. In most courses on partial di�erential equations the problems are treatedseparately; however there is a very elegant link between these problems that can beexplored via Fourier analysis. The Dirichlet-to-Neumannmap takes data for a Dirich-let problem and translates it into data for a Neumann problem without calculating asolution �rst. This map is useful in many areas, particularly in potential theory.

Syst �emes de racines

Maciej Mizerski

McGill University

Un des plus beaux r �esultats de la th �eorie des groupes et alg �ebres de Lie est laclassi�cation des alg �ebres de Lie semi-simples. Cette classi�cation se ram�ene �a l' �etudede la structure des syst �emes de racines. Dans ma pr �esentation, je vais introduireles syst �emes de racines et survoler leur classi�cation en utilisant les diagrammes deDynkin. Aussi je vais tenter de faire le lien avec les groupes et les alg �ebres de Lie.

Cardan's Formulas for Solving Cubic Equations Revisited

Afroze Naqvi

University of Regina

This presentation o�ers a very brief history of Algebra and then proceeds toderive Cardan's Formulas for solving cubic equations. These formulas will be used tosolve some cubic equations.

258

Mathematics of the Ancient Jews and Ancient Israel

Peter Papez

University of Calgary

As mathematicians we are well acquainted with the mathematics of the an-cient Babylonians, Greeks and Egyptians. Even Ancient Chinese mathematicians havegained great notoriety in recent years. But the mathematics of ancient Israel is cer-tainly noteworthy, if not impressive, and has not received the attention it deserves.The purpose of this discussion will be to introduce these impressive achievementsand discuss some problems and the solutions obtained by ancient Jewish scholars.In order to facilitate the discussion, a brief overview of ancient Jewish history willbe given. As well, the Talmud and Talmudic Law will be presented. The discussionwill encompass the very beginning construction of numerals and numeracy within theancient Jewish culture, proceed through the development of arithmetic, touch on im-portant notions concerning science and logic, and conclude with a discussion of somefairly advanced developments in sampling and statistics. Various problems and theirsolutions, as derived by the ancient Jews, will be used to illustrate these develop-ments and the entire discussion will be placed in a historical and Talmudic context.

Computing in the Quantum World

Christian Paquin

Universit �e de Montr �eal

The current computing model is based on the laws of classical physics. But theworld is not classical; it follows the laws of quantummechanics. A quantum computeris a model of computation based on quantummechanics. It has been proved that sucha model is more powerful than its classical counterpart, meaning that it can do thesame computations as a classical computer (in approximately the same time) but thereexist some problems for which the quantum computer is much faster. In this paper Iwill explain what is quantum information, why a quantum computer would be useful,what are the problems to build a quantum computer and how it will work.

Wavelet Compression on Fractal Tilings

Daniel Pich �e

University of Waterloo

Over the last decade, wavelets have become increasingly useful for studyingthe behaviour of functions, and for compression. Though much remains to be inves-tigated in this �eld, certain types of wavelets are fairly easy to construct, namely Haarwavelets. This paper ties together the theory of these wavelets with that of complexbases. An algorithm is proposed for doing wavelet analysis, with wavelets arising inthis fashion. This will enable further study of the properties of these wavelets.

259

Introduction to Representation Theory

Evelyne Robidoux

McGill University

This paper provides an introduction to representation theory, with emphasison representations of �nite groups. The background required will be only a bit ofgroup theory (that is, being familiar with the concepts of groups, homomorphisms,conjugacy classes) and linear algebra (vector spaces, linear transformations, innerproducts, direct sums). A classical reference for this material is the book by Serre,Linear Representations of Finite Groups, which I really recommend.

The E�ect of Impurities in One-Dimensional Antiferromagnets

Alistair Savage

University of British Columbia

A brief overview of a paper to be published shortly with Ian A�eck of theUniversity of British Columbia Department of Physics and Astronomy.

Authentication Codes Without Secrecy

Nelly Sim~oes

Simon Fraser University

Suppose Alice wants to say something to Bob and cares only about the authen-tication of her message. Authentication makes it possible for Bob to receive Alice'smessage and be certain it came from her. Basically an authentication code withoutsecrecy is a process in which a mathematical function transforms what Alice wants tosay to Bob, we call this the source state, into what is called an authentication tag andadds it to the source state to form the message. We also suppose that Alice and Bobmutually trust each other. Alice wants to use an unconditionally secure method ofauthentication. She does not want anyone to be able to modify her message, not evena person with lots of computer power. This is why Alice decides to use a combinator-ial method. We are going to explore an unconditionally secure way of authenticatingAlice's message.

Integer Triangles With a Side of Given Length

Jill Taylor

Mount Allison University

This presentation will give a glimpse into the history of Heronian triangles

(triangles with integer sides and integer area) and one special case of such triangleswhich I have researched this summer. However, the main focus will be a particularproblem involving Heronian triangles with a given perimeter. The solution of thisproblem requires both number theory and basic geometrical concepts.

260

Convergence and Transcendence in the Field of p{adic Numbers

Sarah Sumner

Queen's University

We will �rst explore convergence properties of series in Qp, and then studyinstances when the series

1Xn=0

anpn; an 2 Qp

is transcendental over Qp. We will give a general result describing a large class ofseries of this type which are transcendental over Qp. Our result unfortunately doesnot resolve the transcendence of

P1n=0 n! in Qp. However, our theorem applies to

show transcendence of1Xn=0

�nn!

where �n is a primitive n-th root of unity if (p; n) = 1 and is 1 otherwise.

Random Number Generation

Ren�ee Touzin

Universit �e de Montr �eal

Nowadays, in many scienti�c �elds, random number distributions are needed.Those distributions such as a binomial, a normal, an exponential are built from iiduniform (0; 1) distributions. But this distribution does not really exist. In fact, itconsists of a mathematical and deterministic algorithm that tries to behave stochas-tically.

The building of an e�cient generator requires a strong knowledge of mathe-matics and computer science. A priori, a generator must follow certain theoreticalcriteria. A posteriori, those same generators must pass many statistical tests to besure they look random even though they are deterministic. In this presentation, wewill talk about di�erent kinds of existing generators and the qualities of a good gen-erator. We will give examples of good and bad generators and �nally we will presenta simple implementation in C.

The Probabilistic Method: Proving Existence by Chance

Alexander Yong

University of Waterloo

The Probabilistic Method is based on a simple concept: in order to prove the

existence of some mathematical object, construct an appropriate probability space

and show that the object occurs with positive probability. We will investigate this

powerful proof technique via three examples.

261

THE OLYMPIAD CORNERNo. 199

R.E. Woodrow

All communications about this column should be sent to Professor R.E.

Woodrow, Department of Mathematics and Statistics, University of Calgary,

Calgary, Alberta, Canada. T2N 1N4.

The �rst Olympiad we give this issue is a Danish contest. My thanks goto Ravi Vakil for collecting a copy and forwarding it to me when he was Cana-dian Team Leader to the International Mathematical Olympiad at Mumbai.

GEORG MOHR KONKURRENCEN I MATEMATIK1996

January 11, 9{13

Only writing and drawing materials are allowed.

1. \C in4ABC is a right angle and the legs BC and AC are both oflength 1. For an arbitrary point P on the leg BC, construct points Q, resp.R, on the hypotenuse, resp. on the other leg, such that PQ is parallel to ACand QR is parallel to BC. This divides the triangle into three parts.

Determine positions of the point P on BC such that the rectangularpart has greater area than each of the other two parts.

2. Determine all triples (x; y; z), satisfying

xy = z

xz = y

yz = x .

3. This year's idea for a gift is from \BabyMath", namely a series of9 coloured plastic �gures of decreasing sizes, alternating cube, sphere, cube,sphere, etc. Each �guremay be opened and the succeeding one may be placedinside, �tting exactly. The largest and the smallest �gures are both cubes.Determine the ratio between their side-lengths.

4. n is a positive integer. It is known that the last but one digit in thedecimal expression of n2 is 7. What is the last digit?

5. In a ballroom 7 gentlemen, A, B, C, D, E, F and G are sittingopposite 7 ladies a, b, c, d, e, f and g in arbitrary order. When the gentle-men walk across the dance oor to ask each of their ladies for a dance, one

262

observes that at least two gentlemen walk distances of equal length. Is thatalways the case?

The �gure shows an example. In this example Bb = Ee andDd = Cc.

A B C D E F G

f d b g c e a

s s s s s s s

s s s s s s s

The second two sets of problems come from the St. Petersburg CityMathematical Olympiad. Again my thanks go to Ravi Vakil, Canadian TeamLeader to the InternationalMathematical Olympiad at Mumbai for collectingthem for me.

ST. PETERSBURG CITY MATHEMATICAL OLYMPIADThird Round { February 25, 1996

11th Grade (Time: 4 hours)

1. Serge was solving the equation f(19x� 96=x) = 0 and found 11

di�erent solutions. Prove that if he tried hard he would be able to �nd atleast one more solution.

2. The numbers 1, 2, : : : , 2n are divided into two groups ofn numbers.Prove that pairwise sums of numbers in each group (sums of the form a+ a

included) have identical sets of remainders on division by 2n.

3. No three diagonals of a convex 1996{gon meet in one point. Provethat the number of the triangles lying in the interior of the 1996{gon andhaving sides on its diagonals is divisible by 11.

4. Points A0 and C0 are taken on the diagonal BD of a parallelogramABCD so that AA0kCC0. Point K lies on the segment A0C, and the lineAK meets the line C0C at the point L. A line parallel to BC is drawnthrough K, and a line parallel to BD is drawn through C. These two linesmeet at pointM . Prove that the points D, M , L are collinear.

5. Find all quadruplets of polynomials p1(x), p2(x), p3(x), p4(x)with real coe�cients possessing the following remarkable property: for allintegers x, y, z, t satisfying the condition xy � zt = 1, the equalityp1(x)p2(y)� p3(z)p4(t) = 1 holds.

6. In a convex pentagon ABCDE, AB = BC, \ABE + \DBC =

\EBD, and\AEB+\BDE = 180�. Prove that the orthocentre of triangleBDE lies on diagonal AC.

263

7. Two players play the following game on a 100�100 board. The �rstplayer marks some free square of the board, then the second puts a domino�gure (that is, a 2 � 1 rectangle) covering two free squares one of which ismarked. The �rst player wins if all the board is covered; otherwise the secondwins. Which of the players has a winning strategy?

Selective Round { March 10, 199611th Grade (Time: 5 hours)

1. It is known about real numbers a1, : : : , an+1; b1, : : : , bn that0 � bk � 1 (k = 1, : : : , n) and a1 � a2 � : : : � an+1 = 0. Prove theinequality:

nXk=1

akbk �[Pn

j=1 bj]+1Xk=1

ak .

2. Segments AE and CF of equal length are taken on the sides ABand BC of a triangle ABC. The circle going through the points B, C, Eand the circle going through the points A, B, F intersect at points B andD.Prove that the line BD is the bisector of angle ABC.

3. Prove that there are no positive integers a and b such that for alldi�erent primes p and q greater than 1000, the number ap+bq is also prime.

4. A Young tableau is a �gure obtained from an integral-sided rectangleby cutting out its cells covered by several integral-sided rectangles containingits right lower angle. We call a hook a part of the tableau consisting of somecell and all the cells lying either to the right of it (in the same row) or belowit (in the same column). A Young tableau of n cells is given. Let s be thenumbers of hooks containing exactly k cells. Prove that s(k+ s) � 2n.

5. In a triangle ABC the angle A is 60�. A point O is taken insidethe triangle such that \AOB = \BOC = 120�. A pointD is chosen on thehalf-line CO such that the triangle AOD is equilateral. The perpendicularbisector of the segment AO meets the line BC at point Q. Prove that theline OQ divides the segment BD into two equal parts.

6. There are 120 underground lines in a city; every station may bereached from each other with not more than 15 changes. We say that twostations are distant from each other if not less than 5 changes are neededto reach one of them from the other. What maximum number of pairwisedistant stations can be in this city?

7. a, b, c are integers. It is known that the polynomial x3 + ax2 +

bx + c has three di�erent pairwise coprime positive integral roots, and thepolynomial ax2+ bx+ c has a positive integral root. Prove that the numberjaj is composite.

8. Positive integers 1, 2, : : : , n2 are being arranged in the squares ofan n�n table. When the next number is put in a free square, the sum of the

264

numbers already arranged in the row and column containing this square iswritten on the blackboard. When the table is full, the sum of the numbers onthe blackboard is found. Show an example of this way of putting the numbersin squares such that the sum is minimum.

We next turn to readers' solutions to problems given earlier in the Cor-ner. First some catching up. Over the summer I had some time to sort and �lematerials, and discovered solutions to problems of the 1994 Balkan Olympiadmis�led with 1999 material. So we begin with solutions to the 1994 BalkanOlympiad [1997: 198].

2. [Greece] Show that the polynomial

x4 � 1994x3 + (1993 +m)x2 � 11x+m; m 2 Z

has at most one integral root. (Success rate: 39:66%)

Solutions by Mansur Boase, student, Cambridge, England.Consider

x4 � 1994x3 + (1993 +m)x2 � 11x+m . (�)

Suppose the given polynomial has two integral roots. Then neither can beodd for otherwise

(x4 + 1993x2)� (1994x3) +m(x2 + 1)� 11x

will be odd (as each of the terms in brackets is even) and hence non-zero.

Suppose x1 = 2r1a, r1 > 0 and a odd is a solution.

Considering the polynomial (mod 22r1), we then have thatm � 11x (mod 22r1). Hence m � 2r1(11a) (mod 22r1), and since a isodd, m must be of the form 2r1l1, l1 odd.

If x2 = 2r2b, b odd, is also a solution, then m = 2r2l2, l2 odd, so wemust have r1 = r2.

Thus, if there are two integral roots, both must be of the form 2rk,r > 1, k odd. The product of the two roots must be a multiple of 22r. Thequadratic which has these two roots as zeros is x2 + px + 22rq, where p, qare integers.

Now the given polynomial (*) can be factorized into two quadratics:

(x2 + px+ 22rq)(x2 + sx+ t) .

If s were not integral, then the coe�cient of x3 would not be integral in thequartic, and if t were not integral, the coe�cient of x2 would not be integralin the quartic. Thus s and tmust be integers andm = 22rqt. But the highest

265

power of 2 dividingm is 2r so r � 2r giving r = 0, a contradiction. Hencethe given quartic cannot have more than one integral root.

4. [Bulgaria] Find the smallest number n > 4 such that there is a setof n people with the following properties:

(i) any two people who know each other have no common acquain-tances;

(ii) any two people who do not know each other have exactly two com-mon acquaintances.

Note: Acquaintance is a symmetric relation. (Success rate: 19%)

Solution by Mansur Boase, student, Cambridge, England.

Choose one of the people, A, and suppose A knows x1, x2, : : : , xr.Then by (i) no xi knows an xj for i 6= j. Therefore, by (ii), for each pairfxi; xjg there must exist an Xij who knows both xi and xj in order thatxi and xj have two common acquaintances A andXij. Now A cannot knowany of theXij. Thus by (ii) eachXij can have only two acquaintances amongx1, x2, : : : , xr, namely xi and xj , so all the Xij are distinct.

Any person who is not A, nor an acquaintance of A must by (ii) be anXij. Thus the total number of people must be

�r

2

�+ r + 1.

Now r > 2 and n > 4.

If r = 3; then n = 7 (A) ,If r = 4; then n = 11 (B) ,If r = 5; then n = 16 (C) .

Let us label the people 1; 2; : : : ; n.

Case (A): Without loss of generality suppose 1 knows 2, 3 and 4 and that 5knows 2 and 3, 6 knows 2 and 4 and 7 knows 3 and 4.

Now 5 must have three acquaintances, so he must know one of 6 and7. But he has common acquaintances with both 6 and 7, contradicting (i).

Case (B): Without loss of generality, suppose 1 knows 2, 3, 4 and 5 and 6,7, 8, 9, 10, 11 know pairs f2; 3g, f2; 4g, f2; 5g, f3;4g, f3; 5g and f4; 5grespectively.

Then 6 cannot know 7, 8, 9 or 10 as he has common acquaintances witheach of them. So 6 can only know 2, 3 and 11, while he must know r = 4

people, a contradiction.

Case (C): We claim that n = 16 is the smallest number of people requiredin such a set, by noting that cases (A) and (B) fail and that the followingacquaintance table satis�es (i) and (ii):

266

Person Acquaintances1 2 3 4 5 6

2 1 7 8 9 10

3 1 7 11 12 13

4 1 8 11 14 15

5 1 9 12 14 16

6 1 10 13 15 16

7 2 3 14 15 16

8 2 4 12 13 16

9 2 5 11 13 15

10 2 6 11 12 14

11 3 4 9 10 16

12 3 5 8 10 15

13 3 6 8 9 14

14 4 5 7 10 13

15 4 6 7 9 12

16 5 6 7 8 11

Now we turn to solutions to problems of the Fourth Grade of the 38th

Mathematics Competition of the Republic of Slovenia [1998: 132].

1. Prove that there does not exist a function f : Z ! Z, for whichf(f(x)) = x+ 1 for every x 2 Z.

Solutions by Mohammed Aassila, CRM, Universit �e de Montr �eal,

Montr �eal, Qu �ebec; by Pierre Bornsztein, Courdimanche, France; by Masoud

Kamgarpour, Carson Graham Secondary School, North Vancouver, BC; by

Pavlos Maragoudakis, Pireas, Greece; and by Edward T.H. Wang, Wilfrid

Laurier University,Waterloo, Ontario. We give the solutionbyMaragoudakis.

Suppose that there is such a function. Then f(f(f(x))) = f(x) + 1.Since f(f(x)) = x+ 1, we get f(x+ 1) = f(x) + 1.

By induction f(x + n) = f(x) + n for every n 2 N. Alsof(x) = f(x� n+ n) = f(x� n) + n so

f(x� n) = f(x)� n for every n 2 N .

Finally f(x+ y) = f(x) + y for x; y 2 Z.

For x = 0, f(y) = f(0) + y. For y = f(0), f(f(0)) = f(0) + f(0).

But f(f(0)) = 1; thus 2f(0) = 1, a contradiction.

2. Put a natural number in every empty �eld of the table so that youget an arithmetic sequence in every row and every column.

267

0

74

103

186

Solutions by Pierre Bornsztein, Courdimanche, France; by Masoud

Kamgarpour, Carson Graham Secondary School, North Vancouver, BC;

and by Pavlos Maragoudakis, Pireas, Greece. We give the solution of

Kamgarpour.

Let us say the numbers adjacent to 0 are a0 and a1, with a0 in the rowand a1 the column. We know that in an arithmetic sequence every term isthe arithmetic mean of the term before and after. Therefore, we can putnumbers in the chart as follows:

3a1 74

2a1 a1 � a0 + 103 186

a11

2(a1 + 103) 103 1

2(309� a1) 206� a1

0 a0 2a0 3a0 4a0

Now

1

2(186 + 4a0) = 206� a1 ===) 93 + 2a0 = 206� a1

===) 2a0 + a1 = 113 , (1)

and

74 + 1

2(a1 + 103) = 2(103+ a1 � a0) ===) 3a1 � 4a0 = �161 . (2)

Solving (1) and (2) gives a0 = 50 and a1 = 13, so we can easily put numbersin every �eld as shown:

52 82 112 142 172

39 74 109 144 179

26 66 106 146 186

13 58 103 148 193

0 50 100 150 200

268

3. Prove that every number of the sequence

49 , 4489 , 444889 , 44448889 , : : :

is a perfect square (in every number there are n fours, n � 1 eights and anine).

Solutions by Pavlos Maragoudakis, Pireas, Greece; by Bob Prielipp,

University ofWisconsin{Oshkosh,Wisconsin,USA; and by Edward T.H.Wang,

Wilfrid Laurier University, Waterloo, Ontario. We give Prielipp's solution.

Let 43829 denote the number 444889 and 627 denote the number 667.

We shall show that

(6n�17)2 = 4n8n�19

for each positive integer n.

Because (6n�17)2 =�6(10n�1)

9+ 1

�2, it su�ces to establish that

�6(10n � 1)

9+ 1

�2

= 4n8n�19 for each positive integer n . (�)

Let n be an arbitrary positive integer. Then

�6(10n � 1)

9+ 1

�2=

�6 � 10n + 3

9

�2

=(2 � 10n + 1)2

9

=4 � 102n + 4 � 10n + 1

9

=40n�140n�11

9= 4n8n�19 .

4. Let Q be the mid-point of the side AB of an inscribed quadrilat-eral ABCD and S the intersection of its diagonals. Denote by P and R

the orthogonal projections of S on AD and BC respectively. Prove thatjPQj = jQRj.

269

Solutions by Pierre Bornsztein, Courdimanche, France; and by Toshio

Seimiya, Kawasaki, Japan. We give the solution of Seimiya.

A

B

C D

EF

P

Q

R

S

b

b

Let E be the point symmetric to A with respect to P , and let F be thepoint symmetric to B with respect to R. Then we have

SE = SA; \SEA = \SAE ,

and SF = SB; \SFB = \SBF .

As A, B, C, D are concyclic we get \CAD = \CBD . Thus,

\SEA = \SAE = \SBF = \SFB .

Consequently we have \ASE = \BSF . Thus we get

\BSE = \BSA+ \ASE = \BSA+ \BSF = \FSA .

Since SB = SF and SE = SA, we have

4SEB � 4SAF . (SAS)

Thus we get EB = AF . Since P , Q, R are mid-points of AE, AB, BFrespectively, we have

PQ =1

2EB and QR =

1

2AF .

Therefore we have PQ = QR .

That completes the Corner for this issue. Send me your nice solutionsas well as contest materials.

270

BOOK REVIEWS

ALAN LAW

In Polya's Footsteps, by Ross Honsberger,published by the Mathematical Association of America, 1997,ISBN # 0-88385-326-4, softcover, 328+ pages, $28.95.Reviewed byMurray S. Klamkin.

This is another in a series of books by the author on problems and solutionstaken from various national and international olympiad competitions and very manyof which have appeared previously in the Olympiad Corner of CruxMathematicorum.Also as the author notes, the solutions are his own unless otherwise acknowledged.Otherwise how would it look if everything was copied from elsewhere. I �nd thesesolutions of the author to be a mixed bag, especially in light of the following twoquotations,

A good proof is one that makes us wiser. Yu. I. Manin

Proofs really aren't there to convince you something is true | they're there toshow you why it is true. Andrew Gleason

Some of his solutions are a welcome addition, but there are also quite a num-ber for which the previously published ones are much better and certainly not asoverblown.

I now give some speci�c comments on some of the problems and their solutions.

� page 14. Given any sequence of r digits, is there a perfect square k2 with thesedigits immediately preceding the last digit of k2? The nice solution of the im-possibility is ascribed to Andy Liu, but not from the University of Calgary. As arelated aside, it could be mentioned that in problem 4621, School Science and

Mathematics, it is shown that the given sequence of digits can be the �rst r ormiddle r digits of an in�nite number of squares k2.

� page 26. Here it is shown in an elongated fashion that

nYi=1

�1 +

1

ai

�� (n+ 1)

n,

where

nXi=1

a1 = 1 and ai � 0. This is a well-known inequality appearing in

many places. For a more elegant proof, we can use H�older's Inequality to im-

mediately obtain the known inequality

nYi=1

�1 +

1

ai

� 1

n

� 1 +1

npa1a2 � � � an

.

We can then �nish o� using the AM{GM Inequality

1

n

nXi=1

ai

!n� a1a2 � � �an.

This proof not only leads to learning about the important H�older's Inequality,but also leads to generalizations. For example, let

Pai = A and

Pbi = B.

ThennYi=1

n

r1 +

1

ai+

1

bi� 1 +

n

A+

n

B.

� page 60. Not noted in this nice geometry problem, is that PQRS also has theproperty of having the least perimeter for quadrilaterals inscribed in ABCD.

271

� page 76. Here the author generalizes a Chinese problem by determining theremainder when F (xn+1) is divided by F (x) where F (x) = 1 + x + � � �+xn�1 + xn, and again gives an elongated solution. At the end he noted: \Analternative solution is given in [1992: 102]; additional comment appears in[1994: 46]." Left out is that the additional comment gives a wider generaliza-tion and a solution which is simpler and more compact.

� page 93. Here we are to determine the area of a triangle ABC given threeconcurrent cevians AD;BE and CF intersecting at P , where AD = 20,BP = 6 = PE, CF = 9 and PF = 3. The solution here is almost fourpages long and uses a guess in solving an irrational equation since it is knownthat the solution is an integer. For a much more compact solution with moreunderstanding, consider the following. Let [G] denote the area of a �gure

G. Then immediately, [APB]

[ABC] =3

3+9 = 14 and [APC]

[ABC] =6

6+6 = 12 , so that

[BPC]

[ABC]= 1

4and AP = 15. Now lettinq \BPC = � � �, \CPA = � � �

and \APB = � � , we have 2[BPC] = 6 � 9 sin� = 12 [ABC],

2[CPA] = 9 � 15 sin� = [ABC], and 2[APB] = 15 � 6 sin = 12[ABC].

All that is left is to determine one of the angles �, � and , whose sum is �,and so are angles of some triangle with sides proportional to 15, 12 and 9, re-spectively. Since this is a right triangle, � = �

2and [ABC] = 2 � 6 � 9 = 108.

Note that even if the latter was not a right triangle, we still can determine theangles.

� page 96. Here we have an elementary but long solution to �nd the smallestpositive integer n which makesmn � 1 divisible by 21989 no matter what oddinteger greater than 1 might be substituted for m. For another way of show-ing that n = 21987 is the smallest n, we use the known extension of Euler'sTheorem a�(n) � 1 (mod n), where (a; n) = 1, to the � function. Here,

�(2�) = �(2�) if � = 0, 1, 2;

�(2�) = 1

2�(2�) if � > 2;

�(p�) = �(p

�) if p is an odd prime;

and �(2�p�11 p

�21 � � � p�nn ) is the least common multiple of �(2�), �(p�11 ),

�(p�22 ), : : : , �(p�nn ), where 2, p1, p2, : : : , pn are di�erent primes. Then there

is no exponent � less than �(n) for which the congruence a� � 1 (mod n) issatis�ed for every integer a relatively prime ton. Hence �(21989) = 1

2�(21989)

= 21987. See R.D. Carmichael, The Theory of Numbers Wiley, NY 1914.

� page 106. Here the problem is to determine the acute angle A of a triangle if itis given that vertexA lies on the perpendicular bisector joining the circumcentreO and the orthocentre H. A simpler solution follows almost immediately byusing vectors. Letting A, B and C be vectors from O to the vertices A, B andC, respectively. We have H = A + B + C so that jAj = jB + Cj. Squaringthe latter equation, we get R2 = R2 + R2 + 2B � C = 4R2 � a2. Hencea = R

p3 = 2R sinA so that A = 60�. Note that many metric properties

of a triangle can be determined easily using this vector representation and thefollowing ones for the centroid G = A+B+C

3 and the incentre I = aA+bB+cCa+b+c .

For example, OH2 = (A + B + C)2 = A2 + B2 + C2 + 2B � C + 2C � A+

2A �B = 9R2 � a2� b2� c2, so that we have a simple proof of the inequality9R2 � a2 + b2 + c2, or equivalently, 9

4� sin2 A + sin2 B + sin2C. It also

272

follows immediately that O;H and G are collinear with OH = 3OG. As anexercise for the reader, show that OI2 = R2 � 2Rr.

� page 118. Here AB, CD and EF are chords of a circle that are concurrentat K and are inclined to each other at 60� angles. One has to show thatKA + KD + KE = KB + KC + KF . Again the referred to solutionis neater and shorter. Another easy solution is to use the polar equation ofthe circle with origin atK.

� page 122. Here E is a point on a diameter AC of a circle centre O and one hasto determine the chord BD through E which yields the quadrilateral ABCDof the greatest area. Here the author claims that BD should be perpendic-ular to AC. However, this is correct only if OA � OE

p2. For a com-

plete and simpler solution, let OA = r, OE = a and \BEC = �. Then

BE = �a cos �+pr2 � a2 sin2 � and ED = a cos � +

pr2 � a2 sin2 �, so

that [ABCD] = 2r sin �pr2 � a2 sin2 �. Letting t = sin2 �, we wish to max-

imize F � t(r2 � a2t). Now F is increasing from t = 0 to r22a2

. Consequently,

if r � ap2, F is maximized for t = r2

2a2, and if r > a

p2, F is maximized

for t = 1.

� page 133. Here we have to maximize abcd where a, b, c and d are integers witha � b � c � d and a + b + c + d = 30. Again a previous solution in CRUX

is simpler and more general and points out the use of the widely applicableMajorization Inequality.

� page 134. Here we are given tangents to a circle K from an external point Pmeeting K at A and B. We want to determine the position of C on the mi-nor arc AB such that the tangent at C cuts from the �gure a triangle PQRof maximum area. The author reduces the problem to minimizing the tangentchord QCR or, in his terms, tan�+ tan(90� � �� �) where 2� is the given

angle APB. Then using calculus, he obtains the minimizing angle � as 90���2 ,

which places C at the mid-point of arc AB which is to be expected intuitively.He also justi�es the minimum by examining the second derivative. My crit-icism here is one should eschew calculus methods in these problems. Sincetan�+tan(90�����) = cos�

sin(�+�) cos�= 2 cos �

sin�+sin(�+2�), we have a more

elementary and simpler solution for the minimizing angle.

� page 152. Here we have to show that if for every point P inside a convexquadrilateral ABCD, the sum of the perpendiculars to the four sides (or theirextensions if necessary) is constant, then ABCD is a parallelogram. The solu-tion given in approximately three pages is attributed to me. However, I don'tquite recognize it: my solution in CRUX was considerably shorter.

� page 158. Here one has equilateral triangles drawn outwardly on the sides ofa triangle ABC, so that the three new vertices are D, E and F . Given D, Eand F , one has to construct ABC. Here I have no criticisms of the solution,especially since the author �rst reviews some basic ideas of translations and ro-tations in the plane and then uses these to get a nice solution. I just like to pointout that a solution using complex numbers, although not particularly elegant,is very direct. Let (A; B;C) = (z1; z2; z3) and (D;E; F ) = (w1; w2; w3).Then 2w1 = z2+z3+i

p3(z2�z3). The other two equations follow by a cyclic

change in the subscripts. After some simple algebra, we �nd

2z1 = w1+w21+i

p3

2�w3

1�ip3

2, so that z1 is easy to construct given w1; w2

273

andw3. We can then construct z2 and z3. It also follows, by summing the threeequations, that the centroids of ABC and DEF coincide.

� page 184. Here we have a problem on numbering an in�nite checkerboard withan approximate three page solution and with no attribution. Andy Liu informedme of a prior source of this problem. It occurs with an approximately one pagesolution in A. M. Yaglom and I. M. Yaglom, Challenging Mathematical Prob-

lems with Elementary Solutions II, Holden-Day, San Francisco, 1967, p.129.

� page 203. Here we have to �nd the minimum value of the function f(x) =pa2 + x2 +

p(b� x)2 + c2, where a, b and c are positive numbers. This is

a classic minimum distance problem and is treated in many places. The au-thor repeats the classic re ection solution given in CRUX and also refers to analternate solution given there but does not indicate that this solution gives im-mediate generalizations via Minkowski's Inequality which is another useful toolfor competition contestants: p

pap + xp + yp+ p

pbp + (c� x)p + (d� y)p �

pp(a + b)p + cp + dp where p > 1. If 1 > p > 0, the inequality is reversed.

� page 216. Here we have to show there are in�nitely many non-zero solutionsto the Diophantine equation x2 + y5 = z3. The author, after a page of work,comes up with the solution (x; y; z) = (215a+10; 26a+4; 210a+7). More generalequations of this type have appeared as problems very many times in the last100 years with more general solutions. For example, consider the equation xr+ys + zt = wu where r, s, t and u are given positive integers with u relativelyprime to rst. We now just let x = a(ar +bs+ct)mst; y = b(ar +bs+ct)mtr

and z = c(ar + bs + ct)mrs so that xr + ys + zt = (ar + bs + ct)mrst+1.On setting w = (ar + bs + ct)n, we have to ensure that there are integers mand n such that mrst + 1 = nu. Since u is relatively prime to rst, there isan in�nite set of such pairs of positive integers m and n. A much more di�cultproblem is to �nd relatively prime solutions.

� page 247. Here one has to show that (1 + x1)(1 + x2) � � � (1 + xn) �1+S+ S2

2!+ � � �+ Sn

n!, where the x's are positive numbers and S =

Pxi. In

addition to his solution, the author refers to an alternate solution in CRUX in1989. It should be pointed out that the following is a known stronger inequal-

ity: (1+x1x)(1+x2x) � � � (1+xnx) � 1+xS+ x2S2

2!+� � �+ xnSn

n!(x > 0).

Here the � sign indicates majorization; that is, the coe�cient of xr on the leftis less than or equal to the coe�cient of xr on the right for r = 1, 2, : : : , n.See [1982: 296].

Aside from my criticisms above, the book is a nice collection of problems andsome essays. In a review [1998: 78] of the author's previous book in the same vein,From Erd }os to Kiev, Bill Sands was somewhat critical of the random order of theproblems. While that is true here also, I would not be too critical of this providedthe solutions were given in order; that is, having all the geometry solutions together,the algebra solutions together, and so on. For when one is writing a mathematicscompetition, the order of problems is essentially at random. But when one is learningto solve problems on one's own, it will be more e�ective to have solutions to likeproblems linked together. For a student competitor using this book, I strongly advisethat she or he also look at the corresponding solutions in CRUX with MAYHEM.

Finally, I do not think the book's title, \In P �olya's Footsteps", is appropriate.

274

APROPOS BELL AND STIRLING NUMBERS

Antal E. Fekete

Introduction

In 1877 Dobi ~nski stated [1] that there exist integers qn such that

0n

0!+

1n

1!+

2n

2!+

3n

3!+ � � � = qn

�1

0!+

1

1!+

1

2!+

1

3!+ � � �

�= qne ,

and he calculated their values for n = 1 through 5 (Table 1). Indeed, qnare the Bell numbers, so named in honour of the American mathematicianEric Temple Bell (1883-1960), who was among the �rst to popularize thesenumbers; see [2] and [3], where further references can be found. It may beshown that qn is just the sum of Stirling numbers of the second kind:

qn =

�n

1

�+

�n

2

�+ � � � +

�n

n

�.

Since�n

k

is the number of k{member quotient sets of an n{set, the Bell

number qn is the number of all quotient sets of an n{set. It can also becalculated via recursion in terms of the Stirling numbers of the �rst kind; thefollowing formula is due to G.T. Williams [5]:�

n

n

�qn �

�n

n� 1

�qn�1 +� � � � + (�1)n�1

�n

1

�q1 = 1 ,

where�n

k

�are the coe�cients of the polynomial of degreenwith roots 0 , 1 , 2 ,

: : : , (n� 1):�n

n

�xn�

�n

n� 1

�xn�1+� � � �+(�1)n�1

�n

1

�x = x(x�1)(x�2) � � � (x�n+1) .

R�enyi numbers

Here we show that there exist integers rn such that

0n

0!�

1n

1!+

2n

2!�

3n

3!+� � � � = rn

�1

0!�

1

1!+

1

2!�

1

3!+� � � �

�=

rn

e.

We calculate their values for n = 1 through 4 by bringing to the numeratorthe polynomial with coe�cients

�n

k

�, and splitting it into linear factors which

we can then cancel.

r1 = �1: X0

(�1)nn

n!=X1

(�1)n1

(n� 1)!= �

1

e.

Copyright c 1999 CanadianMathematical Society

275

r2 = 0:

X0

(�1)nn2

n!=

X0

(�1)nn2 � n

n!�

1

e=X0

(�1)n(n� 1)n

n!�

1

e

=X2

(�1)n1

(n� 2)!�

1

e=

1

e�

1

e= 0 .

r3 = 1:

X0

(�1)nn3

n!=

X0

(�1)nn3 � 3n2 + 2n

n!+

0

e+

2

e

=X0

(�1)n(n� 2)(n� 1)n

n!+

2

e

=X2

(�1)n1

(n� 3)!+

2

e= �

1

e+

2

e=

1

e.

r4 = 1:

X0

(�1)nn4

n!=

X0

(�1)nn4 � 6n3 + 11n2 � 6n

n!+

6

e�

6

e

=X0

(�1)n(n� 3)(n� 2)(n� 1)n

n!

=X4

(�1)n1

(n� 4)!=

1

e.

The recursion formula for rn in terms of Stirling numbers of the �rst kind�n

n

�rn �

�n

n� 1

�rn�1 +� � � � + (�1)n�1

�n

1

�r1 = (�1)n

shows that rn is an integer for all n (Table 1). To prove it we write theleft{hand side as

eX0

(�1)k�n

n

�kn �

�n

n�1�kn�1 +� � � � + (�1)n�1

�n

1

�k

k!

= eX0

(�1)k(k� n+ 1) � � � (k� 1)k

k!= e

Xn

(�1)k

(k� n)!= (�1)n .

We call rn R�enyi numbers in honour of the Hungarian mathematician Alfr �edR �enyi (1921-1970) who �rst studied them [4]. They can be expressed as thealternating sum of the Stirling numbers of the second kind:

rn = ��n

1

�+

�n

2

��+ � � � + (�1)n

�n

n

�.

276

Related numbers

We now introduce related numbers exhibitingproperties similar to thoseof Bell and R �enyi numbers. There exist integers an, bn such that

0n

0!+

2n

2!+

4n

4!+ � � � = an cosh(1) + bn sinh(1) ;

and1n

1!+

3n

3!+

5n

5!+ � � � = an sinh(1) + bn cosh(1)

which can be calculated via recursion in terms of the Stirling numbers of the�rst kind:�n

n

�an �

�n

n� 1

�an�1 +� � � � + (�1)n�1

�n

1

�a1 =

�1 if n is even0 if n is odd�

n

n

�bn �

�n

n� 1

�bn�1 +� � � � + (�1)n�1

�n

1

�b1 =

�0 if n is even1 if n is odd

Furthermore, there exist integers cn, dn such that

0k

0!�

2k

2!+

4k

4!�+ � � � = ck cos(1)� dk sin(1) ;

and1k

1!�

3k

3!+

5k

5!�+ � � � = ck sin(1) + dk cos(1) ,

which can be calculated via recursion in terms of the Stirling numbers of the�rst kind:

�n

n

�cn �

�n

n� 1

�cn�1 +� � � � + (�1)n�1

�n

1

�c1 =

8<:

1 if n = 4k

�1 if n = 4k + 2

0 if n is odd

�n

n

�dn �

�n

n� 1

�dn�1 +� � � � + (�1)n�1

�n

1

�d1 =

8<:

0 if n is even1 if n = 4k+ 1

�1 if n = 4k + 3

Table 1: Bell, R �enyi and related numbers

n 0 1 2 3 4 5 6 7 8 9 10

qn 1 1 2 5 15 52 203 877 4140 21147 115975

rn 1 �1 0 1 1 �2 �9 �9 50 267 413

an 1 0 1 3 8 25 97 434 2095 10707 58194

bn 0 1 1 2 7 27 106 443 2045 10440 57781

cn 1 0 �1 �3 �6 �5 33 266 1309 4905 11516

dn 0 1 1 0 �5 �23 �74 �161 57 3466 27361

These related numbers are useful in calculating the number of quotient sets

277

of an n{set with an even or odd number of members:

an =qn + rn

2=

�n

2

�+

�n

4

�+

�n

6

�+ � � � ;

and bn =qn � rn

2=

�n

1

�+

�n

3

�+

�n

5

�+ � � � .

We also have

cn = ��n

2

�+

�n

4

���n

6

�+� � � � ;

and dn =

�n

1

���n

3

�+

�n

5

��+ � � � .

Extended Bell numbers

There exist integers qkn such that

kn

0!+

(k+ 1)n

1!+

(k+ 2)n

2!+

(k+ 3)n

3!+ � � � = e qk;n+1 .

In particular, q1n = qn; for this reason we call qkn the extended Bell num-bers. For each �xed k we have a recursion formula in terms of the Stirlingnumbers of the �rst kind

qkn =

(�1)k�1��k

1

�qn �

�k

2

�qn+1 +

�k

3

�qn+2 �+ � � � + (�1)k�1

�k

k

�qn+k�1

�,

providing the �rst of three methods to calculate the extended Bell numbers(Table 2).

A second method is via the recursion

qkn = kqk;n�1 + qk+1;n�1 .

For a �xed n, the extended Bell numbers q0;n , q1;n , q2;n , : : : are inarithmetic progression of order n � 1. Therefore qkn � Qn�1(k + 1) is apolynomial of degree n� 1 in the variable k. We have the recursion

Qn(k) � (k� 1)Qn�1(k) + Qn�1(k+ 1)

enabling us to calculate these polynomials:

Q0(k) � 1 (constant)Q1(k) � k

Q2(k) � k2 + 1

Q3(k) � k3 + 3k+ 1

Q4(k) � k4 + 6k2 + 4k + 4

Q5(k) � k5 + 10k3 + 10k2 + 20k+ 11

278

Q6(k) � k6 + 15k4 + 20k3 + 60k2 + 66k+ 41

Q7(k) � k7 + 21k5 + 35k4 + 140k3 + 231k2 + 287k+ 162

Q8(k) � k8 + 28k6 + 56k5 + 280k4 + 616k3 + 1148k2 + 1296k+ 715

. . . . . . . . . . . . . . .

A thirdmethod of calculating qkn is furnishedby the di�erence equation�nqk1 = qk�1;n that may be verbalized as follows. In Table 2, the k{th row

can be obtained by calculating the �rst member of each of the higher order

di�erence sequences of the (k+ 1){st row. This property is useful not onlyin calculating the k{th row from the (k + 1){st, quickly and e�ciently, butalso the other way around.

Table 2: Extended Bell Numbers

knn 1 2 3 4 5 6 7 8 n

� � � � � � � � � ��7 1 �6 37 �233 1492 �9685 63581 �421356 ��6 1 �5 26 �139 759 �4214 23711 �134873 ��5 1 �4 17 �75 340 �1573 7393 �35178 ��4 1 �3 10 �35 127 �472 1787 �6855 ��3 1 �2 5 �13 36 �101 293 �848 ��2 1 �1 2 �3 7 �10 31 �21 ��1 1 0 1 1 4 11 41 162 �0 1 1 2 5 15 52 203 877 �1 1 2 5 15 52 203 877 4140 �2 1 3 10 37 151 674 3263 17007 �3 1 4 17 77 372 1915 10481 60814 �4 1 5 26 141 799 4736 29371 190497 �5 1 6 37 235 1540 10427 73013 529032 �6 1 7 50 365 2727 20878 163967 1322035 �7 1 8 65 537 4516 38699 338233 3017562 �k � � � � � � � � qkn

For example, suppose that row 3 is given and we wish to �nd the next, row 4.We enter row 3 as the slanting row indicated by the boxed numbers below,and calculate entries in successive slanting rows as the sum of the adjacenttwo entries in the previous slanting row.

1 5 26 141 799 4736 : : :

4 21 115 658 3937 : : :

17 94 543 3279 : : :

77 449 2736 : : :

372 2287 : : :

1915 : : :

10481 : : :

279

Then row 4 appears as the top line of the calculation. It is clear that, giventhe initial condition qn, we can reconstruct the entire table for qkn as theunique solution to the di�erence equation.

Problems

In passingwe mention some further results, proposed here as problemsthat the reader may wish to solve using various ideas presented above.

(1) Show that

12

1!+

32

3!+

52

5!+ � � � =

22

2!+

42

4!+

62

6!+ � � � .

Are there exponents other than n = 2 for which the equality holds?

(2) Show that

13

1!�

23

2!+

33

3!�+ � � � =

14

1!�

24

2!+

34

3!�+ � � � .

Are there pairs of exponents other than 3, 4 for which an equality ofthis type holds?

(3) Show that

1n

1!�

3n

3!+

5n

5!�+ � � � = �n

�1

1!�

1

3!+

1

5!�+ � � �

�

and0n

0!�

2n

2!+

4n

4!�+ � � � = �n

�1

0!�

1

2!+

1

4!�+ � � �

�

simultaneously hold for n = 3. Find all other n having the same prop-erty. If follows that�

0n

0!�

2n

2!+

4n

4!�+ � � �

�2+

�1n

1!�

3n

3!+

5n

5!�+ � � �

�2

is an integer for n = 3. Clearly, this is also true for n = 1. Find allother n having the same property.

(4) Clearly,�0n

0!+

2n

2!+

4n

4!+ � � �

�2��1n

1!+

3n

3!+

5n

5!+ � � �

�2

is an integer for n = 1. Find all other n having the same property.

(5) Show that there are integers rkn such that

kn

0!�

(k+ 1)n

1!+

(k+ 2)n

2!�

(k+ 3)n

3!+� � � � =

rk;n+1

e.

In particular r1n = rn, in consequence of which we may call rkn theextended R �enyi numbers.

280

(6) Show that there are integers akn , bkn such that

(2k)n

0!+

(2k+ 2)n

2!+

(2k+ 4)n

4!+ � � �

= ak;n+1 cosh(1) + bk;n+1 sinh(1)

and (2k+ 1)n

1!+

(2k+ 3)n

3!+

(2k+ 5)n

5!+ � � �

= ak;n+1 sinh(1) + bk;n+1 cosh(1) .

In particular, a1n = an , b1n = bn.

(7) Show that there are integers ckn , dkn such that

(2k)n

0!�

(2k+ 2)n

2!+

(2k+ 4)n

4!�+ � � �

= ck;n+1 cos(1)� dk;n+1 sin(1)

and (2k+ 1)n

1!�

(2k+ 3)n

3!+

(2k+ 5)n

5!�+ � � �

= ck;n+1 sin(1) + dk;n+1 cos(1) .

In particular, c1n = cn , d1n = dn.

(8) For a �xed k, write a recursion formula in terms of the Stirling numbersof the �rst kind for each of rkn , akn, etc.

(9) Show that, for n �xed, each of the sequences of numbers rkn , akn,etc., is in arithmetic progression of order n� 1. Write polynomials ofdegree n, Rn(k), An(k), etc., de�ned such that rkn � Rn�1(k + 1),akn � An�1(k+ 1), etc.

(10) Write a recursion formula for each of the polynomials Rn(k), An(k).Write the polynomials in explicit form up to n = 8.

(11) Write a recursion formula for each of rkn , akn, etc.

(12) Find and tabulate the values of each of rkn , akn, etc.

(13) Recall that the di�erence equation �nqk1 = qk�1;n uniquely deter-mines the extended Bell numbers qkn under the initial conditionq1n = qn. Write a di�erence equation for the extended R �enyi num-bers rkn under the initial condition r1n = rn. Do the same for each ofakn, bkn, etc.

(14) Let n be �xed; continue the sequence of extended Bell numbers qkn fornegative k (there are at least three ways of doing that) in order to justifyTable 2 for k = �1, �2, �3, : : : . Do the same for each of rkn, akn,etc.

281

(15) Show that for each �xed k we have

qk+1;n+1 = qk1 +

�n

1

�qk2 +

�n

2

�qk3 + � � � +

�n

n

�qk;n+1

and

qk+1;n�1 = qk1��n

1

�qk2+

�n

2

�qk3�+ � � � +(�1)n�1

�n

n

�qk;n+1 .

(16) Show that for each �xed k andm we have

qk+m;n+1 = qk1mn+

�n

1

�qk2m

n�1+

�n

2

�qk3m

n�2+ � � �+�n

n

�qk;n+1 , ;

in particular,

qm;n+1 = q0mn +

�n

1

�q1m

n�1 +

�n

2

�q2m

n�2 + � � � +�n

n

�qn .

(17) Show that�k

1

�q1n +

�k

2

�q2n + � � � +

�k

k

�qkn = qn+k�1 .

(18) Show that

qn � qn�1 +� � � � + (�1)n+1q1 = qn�1(0) = q�1;n+2 .

References

1. G. Dobi ~nski, Summierung Der ReihePnm=n! f �urm = 1, 2, 3, 4, 5, : : : Arch.

f �ur Math. und Physik, vol. 61 (1877), pp. 333-6.

2. Eric Temple Bell, Exponential numbers, Am. Math. Monthly, vol. 41, no. 7(1934), pp. 411-419.

3. Gian-Carlo Rota, The Number of Partitions of a Set, Am. Math. Monthly,vol. 71 (1964), pp. 498-504.

4. Alfr �ed R �enyi, New methods and results in Combinatory Analysis (in Hungar-ian), Magyar Tudom�anyos Akad �emia III Oszt �alya K�ozlem�enyei, vol. 16 (1966),pp. 77-105.

5. G.T. Williams, Numbers generated by the function eex�1, Am.Math. Monthly,

vol. 53 (1945), pp. 323-7.

6. D.E. Knuth, TwoNotes on Notation, Am.Math.Monthly, vol. 99 (1992), pp. 403-22.

7. A.E. Fekete, Apropos Two Notes on Notation, Am. Math. Monthly, vol. 101(1994), pp. 771-8.

Memorial University of NewfoundlandSt.John's, CANADA A1C 5S7e-mail: fekete@math.mun.ca

282

THE SKOLIAD CORNERNo. 39

R.E. Woodrow

This issue we give another example of a team competition with the prob-lems of the 1998 Florida Mathematics Olympiad, written May 14, 1998. Thecontest was organized by Florida Atlantic University. My thanks go to JohnGrant McLoughlin, Memorial University of Newfoundland for sending methe problems.

FLORIDA MATHEMATICS OLYMPIADTEAM COMPETITION

May 14, 1998

1. Find all integers x, if any, such that 9 < x < 15 and the sequence

1 , 2 , 6 , 7 , 9 , x , 15 , 18 , 20

does not have three terms in arithmetic progression. If there are no suchintegers, write \NONE."

2. A sequence a1, a2, a3; : : : is said to satisfy a linear recurrence rela-

tion of order two if and only if there are numbers p and q such that, for allpositive integers n,

an+2 = pan+1 + qan .

Find the next two terms of the sequence

2 , 5 , 14 , 41 , : : : ,

assuming that this sequence satis�es a linear recurrence relation of ordertwo.

3. Seven tests are given and on each test no ties are possible. Eachperson who is the top scorer on at least one of the tests or who is in thetop six on at least four of these tests is given an award, but each person canreceive at most one award. Find the maximum number of people who couldbe given awards, if 100 students take these tests.

4. Some primes can be written as a sum of two squares. We have, forexample, that

5 = 12 + 22; 13 = 22 + 32; 17 = 12 + 42 ,

29 = 22 + 52; 37 = 12 + 62; and 41 = 42 + 52 .

283

The odd primes less than 108 are listed below; the ones that can be writtenas a sum of two squares are boxed in.

3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 ,

37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , 71 ,

73 , 79 , 83 , 89 , 97 , 101 , 103 , 107 .

The primes that can be written as a sum of two squares follow a simple pat-tern. See if you can correctly �nd this pattern. If you can, use this patternto determine which of the primes between 1000 and 1050 can be written asa sum of two squares; there are �ve of them. The primes between 1000 and1050 are

1009 , 1013 , 1019 , 1021 , 1031 , 1033 , 1039 , 1049 .

No credit unless the correct �ve primes are listed.

5. The sides of a triangle are 4, 13, and 15. Find the radius of theinscribed circle.

6. In Athenian criminal proceedings, ordinary citizens presented thecharges, and the 500-man juries voted twice: �rst on guilt or innocence, andthen (if the verdict was guilty) on the penalty. In 399BCE, Socrates (c469{399)was charged with dishonouring the gods and corrupting the youth of Athens.He was found guilty; the penalty was death. According to I.F. Stone's calcu-lations on how the jurors voted:

(i) There were no abstentions;

(ii) There were 80more votes for the death penalty than there were forthe guilty verdict;

(iii) The sum of the number of votes for an innocent verdict and thenumber of votes against the death penalty equalled the number of votes infavour of the death penalty.

a) How many of the 500 jurors voted for an innocent verdict?

b) How many of the 500 jurors voted in favour of the death penalty?

7. Find all x such that 0 � x � � and

tan3 x� 1 +1

cos2 x� 3 cot

��

2� x

�= 3 .

Your answer should be in radian measure.

Last issue we gave the problems of the Newfoundland and LabradorTeachers' Association Mathematics League. Here are the answers.

284

NLTA MATH LEAGUEGAME 1 | 1998{99

1. Find a two-digit number that equals twice the product of its digits.

Solution. Denote the number by ab; We get 10a + b = 2a � b. Try-ing a = 1; 2; : : : ; 9, the only integral solution is with a = 3, b = 6 and36 = 2 � 3 � 6.

2. The degree measures of the interior angles of a triangle are A, B, Cwhere A � B � C. If A, B, and C are multiples of 15, how many possiblevalues of (A;B;C) exist?

Solution. Let A = 15m, B = 15n, and C = 15p. Then 15m+ 15n+

15p = 180 so m + n + p = 12 and m � n � p. Since m, n � 1 we havem+ n � 2 and p � 10. Also 3p � 12 so p � 4. For p �xed, 4 � p � 10 wehave m+ n = 12� p, or m = 12� p� n. This leads to the solutions

p 4 5 5 6 6 6 7 7 8 8 9 10

n 4 4 5 3 4 5 3 4 2 3 2 1

m 4 3 2 3 2 1 2 1 2 1 1 1

There are twelve solutions.

3. Place an operation (+;�;�;�) in each square so that the expres-sion using 1, 2, 3, � � � , 9 equals 100.

1 2 2 2 3 2 4 2 5 2 6 2 7 2 8 2 9 = 100 .

You may also freely place brackets before/after any digits in the expression.Note that the squares must be �lled in with operational symbols only.

Solution. Here is one solution:

(((1 + 2 + 3 + 4 + 5)� 6)� 7) + 8 + 9 = 100 .

How many solutions are there?

4. A, B and C are points on a line that is parallel to another linecontaining pointsD andE, as shown. Point F does not lie on either of theselines.

r r r

r r

r

A B C

F

D E

How many distinct triangles can be formed such that all three of itsvertices are chosen from A, B, C, D, E, and F ?

285

Solution. Any choice of three of the six vertices determines a triangleexcept when they lie on a line; that is, except for the one choice fA;B;Cg.The total is thus

�6

3

�� 1 = 20� 1 = 19.

5. Michael, Jane and Bert enjoyed a picnic lunch. The three of themwere to contribute an equal amount of money toward the cost of the food.Michael spent twice as much money as Jane did buying food for lunch. Bertdid not spend any money on food. Instead, Bert brought $6 which exactlycovered his share. How much (in dollars) of Bert's contribution should begiven to Michael?

Solution. Bert brought $6, which exactly covered his share, so the totalcost of food is 3�$6 = $18. Now Michael spent twice as much as Jane so hespent $12 and she spent $6. The total amount of $6 brought by Bert shouldgo to Michael.

6. Two semicircles of radius 3 are inscribed in a semicircle of radius 6.A circle of radius R is tangent to all three semicircles, as shown. Find R.

3 3

33

RR

R

h

r

r

r r r

r r

Solution. Join the centres of the two smaller semicircles and the centre ofthe circle. This forms an isosceles triangle with equal sides 3+R and base 6units. Call the altitude of this triangle h. The altitude extends to a radius ofthe large semicircle, so h+R = 6. By Pythagoras, h2 + 32 = (R+ 3)2, so

(6� R)2 + 32 = (R+ 3)2 ,

36� 12R+ R2 + 9 = R2 + 6R+ 9 ,

36 = 18R ,

2 = R .

The radius of the small circle is 2.

7. If 5A = 3 and 9B = 125, �nd the value of AB.

Solution. Now 5A = 3, so 52A = 32 = 9 and

52AB = (52A)B = 9B = 125 = 53 ,

so 2AB = 3 and AB = 3

2.

286

8. The legs of a right angled triangle are 10 and 24 cm respectively.

Let A = the length (cm) of the hypotenuse,B = the perimeter (cm) of the triangle,C = the area (cm2) of the triangle.

Determine the lowest common multiple of A, B, and C.

Solution. Then

A =p102 + 242 = 26

B = 10 + 24 + 26 = 60

C = 1

2� 10 � 24 = 120

lcm(26;60; 120) = 3� 8� 5� 13 = 1560.

9. A lattice point is a point (x; y) such that both x and y are integers.For example, (2;�1) is a lattice point, whereas,

�3; 1

2

�and

��1

3; 23

�are not.

How many lattice points lie inside the circle de�ned by x2 + y2 = 20? (DoNOT count lattice points that lie on the circumference of the circle.)

Solution. Since 42 < 20 < 52 we have that �4 � x � 4. For a �xedx in the range we must have �

p20� x2 < y <

p20� x2 for integer x, y

solutions corresponding to interior points of the circle.

x �4 �3 �2 �1 0p20� x2 2

p11

p16

p19

p20

y �1 , 0 , 1 �3 , : : : , 3 �3 , : : : , 3 �4 , : : : , 4 �4 , : : : , 4

Lattice pts. 2 � 3 = 6 2 � 7 = 14 2 � 7 = 14 2 � 9 = 18 9

The total number is then 6 + 14 + 14 + 18 + 9 = 61.

10. The quadratic equation x2 + bx+ c = 0 has roots r1 and r2 thathave a sum which equals 3 times their product. Suppose that (r1 + 5) and(r2+5) are the roots of another quadratic equation x2+ ex+ f = 0. Giventhat the ratio of e : f = 1 : 23, determine the values of b and c in the originalquadratic equation.

Solution. Now r1 + r2 = �b and r1r2 = c, so as r1 + r2 = 3r1r2 weget 3c = �b. Similarly we have

r1 + r2 + 10 = �e ,(r1 + 5)(r2 + 5) = f .

Thus

10� b = �eand b� 10 = e ,

r1r2 + 5(r1 + r2) + 25 = f ,

c� 5b+ 25 = f .

287

From ef= 1

23, 23(b� 10) = c� 5b+ 25. Using b = �3c

23(�3c� 10) = c+ 15c+ 25 ,

so that � 69c� 230 = 16c+ 25 ,

�255 = 85c ,

�3 = c ,

and 9 = b .

The quadratic is x2 + 9x� 3 = 0.

RELAY

R1. Operations � and � are de�ned as follows:

A �B =AB +BA

A+ Band A � B =

AB � BA

A� B.

Simplify N = (3 � 2) � (3 � 2). Write the value of N in Box #1 of the relayanswer sheet.

Solution.

3 � 2 =32 + 23

3 + 2=

9 + 8

5=

17

5,

3 � 2 =32 � 23

3� 2=

9� 8

1= 1 ,

N = (3 � 2) � (3 � 2) =

�17

5� 1�=

�17

5

�1+ (1)17=5

17

5+ 1

= 1 .

R2. A square has a perimeter of P cm and an area of Q sq.cm. Giventhat 3NP = 2Q, determine the value of P . Write the value of P in Box #2of the relay answer sheet.

Solution. From R1,N = 1, so 3P = 2Q. Also the side length is s = P4,

so Q = (P4)2 = P 2

16. So 3P = 2 � P 2

16or 24P = P 2. This gives P = 0 or

P = 24. We use P = 24, assuming the square is not degenerate.

R3. List all two-digit numbers that have digits whose product is P .Call the sum of these two-digit numbers S. Write the value of S in Box #3of the relay answer sheet.

Solution. P = 1 � 24 = 2 � 12 = 4 � 6 = 8 � 3. The two-digit numbersare 46, 64, 38 and 83; so

S = 231 .

288

R4. How many integers between 6 and 24 share no common factorswith S that are greater than 1?

Solution. Now 231 = 11� 21 = 11� 7� 3. The numbers between 6

and 24 that share no common factor with 231 are

8 , 10 , 13 , 16 , 17 , 19 , 20 , 23 .

There are 8 of them.

TIE-BREAKER

Find the maximum value of

f(x) = 14�px2 � 6x+ 25 .

Solution.

f(x) = 14�px2 � 6x+ 25

= 14px2 � 6x+ 9� 9 + 25

= 14�p(x� 3)2 + 16 .

For f(x) to be maximum we want (x�3)2+16 to be minimum. This occurswhen x = 3, and

f(3) = 14�p16 = 14� 4 = 10 .

That completes the Skoliad Corner for this number. Sendme your com-ments, suggestions, and especially suitable material for use in the Corner.

Announcement

The second volume in the ATOM series has just been published. Algebra |

Intermediate Methods by Bruce Shawyer. Contents include:

Mathematical Induction, Series, Binomial Coe�cients,Solutions of Polynomial Equations, and Vectors and Matrices.

For more information, contact the CMS O�ce | address on outside backcover.

289

MATHEMATICAL MAYHEM

Mathematical Mayhem began in 1988 as a Mathematical Journal for and by

High School and University Students. It continues, with the same emphasis,as an integral part of Crux Mathematicorum with Mathematical Mayhem.

All material intended for inclusion in this section should be sent to

Mathematical Mayhem, Department of Mathematics, University of Toronto,

100 St. George St., Toronto, Ontario, Canada. M5S 3G3. The electronicaddress is still

mayhem@math.toronto.edu

The Assistant Mayhem Editor is Cyrus Hsia (University of Western On-tario). The rest of the sta� consists of Adrian Chan (Upper Canada College),Jimmy Chui (University of Toronto), David Savitt (Harvard University) andWai Ling Yee (University of Waterloo).

Shreds and Slices

Another Combinatorial Proof

Dear Cyrus:In the most recent issue of Crux Mathematicorum with Mathematical

Mayhem, Dave Arthur's two-step combinatorial proof of the identity

(n� r)

�n+ r� 1

r

��n

r

�= n

�n+ r � 1

2r

��2r

r

�(1)

was presented, and a one-step combinatorial proof was sought. I have sucha proof, but before I present the polished �nal version, let me explain howI was led to it.

Arthur's solution proceeded in two steps. He counted by two di�erentmethods:

(I) the number of ways of choosing two distinct sets A and B, each with relements, from a set with n+ r � 1 elements, and

(II) the number of ways of choosing two distinct sets C and D, where Chas one element and D has r elements, from a set with n elements.

It occurred to me that both (I) and (II) were ways to �nish o� the selec-tion of three sets A, B, and C, with r, r, and 1 elements respectively, froma set with n+ r elements. We could �rst choose the small set C, and thenchoose sets A and B from the remaining n+ r� 1 elements as in (I), or wecould choose one of the big sets A, and then choose sets C and B = D from

290

the remaining n elements as in (II). Counting both these methods yields theidentity

(n+ r)

��n+ r� 1

2r

��2r

r

��=

�n+ r

r

���n

r

�(n� r)

�: (2)

We can easily see that this identity is equivalent to (1) by multiplying bothsides of (2) by n and using the identity

n

�n+ r

r

�= (n+ r)

�n+ r � 1

r

�(3)

on the right-hand side.

There are two �sthetic problems with this: �rst, we derived the soughtidentity (1) with an extra factor of n + r on both sides; second, we usedthe identity (3) which, though elementary, can be thought of as needing acombinatorial proof of its own (choosing disjoint subsets A andW of sizes rand 1 from a set with n + r elements). The following proof of (1) is free ofthese defects.

Proof of (1). Suppose we have a circular arrangement of n + r boxes,and we have r amber balls, r blue balls, 1 cyan ball, and 1 white ball. Wewant to arrange these balls in the boxes (arrangements that di�er only by arotation being regarded as the same), such that two balls are not allowed tobe in the same box, except that the white ball may be in a box with a blue orcyan ball. We claim that both sides of (1) count the number of ways of doingthis. The two counting methods are:

1. First place the white ball in a box; since rotated arrangements are con-sidered the same, this can be done in only one way, but we have nowused up our freedom to rotate the boxes. Next, place the r amber ballsinto r of the remaining n + r � 1 empty boxes, which can be done in�n+r�1

r

�ways. Then, place the r blue balls into r of the n boxes that

do not contain an amber ball (recall that a blue ball can coexist with awhite ball), which can be done in

�n

r

�ways. Finally, place the cyan ball

in one of the n� r boxes not containing an amber or blue ball (again,the white ball is allowed if it doesn't also contain a blue ball), whichcan be done in n� r ways. The total number of such arrangements isthus �

n+ r� 1

r

��n

r

�(n� r) : (4)

2. This time, place the cyan ball �rst; as before, there is only one way todo this up to rotation. Next, select 2r of the remaining n+r�1 emptyboxes (which we will momentarily �ll with the 2r amber and blue balls),which can be done in

�n+r�1

2r

�ways. Then, place the r amber balls into

r of these selected 2r boxes, and place the r blue balls into the other r

291

selected boxes; this can be done in�2r

r

�ways. Finally, place the white

ball into any of the n boxes that does not contain an amber ball, whichcan be done in n ways. The total number of such arrangements is thus�

n+ r � 1

2r

��2r

r

�n : (5)

Since (4) and (5) are the left- and right-hand sides of (1), respectively,this completes the proof. �

We remark that we could rephrase what we are counting as follows: wewant to select three disjoint subsets A, B, and C, with r, r, and 1 elementsrespectively, from a set S with n+ r elements, and also label one elementW of S that is not in A, except that selections that di�er only by a cyclicpermutation of S are to be considered equal.

Yours sincerely,

Greg MartinDepartment of Mathematics, University of Toronto

Mayhem Problems

The Mayhem Problems editors are:

Adrian Chan Mayhem High School Problems Editor,Donny Cheung Mayhem Advanced Problems Editor,David Savitt Mayhem Challenge Board Problems Editor.

Note that all correspondence should be sent to the appropriate editor |see the relevant section. In this issue, you will �nd only problems | thenext issue will feature only solutions.

We warmly welcome proposals for problems and solutions. With theschedule of eight issues per year, we request that solutions from this issuebe submitted in time for issue 6 of 2000.

High School Problems

Editor: Adrian Chan, 229 Old Yonge Street, Toronto, Ontario, Canada.M2P 1R5 <all238@sprint.com>

We correct problem H253, which �rst appeared in Issue 3.

H253. Find all real solutions to the equationp3x2 � 12x+ 52 +

p2x2 � 12x+ 162 =

p�x2 + 6x+ 280 .

292

H257. Find all integers n such that n2�11n+63 is a perfect square.

H258. Solve in integers for x and y:

6(x! + 3) = y2 + 5 .

H259. Proposed by Alexandre Tritchtchenko, student, Carleton Uni-

versity, Ottawa, Ontario.Solve for x:

2m�n sin(2nx)m�nYi=1

cos(2m�ix) = 1 .

H260. Proposed byMohammed Aassila, CRM,Universit �e deMontr �eal,

Montr �eal, Qu �ebec.Let x1, x2, : : : , xm be real numbers. Prove that

X1�i<j�n

cos(xi � xj) � �n

2.

Advanced Problems

Editor: Donny Cheung, c/o Conrad Grebel College, University of Wa-terloo, Waterloo, Ontario, Canada. N2L 3G6 <dccheung@uwaterloo.ca>

A233. Proposed by Naoki Sato, Mayhem Editor.In C81, we de�ned the following sequence: a0 = 0, a1 = 1, and

an+1 = 4an � an�1 for n = 1, 2, : : : . This sequence exhibits the followingcurious property: For n � 1, if we set (a; b; c) = (an�1; 2an; an+1), thenab + 1, ac + 1, and bc + 1 are always perfect squares. For example, forn = 3, (a; b; c) = (a2; 2a3; a4) = (4; 30; 56), and indeed, 4 � 30 + 1 = 112,4 � 56 + 1 = 152, and 30 � 56 + 1 = 412. Show that this property holds.Generalize, using the sequence de�ned by a0 = 0, a1 = 1, and an+1 =

Nan� an�1, and the triples (a; b; c) = (an�1; (N � 2)an; an+1), where Nis an arbitrary integer.

A234. In triangle ABC, AC2 is the arithmetic mean of BC2 andAB2. Show that cot2B � cotA cotC. (Note: cot � = cos �= sin�.)

(1997 Baltic Way)

A235. ProposedbyMohammed Aassila, CRM, Universit �e deMontr �eal,

Montr �eal, Qu �ebec.The convex polygon A1A2 � � �An is inscribed in a circle of radius R.

Let A be some point on this circumcircle, di�erent from the vertices. Set

293

ai = AAi, and let bi denote the distance from A to the line AiAi+1, i = 1,2, : : : , n, where An+1 = A1. Prove that

a21

b1+a22

b2+ � � �+

a2n

bn� 2nR .

A236. ProposedbyMohammed Aassila, CRM, Universit �e deMontr �eal,

Montr �eal, Qu �ebec.

For all positive integers n and positive reals x, prove the inequality

�2n

1

�x+ 1

+

�2n

3

�x+ 3

+ � � �+�

2n

2n�1�

x+ 2n� 1<

�2n

0

�x

+

�2n

2

�x+ 2

+ � � �+�2n

2n

�x+ 2n

.

Challenge Board Problems

Editor: David Savitt, Department of Mathematics, Harvard University,1 Oxford Street, Cambridge, MA, USA 02138 <dsavitt@math.harvard.edu>

C87. Proposed by Mark Krusemeyer, Carleton College.

Find an example of three continuous functions f(x), g(x), and h(x)

from R to R with the property that exactly �ve of the six composite functionsf(g(h(x))),f(h(g(x))),g(f(h(x))), g(h(f(x))),h(f(g(x))), andh(g(f(x)))are the same function and the sixth function is di�erent.

C88.

(a) Let A be an n�n matrix whose entries are all either +1 or �1. Provethat j detAj � nn=2.

(b) It is conjectured that for there to exist an n�nmatrix A whose entriesare all either +1 or �1 and such that jdetAj = nn=2, it is necessaryand su�cient that n = 1, n = 2, or n is divisible by 4. Prove that thiscondition is necessary.

(c) Can you construct such a matrix, for n equal to a power of 2? Forn = 12?

294

Problem of the Month

Jimmy Chui, student, University of Toronto

Problem. Let a, b, and c be positive real numbers. Prove that

aabbcc � (abc)(a+b+c)=3 .

(1995 CMO, Problem 2)

Solution I. Taking log of both sides, we see that we must prove

a log a+ b log b+ c log c �a+ b+ c

3logabc .

Let f(x) = x logx. Then f 0(x) = logx + 1 and f 00(x) = 1=x. We can seethat f 00(x) > 0 for x > 0. So, f is convex for x > 0. By Jensen's Inequality,

f(a) + f(b) + f(c)

3� f

�a+ b+ c

3

�

===)a loga+ b log b+ c log c

3�

a+ b+ c

3log

�a+ b+ c

3

�.

By the AM-GM Inequality, a+b+c3

� 3pabc . Thus, we have

a log a+ b log b+ c log c � (a+ b+ c) log

�a+ b+ c

3

�

� (a+ b+ c) log(3pabc)

=a+ b+ c

3logabc .

Solution II. Recall Chebychev's Inequality: Let x1, x2, : : : , xn and y1,y2, : : : , yn be two real sequences, either both increasing or both decreasing.Then

x1y1 + x2y2 + � � �+ xnyn

n�

x1 + x2 + � � �+ xn

n�y1 + y2 + � � �+ yn

n.

(In other words, the average of the products is greater than or equal to theproduct of the averages.)

Without loss of generality, let a � b � c. Then loga � log b � log c.By Chebychev's Inequality,

a loga+ b log b+ c log c

3�

a+ b+ c

3�loga+ log b+ log c

3,

which implies that a log a+ b log b+ c log c � a+b+c3

log abc .

295

Solution III. Recall the Weighted AM-GM-HM Inequality: Let x1, x2,: : : , xn be positive real numbers, and let w1, w2, : : : , wn be non-negativereal numbers which sum to 1. Then

w1x1 + w2x2 + � � �+ wnxn � xw1

1 xw2

2 � � �xwn

n �1

w1

x2+ w2

x2+ � � �+ wn

xn

.

Take x1 = a, x2 = b, x3 = c, w1 = a=(a+ b+ c), w2 = b=(a+ b+ c),and w3 = c=(a + b + c). Then using the GM-HM portion of the aboveinequality, we obtain

aa=(a+b+c)bb=(a+b+c)cc=(a+b+c) = (aabbcc)1=(a+b+c)

�1

1

a+b+c+ 1

a+b+c+ 1

a+b+c

=a+ b+ c

3.

By the AM-GM Inequality,

aabbcc ��a+ b+ c

3

�a+b+c� (abc)(a+b+c)=3 .

J.I.R. McKnight Problems Contest 1986

Solutions

3. Prove that the sum of the squares of the �rst n even natural numbersexceeds the sum of the squares of the �rst n odd natural numbers byn(2n+1). Hence, or otherwise, �nd the sum of the squares of the �rstn odd natural numbers.

Partial solution by Luyun Zhong-Qido, Columbia International College,

Hamilton, Ontario, Canada.The sum of the �rst n positive integers is given by

12 + 22 + � � �+ n2 =n(n+ 1)(2n+ 1)

6. (1)

Therefore, 12 + 22 + � � �+ (2n)2 =2n(2n+ 1)(4n+ 1)

6, (2)

and 22 + 42 + � � �+ (2n)2 =4n(n+ 1)(2n+ 1)

6. (3)

296

From (2) and (3),

12 + 32 + � � �+ (2n� 1)2 =2n(2n+ 1)(4n+ 1)

6�

4n(n+ 1)(2n+ 1)

6

=2n(2n+ 1)[(4n+ 1)� 2(n+ 1)]

6

=2n(2n� 1)(2n+ 1)

6. (4)

From (3) and (4),

22 + 42 + � � �+ (2n)2 � [(12 + 32 + � � �+ (2n� 1)2]

=4n(n+ 1)(2n+ 1)

6�

2n(2n� 1)(2n+ 1)

6

=2n(2n+ 1)[2(n+ 1)� (2n� 1)]

6

=3 � 2n(2n+ 1)

6= n(2n+ 1) .

4. (b) Prove that in any acute triangle ABC,

cotA+ cotB + cotC =a2 + b2 + c2

4K,

where K is the area of triangle ABC.

Solution by Luyun Zhong-Qido, Columbia International College, Hamil-

ton, Ontario, Canada.We have the following relations in a triangle:

2R =a

sinA=

b

sinB=

c

sinC, (1)

K =1

2ab sinC =

1

2ac sinB =

1

2bc sinA . (2)

From (1) and (2),

K =1

2bc sinA =

1

2

�a sinB

sinA

��a sinC

sinA

�sinA = a2

sinB sinC

2 sinA,

so a2 =2K sinA

sinB sinC.

Likewise, b2 =2K sinB

sinA sinC, c2 =

2K sinC

sinA sinB.

We note that

cotA+ cotB =cosA

sinA+

cosB

sinB=

cosA sinB + cosB sinA

sinA sinB

=sin(A+B)

sinA sinB=

sinC

sinA sinB.

297

Therefore,

a2 + b2 + c2

4K=

2K sinAsinB sinC

+ 2K sinBsinA sinC

+ 2K sinCsinA sinB

4K

=2K(sin2A+ sin2B + sin2C)

4K sinA sinB sinC

=1

2

�sinA

sinB sinC+

sinB

sinA sinC+

sinC

sinA sinB

�

=1

2(cotB + cotC + cotA+ cotC + cotA+ cotB)

= cotA+ cotB + cotC .

J.I.R. McKnight Problems Contest 1989

PART A

1. A curve has equation y = x3 � 3x. A tangent is drawn to the curve atits relative minimum point. This tangent line also intersects the curveat P . Find the equation of the normal to the curve at P .

2. (a) In a triangle whose sides are 5, 12, and 13, �nd the length of thebisector of the larger acute angle.

(b) This large rectangle has been cut into eleven squares of varioussizes. The smallest square has an area of 81 cm2. Find the dimen-sions of the large rectangle.

3. Solve the following system of equations for x and y, where x, y 2 R:

x3 � y3 = 35 , xy2 � yx2 = 30 .

298

4. The combined volume of two cubes with integral sides is equal to thecombined length of all their edges. Find the dimensions of all cubessatisfying these conditions.

5. A car at an intersection is heading west at 24 m/s. Simultaneously, asecond car, 84 m north of the �rst car, is travelling directly south at10 m/s. After 2 seconds:

(a) Find the rate of change of the distance between the 2 cars.

(b) Using vector methods, determine the velocity of the �rst car rela-tive to the second.

(c) Explain fully why the magnitude of the vector in (b) is not neces-sarily the same as the rate of change in (a).

PART B

1. Prove that

sin 1� + sin3� + sin5� + � � �+ sin97� + sin99� =sin2 50�

sin1�.

2. Determine the nth term and the sum of n terms for the series

3 + 5 + 10 + 18 + 29 + � � � .

3. A circle has equation x2+y2 = 1. Lines with slope 1

2and�1

2are drawn

through C(0; 0) to form a sector of the circle. Find the dimensionsof the rectangle of maximal area which can be inscribed in this sectorgiven that two sides of the rectangle are parallel to the y{axis and therectangle is entirely below the x-axis.

4. In triangle ABC, angles A, B, and C are in the ratio 4 : 2 : 1. Provethat the sides of the triangle are related by the equality 1

a+ 1

b= 1

c.

5. Prove that 7 +

p37

2

!n+

7�

p37

2

!n� 1

is divisible by 3 for all n 2 N.

299

Derangements and Stirling Numbers

Naoki Satostudent, Yale University

In this article, we introduce two important classes of combinatorialnumbers which appear frequently: derangements and Stirling numbers. Wealso hope to emphasize the importance and usefulness of basic counting prin-ciples.

We can think of a permutation on n objects as a 1-1 function � fromthe set f1, 2, : : : , ng to itself. For example, for n = 3, the map � given by�(1) = 1, �(2) = 3, and �(3) = 2 is a permutation on 3 objects, namelythe elements of f1; 2; 3g; a permutation essentially re-arranges the elements.Note that there are n! di�erent permutations on n objects. Then, a derange-ment is a permutation � which has no �xed points; that is, �(i) 6= i for alli = 1, 2, : : : , n. Alternatively, if we think of a permutation on n objects asa distribution of n letters to n corresponding envelopes, then a derangementis a permutation where no letter is inserted into the correct correspondingenvelope. Let Dn denote the number of derangements on n objects. Thenatural question to ask is, what is the formula for Dn?

Problems.

1. Write down all permutations on 2, 3, and 4 elements. How many ofthese are derangements? Is there a systematic way of writing down allpermutations on n elements?

2. What is the total number of �xed points over all permutations on n

elements? You should be able to guess the answer from small cases.

We immediately see thatD1 = 0 andD2 = 1 (by convention, D0 = 0).Assume that n � 2; we will derive a recurrence relation for theDn, by divid-ing all derangements on n elements into two categories, and then countingthe number in each category. Let � be a derangement on the n elements1, 2, : : : , n. Let k = �(1), so k 6= 1 since � is a derangement. There aretwo cases: �(k) = 1 or �(k) 6= 1. Let An be the number of derangementsfor which �(k) = 1, and let Bn be the number of derangements for which�(k) 6= 1.

If �(k) = 1, then � swaps the elements 1 and k, leavingwhat � does tothe remaining elements 2, 3, : : : , k�1, k+1, : : : , n to be considered. Since� re-arranges these elements, it too acts as a derangement on these n � 2

elements, of which there are Dn�2. Going back, there are n � 1 possiblevalues for k, as 1 is omitted, so An = (n� 1)Dn�2.

Copyright c 1999 CanadianMathematical Society

300

If �(k) 6= 1, then let � be the permutation which swaps 1 and k, andleaves everything else �xed. Recall that for maps f and g, f � g denotes thecomposition of the two maps; that is, (f � g)(x) = f(g(x)). In this case, iff and g are permutations, then f � g is the permutation which arises fromapplying g, then applying f . Consider the permutation � � �. We see that(� � �)(1) = �(�(1)) = �(k) 6= 1, (� � �)(k) = �(�(k)) = �(1) = k,and for all other elements i, (� � �)(i) = �(�(i)) = �(i) 6= i, since �is a derangement. Thus, � � � is a permutation which �xes the element k,and which deranges all n� 1 others (this is why we composed � with �), ofwhich there are Dn�1. Again, there are n� 1 possible values of k (again, 1is omitted), so Bn = (n� 1)Dn�1. Therefore,

Dn = An +Bn

= (n� 1)Dn�2 + (n� 1)Dn�1

= (n� 1)(Dn�1 +Dn�2) .

Problems.

3. Show that in general, for permutations � and �, � � � 6= � � �. Canyou determine when � � � = � � �?

4. For a positive integer k, let �k denote the permutation � composedwith itself k times; that is,

�k = � � � � � � � � �| {z }k

.

Prove that for any permutation �, there exists a positive integer k suchthat �k = 1. Moreover, if � is a permutation on n elements, then�n! = 1. Here, 1 stands for the identity permutation, the permutationwhich takes every element to itself.

5. Classify all permutations � on n elements such that � � � = �2 = 1.

6. Prove that for any permutation �, there exist unique permutations �and such that � � � = � � = 1. Problem 5 shows that � = ispossible; is this true in general? Hint: Apply Problem 4!

The next few terms in the sequence are thenD3 = 2,D4 = 9,D5 = 44,etc. We can use this recurrence to derive an explicit formula for Dn. By therelation,

Dn � nDn�1 = �Dn�1 + (n� 1)Dn�2

= � (Dn�1 � (n� 1)Dn�2)

= (�1)2 (Dn�2 � (n� 2)Dn�3)

= � � �= (�1)n�2(D2 � 2D1) = (�1)n .

301

Hence,

Dn

n!=

Dn�1

(n� 1)!+

(�1)n

n!

=Dn�2

(n� 2)!+

(�1)n�1

(n� 1)!+

(�1)n

n!= � � �=

1

0!�

1

1!+

1

2!� � � �+ (�1)n

1

n!

===) Dn = n!

�1

0!�

1

1!+

1

2!� � � �+ (�1)n

1

n!

�.

Remember this expression; as mentioned above, it comes up a lot! Wenote in passing that

1

e=

1Xk=0

(�1)k1

k!=

1

0!�

1

1!+

1

2!� � � � ,

so that

Dn �n!

e

for large n. We also note that it is possible to derive the formula for Dn

using the Principle of Inclusion-Exclusion. We �nish o� derangements withtwo quick problems.

Problem. Let n be a positive integer. Show that

nXk=0

�n

k

�Dk = n! .

Solution. It looks as if we might have to plug away at some algebra,but a combinatorial approach is much more natural and painless. Note thatthe RHS, n!, is simply the number of permutation on n objects, and this isa cue. What is the number of permutations which derange k elements, oralternatively, which �x n� k elements? We �rst choose n�k elements, andderange the rest, which there are Dk ways of doing, for a total of�

n

n� k

�Dk =

�n

k

�Dk .

The result follows from summing over k (since every permutation derangesk elements for some k).

In general, ifx0, x1, x2, : : : and y0, y1, y2, : : : are sequences satisfying

nXk=0

�n

k

�xk = yn ,

302

then it is possible to recover fxng from fyng, via

xn =

nXk=0

(�1)n�k�n

k

�yk .

Indeed, for yn = n!, we obtain that xn = Dn, since

nXk=0

(�1)n�k�n

k

�k! =

nXk=0

(�1)n�kn!

(n� k)!

= n!

�1

0!�

1

1!+

1

2!� � � �+ (�1)n

1

n!

�= Dn .

Problem. Let n be a positive integer. Prove that

nXk=0

k

�n

k

�Dk = (n� 1) � n! .

Solution. Left as an exercise for the reader. (Follow the same type ofreasoning as the previous problem.) Hint: Combinatorially, what does theLHS represent?

We now draw our attention to the Stirling numbers, in particular thoseof the second kind (as opposed to the �rst kind for those in suspense). Thesearise in the following situation: Suppose that we have n distinguishable ballsto distribute among k indistinguishable boxes, and no box can be empty.How many such distributions are there? First, consider the case where theboxes are distinguishable.

If the boxes were allowed to be empty, there would be kn such dis-tributions; assign a box to each ball. Following the Principle of Inclusion-Exclusion, we subtract the number of distributions with at least 1 box empty,which is �

k

k � 1

�(k� 1)n ,

since we must choose out of k � 1 boxes for each ball. We then add thenumber of distributions with at least 2 boxes empty, which is�

k

k � 2

�(k� 2)n ,

and so on. Hence, the total number of distributions is

nXi=0

(�1)i�k

i

�(k� i)n .

303

Then, for the case where the boxes are indistinguishable, we may \removethe labels" of the boxes, and divide by a factor of k!, to obtain the numberof distributions of the original problem:

S(n; k) :=1

k!

nXi=0

(�1)i�k

i

�(k� i)n .

These are the Stirling numbers of the second kind. We list the �rst few here:

n S(n; k); k = 1; 2; : : : ; n

1 1

2 1 1

3 1 3 1

4 1 7 6 1

5 1 15 25 10 1

An important property of these Stirling numbers, one that may be no-ticeable in the table, is the following:

S(n; k) = S(n� 1; k� 1) + kS(n� 1; k) .

This rule easily allows us to generate further rows in the table. We stressthat although we derived a formula for S(n; k) above, it is in general bet-ter to consider the combinatorial signi�cance of this number. So, we give acombinatorial proof here, and leave an algebraic proof for the reader.

Assume that the balls are labelled 1 through n (recall that they are dis-tinguishable). Consider balln. Either it is in a box all by itself, or with others.For the �rst case, the number of distributions is simplyS(n � 1; k � 1), as we can add one box containing ball n to a distributionof n� 1 balls among k � 1 boxes. For the second case, temporarily removeball n. This leaves n � 1 balls among k non-empty boxes, of which thereare S(n� 1; k) distributions. We can then add ball n to any of the k boxes,giving rise to k distinct distributions. Hence,

S(n; k) = S(n� 1; k� 1) + kS(n� 1; k) .

Problem. Show that

S(n; k) =X

1a12a2 � � � kak ,

where the sum is taken over all�n�1k�1

�decompositions of n� k into k non-

negative integers a1, a2, : : : , ak: a1+a2+� � �+ak = n�k. (A decompositionof a non-negative integer N into m parts is a sequence of m non-negativeintegers, the sum of which is N . So, 3 has 4 di�erent decompositions into 2parts: 3 = 0 + 3 = 1 + 2 = 2 + 1 = 3 + 0.)

Solution. The table above makes this virtually obvious, but we will eshout the details. Draw in arrows in the table, joining entries in consecutive

304

rows. An arrow from S(n � 1; k � 1) has weight 1 and an arrow fromS(n � 1; k) has weight k. By virtue of the identity proved above, anentry in the table is equal to the sum of the weights of the paths leadingto it, where the weight of a path is simply the product of the weights of thearrows in it (think of this as a tweaked Pascal's Triangle).

There are�n�1k�1

�paths from entry S(1;1) to S(n; k), and the weight of a

path is precisely the term in the given sum. The result follows from summingover all paths, which clearly gives the given expression.

Problem. Show that

S(n+ 1; k) =

nXi=0

�n+ 1

i

�S(i; k � 1) .

Solution. In a distribution of n+1 balls among k boxes, select one boxas \blue". Let i be the number of balls in the blue box, so 1 � i � n + 1.From the n+1 balls, we may choose i to be in the blue box, leavingn�i+1

to be distributed among k� 1 boxes. There are�n+ 1

i

�S(n� i+ 1; k� 1) =

�n+ 1

n� i+ 1

�S(n� i+ 1; k� 1)

such distributions in this case. The result follows from summing over i.

Problems.

1. Let n be a positive integer. Prove that

nXk=0

k

�n

k

�Dk = (n� 1) � n! .

2. Here is an examplewhich illustrates the Principle of Inclusion-Exclusion:Let A, B, and C be subsets of a set S. Let A denote the complementof A. Prove that

jA\B\Cj = jSj�jAj�jBj�jCj+jA\Bj+jA\Cj+jB\Cj�jA\B\Cj .

For example, let S be the set of all distributions of n distinguishableballs among 3 distinguishable boxes, letA be the subset of distributionswith box 1 empty, and so on. ThenA is the subset of distributions withbox 1 containing at least one ball, and so on, so A\B\C is the subsetof distributions with all three boxes containing at least one ball. Nowdetermine what the formula above turns into. This is a useful formula,because the terms such as jAj and jA \ Bj are easy to compute. Thisformula also generalizes to any number of subsets.

3. By using the derived formula for S(n; k), show that

S(n; k) = S(n� 1; k� 1) + kS(n� 1; k) .

305

4. Let n and k be positive integers, with k < n. Verify that there are�n�1k�1

�decompositions of n� k into k parts. If we restrict the parts to

be positive integers, then how many decompositions are there of n intotal?

5. De�ne a sequence of polynomials fpn(x)g as follows: p1(x) = p(x)

is given, and pn+1(x) = xp0n(x). Find a closed formula for pn(x), interms of p(x) and n.

6. (a) Let n be a positive integer. Prove that the following is an identityin x:

xn =

nXk=0

k!S(n;k)

�x

k

�.

(b) Prove that

1n + 2n + � � �+mn =

nXk=0

k!S(n;k)

�m+ 1

k+ 1

�.

Naoki SatoDepartment of Mathematics

Yale UniversityNew HavenConnecticut

USAsato-naoki@math.yale.edu

306

PROBLEMS

Problem proposals and solutions should be sent to Bruce Shawyer, Departmentof Mathematics and Statistics, Memorial University of Newfoundland, St. John's,

Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution,

together with references and other insights which are likely to be of help to the edi-tor. When a submission is submitted without a solution, the proposer must include

su�cient information on why a solution is likely. An asterisk (?) after a numberindicates that a problem was submitted without a solution.

In particular, original problems are solicited. However, other interesting prob-

lems may also be acceptable provided that they are not too well known, and refer-

ences are given as to their provenance. Ordinarily, if the originator of a problem canbe located, it should not be submitted without the originator's permission.

To facilitate their consideration, please send your proposals and solutions

on signed and separate standard 812"�11" or A4 sheets of paper. These may

be typewritten or neatly hand-written, and should be mailed to the Editor-in-

Chief, to arrive no later than 1 January 2000. They may also be sent by email to

crux-editors@cms.math.ca. (It would be appreciated if email proposals and solu-tions were written in LATEX). Graphics �les should be in epic format, or encapsulated

postscript. Solutions received after the above date will also be considered if there

is su�cient time before the date of publication. Please note that we do not acceptsubmissions sent by FAX.

2446. Correction: Proposed by Catherine Shevlin, Wallsend upon

Tyne, England.A sequence of integers, fang with a1 > 0, is de�ned by

an+1 =

8>><>>:

an2

if a(n) � 0 (mod 4),3an + 1 if a(n) � 1 (mod 4),2an � 1 if a(n) � 2 (mod 4),an+1

4if a(n) � 3 (mod 4).

Prove that there is an integer m such that am = 1.

(Compare OQ.117 in OCTOGON, vol 5, No. 2, p. 108.)

2451. Proposed by Michael Lambrou, University of Crete, Crete,

Greece.Construct an in�nite sequence, fAng, of in�nite subsets of N with the

following properties:

(a) the intersection of any two distinct sets An and Am is a singleton;

(b) the singleton in (a) is a di�erent one if at least one of the distinct setsAn, Am, is changed (so the new pair is again distinct);

(c) every natural number is the intersection of (exactly) one pair of distinctsets as in (a).

307

2452. Proposed by Antal E. Fekete, Memorial University of New-

foundland, St. John's, Newfoundland.Establish the following equalities:

(a)

1Xn=1

(2n+ 1)2

(2n+ 1)!=

1Xn=1

(2n+ 2)2

(2n+ 2)!.

(b)1Xn=1

(�1)n(n+ 1)3

(n+ 1)!=

1Xn=1

(�1)n(n+ 1)4

(n+ 1)!.

(c)

1Xn=1

(�1)n(n+ 1)6

(n+ 1)!=

1Xn=1

(�1)n(n+ 1)7

(n+ 1)!.

2453. Proposed by Antal E. Fekete, Memorial University of New-

foundland, St. John's, Newfoundland.Establish the following equalities:

(a)

1Xn=1

(�1)n(2n+ 1)3

(2n+ 1)!= �3

1Xn=1

(�1)n1

(2n+ 1)!.

(b)

1Xn=1

(�1)n(2n)3

(2n)!= �3

1Xn=1

(�1)n1

(n+ 1)!.

(c)

1Xn=1

(�1)n(2n+ 1)2

(2n+ 1)!

!2

+

1Xn=1

(�1)n(2n)2

(2n)!

!2

= 9 .

2454. Proposed by Gerry Leversha, St. Paul's School, London, Eng-

land.

Three circles intersect each other orthogonally at pairs of points A andA0, B and B0, and C and C0. Prove that the circumcircles of 4ABC and4AB0C0 touch at A.

2455. Proposed by Gerry Leversha, St. Paul's School, London, Eng-

land.

Three equal circles, centred at A, B and C intersect at a common pointP . The other intersection points are L (not on circle centre A), M (noton circle centre B), and N (not on circle centre C). Suppose that Q is thecentroid of 4LMN , that R is the centroid of 4ABC, and that S is thecircumcentre of4LMN .

(a) Show that P , Q, R and S are collinear.

(b) Establish how they are distributed on the line.

2456. Proposed by Gerry Leversha, St. Paul's School, London, Eng-

land.

Two circles intersect orthogonally at P . A third circle touches them atQ andR. LetX be any point on this third circle. Prove that the circumcirclesof 4XPQ and4XPR intersect at 45�.

308

2457. Proposed by Gerry Leversha, St. Paul's School, London, Eng-

land.

In quadrilateral ABCD, we have \A + \B = 2� < 180�, andBC = AD. Construct isosceles triangles DCI, ACJ and DBK, where I,J andK are on the other side of CD from A, such that \ICD = \IDC =

\JAC = \JCA = \KDB = \KBD = �.

(a) Show that I, J andK are collinear.

(b) Establish how they are distributed on the line.

2458. Proposed by Nikolaos Dergiades, Thessaloniki, Greece.Let ABCD be a quadrilateral inscribed in the circle centre O, radius

R, and letE be the point of intersection of the diagonalsAC andBD. Let Pbe any point on the line segment OE and letK, L,M , N be the projectionsof P on AB, BC, CD, DA respectively.

Prove that the linesKL, MN , AC are either parallel or concurrent.

2459. Proposed by Vedula N. Murty, Visakhapatnam, India, modi-

�ed by the editors.Let P be a point on the curve whose equation is y = x2. Suppose that

the normal to the curve at P meets the curve again at Q. Determine theminimal length of the line segment PQ.

2460. Proposed by V�aclav Kone �cn �y, Ferris State University, Big

Rapids, Michigan, USA.

Let y(x) =3

qx+

px2 � 1 +

3

qx�

px2 � 1 for 0 � x � 1.

(a) Show that y(x) is real valued.

(b) Find an in�nite sequence fxng1n=0 such that y (xn) can be expressed interms of square roots only.

2461. Proposed byMohammed Aassila, CRM,Universit �e deMontr �eal,

Montr �eal, Qu �ebec.Suppose that x0, x1, : : : , xn are integers which satisfy x0 > x1 >

: : : > xn. Let

F (x) =

nXk=0

akxn�k , ak 2 R , a0 = 1 .

Prove that at least one of the numbers jF (xk)j, (k = 0, 1, : : : , n) is

greater thann!

2n.

2462. Proposed by Vedula N. Murty, Visakhapatnam, India.If the angles, A, B, C of4ABC satisfy

cosA sinA

2= sin

B

2sin

C

2,

prove that 4ABC is isosceles.

309

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased to

consider for publication new solutions or new insights on past problems.

2324. [1998: 109, 1999: 50] Proposed by Joaqu��n G �omez Rey, IES

Luis Bu ~nuel, Alcorc �on, Madrid, Spain.

Find the exact value of1Xn=1

1

un, where un is given by the recurrence

un = n! +

�n� 1

n

�un�1 ;

with the initial condition u1 = 2.

Solution by Charles R. Diminnie, Angelo State University, San Angelo,

TX, USA; Walther Janous, Ursulinengymnasium, Innsbruck, Austria; and

Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario.

We �rst show that un = n! + (n� 1)!. This is true for n = 1, sinceu1 = 2. Suppose that un = n! + (n� 1)! for some n � 1. Then

un+1 = (n+ 1)! +n

n+ 1(n! + (n� 1)!)

= (n+ 1)! +n

n+ 1(n� 1)!(n+ 1)

= (n+ 1)! + n! ,

completing the induction.

Hence, for allm � 1, we have

mXn=1

1

un=

mXn=1

1

n! + (n� 1)!=

mXn=1

1

(n� 1)!(n+ 1)

=

mXn=1

n+ 1� 1

(n+ 1)!=

mXn=1

�1

n!�

1

(n+ 1)!

�= 1�

1

(m+ 1)!,

from which it follows that

1Xn=1

1

un= lim

m!1

mXn=1

1

un= 1.

Also solved by the proposer.

2339. [1998: 234] Proposed by Toshio Seimiya, Kawasaki, Japan.

A rhombus ABCD has incircle �, and � touches AB at T . A tangentto � meets sides AB, AD at P , S respectively, and the line PS meets BC,CD at Q, R respectively. Prove that

310

(a)1

PQ+

1

RS=

1

BT,

and

(b)1

PS�

1

QR=

1

AT:

Solution by Michael Lambrou, University of Crete, Crete, Greece.

If PS meets � at U we show, slightly generalizing (and correcting) thesituation, that, depending on the position of U , we have

1

PQ�

1

RS= �

1

BT,

1

PS�

1

QR= �

1

AT,

with an appropriate choice of � each time.

Using as coordinate axes the two (perpendicular) diagonals, meeting atO say, we may assume that A, B, C, D have coordinates A(a; 0), B(0; b),C(�a; 0), D(0;�b) respectively, where a, b > 0: The radius of � is OTwhich, being perpendicular to AB is the altitude of OAB. Writing OT = h

we have h � AB = OA � OB = ab, and the coordinates of U are of the form(h cos �; h sin �). As PS ? OU , the equation of PS is clearly

y sin� = �x cos � + h .

Line AB is x

a+ y

b= 1, so the coordinates of P , being on AB and PS, are

easily seen to be

(xP ; yP ) =1

b sin � � a cos �

�a(b sin� � h); b(h� a cos �)

�. (1)

Similarly (or quicker, by replacing a by �a) we �nd the coordinates of Q as

(xQ; yQ) =1

b sin� + a cos �

�� a(b sin� � h); b(h+ a cos �)

�.

(2)

Assume now that the position of U is such that PS cuts, say, AB andBC internally (the rest of the cases follow by a trivial adaptation of ourargument here) and so the coordinates of P are both positive [and Q has anegative x{coordinate and a positive y{coordinate]. By (1) we haveb sin� > h > a cos �. Thus

PQ =

q(xP � xQ)2 + (yP � yQ)2 =

2ab(b sin� � h)

(b sin�)2 � (a cos �)2.

Similarly

RS =2ab(b sin� + h)

(b sin�)2 � (a cos �)2,

311

and so

1

RS�

1

PQ=

(b sin �)2 � (a cos �)2

2ab

�1

b sin � + h�

1

b sin � � h

�

= �(b sin�)2 � (a cos �)2

2ab�

2h

(b sin �)2 � h2.

Writing h2 = a2b2

(AB)2= a2b2

a2+b2, this simpli�es to� (a2+b2)h

ab3which equals� 1

BT

as BT � AB = OB2 = b2, as required.

The proof of 1

PS� 1

QR= � 1

ATis similar and routine.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK.

Bradley and Lambrou were the only readers who recognized that the problem was incor-

rect as stated. There were eleven partial solutions.

2346. [1998: 236] Proposed by Juan-Bosco Romero M�arquez, Uni-

versidad de Valladolid, Valladolid, Spain.

The angles of4ABC satisfy A > B � C. Suppose that H is the footof the perpendicular from A to BC, that D is the foot of the perpendicularfrom H to AB, that E is the foot of the perpendicular from H to AC, thatP is the foot of the perpendicular from D to BC, and that Q is the foot ofthe perpendicular from E to AB.

Prove that A is acute, right or obtuse according as AH �DP �EQ ispositive, zero or negative.

Solution by Nikolaos Dergiades, Thessaloniki, Greece.

As noted by the proposer, this is a generalization of a problem he pro-posed in College Mathematics Journal 28, no. 2, March 1997, 145-6.

Let S = AH �DP � EQ. Then, since \DHA = \PDH = \B, wehave

DP = DH cosB = AH cos2B .

Similarly EQ = AH cos2C, so,

S = AH � AH cos2B � AH cos2C = AH(1� cos2B � cos2C)

= AH(sin2B � cos2C) = AH

�1� cos 2B

2�

1 + cos 2C

2

�= �AH cos(B + C) cos(B �C) ,

or S = AH cosA cos(B � C), where cos(B �C) > 0, and hence

A is acute if S > 0 , A is right if S = 0 , A is obtuse if S < 0 .

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, and MAR �IA

ASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; CHRISTOPHER

312

J. BRADLEY, Clifton College, Bristol, UK; GORAN CONAR, student, Gymnasium Vara�zdin,

Vara�zdin, Croatia; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; VJEKOSLAV

KOVA �C, student, University of Zagreb, Croatia; MICHAEL LAMBROU, University of Crete,

Crete, Greece; GERRY LEVERSHA, St. Paul's School, London, England; TOSHIO SEIMIYA, Ka-

wasaki, Japan; JEREMY YOUNG, student, Nottingham High School, Nottingham, UK; and the

proposer.

2348. [1998: 236] Proposed by D.J. Smeenk, Zaltbommel, the Neth-

erlands.Without the use of trigonometrical formulae, prove that

sin (54�) = 1

2+ sin (18�) :

I. Solution by Richard I. Hess, Rancho Palos Verdes, California, USA.

From Figure 1 below where we assume that AC = 1 we have BC = sin54�,AM = MC = MB = BN = 1

2, \MBN = 36�, \MBQ = \QBN =

18�. ThusMQ = 1

2sin 18�.

Now \CMN = 36� which implies CN = MN = 2MQ = sin18�.Since BC = BN + CN we obtain sin54� = 1

2+ sin18�.

A

BC N

M

Q

54�

54�

72�

72�

36�

36�

18�

18�

Figure 1.

II. Solution by Nikolaos Dergiades, Thessaloniki, Greece.On the circle centred at O with radius r having a diameter AD we take thepoints B and C such that arcAB = arcBC = 36� (see Figure 2 below).Then arcCD = 108�, so that

AB = BC = 2r sin 18� , CD = 2r sin54� .

Since \BOA = \CDA, we see that BO k CD. Choosing E on CD suchthat BE k AD, we see that the quadrilateral BEDO is a rhombus; thusED = r. Since \CEB = \CDA = 36�, we have \CBA = 144�,\OBA = \OBC = 72�, and \EBO = 36�. It then follows that\CBE = 36� and CE = CB = 2r sin 18�. Thus

2r sin 54� = CD = CE + ED = 2r sin 18� + r ,

or sin54� =1

2+ sin18� .

313

AO

D

B

C

E

Figure 2.

III. Solution by Jeremy Young, student, Nottingham High School, Eng-

land.

Set x = sin18� and set y = sin54�. First consider the triangle inFigure 3 below, where we set AC = 1. Then BC = x and CD = 1. ThusAB =

p1� x2 and AD =

p1� x2 + (1 + x)2 =

p2(1 + x). Then

sin54� = y =1+ xp2(1 + x)

=) 2y2� 1 = x .

A

B C D

1p1� x2

x 1

p2(1 + x)

18�!

36�

108�

72�

36�

E

F G H

K

L

1

p1� y2

y 1�y

1

18�

18�54

�

72�

72�

36�

72�

18�

Figure 3. Figure 4.

In Figure 4 above where EG = 1, we have FG = y, EK = 1,GH = 1� y, and EF =

p1� y2 = KH. Then

LG = x =1

2

p(1� y)2 + 1� y2 =

1

2

p2� 2y .

which yields 1� 2x2 = y. Adding this to the previous result (2y2 � 1 = x)gives

2(y+ x)(y� x) = y + x ,

2(y� x) = 1 , since x, y > 0 =) y + x 6= 0,

y = x+1

2.

Also solved by SAM BAETHGE, Nordheim, Texas, USA; FRANCISCO BELLOT ROSADO,

I.B. Emilio Ferrari, and MAR �IA ASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo Cano, Val-

ladolid, Spain; M. BENITO and E. FERN �ANDEZ, Logro ~no, Spain; CHRISTOPHER J. BRADLEY,

314

Clifton College, Bristol, UK; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX,

USA; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA; DIANE and ROY

DOWLING, University ofManitoba, Winnipeg, Manitoba; RICHARDEDEN, student, Ateneo de

Manila University, Philippines; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;

ANGEL JOVAL ROQUET, Spain; GEOFFREY A. KANDALL, Hamden, CT; V �ACLAV KONE �CN �Y,

Ferris State University, Big Rapids, Michigan, USA; VJEKOSLAV KOVA �C, student, University of

Zagreb, Croatia; MICHAEL LAMBROU, University of Crete, Crete, Greece; KEE-WAI LAU, Hong

Kong; GERRY LEVERSHA, St. Paul's School, London, England; VEDULA N. MURTY, Dover,

PA, USA; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; BOB PRIELIPP, University

of Wisconsin{Oshkosh, Wisconsin, USA; TOSHIO SEIMIYA, Kawasaki, Japan; J. SUCK, Essen,

Germany; PANOS E. TSAOUSSOGLOU, Athens, Greece; PAUL YIU, Florida Atlantic University,

Boca Raton, Florida, USA; and the proposer.

2351. [1998: 302] Proposed by Paul Yiu, Florida Atlantic University,

Boca Raton, Florida, USA.

A triangle with integer sides is called Heronian if its area is an integer.

Does there exist a Heronian triangle whose sides are the arithmetic,geometric and harmonic means of two positive integers?

Solution by Michael Lambrou, University of Crete, Crete, Greece.

We show, slightly generalizing the given situation, that no triangle withsides a,

pac, c, with a, c 2 N can have integral area. Indeed, if the area

� 2 N we would have by Heron's formula:

�2 =1

16(a+

pac+ c)(a�

pac + c)(a+

pac� c)(�a+

pac+ c)

=1

16

�4a2c2 � (a2 + c2 � ac)2

�.

If (a; c) = t so that a = pt, c = qt for some p, q 2 N with (p; q) = 1 weobtain

4�

t2=p4p2q2� (p2 + q2� pq)2 .

But the left hand side is rational; so 4p2q2 � (p2 + q2 � pq)2 must be aninteger perfect square, say T 2. But this is impossible because p2 + q2 � pq

is odd (as p, q are not both even) and so

T 2 = 4p2q2 � (p2 + q2 � pq)2 � 0� 1 � 3 (mod 4) ,

giving a contradiction, as no square is congruent to 3 modulo 4. This com-pletes the proof that � =2 N.

Also solved by DUANE BROLINE, Eatsern Illinois University, Charleston, Illinois.

USA; THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; RICHARD

I. HESS, Rancho Palos Verdes, California, USA; JEREMY YOUNG, student, Nottingham High

School, Nottingham, UK; and the proposer.

315

2352. [1998: 302] Proposed by Christopher J. Bradley, Clifton Col-

lege, Bristol, UK.

Determine the shape of4ABC if

cosA cosB cos(A�B) + cosB cosC cos(B �C)

+ cosC cosA cos(C � A) + 2cosA cosB cosC = 1 .

Solution by Nikolaos Dergiades, Thessaloniki, Greece.Since cosA cosB � sinA sinB = cos(A + B) = � cosC, we have

(cosA cosB + cosC)2=�1� cos2A

� �1� cos2B

�, or

cos2A+ cos2B + cos2C + 2 cosA cosB cosC = 1 . (1)

Also,

cos(A+ B) cos(A�B) = cos2A cos2B � sin2A sin2B

= cos2A cos2B ��1� cos2A

� �1� cos2B

�= cos2A+ cos2B � 1 . (2)

Using (2), we have

cosA cosB cos(A�B) = 1

2(cos(A� B) + cos(A+ B)) cos(A� B)

= 1

2

�cos2(A�B) + cos2A+ cos2B � 1

�= 1

2

�cos2A+ cos2B � sin2(A�B)

�.

Hence, the given equality is equivalent to

cos2A+ cos2B + cos2C + 2 cosA cosB cosC

�1

2

�sin2(A� B) + sin2(B � C) + sin2(C � A)

�= 1 ,

which, in view of (1), becomes

sin2(A� B) + sin2(B � C) + sin2(C � A) = 0 .

Hence, A = B = C, which means that 4ABC is equilateral.

Also solved by MICHEL BATAILLE, Rouen, France; THEODORE CHRONIS, student,

Aristotle University of Thessaloniki, Greece; C. FESTRAETS-HAMOIR, Brussels, Belgium;

WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; JUN-HUA HUANG, the Mid-

dle School Attached To Hunan Normal University, Changsha, China; MICHAEL LAMBROU,

University of Crete, Crete, Greece; KEE-WAI LAU, Hong Kong; GERRY LEVERSHA, St. Paul's

School, London, England; VEDULA N. MURTY, Visakhapatnam, India; HEINZ-J �URGEN

SEIFFERT, Berlin, Germany; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel,

the Netherlands; ARAMTANGBOONDOUANGJIT, Carnegie Mellon University, Pittsburgh, PA,

USA; PARAYIOU THEOKLITOS, Limassol, Cyprus; PANOS E. TSAOUSSOGLOU, Athens, Greece;

PAUL YIU, Florida Atlantic University, Boca Raton, Florida, USA; JEREMY YOUNG, student,

Nottingham High School, Nottingham, UK; and the proposer. There was also one incorrect

solution.

316

Most solvers showed that the given equality is equivalent to eitherP

cos2(A�B) = 3,

or toP

cos 2(A � B) = 3, both of which are clearly equivalent toP

sin2(A � B) = 0,

obtained in the solution above.

Four solvers derived the equalityQcos(A � B) = 1, from which the conclusion also

follows immediately. The solution given above is self-contained, and does not use any known

identity (for example,P

cos(2A) = �1�4QcosA, which was used by a few solvers) beyond

the elementary formula transforming product into sum. Lambrou obtained the slightly stronger

result that �XcosA cosB cos(A�B)

�+ 2 cosA cosB cosC � 1 ,

with equality holding if and only if4ABC is equilateral.

2353. [1998: 302] Proposed by Christopher J. Bradley, Clifton Col-

lege, Bristol, UK.

Determine the shape of4ABC if

sinA sinB sin(A�B) + sinB sinC sin(B � C)

+ sinC sinA sin(C � A) = 0 .

Solutionby Gerry Leversha, St. Paul's School, London, England (slightlymodi�ed by the editor).

Since

sinA sinB sin(A� B)

= sin2A sinB cosB � sin2B sinA cosA

= 1

4((1� cos(2A)) sin(2B)� (1� cos(2B)) sin(2A))

= 1

4((sin(2B)� sin(2A)) + sin(2(A� B))) ,

the given equality is equivalent toXsin(2(A� B)) = 0 . (1)

Using elementary formulae transforming sums into products, we have

sin(2(A�B)) + sin(2(B� C)) = 2 sin(A�C) cos(A+ C � 2B) ,

and thus

sin(2(A� B)) + sin(2(B � C)) + sin(2(C � A))

= 2 sin(A� C) cos(A+ C � 2B) + 2 sin(C � A) cos(C �A)

= 2 sin(C �A) (cos(C �A)� cos(A+ C � 2B))

= �4 sin(A� B) sin(B � C) sin(C � A) . (2)

From (1) and (2), we see that the given equality is equivalent to

sin(A�B) sin(B �C) sin(C � A) = 0 ,

317

which holds if and only if at least two of A, B, C are equal to each other.Therefore, the given equality holds if and only if4ABC is isosceles.

Also solved by MICHEL BATAILLE, Rouen, France; THEODORE CHRONIS, student,

Aristotle University of Thessaloniki, Greece; NIKOLAOS DERGIADES, Thessaloniki, Greece;

C. FESTRAETS-HAMOIR, Brussels, Belgium; PETER HURTHIG, Columbia College, Burnaby,

BC; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; JUN-HUA HUANG, the

Middle School Attached To Hunan Normal University, Changsha, China; V �ACLAV KONE �CN �Y,

Ferris State University, Big Rapids, Michigan, USA; MICHAEL LAMBROU, University of

Crete, Crete, Greece; KEE-WAI LAU, Hong Kong; VEDULA N. MURTY, Visakhapatnam,

India; HEINZ-J �URGEN SEIFFERT, Berlin, Germany; TOSHIO SEIMIYA, Kawasaki, Japan;

ARAM TANGBOONDOUANGJIT, Carnegie Mellon University, Pittsburgh, PA, USA; PARAYIOU

THEOKLITOS, Limassol, Cyprus; PAUL YIU, Florida Atlantic University, Boca Raton, Florida,

USA; JEREMY YOUNG, student, Nottingham High School, Nottingham, UK; and the proposer.

There were also two incorrect solutions.

The solution given above is interesting since it shows that the conclusion A = B or

B = C or C = A does not depend on the assumption that A+B +C = �; that is, that A,

B and C are the three angles of a triangle.

2354. [1998: 302] Proposed by Herbert G �ulicher, Westfalische

Wilhelms-Universit�at, M�unster, Germany.In triangle P1P2P3, the line joining Pi�1Pi+1 meets a line �j at the

point Si;j (i; j = 1; 2; 3, all indices taken modulo 3), such that all the pointsSi;j , Pk are distinct, and di�erent from the vertices of the triangle.

1. Prove that if all the points Si;i [note the correction] are non-collinear,then any two of the following conditions imply the third condition:

(a)P1S3;1

S3;2P2

�P2S1;2

S1;3P3

�P3S2;3

S2;1P1

= �1 ;

(b)S1;2S1;1

S1;1S1;3�S2;3S2;2

S2;2S2;1�S3;1S3;3

S3;3S3;2= 1 ;

(c) �1, �2, �3 are either concurrent or parallel.

2. Prove further that (a) and (b) are equivalent if the Si;i are collinear.

Here, AB denotes the signed length of the directed line segment [AB].

Solution by G �unter Pickert, Giessen, Germany.

De�ne Si = Sii and Qi = �iTSi�1Si+1. If �ijjSi�1Si+1 (so that Qi

is at in�nity), then replaceSi+1Qi

QiSi�1by �1.

(a) From Menelaus' Theorem applied to 4Si+1Si�1Pi and the line �i(1 = 1, 2, 3),

Si+1Qi

QiSi�1�Si�1Si�1;i

Si�1;iPi�PiSi+1;i

Si+1;iSi+1

= �1 . (1)

318

De�ne

a =

3Yi=1

PiSi�1;i

Si�1;i+1Pi+1

, b =

3Yi=1

Si;i+1Si

SiSi;i�1, c =

3Yi=1

Si+1Qi

QiSi�1.

The given conditions (a), (b), (c) say (respectively) that a = �1,b = 1, and c = 1. [The value c = 1 is just Ceva's Theorem applied to4S1S2S3, and extended to allow the possibility that one or two of theQi are at in�nity.] From (1) we get

�1 = c �QSi�1Si�1;iQSi+1;iSi+1

�QPiSi+1;iQSi�1;iPi

= c � b � a�1 ,

so that �a = b � c .Two of c, b, �a are equal to 1 if and only if two of (a), (b), (c) hold, inwhich case the third product is 1 and the third condition holds too.

(b) If the Si are collinear thenQi = Si and c = 1; therefore (a) is equivalentto (b).

Also solved by JUN-HUA HUANG, the Middle School Attached To Hunan Normal Uni-

versity, Changsha, China; and the proposer.

Huang's solution is much the same as our featured solution. G �ulicher based his on his

problem 64 in Math. Semesterber. 40 (1992) 91{92.

2355. [1998: 303] Proposed by G.P. Henderson, Garden Hill, Camp-

bellcroft, Ontario.

For j = 1, 2, : : : , m, let Aj be non-collinear points with Aj 6= Aj+1.Translate every even-numbered point by an equal amount to get new pointsA02, A04, : : : , and consider the sequence Bj , where B2i = A02i andB2i�1 = A2i�1. The last member of the new sequence is either Am+1 orA0m+1 according asm is even or odd.

Find a necessary and su�cient condition for the length of the pathB1B2B3 : : : Bm to be greater than the length of the path A1A2A3 : : : Am

for all such non-zero translations.

CRUX 1985 [1994: 250; 1995: 280] provides an example of such a con-�guration. There, m = 2n, the Ai are the vertices of a regular 2n{gon andA2n+1 = A1.

Solution by the proposer.

(Editor's note: In the original submission the paths above were givenas B1B2 � � �Bm+1 and A1A2 � � �Am+1. Since this makes the upper boundson summations simpler we have opted to present the solution with theseendpoints, although logically it makes no di�erence.)

319

We use the same letter for a pointP and a vector�!P . Let the translation

in the problem be�!X . Then

�!B2k =

�!A02k =

�!A2k +

�!X , k = 1, 2, : : : , b(m+ 1)=2c ,

and the lengths of the new segments are����!B2 ��!B1

��� =

����!A02 ��!A1

��� =

����!X +�!A2 �

�!A1

��� ,����!B3 ��!B2

��� =����!A3 �

�!A02

��� =����!X �

��!A3 �

�!A2

���� ,� � � .

The lengths of the paths are

L0 =

mXj=1

�����!Bj+1 ��!Bj

��� =

mXj=1

����!X � (�1)j���!Aj+1 �

�!Aj

����and L =

mXj=1

�����!Aj+1 ��!Aj

��� .Now L0 is to be a minimum at

�!X =

�!0 . If we form the partial derivatives

of L0 with respect to the components of�!X and set

�!X =

�!0 , we get

mXj=1

(�1)j���!Aj+1 �

�!Aj

������!Aj+1 �

�!Aj

��� =�!0 . (1)

The minimum of a sum like L0 does not always occur at a point where thederivatives are zero. However, in this case we will prove that (1) actually isthe required condition.

That is, the sum of unit vectors along the odd segments�!A2 �

�!A1,�!

A4 ��!A3, : : : , is equal to the sum of unit vectors along the even segments.

The lengths of the segments are arbitrary provided they are greater than zero.It is only their directions that matter. In the case of the regular 2n{gon, both

sums are�!0 because each consists of unit vectors parallel to the sides of a

regular n{gon.

Set dj =�����!Aj+1 �

�!Aj

��� > 0 and�!Cj = (�1)j

���!Aj+1 �

�!Aj

�=dj ,

a unit vector parallel to the jth segment. Equation (1) becomes

mXj=1

�!Cj =

�!0 . (2)

The lengths of the paths are L0 =

mXj=1

����!X � dj�!Cj

��� and L =

mXj=1

dj .

320

To prove the necessity of (2), assume L0 � L for all�!X .

Set yj =����!X � dj

�!Cj

��� � dj . ThenmXj=1

yj � 0 . Squaring both sides of

����!X � dj�!Cj

��� = dj + yj and dividing by dj , we get

�!X2 =dj � 2

�!X �

�!Cj = 2yj + y2j=dj

�!X2

mXj=1

(1=dj)� 2�!X �

mXj=1

�!Cj = 2

mXj=1

yj +

mXj=1

(y2j=dj) � 0 .

When we set�!X =

mXj=1

�!Cj

� mXj=1

(1=dj) , this becomes

0@ mXj=1

�!Cj

1A

2

� 0 ,

and we have (2).

Given (2) we are to prove that L0 > L for all�!X 6=

�!0 . We have����!X � dj

�!Cj

��� =����!X � dj

�!Cj

��� ����!Cj��� � �����!X � dj�!Cj

���!Cj

��� ; (3)

L0 �mXj=1

�����!X � dj�!Cj

���!Cj

��� �������mXj=1

��!X � dj

�!Cj

���!Cj

������=

�������!X �

mXj=1

�!Cj �

mXj=1

dj

������ = L .

If L0 = L for some�!X 6=

�!0 , there is equality in (3) for all j. Then�!

X � dj�!Cj = cj

�!Cj , where cj 6= �dj ; thus

�!Cj =

�!X =(cj + dj) , and the

points are collinear.

All of the above is valid in a Euclidean space of any number of dimen-sions.

There were no other solutions submitted.

Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell

Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil

Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,

J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

321

THE ACADEMY CORNERNo. 26

Bruce ShawyerAll communications about this column should be sent to BruceShawyer, Department of Mathematics and Statistics, Memorial Universityof Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

Once again, we are pleased to present the Bernoulli Trials. Many thanks toChristopher Small for sending them to us.

The Bernoulli Trials 1999

Christopher G. Small & Ravindra MaharajUniversity of Waterloo

The Bernoulli Trials is an undergraduate mathematics competition heldannually at the University of Waterloo. Students competing in the BernoulliTrials proceed through a sequence of rounds, with a new mathematical prob-lem for each round. The mathematical problems are presented as proposi-tions, and the students are given ten minutes to decide upon the correctnessof each proposition. The double-knockout format means that students maymake one mistake, but are eliminated after two mistakes.

For the second year in a row, the competition was won by Fr �ed �ericLatour, who survived eleven roundswith only one mistake. Dennis The lastedten rounds for a second place showing. Students ranked third through �fthwere eliminated in the tenth round, and entered a playo� round to break thetie. After the tiebreaker, Byung Kyu Chun and Derek Kisman were still tiedfor third and fourth, while Sabin Cautis was ranked �fth.

In keeping with the spirit of the competition, Fr �ed �eric received 100toonies (Can $200) in laundry money, a prize awarded by the Dean of Mathe-matics. The Dean also awarded 100 loonies (Can $100) to Dennis, 80 quarterseach to Byung and Derek, and 40 quarters to Sabin. A total of 30 studentsparticipated.

The contest organiser this year was Christopher Small. He was ably as-sisted by Ravindra Maharaj and Ken Davidson. Kristin Lord was the contestphotographer.

322

The Problems

1. Let x and y be any positive integers satisfying x y = 1999x+ 1999 y .

TRUE OR FALSE? It follows that x � 3998000.

2. TRUE OR FALSE? The equation sin(cosx) = cos(sinx) has no solu-tions in real values x.

3. In the equation

13107933121311959518748AB = CD � (50833� 5083)2

four of the digits are omitted as shown. (The digits are not necessarilydistinct.)

TRUE OR FALSE? B = 8.

4. Circles with radii 12 and 4 are tangent as shown. A square is drawninside the larger circle touching it and the smaller one as shown.

B

O CA

The length of a side of the square is expressed asa+

pb

c, where a, b

and c are integers, and the ratio is expressed in lowest form.

TRUE OR FALSE? b = 2734.

5. Suppose the 64 squares of an 8 � 8 chessboard of unit squares are tobe coloured either black or white.

TRUE OR FALSE? There are exactly 215 ways of colouring the squaresso that every 2� 2 square of adjacent unit squares has exactly 3 of the4 squares the same colour.

323

6. Consider the equation f(x)f(x+ 1) = f�f(x) + x

�.

TRUE OR FALSE? There exists a non-constant di�erentiable functionf : R! R satisfying this equation for all real x.

7. Suppose f : R ! R has derivatives of all orders and that the doublyin�nite series

� � �+ f 00(x) + f 0(x) + f(x) +

Z x

0

f(t)dt +

Z x

0

Z y

0

f(t)dt dy + � � �

converges absolutely and uniformly to some function �(x) for allx 2 [�1999;+1999]. Suppose also that �(0) = 1999.

TRUE OR FALSE? �(�1999)< 0.

8. TRUE or FALSE?5

4<

Z 1

0

x�x dx <3

2.

9. Three points A, B and X are randomly chosen from the interior of acircle. An additional point X0 is randomly chosen on the boundary ofthe circle.

X0

A

X

B

Let � be the probability that4ABX is acute, and let �0 be the prob-ability that 4ABX0 is acute.

TRUE OR FALSE? � < �0.

10. In 3{dimensional tic-tac-toe on a 4� 4 � 4 board, a winning line is asequence of four counters in a straight line. In the �gure, the X's forma winning line, as do the Y's and the Z's.

XXXX

YY

YY

Z

Z

Z

Z

--

-

324

Note that the X's form a line parallel to an edge of the cube, the Y'sform a line parallel to a face but not to any edge, and the Z's are notparallel to any face or edge.

TRUE OR FALSE? It is possible to distribute 16 counters among the 64cells so that every winning line of the opponent parallel to a face oredge is blocked.

11. Four men and four women decide to play a mixed-doubles tennis tour-nament on two adjacent courts at a tennis club so that each person playsin one mixed-doubles match per day.

The tournament is arranged so that

� a man and a woman always play against a man and a woman;

� no person ever plays with anyone else more than once;

� no person ever plays against anyone else more than once.

TRUE OR FALSE? Such a tournament cannot be played on the two courtson three days.

325

THE OLYMPIAD CORNERNo. 200

R.E. Woodrow

All communications about this column should be sent to Professor R.E.Woodrow, Department of Mathematics and Statistics, University of Calgary,Calgary, Alberta, Canada. T2N 1N4.

As a set of problems for this issue, we give problems of the XL Math-ematical Olympiad of the Republic of Moldova. My thanks go to Ravi Vakilfor collecting them when he was Team Leader for Canada at the InternationalMathematical Olympiad at Mumbai.

REPUBLIC OF MOLDOVAXL MATHEMATICAL OLYMPIAD

Chi�sin�au, 17{20 April, 1996First Day (Time: 4 hours)

10 Form

1. Let n = 213 � 311 � 57. Find the number of divisors of n2 which areless than n and are not divisors of n.

2. Distinct square trinomials f(x) and g(x) have leading coe�cient 1.It is known that f(�12)+f(2000)+f(4000) = g(�12)+g(2000)+g(4000).Find all the real values of x which satisfy the equation f(x) = g(x).

3. Through the vertices of a triangle tangents to the circumcircle areconstructed. The distances of an arbitrary point of the circle to the straightlines containing the sides of the triangle are equal to a, b and c and to thetangents are equal to x, y and z. Prove that a2 + b2 + c2 = xy + xz + yz.

4. Two brothers sold n lambs at a price n dollars. They divide themoney as follows: the elder brother took 10 dollars, the younger one took10 dollars, the elder one took again 10 dollars and so on. At the end itturned out that the sum for the younger brother was less than 10. He tookthe remainder and the pen-knife of his brother. The brothers agreed that thedivision was correct. What is the cost of the pen-knife?

11{12 Form

1. Prove the equality

1

666+

1

607+� � �+ 1

1996= 1+

2

2 � 3 � 4+2

5 � 6 � 7+� � �+2

1994 � 1995 � 1996 .

326

2. Prove that the product of the roots of the equationp1996 � xlog1996 x = x6

is an integer and �nd the last four digits of this integer.

3. Two disjoint circles C1 and C2 with centres O1 and O2 are given. Acommon exterior tangent touches C1 and C2 at points A andB respectively.The segment O1O2 cuts C1 and C2 at points C and D respectively. Provethat:

(a) the points A, B, C and D are concyclic;(b) the straight lines (AC) and (BD) are perpendicular.

4. Among n coins, identical by form, less than half are false and di�erby weight from the true coins. Prove that with the help of scales (withoutweights) using no more than n � 1 weighings one can �nd at least one truecoin.

Second Day (Time: 4 hours)

10 Form

5. Prove that for all natural numbers m � 2 and n � 2 the smallestamong the numbers n

pm and m

pn does not exceed the number 3

p3.

6. Prove the inequality 2a1+2a2+� � �+2a1996 � 1995+2a1+a2+���+a1996

for any real non-positive numbers a1, a2, : : : , a1996.

7. The perpendicular bisector to the side [BC] of a triangle ABCintersects the straight line (AC) at a pointM and the perpendicular bisectorto the side [AC] intersects the straight line (BC) at a pointN . Let O be thecentre of the circumcircle to the triangle ABC. Prove that:

(a) points A, B, M , N and O lie on a circle S;(b) the radius of S equals the radius of the circumcircle to the triangle

MNC.

8. Among 1996 coins, identical by form, two are false. One is heavierand the second is lighter than a true coin. With the help of scales (with-out weights) using four weighings, what is the way to show if the combinedweight of these two coins is equal, greater than or less than, the combinedweight of two true coins?

11{12 Form

5. Let p be the number of functions de�ned on the set f1; 2; : : : ; mg,m 2 N�, with values in the set f1, 2, : : : , 35, 36g and q be the number offunctions de�ned on the set f1, 2, : : : , ng, n 2 N�, with values in the setf1, 2, 3, 4, 5g. Find the least possible value for the expression jp� qj.

6. Solve in real numbers the equation

2x2 � 3x = 1+ 2xpx2 � 3x .

327

7. On a sphere distinct points A, B, C and D are chosen, so thatsegments [AB] and [CD] cut each other at point F , and points A, C and Fare equidistant to a point E. Prove that the straight lines (BD) and (EF )are perpendicular.

8. 20 children attend a rural elementary school. Every two childrenhave a grandfather in common. Prove that some grandfather has not lessthan 14 grandchildren in this school.

Next, we give solutions by our readers to problems of the VIII NordicMathematical Contest [1998: 133].

1. Let O be a point in the interior of an equilateral triangle ABC withside length a. The lines AO, BO and CO intersect the sides of the triangleat the points A1, B1 and C1 respectively. Prove that

jOA1j+ jOB1j+ jOC1j < a .

Solutions by Mohammed Aassila, CRM, Universit �e de Montr �eal,Montr �eal, Qu �ebec; by Pierre Bornsztein, Courdimanche, France; and byToshio Seimiya, Kawasaki, Japan. We give the solution by Seimiya.

B A1 O0 A0 C

A

B1

C1

O

Since AB = AC we have, in4AA1C

\ABC = \ACB = \ACA1 ,

so that AA1 < AC = a.

Similarly we have that BB1 < a and CC1 < a. We denote the area of4PQR by [PQR]. Let A0, O0 be the feet of the perpendiculars from A, Oto BC respectively. There, we have

OA1

AA1

=OO0

AA0=

[OBC]

[ABC];

OB1

BB1

=[OCA]

[ABC];

OC1

CC1

=[OAB]

[ABC].

328

Thus we get

OA1

AA1

+OB1

BB1

+OC1

CC1

=[OBC] + [OCA] + [OAB]

[ABC]= 1 .

Since AA1 < a, BB1 < a, CC1 < a we have

OA1

AA1

+OB1

BB1

+OC1

CC1

>OA1

a+OB1

a+OC1

a.

Thus we have 1 >OA1

a+OB1

a+OC1

a.

This implies that OA1 +OB1 + OC1 < a .

2. A �nite set S of points in the plane with integer coordinates is calleda two-neighbour set, if for each (p; q) inS exactly two of the points (p+1; q),(p; q+1), (p�1; q), (p; q� 1) are in S. For which n does there exist a two-neighbour set which contains exactly n points?

Solution by Pierre Bornsztein, Courdimanche, France.

On va prouver que les valeurs de n cherch �ees sont tous les entierspairs sup �erieurs ou �eqaux �a 4 sauf 6. On v �eri�e facilement que n � 4 estn �ecessaire.

Soit S un tel ensemble avec card S = n. PourM(p; q) 2 S, on dit quele point P est voisin de M lorsque P 2 S et P 2 f(p + 1; q), (p; q + 1),(p�1; q), (p; q�1)g. Alors tout point de S admet exactement deux voisins.

Soit M1(p; q) 2 S.

Soit M2 un voisin deM1.

On poseM3 est le voisin de M2 avec M3 6= M1, etc.

SiMk et Mk+1 sont construits (k � 1) on pose Mk+2 est le voisin deMk+1 avec Mk+2 6=Mk.

On construit ainsi, �a partir de M1, une suite de points de S. Mais Sest �ni. Donc il existe i; j 2 N

� avec i < j tels que Mi = Mj. Soit alorsE = fk 2 N� j 9 i 2 N�; i < k et Mi = Mkg. On a E 6= �, E � N

�, doncE admet un plus petit �el �ement. On pose p = minE.

Soit alors i < p, i 2 N� tel que Mi = Mp. On a alors i = 1 : en e�et,si on suppose i > 1 alorsMi�1,Mi+1, Mp�1 sont des voisins de Mi =Mp

avec

�Mi�1 6=Mi+1 par constructionMi�1 6=Mp�1 car p = minE

d'o �u Mi+1 =Mp�1, etcomme p = minE, on a i+ 1 = p � 1 (on a p� 1 � p et i+ 1 � p) d'o �up = i+ 2, et Mi = Mi+2, ce qui est impossible par construction.

329

D�e�nition : Une cha�ne C, de longueur a (a 2 N�; a � 2) est un ensemblefM1, : : : , Mag de points de S tels que pour tous i, j de f1, : : : , ag

� Mi 2 S,� Mi+1 et Mi sont voisins,� Ma et M1 sont voisins,� si i 6= j, alors Mi 6= Mj .

Sans perte de g �en �eralit �e on peut toujours supposer que pour touti 2 f1; : : : ; ag, Mi(pi; qi) avec

� pi � 1� si i > 1 et pi = p1 alors qi > q1� M2(p1 + 1; q1)

Cela entra�ne Ma(p1; q1 + 1) et permet d' �eviter de consid �erer deux foix lameme cha�ne si l'on change de \sens de parcours" ou de \point de d �epart".

On veri�e ais �ement que a � 4.

Propri �et �e : Si C = fM1, : : : , Mag et C0 = fP1, : : : , Pbg sont deux cha�nesalors C \C0 = � ou C = C0.

Preuve : Par l'absurde. Si C \C0 6= � et C 6= C0, alors il existeM 2 C \C0

tel que :

� il existe i 2 f1, : : : , ag, j 2 f1, : : : , bg,M =Mi = Pj

� card(Mi�1;Mi+1; Pj�1; Pj+1) � 3 (On pose M�1 =Ma, P�1 = Pb).

Mais alors M 2 S poss �ede au moins trois voisins, qui est un contra-diction.

D'apr �es ce qui pr �ec �ede, 8M 2 S 9 chaine C tel que M 2 C, doncS � S

cha�neC.

Par d �e�nition, toute cha�ne est contenue dans S, d'o �u S =Scha�neC.

Toutes ces cha�nes �etant deux �a deux disjointes, alors

n = card S =X

cha�ne

card C . (1)

Soit une cha�ne C = fM1, : : : , Mag.On pose h = nombre de \sauts" vers le haut (c. �a.d. le nombre d'indices

i tel que pi+1 = pi et qi+1 = qi + 1)

b = nombre de \sauts" vers le bas (idem avec pi+1 = pi, qi+1 = qi� 1),

g = nombre de \sauts" vers la gauche (idem avec pi+1 = pi� 1, qi+1 = qi),

d = nombre de \sauts" vers la droite (idem avec pi+1 = pi + 1, qi+1 = qi).

330

Le nombre de sauts vers la gauche est �egal au nombre de sauts vers ladroite, et le nombre de sauts vers le haut est �egal au nombre de sauts versle bas plus un (rappel qa = q1 + 1).

C. �a.d. h = b+ 1, g = d ;or a = nombre de points = nombre de sauts plus un = d+ g + h+ b+ 1 .Donc a = 2(g+ b+ 1), pair.

Toute cha�ne est donc de longeur paire.

D'apr �es (1), on en d �eduit que n est pair.

Pour n = 4 :r

r

r

rM3

M2

M4

M1

convient.

Pour n = 6 : Comme a � 4, un tel ensemble S, s'il existe, n'utilise q'uneseule cha�ne, et g + b = 3, avec g, b 2 N�, d'o �u g = b = 1 et d = 1, h = 2.

SiM1 est choisis,M2 et M6 sont impos �e par construction.

Car d = 1, il y a deux places possibles pour M3 :

1ercas

r

r

r

r

r

r

M3

M2

M4

M1

M6

2i �emecas

r

r

r

r

r

M2

M3

M4

M1

M6

Dans les deux cas, la place pour M4 est impos �e car d = 1 = b. Dansle 1er cas, M1 a donc 3 voisins, une contradiction. Dans le 2i�eme cas, M3 atrois voisins, contradiction.

Donc n = 6 est impossible.

Pour n = 2p, p � 42z}|{

p� 1

8>>>>>><>>>>>>:

� � �

� �

� �

.

.

.� � �

convient .

D'o �u le r �esultat annonc �e.

331

3. A square piece of paper ABCD is folded by placing the corner Dat some point D0 on BC (see �gure). Suppose AD is carried into A0D0,crossing AB at E. Prove that the perimeter of triangle EBD0 is half as longas the perimeter of the square.

D C

A B

D0

E

A0

Solutions by Mohammed Aassila, CRM, Universit �e de Montr �eal,Montr �eal, Qu �ebec; by Pierre Bornsztein, Courdimanche, France; and byToshio Seimiya, Kawasaki, Japan. We give the solution by Aassila.

D C

A B

D0

E

A0

qF

AB +BC = AE + EB + BD0 +D0C

= EF +EB + BD0 + FD0 = ED0 +EB + BD0 .

[Editor's comment: it is easy to show thatED0 is tangent to the circle, centreD, radius DC; see More Mathematical Morsels by Ross Honsberger, MAADoliciani Mathematical Expositions, 1991, p. 11.]

4. Determine all positive integers n < 200 such that n2 + (n+ 1)2 isa perfect square.

Solutions by Mohammed Aassila, CRM, Universit �e de Montr �eal,Montr �eal, Qu �ebec; by Pierre Bornsztein, Courdimanche, France; by PanosE. Tsaoussoglou, Athens, Greece; and by Edward T.H. Wang, Wilfrid LaurierUniversity, Waterloo, Ontario. We give Wang's solution and remark.

There are exactly three solutions, n = 3, 20, 119. Suppose thatn2 + (n + 1)2 = k2 for some natural number k. Then (n;n + 1; k) is aPythagorean triple. It is in fact a primitive Pythagorean triple since(n; n+ 1) = 1. By a well known result, there exist natural numbers s and twith opposite parities and s > t such that either

332

(i) n = 2st, n+ 1 = s2 � t2, k = s2 + t2; or

(ii) n = s2 � t2, n+ 1 = 2st, k = s2 + t2.

In case (i) we have s2 � 2ts � (t2 + 1) = 0. Solving for s we �nds = t +

p2t2 + 1. Clearly s 2 N if and only if 2t2 + 1 is a perfect square.

Since n < 200 implies st < 100, we have t2 < 100 or t < 10. Substitutingt = 1, 2, : : : , 9 reveals that 2t2 + 1 is a square only when t = 2, in whichcase s = 5, n = 20, n+ 1 = 21 and k = 29.

In case (ii) we have s2 � 2ts � (t2 � 1) = 0. Solving for s, we �nds = t+

p2t2 � 1. (Note that t�p2t2 � 1 < 0 for all t � 1). As in case (i),

n + 1 < 201 implies st � 100 and thus t2 < 100 or t < 10. Substitutingt = 1, 2, : : : , 9 reveals that 2t2 � 1 is a perfect square exactly when t = 1and t = 5. When t = 1, we get s = 2, n = 3, n + 1 = 4, k = 5. Whent = 5, we get s = 12, n = 119, k = 169.

Remark. In fact, the characterization of all (primitive) Pythagoreantriples of the form (n;n+1; k) is known; for example, in Chapter 2.4 of Ele-mentary Theory of Numbers byW. Sierpinski, 2nd ed., 1985, it is proved thatif the sequences fxng, fyng, and fzng are de�ned recursively byxn+1 = 3xn + 2zn + 1, yn+1 = xn+1 + 1, zn+1 = 4xn + 3zn + 2 forall n � 1, with initial values x1 = 3, y1 = 4 and z1 = 5, then (xn; yn; zn),n = 1, 2, 3, : : : would be all the (primitive) Pythagorean triples for whichyn = xn +1. Using these iterations one can easily verify that the triples weobtained in the solution do indeed give the �rst three such triples. The nextfour values of n are 696, 4059, and 23660.

We �nish this number of the Corner by giving readers' solutions toproblems of the 44th Lithuanian Mathematical Olympiad [1998: 196{197].

GRADE XI

1. You are given a set of 10 positive integers. Summing nine of themin ten possible ways we get only nine di�erent sums: 86, 87, 88, 89, 90, 91,93, 94, 95. Find those numbers.

Solutions by Mohammed Aassila, CRM, Universit �e de Montr �eal,Montr �eal, Qu �ebec; by Pierre Bornsztein, Courdimanche, France; byChristopher J. Bradley, Clifton College, Bristol, UK; by Murray S. Klamkin,University of Alberta, Edmonton, Alberta; and by Edward T.H. Wang, WilfridLaurier University, Waterloo, Ontario. We give Bradley's write-up.

Let S be the sum of all ten positive integers and suppose x is therepeated sum. Call the elements a1, a2, : : : , a10. Then we have

S � a1 = 86 , S � a2 = 87 , : : : , S � a9 = 95 , S � a10 = x .

333

Adding, 9S = 813 + x. The only value of x from 86, 87, 88, : : : , 95 whichmakes 813 + x divisible by 9 is x = 87 and then S = 100. It follows thatthe ten numbers are respectively 14, 13, 12, 11, 10, 9, 7, 6, 5 and 13.

2. What is the least possible number of positive integers such that thesum of their squares equals 1995?

Solutions by Mohammed Aassila, CRM, Universit �e de Montr �eal,Montr �eal, Qu �ebec; by Pierre Bornsztein, Courdimanche, France; by MurrayS. Klamkin, University of Alberta, Edmonton, Alberta; by Bob Prielipp, Uni-versity of Wisconsin{Oshkosh, WI, USA; and by Edward T.H. Wang, WilfridLaurier University, Waterloo, Ontario. We give Klamkin's solution.

First note that 1995 = 3 � 5 � 7 � 19. We now use some known theorems[1] on representations of a number as sums of squares.

1. A natural numbern is the sumof two squares if and only if the factorizationof n does not contain any prime of the form 4k+3 that has an odd exponent.

2. A natural number n can be the sum of three squares if and only if it is notof the form 4l(8k+ 7), where k, l are non-negative integers.

3. Each odd natural number is the sum of the squares of four integers, twoof which are consecutive numbers.

In view of the above theorems, the minimum number is three and arepresentation is given by 1995 = 12 + 252 + 372.

Comment. A number for which the minimum number of squares is fouris given by 1992 = 3 � 8 � 83 = 102+182 +282+282 = 22 +42 +62 +442.Also 1995 = 132 + 242 + 252 + 252.

Reference

[1] W. Sierpinski, Elementary Theory of Numbers, Hafner, NY, 1964, pp.351, 363, 367.

3. Replace the asterisks in the \equilateral triangle"

� � � � � � � � �� � � � � � � �� � � � � � �� � � � � �� � � � �� � � �� � �� ��

by the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 so that, starting from the second line,each number is equal to the absolute value of the di�erence of the nearesttwo numbers in the line above.

334

Is it always possible to inscribe the numbers 1, 2, : : : , n, in the wayrequired, into the equilateral triangle with the sides having n asterisks?

Solutions by Mohammed Aassila, CRM, Universit �e de Montr �eal,Montr �eal, Qu �ebec; and by Christopher J. Bradley, Clifton College, Bristol,UK. We give Bradley's solution.

The case n = 9 can be done as follows:

1 7 8 1 9 8 1 7 66 1 7 8 1 7 6 1

5 6 1 7 6 1 51 5 6 1 5 4

4 1 5 4 13 4 1 3

1 3 22 1

1

For the case n = 6m+ 3 one starts with the row

1 4m+3 : : : 1 6m+1 6m+21

centre squarez }| {

6m+ 3 6m+2 16m+1 6m 1 : : : 4m+34m+2 .

Then if one works down six rows you get

1 4m�1 : : : 1 6m�5 6m�4 1 6m�3 6m�4 1 6m�5 6m�6 1 : : : 4m�1 4m�2

and the result then follows by induction onm, with the pattern up tom = 1as in the �rst triangle.

4. A function f : N! N is such that f(f(m)+ f(n)) = m+n for allm, n 2 N (N = f1, 2, : : : g denotes the set of all positive integers). Find allsuch functions.

Solutions by Mohammed Aassila, CRM, Universit �e de Montr �eal,Montr �eal, Qu �ebec; by Pierre Bornsztein, Courdimanche, France; and byChristopher J. Bradley, Clifton College, Bristol, UK. We give the solutionby Bornsztein.

Pour m, n 2 N�, d'apr �es

f(f(m)+ f(n)) = m+ n (1)

f(f(m)+ f(n))+ f(f(m)+ f(n)) = 2(m+ n)

et donc

f [f(f(m)+ f(n))+ f(f(m)+ f(n))] = f(m) + f(n)+ f(m) + f(n)

= 2(f(m)+ f(n))

etf [f(f(m)+ f(n))+ f(f(m)+ f(n))] = f(2m+ 2n) .

335

Ainsi

2f(m)+ 2f(n) = f(2m+ 2n) . (2)

Pour m = n, 4f(n) = f(4n).

Pour m = 2p+ 1, n = 2p � 1, 2f(2p+ 1) + 2f(2p� 1) = f(8p) =4f(2p) d'o �u pour p � 1,

f(2p+ 1) = 2f(2p)� f(2p� 1) . (3)

De meme pour m = 2p+ 2, n = 2p� 2, (p � 2) on obtient

f(2p+ 2) = 2f(2p)� f(2p� 2) . (4)

On pose f(1) = a, f(2) = b.

En utilisant (3) et (4) on trouve

f(3) = 2b� a ,

f(4) = f(4� 1) = 4f(1) = 4a ,

f(5) = 9a� 2b ,

f(6) = 8a� b .

Mais pour m = 2, n = 1, (2) conduit �a f(6) = 2f(2) + 2f(1) = 2a + 2b.Donc 8a � b = 2a + 2b ; c. �a.d. b = 2a. On en d �eduit que pour n 2 f1,: : : , 6g, f(n) = an. Une r �ecurrence imm�ediate en utilisant (3) et (4) permetd'obtenir :

f(n) = an; pour tout n 2 N� .Alors dans (1), pour tous m;n 2 N�

m+ n = f(f(m)+ f(n)) = f(a(m+ n)) = a2(m+ n)

d'o �u a = 1 et f = idN�. R �eciproquent, f = idN� convient.

5. In the trapezium ABCD, the bases are AB = a, CD = b, andthe diagonals meet at the point O. Find the ratio of the areas of the triangleABO and trapezium.

Solutions by Mohammed Aassila, CRM, Universit �e de Montr �eal,Montr �eal, Qu �ebec; byMiguel Amengual Covas, Cala Figuera,Mallorca, Spain;by Pierre Bornsztein, Courdimanche, France; and by Christopher J. Bradley,Clifton College, Bristol, UK. We give the solution by Amengual.

D C

BA

b

a

O

336

Since triangles with the same height have areas in proportion to their basesand since4AOB � 4COD, we have

[DOA]

[AOB]=

OD

OB=

b

aor [DOA] =

b

a[AOB] ,

where [P ] denotes the area of polygon [P ].Also, since the areas of similar triangles are proportional to the squares

on corresponding sides,

[COD]

[AOB]=

b2

a2or [COD] =

b2

a2[AOB] .

Finally, since 4ABD and 4ABC have the same base and equal alti-tude [ABD] = [ABC], and since

[ABD] = [AOB] + [DOA]; [ABC] = [AOB] + [BOC];

it follows that [DOA] = [BOC] .Consequently,

[ABCD] = [AOB] + [BOC] + [COD] + [DOA]

= [AOB] + 2[DOA] + [COD] ;

that is, [ABCD] =

�1 + 2 � b

a+b2

a2

�[AOB] ,

giving [ABCD] =

�a+ b

a

�2[AOB] ,

and[AOB]

[ABCD]=

�a

a+ b

�2.

Remark. The following related problem appears in Solving Problemsin Geometry by V. Gusev, V. Litvinenko and A. Mordkovich, Mir Publishers,Moscow 1988, page 60: The diagonals of a trapezoid ABCD (ADkBC) in-tersect at the point O. Find the area of the trapezoid if it is known that thearea of the triangle AOD is equal to a2, and the area of the triangle BOCis equal to b2.

GRADE XII

1. Consider all pairs of real numbers satisfying the inequalities

�1 � x+ y � 1 , � 1 � xy + x+ y � 1 .

Let M denote the largest possible value of x.

337

(a) Prove that M � 3.

(b) Prove that M � 2.

(c) FindM .

Solutions by Mohammed Aassila, CRM, Universit �e de Montr �eal,Montr �eal, Qu �ebec; by Pierre Bornsztein, Courdimanche, France; byChristopher J. Bradley, Clifton College, Bristol, UK; by Murray S. Klamkin,University of Alberta, Edmonton, Alberta; and by Edward T.H. Wang, WilfridLaurier University, Waterloo, Ontario. We use the solution by Klamkin.

Letting u = x+ 1 and v = y+ 1, the given inequalities become

0 � uv � 2 , 1 � u+ v � 3 .

Clearly u and v must each be non-negative and so the maximum value of uis 3. Finally, xmax = 2.

2. A positive integer n is called an ambitious number if it possessesthe following property: writing it down (in decimal representation) on theright of any positive integer gives a number that is divisible by n. Find:

(a) the �rst 10 ambitious numbers;

(b) all the ambitious numbers.

Solutions by Mohammed Aassila, CRM, Universit �e de Montr �eal,Montr �eal, Qu �ebec; by Christopher J. Bradley, Clifton College, Bristol, UK;by Murray S. Klamkin, University of Alberta, Edmonton, Alberta; and byEdward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We giveBradley's solution.

We assume in this solution that n cannot beginwith a zero (so that 04 isnot acceptable). For n to be ambitious it must be a k{digit number dividing10k �m for allm, which is so if and only if n j 10k. The ambitious numbersare thus 1, 2, 5, 10, 20, 25, 50, 100, 125, 200, : : : .

3. The area of a trapezium equals 2; the sum of its diagonals equals 4.Prove that the diagonals are orthogonal.

Solutions by Mohammed Aassila, CRM, Universit �e de Montr �eal,Montr �eal, Qu �ebec; byMiguel Amengual Covas, Cala Figuera,Mallorca, Spain;by Pierre Bornsztein, Courdimanche, France; by Christopher J. Bradley,Clifton College, Bristol, UK; and by Murray S. Klamkin, University of Al-berta, Edmonton, Alberta. We give the solution of Klamkin.

If a and b are the diagonal lengths and � the angle between them, thenthe area 2 = 1

2ab sin �. Since 4 = a+ b � 2

pab, the maximum value of ab

is 4. Hence sin � = 1, so that � = �=2.

4. 100 numbers are written around a circle. Their sum equals 100. Thesum of any 6 neighbouring numbers does not exceed 6. The �rst number is6. Find the remaining numbers.

338

Solutions by Mohammed Aassila, CRM, Universit �e de Montr �eal,Montr �eal, Qu �ebec; and by Pierre Bornsztein, Courdimanche, France. Wegive the solution of Bornsztein.

On appelle x1, x2, : : : , x100 ces nombres dans le sens horaire, et onpose x100+i = xi. On a, pour i = 1, 2, : : : , 100

xi + xi+1 + � � �+ x100 + x1 + : : : xi�1 + xi + xi+1 = 100 + xi + xi+1

et comme il y a 102 = 6� 17 termes dans cette somme, en les groupant par6 cons �ecutifs, on a 100 + xi + xi+1 � 102

c. �a.d: xi + xi+1 � 2 .

Alors

100 = (x1 + x2) + (x3 + x4) + � � �+ (x99 + x100) � 2� 50

avec �egalit �e ssi pour tout i � 1, x2i�1 + x2i = 2. Donc pour i � 1,x2i�1 + x2i = 2.

De meme

100 = (x2 + x3) + (x4 + x5) + � � �+ (x100 + x1) � 2� 50

avec �eqalit �e ssi x2i + x2i+1 = 2 pour i � 1.

Finalement, pour tout i � 1, xi + xi+1 = 2 . Or x1 = 6 doncx2 = �4 et par r �ecurrence imm�ediate pour i = 1, 2, : : : , 50, x2i�1 = 6,x2i = �4. Cette famille v �eri�e bien les conditions de l' �enonc �e.

5. Show that, at any time, moving both the hour-hand and the minute-hand of the clock symmetrically with respect to the vertical (6 � 12) axisresults in a possible position of the clock-hands. How many straight linescontaining the centre of the clock-face possess the same property?

Solution by Mohammed Aassila, CRM, Universit �e de Montr �eal,Montr �eal, Qu �ebec.

Eleven,360�n

11(n = 0, 1, 2, : : : , 10) .

That completes the Corner for this issue. Send me your nice solutionsas well as contest materials.

339

BOOK REVIEWS

ALAN LAW

Problem-Solving Strategies for E�cient and Elegant Solutions by Alfred S.Posamentier and Stephen Krulik,published by Corwin Press, 1998.ISBN # 0-8039-6698-9, softcover, 249+ pages, $29.95 (U.S.).Reviewed by Ian VanderBurgh, University of Waterloo, Waterloo,Ontario.

This new book from well-known authors Alfred Posamentier andStephen Krulik is an excellent introduction to problem solvingand the thoughtprocesses behind problem solving.

\Problem Solving Strategies : : : " devotes each of ten chapters to atechnique useful in solving problems. These techniques range from \WorkingBackwards" to \Finding a Pattern" to \Making a Drawing". Each chaptershows �rst how each technique can be used in everyday life, and then showsits use in a mathematical example. Posamentier and Krulik then presentroughly 20 problems with solutions in each chapter.

There are de�nite advantages to their presentation. The layout of eachchapter is consistent, which makes their book very easy to read. I liked theidea of showing clear applications in everyday life | this gives the readersomething to relate to if he or she has never seen a particular techniquepreviously in a mathematical setting. Also, with each problem, two solutionsare usually given, one \brute force" and the other using the technique of thechapter, giving a clear contrast between a potentially ugly solution and anelegant one. The exposition does, however, tend to get a little wordy, whichcan be either a plus or a minus.

There are some di�culties that I encountered in working through thisbook. A couple of the problems and solutions are incomprehensible, andother problems are really overly simplistic. My major concern arose whensome of the problems said \Show that : : : " and the authors in their solutionfound a pattern or considered a couple of extreme cases, thus considering theproblem �nished without making any attempt to prove the required result.This is a �ne way to demonstrate the use of their techniques, but could bedangerous for a student reader unfamiliar with the concept of a proof. Myother concern is that the book is lacking a conclusion to tie together all ofthe ideas presented.

That being said, this book can serve as a good resource for teachers whowant to brush up on their problem solving skills or teach them to youngerhigh school students. The vast majority of the problems would be accessibleto students in Grades 9{11. The techniques presented are good as general

340

problem solving skills, but are also of excellent value for use on multiple-choice or answer-only contests, as they generally lead one to the correct re-sult fairly quickly. If nothing else, this book is a good collection of problems,and thus could serve as a good resource.

The art of problem solving, Volume 2, by S. Lehoczky and R. Rusczyk.Published by Greater Testing Concepts, P.O. Box A-D, Stanford, CA 94309,1994. Paperback, 390+ pages, solution manual 212 pages, without ISBNnumber, US $27 without solution manual or $35 with. (May be ordered incombination with Volume 1, which has been reviewed in Crux [1994: 135-136], US $47 without solution manuals or $60 with.)Reviewed by Andy Liu, University of Alberta, Edmonton, Alberta

This volume consists of 27 chapters, 7 on algebra, 4 on geometry, 2 onanalytic geometry, 2 on trigonometry, 2 on vectors, 5 on combinatorics, 3on number theory, an introductory chapter on proof techniques and a �nalchapter of further problems. In all, there are 237 examples, 412 exercises and509 problems. The companion manual contains solutions to all the exercisesand problems.

Much that was said enthusiastically about Volume 1 applies here also.The only slight disappointment is that while the topics covered are moreadvanced, the treatment is at essentially the same level as that in Volume 1.Nevertheless, this set provides excellent preparation for introductory levelmathematics competitions, and important stepping stones towards furtherstudy in solving problems of Olympiad calibre.

341

THE SKOLIAD CORNERNo. 40

R.E. Woodrow

In this issue, we give the Final Round Parts A and B of the 1998 BritishColumbia Colleges Junior High School Mathematics Contest. My thanks goto Jim Totten, University College of the Cariboo, one of the organizers, forforwarding the materials for use in the Corner.

BRITISH COLUMBIA COLLEGES JR. HIGH SCHOOLMATHEMATICS CONTEST

Final Round 1998Part A

1. Each edge of a cube is coloured either red or black. If every face ofthe cube has at least one black edge, the smallest possible number of blackedges is:

(a) 6 (b) 5 (c) 4 (d) 3 (e) 2

2. Line AE is divided into four equal parts by the points B, C and D.Semicircles are drawn on segments AC, CE, AD and DE creating semicir-cular regions as shown. The ratio of the area enclosed above the line AE tothe area enclosed below the line is:

r r r r rAB C

DE

(a) 4 : 5 (b) 5 : 4 (c) 1 : 1 (d) 8 : 9 (e) 9 : 8

3. A container is completely �lled from a tap running at a uniform rate.The accompanying graph shows the level of the water in the container at anytime while the container is being �lled. The segment PQ is a straight line.The shape of the container which corresponds with the graph is:

342

Full

WaterLevel

EmptyTime

q

q

Q

P

(a) (b) (c) (d) (e)

4. The digits 1, 9, 9, and 8 are placed on four cards. Two of the cardsare selected at random. The probability that the sum of the numbers on thecards selected is a multiple of 3 is:

(a) 14

(b) 13

(c) 12

(d) 23

(e) 34

5. The surface areas of the six faces of a rectangular solid are 4, 4, 8, 8,18 and 18 square centimetres. The volume of the solid, in cubic centimetresis:

(a) 24 (b) 48 (c) 60 (d) 324 (e) 576

6. The area of the small triangle in the diagram is 8 square units. Thearea of the large triangle, in square units, is:

2b 3b

2a

a

(a) 18 (b) 20 (c) 24 (d) 28 (e) 30

7. At 6:15 the hands of the clock form two positive angles with a sumof 360�. The di�erence of the degree measures of these two angles is:

(a) 165 (b) 170 (c) 175 (d) 180 (e) 185

8. The last digit of the number 826 is:

(a) 0 (b) 2 (c) 4 (d) 6 (e) 8

343

9. For the equation Ax+3

+ Bx�3

= �x+9x2�9

to be true for all values of x

for which the expressions in the equation make sense, the value of AB is:

(a) 2 (b) �1 (c)�2 (d)�3 (e) �610. A hungry hunter came upon two shepherds, Joe and Frank. Joe had

three small loaves of bread and Frank �ve loaves of the same size. The loaveswere divided equally among the three people, and the hunter paid $8 for hisshare. If the shepherds divide the money so that each gets an equitable sharebased on the amount of bread given to the hunter, the amount of money thatJoe receives is:

(a) $1 (b) $1:50 (c) $2 (d) $2:50 (e) $3

Part B

1. Four positive integers sum to 125. If the �rst of these numbers isincreased by 4, the second is decreased by 4, the third is multiplied by 4 andthe fourth is divided by 4, you produce four equal numbers. What are thefour original numbers?

2. A semi-circular piece of paper of radius 10 cm is formed into a conicalpaper cup as shown (the cup is inverted in the diagram):

V 10 cm

V

10 cm

VN N

Find the height of the paper cup; that is, the depth of water in the cupwhen it is full.

3. In the diagram a quarter circle is inscribed in a square with sidelength 4, as shown. Find the radius of the small circle that is tangent to thequarter circle and two sides of the square.

4

4

344

4. Using the digits 1, 9, 9 and 8 in that order create expressions equalto 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. Youmay use any of the four basic operations(+, �, �, �), the square root symbol (p) and parentheses, as necessary.For example, valid expressions for 25 and 36 would be

25 = �1 + 9 + 9 + 8

36 = 1 + 9�p9 + 8

Note: You may place a negative sign in front of 1 to create �1 if you wish.

5. At 6 am, one Saturday, you and a friend begin a recreational climb ofMt. Mystic. Two hours into your climb, you are overtaken by some scouts.As they pass, they inform you that they are attempting to set a record forascending and descending the mountain. At 10 am, they pass you again ontheir way down, crowing that they had not stopped once to rest, not even atthe top.

You �nally reach the summit at noon. Assuming that both you and thescouts travelled at a constant vertical rate, both climbing and descending,when did the scouts reach the top of Mt. Mystic?

In the last number, we gave the problems of the Florida MathematicsOlympiad, Team Competition for 1998. Next we give the solutions. Mythanks go to John Grant McLoughlin,Memorial University of Newfoundland,for forwarding the contest for our use.

FLORIDA MATHEMATICS OLYMPIADTEAM COMPETITION

May 14, 1998

1. Find all integers x, if any, such that 9 < x < 15 and the sequence

1 , 2 , 6 , 7 , 9 , x , 15 , 18 , 20

does not have three terms in arithmetic progression. If there are no suchintegers, write \NONE."

Answer: x = 14.The following chart allows us to eliminate 10, 11, 12, and 13:

x Possible arithmetic progression(s)if x is in the sequence

10 2, 6, 10; 2, 10, 18; 10, 15, 20.11 2, 11, 20; 7, 9, 11; 7, 11, 15.12 2, 7, 12; 6, 9, 12; 6, 12, 18; 9, 12, 15.13 1, 7, 13; 6, 13, 20.

345

As can be veri�ed, the sequence

1 ,2 ,6 ,7 ,9 ,14 ,15 ,18 ,20

does not have three terms in arithmetic progression.

2. A sequence a1, a2, a3, : : : is said to satisfy a linear recurrencerelation of order two if and only if there are numbers p and q such that, forall positive integers n,

an+2 = pan+1 + qan .

Find the next two terms of the sequence

2 ,5 ,14 ,41 , : : :

assuming that this sequence satis�es a linear recurrence relation of ordertwo.

Answer: The next two terms are 122 and 365.Let a1, a2, a3, a4, : : : be the sequence

2 ,5 ,14 ,41 , : : : .

Since this sequence satis�es a linear recurrence of order two, we know that,in particular, there are numbers p and q such that

a3 = pa2 + qa1 and a4 = pa3 + qa2 .

That is,14 = 5p+ 2q and 41 = 14p+ 5q .

Solving this system of equations in the usual way, we see that

p = 4 and q = �3 .

Thus, for all positive integers n,

an+2 = 4an+1 � 3an .

Hence the next two terms are

a5 = 4a4 � 3a3 = 4(41)� 3(14) = 122 ,

anda6 = 4a5 � 3a4 = 4(122)� 3(41) = 365 .

3. Seven tests are given and on each test no ties are possible. Eachperson who is the top scorer on at least one of the tests or who is in thetop six on at least four of these tests is given an award, but each person canreceive at most one award. Find the maximum number of people who couldbe given awards if 100 students take these tests.

346

Answer: 15.The maximum number of people who can receive an award occurs when

(i) no person is the top scorer more than once, and(ii) the 35 positions second through sixth in the seven competitions are �lledby each of 8 non-top-scorers at least four times.

This can happen as follows:

TestPosition 1 2 3 4 5 6 7

1 T U V W X Y Z2 A B C D E F G3 H A B C D E F4 G H A B C D E5 F G H A B C D6 E F G H { { {

4. Some primes can be written as a sum of two squares. We have, forexample, that

5 = 12 + 22 , 13 = 22 + 32 , 17 = 12 + 42 ,

29 = 22 + 52 , 37 = 12 + 62 , and 41 = 42 + 52 .

The odd primes less than 108 are listed below; the ones that can be writtenas a sum of two squares are boxed in.

3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 ,

37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , 71 ,

73 , 79 , 83 , 89 , 97 , 101 , 103 , 107 .

The primes that can be written as a sum of two squares follow a simple pat-tern. See if you can correctly �nd this pattern. If you can, use this patternto determine which of the primes between 1000 and 1050 can be written asa sum of two squares; there are �ve of them. The primes between 1000 and1050 are

1009 , 1013 , 1019 , 1021 , 1031 , 1033 , 1039 , 1049 .

No credit unless the correct �ve primes are listed.

Answer: 1009, 1013, 1021, 1033, 1049.Remark: An odd prime can be written as a sum of two squares if and

only if when the prime is divided by 4 the remainder is 1.

5. The sides of a triangle are 4, 13, and 15. Find the radius of theinscribed circle.

Answer: 1:5.

347

Let A, B, C be the vertices of the triangle. The area of 4ABC is, byHero's Formula, p

16(12)3(1) = 24 .

Let O be the centre of the inscribed circle and let r be its radius. Thus

24 = area of4AOB + area of 4BOC + area of4AOC= 0:5r(AB) + 0:5r(BC) + 0:5r(AC)

= 0:5r(4 + 13 + 15) = 16r ,

we see that r = 1:5.

6. In Athenian criminal proceedings, ordinary citizens presented thecharges, and the 500-man juries voted twice: �rst on guilt or innocence, andthen (if the verdict was guilty) on the penalty. In 399 BCE, Socrates (c. 469{399) was charged with dishonouring the gods and corrupting the youth ofAthens. He was found guilty; the penalty was death. According to I.F.Stone's calculations on how the jurors voted:

(i) There were no abstentions;

(ii) There were 80 more votes for the death penalty than there were forthe guilty verdict;

(iii) The sum of the number of votes for an innocent verdict and thenumber of votes against the death penalty equalled the number of votes infavour of the death penalty.

a) How many of the 500 jurors voted for an innocent verdict?

b) How many of the 500 jurors voted in favour of the death penalty?

Answer: a) 220 voted for an innocent verdict. b) 360 voted in favour ofthe death penalty.

Let x be the number of jurors who voted innocent and let y be thenumber of jurors who voted in favour of the death penalty. Since there wereno abstentions,

500� x jurors voted guilty

and

500� y jurors voted against the death penalty.

From (ii) we see that

y � (500� x) = 80

and from (iii) we see that

x+ (500� y) = y .

Simplifying we see that

x+ y = 580 and x� 2y = �500 .

348

Solving this system of equations we see that

3y = 1080

y = 360 .

Hence x = 220.

7. Find all x such that 0 � x � � and

tan3 x� 1 +1

cos2 x� 3 cot

��

2� x

�= 3 .

Your answer should be in radian measure.

Answer: �3, 2�

3, and 3�

4.

Since

�1 + 1

cos2 x= �1 + sec2 x = tan2 x

and

cot

��

2� x

�= tanx ,

we have the following chain of equations:

tan3 x� 1 +1

cos2 x� 3 cot

��

2� x

�= 3 ,

tan3 x+ tan2 x� 3 tanx� 3 = 0 ,

(tan2 x)(tanx+ 1)� 3(tanx+ 1) = 0 ,

(tan2 x� 3)(tanx+ 1) = 0 ,

tanx =p3 , tanx = �

p3 , or tanx = �1 ,

x =�

3,2�

3, or

3�

4.

That completes the Skoliad Corner for this number. I need your suit-able level contests and welcome your comments for future directions of theCorner.

349

MATHEMATICAL MAYHEM

Mathematical Mayhem began in 1988 as a Mathematical Journal for and byHigh School and University Students. It continues, with the same emphasis,as an integral part of Crux Mathematicorum with Mathematical Mayhem.

All material intended for inclusion in this section should be sent toMathematical Mayhem, Department of Mathematics, University of Toronto,100 St. George St., Toronto, Ontario, Canada. M5S 3G3. The electronicaddress is still

mayhem@math.toronto.edu

The Assistant Mayhem Editor is Cyrus Hsia (University of Western On-tario). The rest of the sta� consists of Adrian Chan (Upper Canada College),Jimmy Chui (University of Toronto), David Savitt (Harvard University) andWai Ling Yee (University of Waterloo).

Mayhem Problems

The Mayhem Problems editors are:

Adrian Chan Mayhem High School Problems Editor,Donny Cheung Mayhem Advanced Problems Editor,David Savitt Mayhem Challenge Board Problems Editor.

Note that all correspondence should be sent to the appropriate editor |see the relevant section. In this issue, you will �nd only solutions | thenext issue will feature only problems.

We warmly welcome proposals for problems and solutions. With theschedule of eight issues per year, we request that solutions from the lastissue be submitted in time for issue 6 of 2000.

High School Solutions

Editor: Adrian Chan, 229 Old Yonge Street, Toronto, Ontario, Canada.M2P 1R5 <all238@sprint.com>

H241. Find the integer n that satis�es the equation

1 � 1998 + 2 � 1997 + 3 � 1996 + � � �+ 1997 � 2 + 1998 � 1 =

�n

3

�.

Solution by Shawn Godin, Cairine Wilson S.S., Orleans, Ontario; PennyNom, University of Regina, Regina, Saskatchewan; and Edward T.H. Wang,

350

Wilfrid Laurier University, Waterloo, Ontario. (All three had virtually iden-tical solutions.)

Consider a set of three distinct numbers fa, b, cg chosen from the setf1, 2, : : : , ng such that a < b < c. Clearly, b = k for some k = 2, 3, : : : ,n� 1. Now, having chosen b, there are k � 1 ways to choose a, and n� kways to choose c. Hence,�

n

3

�=

n�1Xk=2

(k� 1)(n� k) .

Our original equation is the case n = 2000 of this equation, so therequired value of n is 2000.

Also solved byMasoudKamgarpour, Carson Graham Secondary School,North Vancouver, BC; and D.J. Smeenk, Zaltbommel, the Netherlands.

Remark. Three other solutions by Godin were received.

H242. Let a and b be real numbers that satisfy a2 + b2 = 1. Provethe inequality

ja2b+ ab2j �p2

2.

Determine the values of a and b for which equality occurs. (See how manyways you can solve this!)

Solution I by Masoud Kamgarpour, Carson Graham Secondary School,North Vancouver, BC.

Since a2+b2�2ab = (a�b)2 � 0 for all real a and b, 2ab � a2+b2 = 1,and so ab � 1=2. Since a2 + b2 + 2ab = (a + b)2 � 0 for all real aand b, 2ab � �(a2 + b2) = �1, and so ab � �1=2. Combining these twoinequalities, we have jabj � 1=2.

Furthermore, (a+b)2 = a2+2ab+b2 = (a2+b2)+2ab � 1+1 = 2,and so ja+ bj � p2. Multiplying these together, we get

ja2b+ ab2j = jab(a+ b)j = jabjja + bj �p2

2,

which leads to the desired result.

Equality occurs if and only if jabj = 1=2 and ja + bj = p2, which

implies (a� b)2 = a2 � 2ab+ b2 = 1� 2ab = 0, so a = b = �p2=2.

Solution II. By the QM-AM-GM inequality, we haver1

2=

sa2 + b2

2� jaj+ jbj

2�pjabj .

Thus, jabj � 1=2 and jaj + jbj � p2, with equality if and only if

a = b. But a2 + b2 = 1, so this implies that equality occurs if and only ifa = b = �p2=2.

351

By the Triangle Inequality, ja+ bj � jaj+ jbj � p2, so ja+ bj � p2.Hence, ja2b+ ab2j = jab(a+ b)j = jabjja + bj � p

2=2, as required.

Solution III. Since a and b are real numbers with a2 + b2 = 1, thereexists a � such that a = sin � and b = cos �, 0 � � < 2�. Then we have that��a2b+ ab2

�� =��sin2 � cos � + cos2 � sin �

��=

��� sin� cos �(sin � + cos �)��� =

1

2

���sin2�(sin � + cos �)���

=

p2

2

���sin2�(sin � � cos �4 + cos � � sin �4)���

=

p2

2

���sin2� sin(�+ �4)��� .

Since j sin2�j � 1 and j sin(�+ �4)j � 1, the desired inequality follows

immediately. Equality occurs if and only if j sin2�j = 1 and j sin(�+ �4)j = 1,

which occurs if and only if � = �=4 or 5�=4, so a = b = �p2=2.

Also solved by Edward T.H. Wang, Wilfrid Laurier University, Water-loo, Ontario.

H243. For a positive integer n, let f(n) denote the remainder ofn2 +2 when divided by 4. For example, f(3) = 3 and f(4) = 2. Prove thatthe equation

x2 + (�1)yf(z) = 10y

has no integer solutions in x, y, and z.

Solution by Masoud Kamgarpour, Carson Graham Secondary School,North Vancouver, BC; and Edward T.H. Wang, Wilfrid Laurier University,Waterloo, Ontario.

Since (2k)2 � 0 (mod 4) and (2k+1)2 = 4k2 +4k+ 1 � 1 (mod 4),every perfect square leaves a remainder of 0 or 1 when divided by 4. There-fore, when n2 + 2 is divided by 4, the remainder must be 2 or 3. Hence,f(z) = 2 or 3 for all z. So, (�1)yf(z)� 2, 3, 7, or 8 modulo 10.

But, x2 must be 0, 1, 4, 5, 6, or 9modulo 10. Therefore, x2+(�1)yf(z)can never be 0 modulo 10, and the proof is complete.

H244. For a positive integer n, let P (n) denote the sum of the digitsof n. For example, P (123) = 1 + 2 + 3 = 6. Find all positive integers nsatisfying the equation P (n) = n=74.

Solution by Keon Choi, student, A.Y. Jackson Secondary School, NorthYork, Ontario.

Let n be a k{digit number, so 10k�1 � n < 10k. Then P (n) � 9k,since each of the k digits is at most 9, so

n� 74P (n) � 10k�1 � 74 � 9k = 10k�1 � 666k .

352

If k = 5, then n� 74P (n)� 104� 666 � 5 > 0, so P (n) cannot equaln=74. If k > 5, then n � 74P (n) > 0, since 10k�1 increases much fasterthan 666k (a simple result by induction). So, in order for P (n) = n=74, werequire k � 4.

By the rule of divisibility by 9, P (n) � n (mod 9). Hence, 74P (n) =n � P (n) =) 73P (n) � 0 (mod 9). Since 9 is relatively prime to 73,P (n) � 0 (mod 9), and P (n) = n=74 must be a multiple of 9. Also, sincen has at most four digits, P (n) � 9 � 4 = 36.

If n=74 = 9, then n = 666, but P (666) = 18 6= 9.

If n=74 = 18, then n = 1332, but P (1332) = 9 6= 18.

If n=74 = 27, then n = 1998, and indeed P (1998) = 27.

If n=74 = 36, then n = 2664, but P (2664) = 18 6= 36.

Hence, the only solution is n = 1998.

Also solved by Mara Apostol, student, A.Y. Jackson Secondary School,North York, Ontario; Nick Harland, student, Vincent Massey C.I., Win-nipeg, Manitoba; Kenneth Ho, student, Don Mills C.I., Don Mills, Ontario;Masoud Kamgarpour, Carson Graham Secondary School, North Vancouver,BC; Condon Lau, student, David Thompson S.S., Vancouver, BC; EdwardT.H. Wang, Wilfrid Laurier University, Waterloo, Ontario; Dale Whitmore,student, Prince of Wales C.I., St. John's, Newfoundland; Wendy Yu, student,Danforth College and Tech. Inst., Toronto, Ontario.

Advanced Solutions

Editor: Donny Cheung, c/o Conrad Grebel College, University ofWaterloo, Waterloo, Ontario, Canada. N2L 3G6 <dccheung@uwaterloo.ca>

A217. Proposed by Alexandre Trichtchenko, OAC student, Brook�eldHigh School, Ottawa.

Show that for any odd prime p, there exists a positive integer n suchthat n, nn, nn

n

, : : : all leave the same remainder upon division by p wheren does not leave a remainder of 0 or 1 upon division by p.

Solution. We claim that n = 2p � 1 will satisfy the given conditions.First, n = 2p� 1 is odd and n = 2p� 1 � �1 (mod p). So for any tower ofns, we have

nn:::n

� (�1)n:::n

� �1 (mod p) .

Therefore, all the numbers n, nn, nnn

, : : : leave a remainder of p� 1 whendivided by p. Finally, since p is an odd prime, 2p� 1 cannot be 0 or 1.

353

A218. Proposed by Mohammed Aassila, Strasbourg, France.

(a) Suppose f(x) = xn+qxn�1+t, where q and t are integers, and supposethere is some prime p such that p divides t but p2 does not divide t.Show, by imitating the proof of Eisenstein's Theorem, that either f isirreducible or f can be reduced into two factors, one of which is linearand the other irreducible.

(b) Deduce that if both q and t are odd then f is irreducible.

(Generalization of Question 1, IMO 1993)

[Ed.: Here \irreducible" means \irreducible over the integers".]

Solution by the proposer. (a) Suppose on the contrary that f(x) =g(x)h(x) with

g(x) = xr + cr�1xr�1 + � � �+ c0 ,

h(x) = xs + bs�1xs�1 + � � �+ b0 ,

0 < r, s < n and r + s = n. Since p divides t and p2 does not divide t, wehave p dividing one of c0 or b0, but not both. Say p divides c0, but it doesnot divide b0. Letm be the largest i such that p divides the ith coe�cient ci.Then 0 � m < r and p does not divide cm+1. The coe�cient of xm+1 inf(x) is

cm+1b0 + cmb1 + � � � � cm+1b0 (mod p) .

Further, this is not congruent to 0modulo p. So n > r � m+1 � n�1 � r.Hence n� 1 = r, s = 1, and so h(x) is a linear factor. It remains to showthat g(x) is irreducible, but since m = r � 1, we have that p divides c0, c1,: : : , cr�1, and p does not divide cr. So it follows that g(x) is irreducible byEisenstein's criterion.

(b) If q and t are odd, then f(x) is odd for any integer x, and so it hasno integer roots. Thus it cannot have a linear factor and so by part (a) isirreducible.

A219. Proposed by Mohammed Aassila, Strasbourg, France.Solve the following system:

3

�x+

1

x

�= 4

�y +

1

y

�= 5

�z +

1

z

�; xy + yz + zx = 1 .

Solution by D.J. Smeenk, Zaltbommel, the Netherlands.The values x, y, and z are all positive or all negative. If (x0; y0; z0) is

a solution of the equations, then (�x0;�y0;�z0) must also be a solution.

In any triangle ABC, with angles A, B, and C, we have

tanA

2tan

B

2+ tan

B

2tan

C

2+ tan

C

2tan

A

2= 1 .

354

We denote x = tan(A=2), y = tan(B=2), and z = tan(C=2). Thenwe have x+ 1=x = 2= sinA, y + 1=y = 2= sinB, and z + 1=z = 2= sinC.So the equation to be solved can be rewritten as

3

sinA=

4

sinB=

5

sinC.

So triangle ABC has a right angle at C. We �nd that sinA = 3=5,cosA = 4=5, sinB = 4=5, cosB = 3=5, and C = �=2, and

x = tanA

2=

1� cosA

sinA=

1

3,

y = tanB

2=

1� cosB

sinB=

1

2,

z = tanC

2= 1 .

The solutions (x; y; z) are (1=3; 1=2;1) and (�1=3;�1=2;�1).A220. Proposed by Waldemar Pompe, student, University of War-

saw, Poland.P is an interior point of triangle ABC. D, E, and F are the feet of the

perpendiculars from P to the lines BC, CA, and AB, respectively. Let Qbe the interior point of triangle ABC such that

\ACP = \BCQ and \BAQ = \CAP .

Prove that \DEF = 90� if and only ifQ is the orthocentre of triangleBDF .

Solution. First for some notation. Let x = \BAQ = \CAP ,y = \ACP = \BCQ. Let AQ intersect FP at U , CQ and DP at V ,AQ and FE at S, and CQ and DE at T .

A

B CD

E

F

PU

S

V T

Q

q

q qq

q

q

q q

q

q

qq

Notice that AQ and CQ are perpendicular to EF and ED respec-tively. To see this, consider triangle FAU . Then \UFS = \PFE =\PAE = x since PFAE is a cyclic quadrilateral. Now \AUF = 90� � x,so that \FSU = 90�. Thus EF is perpendicular to AQ. Likewise, in tri-angle DCV , we see that \VDT = \PDE = \PCE = y. Thus ED isperpendicular to CQ.

355

Assume that \DEF = 90�. We show thatDQ is parallel to PF . SincePF is perpendicular to AB, DQ extended to AB will be the altitude oftriangle BDF from vertex D. Likewise, it can be shown that FQ is parallelto PD and it extended is the altitude of triangle BDF from vertex F . Thiswill mean that Q, the intersection of these two altitudes, is the orthocentreof triangle BDF .

First, consider the quadrilateral QSET . Then \QSE = \QTE = 90�

since we showed above that CQ is perpendicular to ED. Recall \SET =\FED = 90� by assumption. Thus \SQT = 90�.

Now trianglesDPC and QAC are similar since they both have a rightangle and \QCA = \DCP . Thus DC=PC = QC=AC. Using this factand knowing \DCQ = \PCA implies that triangles QDC and APC aresimilar. Thus \DQC = \PAC = x.

Lines FE and QC are parallel since they are extended line segments ofopposite sides of the rectangle QSET . Line FP makes an angle of x withFE since \PFE = \PAE. Line QD makes an angle of x also withQC andin the same con�guration as shown in the diagram. Thus FP and QD areparallel.

Challenge Board Problems

Editor: David Savitt, Department of Mathematics, Harvard University,1 Oxford Street, Cambridge, MA, USA 02138 <dsavitt@math.harvard.edu>

C79. Proposed by Mohammed Aassila, Strasbourg, France.Let f : R+ ! R

+ be a non-increasing function, and assume that thereare two constants p > 0 and T > 0 such thatZ

1

t

f(s)p+1 ds � Tf(0)pf(t)

for all t 2 R+. Prove that

f(t) � f(0)

�T + pt

T + pT

��1=pfor all t � T .

Solution by the proposer. If f(0) = 0, then f � 0 and there is nothingto prove. Otherwise, replacing f(x) by f(x)=f(0), we may assume thatf(0) = 1, and our hypothesis becomesZ

1

t

f(s)p+1 ds � Tf(t)

356

for all t 2 R+. Introduce the auxiliary function F : R+ ! R+ de�ned as

F (t) =

Z1

t

f(s)p+1 ds ,

so that F (t) � Tf(t). Then F is non-increasing and locally absolutely con-tinuous, and therefore

F 0(t) = �f(t)p+1 � �T�p�1F (t)p+1

almost everywhere in (0;1).

(Readers who are unfamiliar with the terms `locally absolutely contin-uous' and `almost everywhere' may freely ignore them: they are technicalconditions which allow us to solve the problem in greater generality. Ifinstead we add the hypothesis that f is continuous, then we may replace`locally absolutely continuous' with `di�erentiable' and `almost everywhere'with `everywhere', and the proof carries through with no other changes.)

Letting B be the (possibly in�nite) supremum of the set ft j F (t) > 0g,we �nd that

(F�p)0 = �pF�p�1F 0 � pT�p�1

almost everywhere in (0;B). It follows by integration that

F (t)�p� F (0)�p � pT�p�1t

for all t 2 [0; B), whence

F (t) � �F (0)�p + pT�p�1t

��1=pfor all t 2 [0; B). Since F (0) � T , it follows further that

F (t) � (T�p + pT�p�1t)�1=p = T (p+1)=p(T + pt)�1=p .

Since t � B by de�nition implies F (t) = 0, the above inequality evidentlyholds for all t 2 R+.

We can estimate the left-hand side of this inequality as follows:

F (t) =

Z1

t

f(s)p+1 ds

�Z T+(p+1)t

t

f(s)p+1 ds

� (T + pt)f(T + (p+ 1)t)p+1 ,

where the right-most inequality uses the fact that f is non-increasing. Con-sequently,

(T + pt)f(T + (p+ 1)t)p+1 � T (p+1)=p(T + pt)�1=p ,

357

which reduces immediately to

f(T + (p+ 1)t) � T 1=p(T + pt)�1=p .

Writing t = T + (p+ 1)t0 for t � T , we �nd

f(t) ��T + pt0

T

��1=p=

�T + pt

T + pT

��1=p

,

as desired.

C80. Suppose a1, a2, : : : , am are transpositions in Sn (the symmetricgroup on n elements) such that a1a2 � � � am = 1. Show that if the ai generateSn, thenm � 2n� 2.

Solution. We take our Sn to be the symmetric group on the n integers1, 2, : : : , n. Recall that when we say that the permutation ai �xes the integerj, we mean that ai(j) = j or, equivalently, that j does not appear in thecycle decomposition of ai. (We will use cycle-decomposition notation forthese transpositions; that is, ai = (b c) is the permutation which swaps band c and �xes everything else. Henceforth, we will also denote the identitypermutation by e, so as not to confuse it with the integer 1 which is beingpermuted.)

To begin with, we will argue that unless none of the transpositions ai�x 1 (that is, unless all the ai are of the form (1 bi) already), it is possible toreplace the product a1a2 � � � am = e with a new product a01a

0

2 � � � a0m = e ofthe same length, such that the a0i still generate Sn and such that fewer of thea0i �x 1 than do the ai. To show this, observe that either none of the ai �x 1,or else there must exist some pair of transpositions of the form aj = (1 b) andak = (b c)with 1, b, and c all di�erent: if not, this would certainly contradictthe assumption that the ai generate Sn. Choose such an aj and ak with jk�jjas small as possible, and without loss of generality assume that k > j. Thenby the minimality of jk � jj, all of the transpositions aj+1, : : : , ak�1 must�x b and c, so the product aj+1 � � � ak�1(b c) = (b c)aj+1 � � � ak�1. Thus

aj � � � ak = (1 b)aj+1 � � � ak�1(bc)= (1 b)(b c)aj+1 � � � ak�1= (1 c)(1 b)aj+1 � � � ak�1 .

So,

e = a1 � � � am= a1 � � � aj�1(1 c)(1 b)aj+1 � � � ak�1ak+1 � � � am ,

and the latter product is certainly still of the same length, its transpositionsstill generate Sn (because (b c) = (1 c)(1 b)(1 c)), and fewer of the transpo-sitions �x 1, as desired.

358

Evidently, by repeated application of the above procedure, we eventu-ally arrive at a product a01 � � � a0m = e of the same length m, and such thatnone of the a0i �x 1; that is, all of the a

0

i are of the form (1 bi) for some integerbi between 2 and n. Since the a0i generate Sn, each of (1 2), : : : , (1 n)mustappear in the product at least once. However, if (1 b) were to appear in theproduct a01 � � � a0m = e exactly once, then certainly the product could not �xb; so, each of (1 2), : : : , (1 n)must appear in the product at least twice, andtherefore m � 2(n� 1), as desired.

Problem of the Month

Jimmy Chui, student, University of Toronto

Problem. Let a be the length of a side and b bethe length of a diagonal in the regular pentagonPQRST . Prove that

b

a� a

b= 1 .

(1998 Descartes, D1)

Q

R

S

P

T

a

b

Solution I. The internal angle at each vertex of a regular pentagon is108�. Note that a and b are positive, and by the Cosine Law, we have

b2 = a2 + a2 � 2 � a � a � cos 108�= 2a2(1 + cos 72�) = 4a2 cos2 36� ,

and so b = 2a cos 36�.

Let x = cos 18�. We know that 0 < x < 1 and that cos(5 � 18�) = 0.Setting � = 18� and using De Moivre's Theorem, we have cis 5� = (cis �)5.Comparing real terms on both sides, we get

cos 5� =

�5

0

�(cos �)5 �

�5

2

�(cos �)3(sin�)2 +

�5

4

�(cos �)(sin �)4

= x5 � 10x3(1� x2) + 5x(1� x2)2 = 16x5 � 20x3 + 5x ,

so 16x4 � 20x2 + 5 = 0, since x 6= 0.

By the quadratic formula, x2 =20�p400� 320

32=

5�p5

8,

with x positive. Since

s5�p5

8<

r5� 1

8=

r1

2= cos 45� , we must

359

have

cos 18� = x =

s5 +

p5

8.

Then,

cos 36� = 2x2 � 1 =1 +

p5

4.

Evaluating b, we obtain

b =1+

p5

2� a .

Finally,

b

a� a

b=

1+p5

2� 2

1 +p5

=(1 +

p5)2 � 4

2(1 +p5)

=2 + 2

p5

2(1 +p5)

= 1 .

Solution II. Draw the line PS. This has a length of b. By symmetry,\QPR = \TPS = 36�, and so \RPS = 36�. From triangle PQR, we get

cos 36� =a2 + b2 � a2

2ab=

b

2a.

From triangle PRS, we get

cos 36� =b2 + b2 � a2

2b2=

2b2 � a2

2b2.

Now, a and b are both positive and cannot be equal, and so we have

b

2a=

2b2 � a2

2b2

===) b3 = 2ab2 � a3

===) 0 = a3 � 2ab2 + b3

= (a� b)(a2 + ab� b2)===) 0 = a2 + ab� b2

===) 1 =b

a� a

b.

Solution III. Construct a point A on PR such that SA bisects the an-gle PSR. Chasing angles, we observe that triangle SRA is isosceles, withRS = AS = a, and that triangle SPA is isosceles, with SA = PA = a.We can also see that triangle SAR is similar to triangle PRS, and from this,we have

AR

AS=

RS

RP===) b� a

a=

a

b===) b

a� a

b= 1 .

360

J.I.R. McKnight Problems Contest 1990

1. A certain numberN consists of three digits which are consecutive termsof an arithmetic sequence. If N is divided by the sum of its digits thequotient is 48. Also, if 198 is subtracted from N , the resulting numbercomprises the digits of N in reversed order. FindN .

2. Two chords on a circle AB and CD intersect at right angles at E suchthat AE = 2, EB = 6 and DE = 3. Find the area of the circle.

3. Solve for x:

log10

�px+

2px

�=

1

2+ log10 2 .

Leave your answer in simplest radical form.

4. A point P is taken on the curve y = x3. The tangent at P intersectsthe curve at Q. Prove that the slope of the curve at Q is four times theslope at P .

5. The sum of the length of the hypotenuse and another side of a rightangled triangle is 12. Prove that the area of the triangle is maximumwhen the angle between the two sides is 60�.

6. A curve of intersection of a sphere with a plane through the centre ofthe sphere is called a great circle. One great circle divides a sphere'ssurface into 2 regions; 2 great circles divide the sphere into 4 regions.If no 3 great circles intersect in a common point, �nd the number ofregions into which the surface of a sphere is divided by: 3, 4, 5, n greatcircles.

7. The sum of the �rst three terms of a geometric sequence is 37 and thesum of their squares is 481. Find the �rst three terms of all such se-quences.

8. The pointD is on the sideBC of an equilateral triangleABC such thatBD = 1

3BC and E is a point on AB equidistant from A andD. Prove

that CE = EB + BD.

9. Given that

tanA+ tanB = a ,

cotA+ cotB = b ,

tan(A+B) = c ,

�nd c in terms of a and b.

10. Determine the number of factors of 2 in the �rst positive integer greaterthan (4 + 2

p3)1990.

361

Waiting in Wonderland

Cyrus Hsiastudent, University of Toronto

Jack Wang, a student at Cedarbrae Collegiate Institute in Toronto, sentme the following fabulous problem. It was originally a problem in the 1998J.I.R. McKnight Problem Solving Competition held annually in Scarboroughand whose problems we are currently publishing in CRUX with MAYHEM.The problem stated here is modi�ed from the original problem in the paper.

Jack's Meeting Problem

A teacher at Cedarbrae, Mr. Iacobucc, and his girlfriend want to meetin Canada's Wonderland between 1:00 pm and 2:00 pm. [Ed: Whether ornot his girlfriend's name is Alice is uncon�rmed at the time this article waswritten.] Each of them agrees to wait for each other for 15minutes. Assumeeach of them arrives between the designated times randomly. What is theprobability that they will meet each other?

The reader is encouraged to �nd the answer to this problem on his/herown before reading the solution. The answer turns out to be a relativelynice rational number. If you solve it, you may wish to skip the next sectionwhich deals with some simple properties of probability. Otherwise, sit backand absorb some motivating facts that may be useful in solving the aboveproblem.

An Aside on Simple Probability

Finite OutcomesReaders will be familiar with the examples of tossing coins or picking

coloured marbles from a bag. In any experiments similar to these, there arecertain outcomes (tossing 2 heads or picking a red marble) whose likelihoodof occurring we want to know. The desired outcome is called an event. Theset of all possible outcomes in a given experiment is called a sample space.

Now to �nd the probability, we need to know the number of ways thatthe desired outcome (event) occurs compared to the number of ways of allpossible outcomes (size of the sample space). In fact, the probability of anevent, E, in a given sample space, S, is given by

P (E) =N(E)

N(S),

where N(A) is the number of ways of event A occurring.

Copyright c 1999 CanadianMathematical Society

362

For example, suppose you wish to know the probability of tossing 2heads in 2 coin tosses. Then N(tossing 2 heads) = 1 since there is only oneway of doing this, when both tosses are heads. ButN(all possible outcomes)equals 4, since we could have (tail, tail), (tail, head), (head, tail) or (head,head). The probability is then 1=4.

In�nite Outcomes

Often the number of ways for the desired outcome or the sample spaceis in�nite. In this case, the formula for probability given above is no longervalid. For example, suppose the real line from 0 to 4 is drawn on a piece ofpaper in black ink with total length 4 centimetres. Now arbitrarily highlight aline segment of length 1 centimetre. What is the probability that a randomlyselected real number from 0 to 4 lies within the highlighted segment?

Now if we used the formula, we have to check the number of all possibleoutcomes. The number of ways of selecting a real number from 0 to 4 isin�nite. In fact, the number of ways of selecting a real number from anyarbitrary line segment of positive length is in�nite.

However, it is intuitive that a randomly selected real number will lieon the non-highlighted sections three times as likely than it would on thehighlighted section, since the length of non-highlighted to highlighted regionsis three to one. The probability should then be 1=4.

We could apply similar reasoning to selecting points in higher dimen-sions. Suppose a circular archery target has radius 2 feet in length and thebull's-eye region is a circle of radius 1 foot in its centre. What is the proba-bility of an arrow landing in the bull's-eye if the arrow lands randomly withinthe target? By calculating the areas, we have a � square feet region of thebull's-eye and a 4� square feet region of the whole target. The probabilitywould then be 1=4. [Ed: In case the reader is wondering, not every proba-bility is 1=4. The reader is strongly urged to verify this.]

So what does all of this have to do with the problem? Let us go on tothe solution.

Solution to Jack's Meeting Problem

Consider Mr. Iacobucc's time line onthe x{axis and his girlfriend's on they{axis. Say the origin is at 1:00 forboth people and time progresses inthe positive direction as shown in thefollowing graph.

2:00

1:00

1:00 2:00

363

Now if Mr. Iacobucc is there from[t; t + 15], 0 � t � 60 minutes after1:00, draw in the two lines x = t andx = t + 15. Similarly, his girlfriendgets there for [s; s+ 15] and so drawin the two lines y = s and y = s+15.The graph is now as follows.

s

s+15

t t+15

So, for what cases of s and t do they meet? If they meet, then there issome point in time when they are both there. In other words, there is sometime k in which t � k � t+15 and s � k � s+15. This time k exists if andonly if the square formed by the four lines above intersect the line y = x.(Why?)

The question then becomes: What isthe probability of a square of sides 15units (minutes) intersecting the liney = x in the square of sides 60 units(minutes)? Another way to look atthis is to �gure out all possible posi-tions to place the lower left corner ofthe small square in the large squareand the positions that would makethe square intersect the line y = x.

15

15

15

15

-� 60

6

?

60

It turns out that if we choose the lower left corner of the small square any-where in region shown, then intersection occurs. Note: We allow for thesmaller square to extend beyond the bounds of the larger square since weallow for either of the couple to come within the last 15 minutes.

By simply calculating the area of the region to the area of the wholesquare, we �nd that the probability of meeting is then

602 � 452

602=

1575

3600=

7

16.

In general, Jack pointed out the result for two people meeting in acommon interval m and waiting for a time interval n. It simpli�es to thefollowing: The probability of meeting is

2n

m��n

m

�2,

with of course n � m being real numbers.

Another Aside on Probability

Jack also notes the following interesting similarity: Suppose A and Bare two events. The probability of either one or the other happening is given

364

by the equation

P (A or B) = P (A) + P (B)� P (A and B) .

Suppose A and B are two independent events and that P (A) = P (B)= n=m. For example, suppose A is the event of throwing a heads and Bis the event of getting an even number in a die roll. Now in one toss of acoin and a die roll, what is P (A or B), the probability of getting heads or aneven number? To get a heads or an even number, we can get a heads and rollany number, roll an even number and toss heads or tails, then remove thecase were we counted tossing heads and rolling an even number twice. SoP (A or B) = P (A)+P (B)�P (A andB). Since tossing a heads and rollingan even number are independent events, P (A and B) = P (A)�P (B). Theresult is that P (A or B) = 2n=m� (n=m)2.

This answer makes us very suspicious that Jack's problem could besolved another way. The question to ask then, is whether or not we canapply this method to solving Jack's problem more elegantly.

Some Generalizations

Two People with Di�erent Amounts of Patience

Unfortunately, not everyone has thesame amount of patience. SupposeMr. Iacobucc is only willing to waitfor a minutes while his girlfriend willwait bminutes. What is the probabil-ity of meeting within a d minute in-terval?

b

a

a

b

-� d

6

?

d

Three's Company

Now suppose that Jack was also to meet the couple at Wonderland.Again, all three people are willing to wait 15 minutes and it is within thetime from 1:00 pm to 2:00 pm. Now what is the probability that all threewill meet?

Here, using the same technique as above, we would be trying to �nd theratio of a certain volume to the volume of a cube. The group would all meetat a certain point in time if a cube of sides 15 intersects the line x = y = zin space.

Another Probability Problem

Problem.

Take a line segment of length 1. What is the probability that makingtwo random cuts along the line will produce three line segments that are thesides of a non-degenerate triangle?

365

Solution.First let us set up the problem. Sup-pose the three line segments havelengths x, y, and z. The constraintson them are x + y + z = 1, and x,y, z � 0. This de�nes the portionof a plane in the �rst quadrant of thethree dimensional cartesian space asshown:

��

��

��

-

6(0; 0; 1)

(0; 1; 0)

(1; 0; 0)

z

y

x

Now for the three lengths to form a non-degenerate triangle the followingtriangle inequalities must hold: x + y > z, x + z > y, and y + z > x.Each of these inequalities de�nes the space on one side of a given plane. Forexample, x+y > z de�nes the space below the plane given by x+y�z = 0.

Now where do the three space do-mains intersect on the plane x+ y+z = 1? The plane x + y + z = 1intersects the �rst quadrant to forma triangle. The three space domainsintersect in a triangle whose verticesare the mid-points of the sides of the�rst triangle.

��

��

��

-

6z

y

x

a

a

a

a

a

a

The probability is then 1=4.

Exercises

1. What is the answer to the probability of three people meeting as givenabove?

2. Exploration: Investigate a similar problem for four or more people.

3. Again consider three people. We assumed that each person will remainexactly 15minutes. Usually, however, when one person meets anotherthey will both wait and leave together so that either the �rst personurges the second to both leave or else the second will force the �rst tocontinue another 15 minutes. In either case, what is the probability ofall three meeting?

AcknowledgementsJack Wang, grade 12 student at Cedarbrae Collegiate Institute in Toronto,

Ontario, Canada, for his interesting account of the problem.

Cyrus Hsia

21 Van Allan Road

Scarborough, Ontario

M1G 1C3

hsia@math.toronto.edu

366

PROBLEMS

Problem proposals and solutions should be sent to Bruce Shawyer, Department

of Mathematics and Statistics, Memorial University of Newfoundland, St. John's,

Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution,

together with references and other insights which are likely to be of help to the edi-

tor. When a submission is submitted without a solution, the proposer must include

su�cient information on why a solution is likely. An asterisk (?) after a number

indicates that a problem was submitted without a solution.

In particular, original problems are solicited. However, other interesting prob-

lems may also be acceptable provided that they are not too well known, and refer-

ences are given as to their provenance. Ordinarily, if the originator of a problem canbe located, it should not be submitted without the originator's permission.

To facilitate their consideration, please send your proposals and solutions

on signed and separate standard 81

2"�11" or A4 sheets of paper. These may

be typewritten or neatly hand-written, and should be mailed to the Editor-in-

Chief, to arrive no later than 1 April 2000. They may also be sent by email to

crux-editors@cms.math.ca. (It would be appreciated if email proposals and solu-

tions were written in LATEX). Graphics �les should be in epic format, or encapsulated

postscript. Solutions received after the above date will also be considered if there

is su�cient time before the date of publication. Please note that we do not accept

submissions sent by FAX.

2463?. Proposed by V�aclav Kone �cn �y, Ferris State University, BigRapids, MI, USA.

Suppose that k 2 Z. Prove or disprove that

�tan

�3n�

11

�+ 4sin

�2n�

11

��2

=

�tan

�5n�

11

�� 4 sin

�4n�

11

��2=

8><>:

11 for n 6= 11k ,

0 for n = 11k .

2464. Proposed by Michael Lambrou, University of Crete, Crete,Greece.

Given triangle ABC with circumcircle �, the circle �A touches AB andAC at D1 and D2, and touches � internally at L. De�ne E1, E2, M , andF1, F2, N in a corresponding way. Prove that

(a) AL, BM , CN are concurrent;

(b) D1D2, E1E2, F1F2 are concurrent, and that the point of concurrency isthe incentre of4ABC.

367

2465?. Proposed by Albert White, St. Bonaventure University,St. Bonaventure, NY, USA.

For n � 1, prove that

n�1Xi=0

�n

i

�n�1�iXj=0

�n� 1

j

�= 4n�1 .

2466. Proposed by Victor Oxman, University of Haifa, Haifa, Israel.Given a circle (but not its centre) and two of its arcs, AB and CD,

and their mid-points M and N (which do not coincide and are not the endpoints of a diameter), prove that all the unmarked straightedge and compassconstructions that can be carried out in the plane of the circle can also bedone with an unmarked straightedge alone.

2467. Proposed by Walther Janous, Ursulinengymnasium, Inns-bruck, Austria.

Given is a line segment UV and tworays, r and s, emanating from V suchthat \(UV; r) = \(r; s) = 60�,and two lines, g and h, on U suchthat \(UV; g) = \(g;h) = �,where 0 < � < 60�. U V

�

�60�

60�

rs

h

g

The quadrilateral ABCD is determined by g, h, r and s. Let P be thepoint of intersection of AB and CD.

Determine the locus of P as � varies in (0; 60�).

2468. Proposed by Walther Janous, Ursulinengymnasium, Inns-bruck, Austria.

For c > 0, let x, y, z > 0 satisfy

xy + yz + zx+ xyz = c . (1)

Determine the set of all c > 0 such that whenever (1) holds, then wehave

x+ y+ z � xy + yz+ zx .

2469. Proposed by Paul Yiu, Florida Atlantic University, Boca Ra-ton, FL, USA.

Given a triangle ABC, consider the altitude and the angle bisector ateach vertex. Let PA be the intersection of the altitude from B and the bisec-tor at C, and QA the intersection of the bisector at B and the altitude at C.These determine a line PAQA. The lines PBQB and PCQC are analogouslyde�ned. Show that these three lines are concurrent at a point on the linejoining the circumcentre and the incentre of triangleABC. Characterize thispoint more precisely.

368

2470. Proposed by Paul Yiu, Florida Atlantic University, Boca Ra-ton, FL, USA.

Given a triangle ABC, consider the median and the angle bisector ateach vertex. LetPA be the intersection of the median fromB and the bisectoratC, andQA the intersection of the bisector atB and themedian atC. Thesedetermine a linePAQA. The linesPBQB andPCQC are analogously de�ned.Show that these three lines are concurrent. Characterize this intersectionmore precisely.

2471. Proposed by Vedula N. Murty, Dover, PA, USA.

For all integers n � 1, determine the value of

nXk=1

(�1)k�1kk+ 1

�n+ 1

k

�.

2472. Proposed by V�aclav Kone �cn �y, Ferris State University, BigRapids, MI, USA.

If A, B, C are the angles of a triangle, prove that

cos2�A�B

2

�cos2

�B � C

2

�cos2

�C �A

2

�

��8 sin

�A

2

�sin

�B

2

�sin

�C

2

��3.

2473. Proposed byMiguel Amengual Covas, Cala Figuera, Mallorca,Spain.

Given a point S on the side AC of triangle ABC, construct a linethrough S which cuts lines BC and AB at P and Q respectively, such thatPQ = PA.

2474. Proposed by Mohammed Aassila, CRM, Universit �e deMontr �eal, Montr �eal, Qu �ebec.

Let f : R+ ! R+ be a decreasing continuous function satisfying, for

all x, y 2 R+:

f(x+ y) + f�f(x) + f(y)

�= f

�f�x+ f(y)

�+ f

�y+ f(x)

��.

Obviously f(x) = c=x (c > 0) is a solution. Determine all other solutions.

2475. Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,Spain.

Prove that

nXj=0

nXk=1

(�1)j+k�2n

2j

��2n

2k� 1

�= 0 .

369

SOLUTIONSNo problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.

The names of MICHEL BATAILLE, Rouen, France (2324), WALTHER JANOUS, Ursulinen-gymnasium, Innsbruck, Austria (2324) and ARAM TANGBOONDOUANGJIT, Carnegie MellonUniversity, Pittsburgh, PA, USA (2352) were inadvertently omitted from the lists of solvers inthe last issue.

2356. [1998: 303] Proposed by Victor Oxman, University of Haifa,Haifa, Israel.

Five points, A, B, C,K, L, with whole number coordinates, are given.The points A, B, C do not lie on a line.

Prove that it is possible to �nd two points, M , N , with whole numbercoordinates, such that M lies on the line KL and 4KMN is similar to4ABC.

Solutionby Jeremy Young, NottinghamHigh School,Nottingham, Eng-land.

Take A as the origin of the coordinate system, such that the points A,B, C have position vectors

�00

�,�x1y1

�,�x2y2

�respectively. Let � be the angle

through which we must rotate AB counterclockwise to make it parallel toKL. Since AB and KL each pass through two points with whole numbercoordinates, both have rational slopes. Therefore, tan� = p

q, some p; q 2 Z,

if � 6= 90�.

The images of B and C under rotation through � 6= 90� are

B0 =

�x1 cos�� y1 sin�x1 sin�+ y1 cos�

�=

cos�

q

�qx1 � py1px1 + qy1

�and

C0 =cos�

q

�qx2 � py2px2 + qy2

�.

De�ne B00 and C00 by enlargement with scalar factor qcos�

; giving

B00 =

�qx1 � py1px1 + qy1

�and C00 =

�qx2 � py2px2 + qy2

�.

If � = 90� take B00 = B0 =

� �y1x1

�and C00 = C0 =

� �y2x2

�.

Since p, q, x1, x2, y1 and y2 are whole numbers, these are points with wholenumber coordinates and4KMN is simply a translation of 4AB00C00 withA going to K.

Also solved by RICHARD I. HESS, RanchoPalos Verdes, CA, USA;MICHAEL LAMBROU,University of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul's School, London, England; andthe proposer. There was one incomplete solution.

370

2357. [1998: 303] Proposed by Gerry Leversha, St. Paul's School,London, England.

An unsteady man leaves a place to commence a one-dimensional ran-dom walk. At each step he is equally likely to stagger one step to the eastor one step to the west. Let his expected absolute distance from the startingpoint after 2n steps be a. Now consider 2n unsteady men each engaging inindependent random walks of this type. Let the expected number of men atthe starting point after 2n steps be b. Show that a = b.

Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Austria.

We label the points that it is possible to be stepped upon by : : : , �2,�1, 0, 1, 2, : : : ; that is, Z. We declare 0 to be the starting point of all walks.

Since there are 2n random steps, the only eligible �nal points are 2j,where �n � j � n.

Let a(2n)j be the number of all possible 2n{walks ending in 2j. Then,

clearly, a(0)0 = 1, a

(2)�1 = a

(2)1 = 1, a

(2)0 = 2, and a

(0)j = a

(2)j = 0 for all

other values of j. Furthermore, a(2n+2)k = a

(2n)k�1 + 2a

(2n)k + a

(2n)k+1 , where

n � 1 and k 2 Z.This relation is just the \doubled" Pascal Triangle Iteration, having

a(2n)0 =

�2nn

�as its central element.

Therefore, a(2n)k =

� 2nn+k

�, �k � n � k, and further,

P (X=2k) =

8>>><>>>:

2� 2nn+k

�22n

, k > 0 ,�2nn

�22n

, k = 0 .

Here, X denotes the absolute distance from the starting point after 2n steps.

Now,

E(X) =nX

k=0

2kP (X=2k) =1

22n�1

nXk=1

2k

�2n

n+ k

�=

sn

22n�2,

where

sn =

nXk=1

k

�2n

n+ k

�=

nXk=1

(n+ k)

�2n

n+ k

�� n

nXk=1

�2n

n+ k

�

= 2n

nXk=1

�2n� 1

n+ k� 1

�� n

�1

2

��22n �

�2n

n

��

= 2n

�1

2

�22n�1 � n 22n�1 +

n

2

�2n

n

�=

n

2

�2n

n

�.

371

Therefore, E(X) =n

22n�1

�2n

n

�= a.

Let Y be the number of men (out of 2n possibilities) being at the start-ing point after 2n steps.

Then, Y is binomial p{distributedwith p = P (X=0), yielding, at once,that

E(Y ) = 2np =n

22n�1

�2n

n

�= b .

Thus, E(X) = E(Y ); that is a = b.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; RICHARDI. HESS, Rancho Palos Verdes, CA, USA; KATHLEEN E. LEWIS, SUNY Oswego, Oswego, NY,USA; JEREMY YOUNG, student, Nottingham High School, Nottingham, UK; and the proposer.

Janous comments that this is one of his favourite problems from CRUX with MAYHEM

in 1998! He also calculated the variances, and found that

V (X) � 2n

�1 � 2

�

�and V (Y ) � 2

�

�pn� � 1

�.

The proposer comments: This problem arose out of an English A level statistics course inwhich one pupil mistakenly answered the wrong question but got the `right' answer. It is not toodi�cult to prove, by combinatorial arguments, that the two expectations are equal, but the realchallenge is to show why they are. Can one problem be reduced to the other? Kathleen Lewiscomes closest to achieving this by showing that the two satisfy parallel recurrence relations, buteven so, I do not feel really satis�ed that we are dealing with two aspects of the same thing. Iawait some marvellous illumination which will dispose of the matter in a nutshell so that I cansay: There! It was obvious all the way along! [Ed. Can any reader help?]

2358. [1998: 303] Proposed by Gerry Leversha, St. Paul's School,London, England.

In triangle ABC, let the mid-points of BC, CA, AB be L, M , N ,respectively, and let the feet of the altitudes from A, B, C be D, E, F ,respectively. Let X be the intersection of LE and MD, let Y be the inter-section ofMF andNE, and let Z be the intersection ofND and LF . Showthat X, Y , Z are collinear.

Editor's comment. Most solvers noted that points L, M , N , D, Eand F all lie on the nine-point circle (a.k.a. Euler's circle or Feuerbach'scircle), so by Pascal's Theorem (see, for example: Roger A. Johnson,ModernGeometry, Houghton-Mi�en Co., 1929, Theorem 385)X, Y , Z are collinear.

Bellot Rosado, Bataille, Smeenk and the proposer all note that X, Yand Z lie on the Euler line of 4ABC.

Solved by CHETAN BALWE, student, University of Michigan, Ann Arbor, MI, USA;MICHEL BATAILLE, Rouen, France (2 solutions); FRANCISCO BELLOT ROSADO, I.B. EmilioFerrari, Valladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK (2solutions); NIKOLAOS DERGIADES, Thessaloniki, Greece; C. FESTRAETS-HAMOIR, Brussels,Belgium; WALTHER JANOUS,Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU,University of Crete, Crete, Greece; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria;TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel, the Netherlands(2 solutions);

372

PAUL YIU, Florida Atlantic University, Boca Raton, FL, USA(2 solutions); JEREMY YOUNG,student, Nottingham High School, Nottingham, UK; NO NAME on the submission; and theproposer.

2359. [1998: 303] Proposed by Vedula N. Murty, Visakhapatnam,India.

Let PQRS be a parallelogram. Let Z divide PQ internally in the ratiok : l. The line through Z parallel to PS meets the diagonal SQ at X. Theline ZR meets SQ at Y .

Find the ratio XY : SQ.

I. Solution by Michael Lambrou, University of Crete, Crete, Greece.We clearly have

SX

XY=

SX

SQ� SQXY

=PZ

PQ� SQXY

=k

k + l� SQXY

.

Also from the similarity of4XY Z and 4QY R, we have

Y Q

XY=

QR

XZ=

PS

XZ=

PQ

ZQ=

k + l

l.

Hence

SQ

XY=

SX +XY + Y Q

XY=

SX

XY+1+

Y Q

XY=

k

k+ l� SQXY

+1+k+ l

l

so thatl

k+ l� SQXY

=k+ 2l

l,

from whichXY : SQ =l2

(k+ l)(k+ 2l), giving the required ratio.

II. Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Aus-tria.

We shall use the method of \closed vector-chains". Let�!PQ =

�!a and�!

PS =�!b . Then

�!a and

�!b are linearly independent. Set � = l=(k+l). Then�!

ZR = ��!a +

�!b , whence it follows that

�!ZY = �

�!ZR for a certain � 2 R.

Furthermore�!QS =

�!b ��!a and

��!Y X = �

�!QS for a certain � 2 R. (We must

�nd the value of �.) Finally�!ZX = �

�!b (because 4ZQX and 4PQS are

similar). Therefore, the \chain-relation"�!ZX =

�!ZY +

��!Y X reads:

��!b = �(�

�!a +

�!b ) + �(

�!b ��!a ) ;

that is,�!b (�� �� �) =

�!a (��� �)

373

whence (due to linear independence)

�� �� � = 0 and ��� � = 0 ,

implying

� =�2

�+ 1=

l2

(k+ l)(k + 2l).

III. Solutionby Francisco Bellot Rosado, I.B. Emilio Ferrari, Valladolid,Spain.

We will give a solution by way of coordinates. Suppose the points arecoordinatized as S(0;0), P (1; v), Z(1 + k; v), Q(1 + k+ l; v), R(k + l; 0),and K(k;0). The equations of the concerned lines are:

SQ : y =v

1 + k+ lx ;

ZR :x� k� l

1� l=y

v;

ZX = ZK : y = v(x� k) .

Then we have

SQ2 = v2 + (1 + k + l)2 .

The coordinates of the points X and Y are:

X = ZK \ SQ =

�k(1 + k+ l)

k+ l;vk

k+ l

�,

Y = ZR \ SQ =

�(k+ l)(1 + k + l)

k + 2l;v(k+ l)

k + 2l

�.

A straightforward computation gives us

XY

SQ=

l2

(k+ l)(k + 2l).

Also solved by SAM BAETHGE, Nordheim, TX, USA; CHRISTOPHER J. BRADLEY,Clifton College, Bristol, UK; NIKOLAOS DERGIADES, Thessaloniki, Greece; MASOUDKAMGARPOUR, student, Carson Graham Secondary School, North Vancouver, BritishColumbia; GEOFFREY A. KANDALL, Hamden, CT, USA; MITKO KUNCHEV, Baba TonkaSchool of Mathematics, Rousse, Bulgaria; GERRY LEVERSHA, St. Paul's School, London,England; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; HEINZ-J �URGEN SEIFFERT,Berlin, Germany; TOSHIO SEIMIYA, Kawasaki, Japan; MAX SHKARAYEV, Tucson, AZ,USA; D.J. SMEENK, Zaltbommel, the Netherlands; J. SUCK, Essen, Germany; PARAYIOUTHEOKLITES, Limassol, Cyprus; UNIVERSITY OF ARIZONA PROBLEM SOLVING GROUP, Tuc-son, AZ, USA; JEREMY YOUNG, student, Nottingham High School, Nottingham, UK; and theproposer. There was one incorrect solution submitted.

374

2360. [1998: 304] Proposed by K.R.S. Sastry, Dodballapur, India.In triangle ABC, let BE and CF be internal angle bisectors, and let

BQ and CR be altitudes, where F and R lie on AB, and Q and E lie onAC. Assume that E, Q, F and R lie on a circle that is tangent to BC.

Prove that triangle ABC is equilateral.

Solution by Paul Yiu, Florida Atlantic University, Boca Raton, FL, USA.The lengths of the various segments are

AR = b cosA , AQ = c cosA , AF =bc

a+ b, AE =

bc

a+ c.

Since E, Q, F , and R lie on a circle, AE � AQ = AF � AR, so thatc

a+ c=

b

a+ b. From this we have b = c.

Note that CQ = a cosC and CE =ab

a+ c=

ac

a+ c. Since the

given circle is tangent to BC and (by symmetry) that point of tangency is the

mid-point ofBC, we haveCQ�CE =

�a

2

�2. From this, cosC =

a+ c

4c.

Since cosC =a=2

cin an isosceles triangle, we deduce that a = c so that

the given triangle is equilateral.

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; NIKOLAOS DERGIADES, Thessa-loniki, Greece; MICHAEL LAMBROU, University of Crete, Crete, Greece; GERRY LEVERSHA,St. Paul's School, London, England; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zalt-bommel, the Netherlands; JEREMY YOUNG, student, Nottingham High School, Nottingham,UK; and the proposer.

2361. [1998: 304] Proposed by K.R.S. Sastry, Dodballapur, India.The lengths of the sides of triangle ABC are given by relatively prime

natural numbers. Let F be the point of tangency of the incircle with sideAB. Suppose that \ABC = 60� and AC = CF . Determine the lengths ofthe sides of triangle ABC.

Almost identical solutions by Sam Baethge, Nordheim, TX, USA, andby Paul Yiu, Florida Atlantic University, Boca Raton, FL, USA.

LetD andE be the points of tangency of the incircle with sidesAC andBC respectively. LetBF = BE = x, CD = EC = y, andAD = AF = z.Applying the Cosine Law to angle B of triangles BCF and BCA yields

FC2 = (y+ z)2 = (x+ y)2 + x2 � x(x+ y) (1)

AC2 = (y+ z)2 = (x+ y)2 + (x+ z)2 � (x+ y)(x+ z) . (2)

375

Equating the right members of (1) and (2) produces z(z+ x� y) = 0. Sincez > 0, y = x + z. Substituting in (2) yields (2x � 3z)(x + z) = 0, which

implies that x =3z

2. Because the triangle's sides are relatively prime,

z = 2, x = 3; and y = 5. Consequently, AB = 5, BC = 8, andAC = CF = 7.

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; NIKOLAOS DERGIADES, Thes-saloniki, Greece; C. FESTRAETS-HAMOIR, Brussels, Belgium; RICHARD I. HESS, Ran-cho Palos Verdes, CA, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;GEOFFREY A. KANDALL, Hamden, CT, USA; MICHAEL LAMBROU, University of Crete, Crete,Greece; GERRY LEVERSHA, St. Paul's School, London, England; GOTTFRIED PERZ, Pestalozzi-gymnasium, Graz, Austria; HEINZ-J �URGEN SEIFFERT, Berlin, Germany; TOSHIO SEIMIYA,Kawasaki, Japan; MAX SHUARAYEV, Tuscon, AZ, USA; D.J. SMEENK, Zaltbommel, the Neth-erlands; PANOS E. TSAOUSSOGLOU, Athens, Greece; JEREMY YOUNG, student, NottinghamHigh School, Nottingham, UK; and the proposer (who provided two solutions).

Only Seimiya took the trouble to verify that these necessary conditions are also su�cient;in particular, 8 < 7 + 5 so that the desired 5{7{8 triangle does, in fact, exist.

2362. [1998: 304] Proposed by Mohammed Aassila, Universit �eLouis Pasteur, Strasbourg, France.

Suppose that a, b, c > 0. Prove that

1

a(1 + b)+

1

b(1 + c)+

1

c(1 + a)� 3

1 + abc.

Solution by Jun-hua Huang, the Middle School Attached To HunanNormal University, Changsha, China.

We use the well-known inequality t+ 1t� 2 for t > 0. Equality occurs

if and only if t = 1. Note that

1 + abc

a(1 + b)=

1 + a

a(1 + b)+b(1 + c)

1 + b� 1 ,

1 + abc

b(1 + c)=

1 + b

b(1 + c)+c(1 + a)

1 + c� 1 ,

and1 + abc

c(1 + a)=

1 + c

c(1 + a)+a(1 + b)

1 + a� 1 .

Then

1 + abc

a(1 + b)+

1 + abc

b(1 + c)+

1 + abc

c(1 + a)� 2 + 2 + 2� 3 = 3 ,

by the above inequality. Equality holds when

1 + a

a(1 + b)=

1 + b

b(1 + c)=

1 + c

c(1 + a)= 1 ;

376

that is, when a = b = c = 1.

Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia andHerzegovina; NIKOLAOS DERGIADES, Thessaloniki, Greece; WALTHER JANOUS, Ursulinen-gymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete, Greece;KEE-WAI LAU, Hong Kong, China; PHIL McCARTNEY, Northern Kentucky University, KY,USA; VEDULA N. MURTY, Dover, PA, USA; M. PAROS ALEXANDROS, Paphos, Cyprus, andPARAYIOU THEOKLITOS, Limassol, Cyprus; PANOS E. TSAOUSSOGLOU, Athens, Greece;JEREMY YOUNG, student, NottinghamHigh School, Nottingham, UK; and the proposer. Therewas also one incomplete solution submitted.

2363. [1998: 304] Proposed by Walther Janous, Ursulinengymnas-ium, Innsbruck, Austria.

For natural numbers a; b; c > 0, let

q(a; b; c) := a+

a+a+

a+a+:::b+:::

b+c+:::a+:::

b+c+

a+:::b+:::

a+c+:::a+:::

b+c+

a+a+:::b+:::

b+c+:::a+:::

a+c+

a+:::b+:::

a+c+:::a+:::

(in the nth \column" above, from the third one onwards, we have, from topto bottom, the sequence a, b, c, a repeated 2n�3 times), where it is assumedthat the right side (understood as an in�nite process) yields a well-de�nedpositive real number.

The original, a Talent Search Problem, asked to determine q(1;3; 5).The value is 3

p2 (seeMathematics and Informatics Quarterly, 7 (1997), No. 1,

p. 53).

Determine whether or not there exist in�nitely many triples (a; b; c)such that q(a; b; c) is the cube root of a natural number.

Solution by Nikolaos Dergiades, Thessaloniki, Greece.Let x1 = a=b, y1 = c=a, and

xn =a+ xn�1

b+ yn�1; yn =

c+ xn�1

a+ yn�1; (1)

then q(a; b; c) = limn!1(a + xn�1), and from this we conclude that thesequence xn converges to a positive real number, so we can put

x = limn!1

xn = limn!1

xn�1 ,

and from (1) we conclude that the limit of yn exists and we can put

y = limn!1

yn = limn!1

yn�1 .

377

So q(a; b; c) = a+ x, and from (1) we get

x =a+ x

b+ yand y =

c+ x

a+ y. (2)

From (2) we have

y =a

x+ 1� b

and thus �a

x+ 1� b

��a+

a

x+ 1� b

�= c+ x .

or

x3 + cx2 = (a+ x� bx)(ax+ a+ x� bx) .

or [after some rearrangement]

(a+ x)3 + (c� 4a+ 2b+ ba� b2 � 1)x2 + (2ba� 4a2 � 2a)x

= a3 + a2 . (3)

Taking b = 2a + 1 [to make the coe�cient of x equal to zero] and thenc = a(2a+ 3) [to make the coe�cient of x2 equal to zero], we get from (3)that (a+ x)3 = a3 + a2 or

q(a; b; c) =3pa3 + a2 .

where a can be any of an in�nity of natural numbers.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; V �ACLAVKONE �CN �Y, Ferris State University, Big Rapids, MI, USA; MICHAEL LAMBROU, University ofCrete, Crete, Greece; GERRY LEVERSHA, St. Paul's School, London, England; STAN WAGON,Macalester College, St. Paul, MN, USA; JEREMY YOUNG, student, Nottingham High School,Nottingham, UK; and the proposer.

All answers were the same. In fact, Lambrou notes that the above solutions are the onlyones possible.

2364. [1998: 363] Proposed by Walther Janous, Ursulinengymnas-ium, Innsbruck, Austria.

A sequence fxng is given by the recursion: x0 = p, xn+1 = qxn+q�1(n � 0), where p is a prime and q � 2 is an integer.

(1) Suppose that p and q are relatively prime. Prove that the sequencefxng does not consist of only primes.

(2)? Suppose that pjq. Prove that the sequence fxng does not consist ofonly primes.

378

I. Solution to both parts by Manuel Benito and Emilio Fernandez, I.B.Praxedes Mateo Sagasta, Logro ~no, Spain.

The sequence is

p , q(p+ 1)� 1 , q2(p+ 1)� 1 , : : : , qn(p+ 1)� 1 , : : :

[as can easily be proved by induction | Ed.] If x1 = q(p+ 1)� 1 is primethen by the Fermat Theorem, qx1 � q (mod x1) and so

xx1 = qx1(p+ 1)� 1 � q(p+ 1)� 1 � 0 (mod x1) .

Hence xx1 is not a prime number. [And if x1 is not prime, we are done.Note that this proof does not use that p is prime. | Ed.]

II. Solution to both parts by Heinz-J �urgen Sei�ert, Berlin, Germany.

More generally, we consider the sequence fxng de�ned by the recur-sion

x0 = p , xn+1 = qxn + r , n � 0 ,

where p, q and r are any positive integers. We shall prove that this sequencecontains in�nitely many composite natural numbers.

A simple induction argument shows that

xn = pqn + r

n�1Xj=0

qj , n � 0 . (1)

First of all we note that fxng is a strictly increasing sequence of positiveintegers. Let k be a positive integer such that xk = s is a prime; if such kdoes not exist, then we are done.

[Editorial note. Since k � 1, it is clear that s = xk > q. This slightlysimpli�es the proposer's original proof.]

Case 1: q = 1. Equation (1) with n = k gives s = xk = p + rk. Itfollows that for all u � 0,

xus+k = p+ r(us+ k) � p+ rk � 0 (mod s) .

Case 2: q > 1. Since qs�1 � 1 (mod s) by Fermat's Little Theorem,and since gcd(q� 1; s) = 1, we then have

qu(s�1)� 1

q � 1� 0 (mod s) for all integers u � 0 .

379

Hence, for all u � 0,

xu(s�1)+k = pqu(s�1)qk + r

0@k�1Xj=0

qj + qku(s�1)�1X

j=0

qj

1A

� pqk + r

k�1Xj=0

qj + rqku(s�1)�1X

j=0

qj (mod s)

= xk + rqk

qu(s�1)� 1

q � 1

!� xk = s � 0 (mod s) ,

where we have used (1) twice and the closed form expression for �nite geo-metric sums. This completes the solution, which is an extension of the proofof the following theorem found in [1]: Let a, b and q be integers such thata � 1, q � 2, and gcd(aq; b) = 1. Then the sequence aqn + b, n � 0, doesnot consist of only primes.

The sequence considered in the proposal is of the form xn = (p+ 1)qn� 1,n � 0, so that the statements of both parts of the proposal are obtainedfrom the above theorem with a = p+ 1 and b = �1.

Reference:

[1] A. Aigner, Zahlentheorie, de Gruyter Verlag, 1975, S. 190, Satz 105.

Both parts also solved by THEODORE CHRONIS, Athens, Greece; NIKOLAOSDERGIADES, Thessaloniki, Greece; MICHAEL LAMBROU, University of Crete, Crete, Greece;and JEREMY YOUNG, student, Nottingham High School, Nottingham, UK. Part (1) only solvedby CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; RICHARD I. HESS, Rancho PalosVerdes, CA, USA; GERRY LEVERSHA, St. Paul's School, London, England; and the proposer.

Lambrou also gave a generalization, which included the observation that the givensequence contains in�nitely many composites.

2365. [1998: 363] Proposed by Victor Oxman, University of Haifa,Haifa, Israel.

Triangle DAC is equilateral. B is on the line DC so that \BAC =70�. E is on the line AB so that \ECA = 55�. K is the mid-point ofED. Without the use of a computer, calculator or protractor, show that60� > \AKC > 57:5�.

SolutionbyManuel Benito and Emilio Fernandez, I.B. Praxedes Mateo

Sagasta, Logro ~no, Spain [slightly modi�ed by the editor].Let us consider the circle with centre A and radiusAC. From the given

information, \AEC = 55� and4AEC is isosceles. It follows that this circlepasses through points D and E. Since K is the mid-point of the chord EDof that circle, we have \KAC = 65�. The radial line AK meets the circle atthe point H, say, and the chord CE at the point L, say, and we obviouslyhave that

\ALC > \AKC > \AHC:

380

But \ALC = 60� because \ACL = 55�, and \AHC = 57:5� because4AHC is isosceles. This completes the proof.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; NIKOLAOSDERGIADES, Thessaloniki, Greece; HIDETOSHI FUKAGAWA, Gifu, Japan; RICHARD I. HESS,Rancho Palos Verdes, CA, USA; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, MI,USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; KEE-WAI LAU, Hong Kong,China; GERRY LEVERSHA, St. Paul's School, London, England; GOTTFRIED PERZ, Pestalozzi-gymnasium, Graz, Austria; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel,the Netherlands; ECKARD SPECHT, Magdeburg, Germany; and the proposer.

2366. [1998: 364] Proposed by Catherine Shevlin, Wallsend-upon-Tyne, England.

Triangle ABC has area p, where p 2 N. Let

� = min�AB2 + BC2 + CA2

�;

where the minimum is taken over all possible triangles ABC with area p,and where � 2 N.

Find the least value of p such that � = p2.

Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Austria.

We have p = 12ab sin(C); that is ab =

2p

sin(C). Furthermore,

a2 + b2 + c2 = 2�a2 + b2 � ab cos(C)

�� 2 (2ab� ab cos(C))

= 2 � 2p�

2

sin(C)� cot(C)

�.

De�ne f(C) =2

sin(C)� cot(C). Hence f 0(C) =

1� 2 cos(C)

sin2(C).

It follows that f(C) attains its minimum at C = �3, yielding

minfa2 + b2 + c2 2 Rg = 4pf��3

�= 4p

p3.

Hence, minfa2+b2+c2 2 Ng = p2 � 4pp3, implying that p � 4

p3;

that is, pmin = 7.

Also solved by MANUEL BENITO and EMILIO FERNANDEZ, I.B. Praxedes Ma-teo Sagasta, Logro ~no, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;C. FESTRAETS-HAMOIR, Brussels, Belgium; MICHAEL LAMBROU, University of Crete, Crete,Greece; KEE-WAI LAU, Hong Kong, China; and the proposer. There were two incorrect solu-tions.

All the solutions were quite di�erent in approach. Benito and Fern �andez also show thatthe solution is, in fact, unique.

The proposer comments that this is based on a problem in the New Scientist, but thatthe answer published there was wrong!

381

2367. [1998: 364] Proposed by K.R.S. Sastry, Dodballapur, India.In triangle ABC, the Cevians AD, BE intersect at P . Prove that

[ABC]� [DPE] = [APB]� [CDE] .

(Here, [ABC] denotes the area of4ABC, etc.)

Solution by Toshio Seimiya, Kawasaki, Japan.

SincePD

AD=

[DPE]

[DEA]and

AE

CE=

[DEA]

[CDE], we have

[DPE]

[CDE]=

[DPE]

[DEA]� [DEA][CDE]

=PD

AD� AECE

. (1)

SincePD

AD=

[PBC]

[ABC]and

AE

CE=

[APB]

[PBC], we have

[APB]

[ABC]=

[APB]

[PBC]� [PBC][ABC]

=AE

CE� PDAD

. (2)

From (1) and (2) we have

[DPE]

[CDE]=

[APB]

[ABC].

This implies [ABC]� [DPE] = [APB]� [CDE].

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain;SAM BAETHGE, Nordheim, TX, USA (2 solutions); MICHEL BATAILLE, Rouen, France;FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; MANUELBENITO and EMILIO FERNANDEZ, I.B. Praxedes Mateo Sagasta, Logro ~no,Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; NIKOLAOSDERGIADES, Thessaloniki, Greece; C. FESTRAETS-HAMOIR, Brussels, Bel-gium; HIDETOSHI FUKAGAWA, Gifu, Japan; RICHARD I. HESS, Rancho PalosVerdes, CA, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;GEOFFREY A. KANDALL, Hamden, CT, USA; MICHAEL LAMBROU, University of Crete, Crete,Greece; GERRY LEVERSHA, St. Paul's School, London, England; VICTOR OXMAN, University ofHaifa, Haifa, Israel; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; HEINZ-J �URGENSEIFFERT, Berlin, Germany; D.J. SMEENK, Zaltbommel, the Netherlands; PARAGIOUTHEOKLITOS, Limassol, Cyprus; PANOS E. TSAOUSSOGLOU, Athens, Greece; JEREMYYOUNG, student, Nottingham High School, Nottingham, UK; and the proposer.

The proposer implied (and all solvers assumed) both (a) that P is inside4ABC, and (b)that all areas and lengths are positive. Bradley points out that had signed areas and segmentsbeen employed, then [ABC] � [PDE] = [ABP ] � [CED] holds for any P in the plane of4ABC.

Three years ago Kandall deduced Sastry's equality as the �rst step of his featured solutionto problem 1469 in Mathematics Magazine, 69:2 (April, 1996) 144-146. His solution thereinvokes Menelaus' Theorem (which is implicit in Seimiya's solution above). Indeed, many ofthe submitted solutions used Menelaus' Theorem; the use of either a�ne or areal coordinatesalso leads to nice solutions.

382

2368. [1998: 364] Proposed by Iliya Bluskov, Simon Fraser Univer-sity, Burnaby, BC.

Let (a1; a2; :::; an) be a permutation of the integers from 1 to nwith theproperty that ak+ak+1+ :::+ak+s is not divisible by (n+1) for any choiceof k and s where k � 1 and 0 � s � n� k� 1. Find such a permutation

(a) for n = 12 ;

(b) for n = 22 .

Solutionby Jeremy Young, student, NottinghamHigh School, Notting-ham, UK (modi�ed slightly by the editor).

(a) Set ak � 2k (mod 13), 1 � k � 12, 1 � ak � 12. Then we obtain thefollowing permutation of 1, 2, : : : , 12 ;

k = 1 2 3 4 5 6 7 8 9 10 11 12ak = 2 4 8 3 6 12 11 9 5 10 7 1

For any integers k and s with k � 1, 0 � s � 11� k, we have

ak+ ak+1 + � � �+ ak+s � 2k+2k+1+ � � �+2k+s = 2k�2s+1 � 1

�.

Clearly, 2k 6� 0 (mod 13). Furthermore, since s+1 � 12�k � 11, wesee from the table above that 2s+1 6� 1 (mod 13). Hence, ak+ak+1+� � �+ ak+s is not divisible by 13.

(b) Set ak � 5k (mod 23), 1 � k � 22, 1 � ak � 22. Then we obtain thefollowing permutation of 1, 2, : : : , 22 ;

k = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22ak = 5 2 10 4 20 8 17 16 11 9 22 18 21 13 19 3 15 6 7 12 14 1

For any integers k and s with k � 1, 0 � s � 21� k, we have

ak + ak+1 + � � �+ ak+s � 5k + 5k+1 + � � �+ 5k+s

=5k�5s+1 � 1

�4

.

Clearly, 5k 6� 0 (mod 23). Furthermore, since s + 1 � 22� k � 21,the table above reveals that 5s+1 6� 1 (mod 23).

Thus5k�5s+1 � 1

�4

6� 0 (mod 23). Hence, ak + ak+1 + � � �+ ak+s is

not divisible by 23.

Also solved by MANUEL BENITO MU ~NOZ and EMILIO FERN �ANDEZ MORAL, I.B.Sagasta, Logro ~no, Spain; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; MICHAELLAMBROU, University of Crete, Crete, Greece; KATHLEEN E. LEWIS, SUNY Oswego, Oswego,NY, USA; and the proposer. Part (a) only was solved by CHARLES ASHBACHER, Cedar Rapids,IA, USA.

383

The solutions given by Hess and the proposer were essentially the same as theone above. Ashbacher, Benito and Fern �andez, and Lewis all gave the same permutation(1; 2; 3; 4; 5; 6; 10; 7; 12; 11; 8; 9) for n = 12. On the other hand, their permutations forn = 22 were all di�erent.

The proposer actually showed in general that if p is a prime and if q is a primitive rootmod p (that is, (q; p) = 1 and p � 1 is the smallest positive integer such that qp�1 � 1(mod p)| Ed.), then

�q0; q1; q2; : : : ; qp�2

�has the desired property. The proposed problem

is the special case when p = 13 and 23. The proof of this general fact is quite easy and isessentially contained in Young's solution.

2369?. [1998: 364] Proposed by Federico Arboleda, student, Bo-

got �a, Colombia (age 11).Prove or disprove that for every n 2 N, there exists a 2n{omino such

that every n{omino can be placed entirely on top of it.

(Ann{omino is de�ned as a collection ofn squares of equal size arrangedwith coincident sides.)

SolutionbyMichael Lambrou, University of Crete, Crete, Greece, mod-i�ed by the editors.

We show that the answer is negative for each large enough n. For sim-plicity we demonstrate this for n = 11, but the technique easily generalizes(Ed. for n odd).

In order to cover the \staircase", , we need 11 squares.

To cover the \ladder", , we need a further 9 squaressince the staircase has only 2 squares in each row and column. Note thatthe staircase has 6 rows and 6 columns, and the ladder can overlap at mostone row if it is placed horizontally and at most one column if it is placedvertically. So either 5 rows or 5 columns of the staircase are not covered

by the ladder. Consider now the \el" shape, , which has a row

of 6 and a column of 6. There will be at least 3 squares on it uncovered byeither the staircase or the ladder, since it can overlap at most 2 squares ofthe staircase which are not on the ladder if it has a maximum overlap withthe ladder (if it does not have a maximum overlap with the ladder then therewill be at least 6 squares on it uncovered by the ladder and the staircase).To sum up we need at least 11 + 9 + 3 = 23 > 22 = 2(11) squares to coverjust these three 11{ominoes, completing the demonstration. (Generally, forn = 2k+ 1, the numbers 11, 9, 3 are respectively 2k+1, 2k� 1, k� 2 and(2k+ 1) + (2k� 1) + (k� 2) > 2(2k+ 1) if k � 5.)

384

Also solved by MANUEL BENITO and EMILIO FERNANDEZ, I.B. Praxedes MateoSagasta, Logro ~no, Spain.

Benito and Fern �andez give a set of eight 7{ominoes which require at least 15 squares to cover:

Editor's note: It is interesting that, although the above example, as well as Lambrou'ssolution, considers only odd n, it certainly generalizes to even n as well.

The proposer observes that the statement to be proved or disproved \is trivially true forn = 1, 2, 3 and 4. For n = 5 the following decamino shows that it is also true:

We believe the following dodecomino shows it is true for n = 6:

(end of quote). If we do not distinguish between re ections of 6{ominoes, then the above12{omino seems to work; but it fails if we distinguish among the re ections; for example, thefollowing cannot be covered without re ecting it:

Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell

Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil

Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,

J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

385

THE ACADEMY CORNERNo. 28

Bruce ShawyerAll communications about this column should be sent to BruceShawyer, Department of Mathematics and Statistics, Memorial Universityof Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

A Trial Balloon

Vedula N. Murty

The following problem on heights and distances was set in the I.I.T.Entrance Exam in 1979.

A balloon is observed simultaneously from three points,A, Band C, on a straight road directly beneath it. The angular ele-vation at B is twice that at A, and the angular elevation at C isthree times that at A. If the distance between A and B is a, andthe distance between B and C is b, �nd the height of the balloonin terms of a and b.

Subsequently, this problem appeared in many textbooks in India with a so-lution which is straightforward. The height of the balloon is

y =ap3b2 + 2ab� a2

2b.

The conditions on a and b under which this solution is valid are not given inany of the solutions printed in the textbooks.

Professor M. Perisastry, a retired Professor of Mathematics at M.R.College, Viziahagaram, Andhra Pradesh, India, noted that

y > 0 =) 3b2 + 2ab� a2 > 0

=) 4b2 � (a� b)2 > 0

=) ja� bj < 2b

=) 0 < a < 3b .

Moreover, it is easily seen that

a

b=

sin(3�)

sin�= 3� 4 sin2 � ,

386

where � is the angle of elevation at A.

Since the angle of elevation at C is 3� < �=2, this implies that

0 < � <�

6=) 0 < sin� <

1

2

=) a

b> 2

=) a > 2b .

Hence,a

3< b <

a

2.

Readers of CRUX withMAYHEMmay be interested in the above prob-lem, and teachers should pay attention to the conditions under which a givensolution is valid.

BAD CANCELLATIONS

�����tan�1 �2n+1�

tan�1 (2n)� n3=2

(n+ 1)3=2

����� = 2

QUESTIONS on MATHEMATICIANS

The year 1796 was the turning point in a (future) mathematician'scareer. Who was he?

And which mathematician was born in 1796?

What do Wilhelm Ackermann, Pavel Sergeevich Aleksandrov, Lester R.Ford, Ronald Martin Foster, Valeri�i Ivanovich Glivenko, KazimierzKurtowski, and Carl Ludwig Seigel have in common?

What is the di�erence between Gustav Magnus Mittag-Le�er andGaspard Monge?

387

THE OLYMPIAD CORNERNo. 201

R.E. Woodrow

All communications about this column should be sent to Professor R.E.Woodrow, Department of Mathematics and Statistics, University of Calgary,Calgary, Alberta, Canada. T2N 1N4.

As a �rst contest this number we give the Thirty-�rst Canadian Math-ematical Olympiad 1999. The contest was held March 31, 1999 with 81 com-petitors from 48 schools in �ve Canadian provinces participating. Competi-tors are invited on the basis of their performance on other contests. Eachquestion was marked out of 7 marks for a total possible score of 35.

First prize went to Jimmy Chui, Second to Adrian Chan, Third to DavidPritchard, and Honourable Mentions go to Edmond Choi, MasoudKamgarpour, Jessie Lei, Pierre LeVan, Dave Nicholson, and Yannick Solari.Congratulations!

My thanks go to Daryl Tingley, Chair of the Canadian MathematicalOlympiad Committee for furnishing us with the contest as well as selectedsolutions by the contestants, which appear at the end of this number of theCorner. But try them �rst!

THE THIRTY-FIRST CANADIANMATHEMATICAL OLYMPIAD 1999

March 31, 1999

1. Find all real solutions to the equation 4x2� 40[x] + 51 = 0. Here,if x is a real number, then [x] denotes the greatest integer that is less thanor equal to x.

2. Let ABC be an equilateral triangle of altitude 1. A circle withradius 1 and centre on the same side of AB as C rolls along the segmentAB. Prove that the arc of the circle that is inside the triangle always has thesame length.

3. Determine all positive integersnwith the property thatn = (d(n))2.Here d(n) denotes the number of positive divisors of n.

4. Suppose a1, a2, : : : , a8 are eight distinct integers from f1, 2,: : : , 16, 17g. Show that there is an integer k > 0 such that the equationai�aj = k has at least three di�erent solutions. Also, �nd a speci�c set of 7distinct integers from f1, 2, : : : , 16, 17g such that the equation ai� aj = kdoes not have three distinct solutions for any k > 0.

388

5. Let x, y, and z be non-negative real numbers satisfyingx+ y + z = 1. Show that

x2y + y2z + z2x � 4

27,

and �nd when equality occurs.

Next we give the 28th United States of America Mathematical Olympiad.These problems are copyrighted by the committee on the American Mathe-matical Competitions of the Mathematical Association of America and maynot be reproduced without permission. Solutions, and additional copies ofthe problems may be obtained from Professor Titu Andreescu, AMCDirector, University of Nebraska, Lincoln, NE, USA 68588{0658. As always,we welcome your original, \nice" solutions and generalizations which di�erfrom the published solutions.

28th UNITED STATES OF AMERICAMATHEMATICAL OLYMPIAD

Part I 9 a.m. | 12 noonApril 27, 1999

1. Some checkers placed on an n�n checkerboard satisfy the followingconditions:

(a) every square that does not contain a checker shares a side with one thatdoes;

(b) given any pair of squares that contain checkers, there is a sequence ofsquares containing checkers, starting and ending with the given squares, suchthat every two consecutive squares of the sequence share a side.

Prove that at least (n2�2)=3 checkers have been placed on the board.

2. Let ABCD be a cyclic quadrilateral. Prove that

jAB � CDj+ jAD � BCj � 2jAC � BDj .

3. Let p > 2 be a prime and let a, b, c, d be integers not divisible by p,such that

fra=pg+ frb=pg+ frc=pg+ frd=pg = 2

for any integer r not divisible by p. Prove that at least two of the numbersa+b, a+c, a+d, b+c, b+d, c+d are divisible by p. (Note: fxg = x�bxcdenotes the fractional part of x.)

389

Part II 1 p.m. | 4 p.m.April 27, 1999

4. Let a1, a2, : : : , an (n > 3) be real numbers such that

a1 + a2 + � � �+ an � n and a21 + a22 + � � �+ a2n � n2 .

Prove that max(a1, a2, : : : , an) � 2.

5. The Y2K Game is played on a 1� 2000 grid as follows. Two playersin turn write either an S or an O in an empty square. The �rst player whoproduces three consecutive boxes that spell SOS wins. If all boxes are �lledwithout producing SOS then the game is a draw. Prove that the secondplayer has a winning strategy.

6. Let ABCD be an isosceles trapezoid withAB k CD. The inscribedcircle ! of triangle BCD meets CD at E. Let F be a point on the (internal)angle bisector of \DAC such that EF ? CD. Let the circumscribed circleof triangle ACF meet line CD at C and G. Prove that the triangle AFG isisosceles.

We next give selected problems of theUkrainianMathematical Olympiadof March 26{27, 1996. My thanks go to J.P. Grossman, Team Leader ofthe Canadian International Olympiad Team at Mumbai, India, for collect-ing these problems.

UKRAINIANMATHEMATICAL OLYMPIADMarch 26{27, 1996Selected Problems

1. (8th grade) A regular polygon with 1996 vertices is given. Whatminimal number of vertices can we delete so that we do not have four verticesremaining, which form: (a) a square? (b) a rectangle?

2. (9th grade) Ivan has made the models of all triangles with integerlengths of sides and perimeters 1993 cm. Peter has made the models of alltriangles with integer lengths of sides and perimeters 1996 cm. Who hasmore models?

3. (10th grade) Prove that sin(�=20)+sin(2�=20)+� � �+sin(9�=20) <99=10�(2=�) arcsin(1=10)�(2=�) arcsin(2=10)�� � ��(2=�) arcsin(9=10).

4. (10th grade) Let S be the set of all points of the coordinate planewith integer coordinates. We shall say that a one-to-one correspondence of Spreserves a distance x if any two points in S at distance x have the images atdistance x. Is it true that a one-to-one correspondence necessarily preservesall positive distances if:

390

(a) it preserves the distance 1?

(b) it preserves the distance 2?

(c) it preserves the distance 2 and the distance 3?

5. (10th grade) Let O be the centre of the parallelogram ABCD with\AOB > �=2. We take the points A1, B1 on the half-lines OA, OB,respectively so that A1B1 k AB and \A1B1C = \ABC=2. Prove thatA1D ? B1C.

6. (11th grade) The sequence fang, n � 0, is such that a0 = 1,a499 = 0 and for n � 1, an+1 = 2a1an � an�1.(a) Prove that ja1j � 1.

(b) Find a1996.

7. (11th grade) Does a function f : R ! R exist which is not a poly-nomial and such that for all real x

(x� 1)f(x+ 1)� (x+ 1)f(x� 1) = 4x(x2 � 1) ?

8. (11th grade) LetM be the number of all positive integers which haven digits 1, n digits 2 and no other digits in their decimal representations. LetN be the number of all n{digit positive integers with only digits 1, 2, 3, 4 inthe representation where the number of 1's equals the number of 2's. Provethat M = N .

As another problem set we give the problems of the XII Italian Mathe-matical Olympiad, Cesenatico, 3May, 1996. Again thanks for collecting thesego to J.P. Grossman, Team Leader of the Canadian International OlympiadTeam at Mumbai, India.

XII ITALIANMATHEMATICAL OLYMPIADCesenatico, 3 May, 1996

1. Among the triangles with an assigned side l and with given area S,determine all those for which the product of the three altitudes is maximum.

2. Prove that the equation a2+b2 = c2+3 has in�nitely many integersolutions (a; b; c).

3. Let A and B be opposite vertices of a cube with side 1. Find theradius of the sphere with centre interior to the cube, tangent to the threefaces meeting in A and tangent to the three edges meeting in B.

4. Given an alphabet with three letters a, b, c, �nd the number ofwords of n letters which contain an even number of a's.

391

5. Let a circle C and a pointA exterior to C be given. For each point Pon C construct the square APQR, with anticlockwise ordering of the lettersA, P , Q, R. Find the locus of the point Q when P runs over C.

6. What is the minimum number of squares that one needs to draw ona white sheet in order to obtain a full grid of size n? (The picture shows a fullgrid of size 6).

As a relative newcomer to the Corner, we next give the problems ofthe South African Mathematics Olympiad, Third Round, 7 September 1995,Section A and B. Again thanks go to J.P. Grossman for collecting these whileat the IMO at Mumbai, India as Canadian Team Leader.

SOUTH AFRICAN MATHEMATICS OLYMPIADThird Round | 7 September 1995

SECTION A

1. Prove that there are no integers m and n such that

419m2 + 95mn+ 2000n2 = 1995 .

2. ABC is a triangle with \A > \C, and D is the point on BC suchthat \BAD = \ACB. The perpendicular bisectors ofAD andAC intersectin the point E. Prove that \BAE = 90�.

3. Suppose that a1, a2, a3, : : : , an are the numbers 1, 2, 3, : : : , n butwritten in any order. Prove that

(a1 � 1)2 + (a2 � 2)2 + (a3 � 3)2 + � � �+ (an � n)2

is always even.

4. Three circles, with radii p, q, r, and centres A, B, C, respectively,touch one another externally at points D, E, F . Prove that the ratio of theareas of4DEF and 4ABC equals

2pqr

(p+ q)(q+ r)(r+ p).

392

SECTION B

1. The convex quadrilateral ABCD has area 1, and AB is producedto E, BC to F , CD to G and DA to H, such that AB = BE, BC = CF ,CD = DG and DA = AH. Find the area of the quadrilateral EFGH.

2. Find all pairs (m;n) of natural numbers with m < n such thatm2 + 1 is a multiple of n and n2 + 1 is a multiple of m.

3. The circumcircle of 4ABC has radius 1 and centre O, and P is apoint inside the triangle such that OP = x. Prove that

AP � BP � CP � (1 + x)2(1� x) ,

with equality only if P = O.

The next problems are those of the Taiwan Olympiad, 1996. Thanks goto J.P. Grossman, Team Leader for Canada at the IMO at Mumbai, India, forcollecting them.

TAIWANMATHEMATICAL OLYMPIAD 1996

1. Let the angles �, �, be such that 0 < �, �, < �2

and�+ � + = �

4. Suppose that

tan� =1

a; tan� =

1

b, tan =

1

c,

where a, b, c are positive integers. Determine the values of a, b, c.

2. Let a be a real number such that 0 < a � 1 and a � aj � 1a, for

j = 1, 2, : : : , 1996. Show that for any non-negative real numbers �j (j = 1,2, : : : , 1996), with

1996Xj=1

�j = 1 ,

one has 1996Xi=1

�iai

!0@1996Xj=1

�ja�1j

1A � 1

4

�a+

1

a

�2.

3. Let A and B be two �xed points on a �xed circle. Let a point Pmove on this circle and let M be a corresponding point such that either Mis on the segment PA with AM =MP + PB orM is on the segment PBwith AP +MP = PB. Determine the locus of such points P .

393

4. Show that for any real numbers a3, a4, : : : , a85, the roots of theequation

a85x85 + a84x

84 + � � �+ a3x3 + 3x2 + 2x+ 1 = 0

are not all real.

5. Find 99 integers a1, a2, : : : , a99 = a0, satisfying

jak�1 � akj � 1996 for all k = 1, 2, : : : , 99,

so that the number

m = maxfjak�1 � akj; k = 1, 2, : : : , 99g

is as small as possible, and determine the minimum valuem� ofm.

6. Let q0, q1, q2, : : : be a sequence of integers such that

(a) for any m > n, m� n is a factor of qm � qn, and

(b) jqnj � n10 for all integers n � 0.

Show that there exists a polynomial Q(x) satisfyingQ(n) = qn for all n.

The next problems are those of the Croatian National MathematicsCompetition, Kraljevica, May 16{19, 1996, IV Class and IMO Team Selec-tion Competition problems. Thanks go to J.P. Grossman, Team Leader forCanada at the IMO at Mumbai, India, for collecting the problem set.

CROATIAN NATIONALMATHEMATICSCOMPETITION

Kraljevica, May 16{19, 1996IV CLASS

1. Is there any solution of the equation

[x] + [2x] + [4x] + [8x] + [16x] + [32x] = 12345?

([x] denotes the greatest integer which does not exceed x.)

2. Determine all pairs of numbers �1, �2 2 R for which every solutionof the equation (x+ i�1)

n + (x+ i�2)n = 0 is real. Find the solutions.

3. Determine all functions f : R �! R continuous at 0, which satisfythe following relation f(x)�2f(tx)+f(t2x) = x2 for all x 2 R , wheret 2 (0; 1) is a given number.

394

4. Let � and � be positive irrational numbers such that 1�+ 1

�= 1

and A = f[n�] j n 2 Ng, B = f[n�] j n 2 Ng. Prove that A [ B = N andA \ B = ;.Remark: You can prove the following equivalent assertion: For a function� : N! N de�ned by

�(m) = Cardfk j k 2 N, k � m, k 2 Ag+Cardfk j k 2 N, k � m, k 2 Bg

one has �(m) = m, 8m 2 N. ([x] denotes the greatest integer which doesnot exceed x.)

ADDITIONAL COMPETITION FOR SELECTION OF THE IMO TEAMMay 18, 1996

1. (a) n = 2k + 1 points are given in the plane. Construct an n{gonsuch that these points are mid-points of its sides.

(b) Arbitrary n = 2k, k > 1, points are given in the plane. Prove thatit is impossible to construct an n{gon, in each case, such that these pointsare mid-points of its sides.

2. The side-length of the squareABCD equals a. Two pointsE and Fare given on sidesBC and AB such that the perimeter of the triangle BEFequals 2a. Determine \EDF .

3. Find all pairs of consecutive integers the di�erence of whose cubesis a full square.

4. Let A1, A2, : : : , An be a regular n{gon inscribed in the circle ofradius 1 with the centre at O. A point M is given on the ray OA1 outsidethe n{gon. Prove that

nXk=1

1

jMAkj� n

jOM j .

We next turn to solutions. First an alternative solution to that givenearlier this year to problem 5 of the Iranian Olympiad (1994) [1999: 142{143]

5. [1998: 6{7] [1999: 142{143] IranianMathematical Olympiad (1994).Show that if D1 and D2 are two skew lines, then there are in�nitely

many straight lines such that their points have equal distance fromD1 andD2.

Comments by J. Chris Fisher, University of Regina, Regina, Saskatche-wan; with alternative solution by Aart Blokhuis, Mathematics Department,Eindhoven University of Technology, the Netherlands.

395

Rename the lines l and m. Fix points A and B on l a unit distanceapart. For each A0 on m there are two points B0 and B00 on m that are aunit distance from A0. There is a unique rotation that takes A and B to A0

andB0, and another taking them toA0 andB00; the points of the axes of thesetwo rotations are equidistant from l and m since the perpendicular from anaxis point to l is taken by the rotation to the perpendicular from that point tom. Each A0 leads to a di�erent pair of lines. (To see that the rotation existsas claimed, take as mirror 1 the plane of points equidistant from A and A0;if B� is the image of B under re ection in mirror 1 then take mirror 2 to bethe perpendicular bisector (necessarily through A0) of B� and either B0 orB00. The product of re ections in these two mirrors is a rotation about theirline of intersection.)

Comments. (1) The locus of points equidistant from the skew lines l andm is a ruled surface, namely the hyperbolic paraboloid. To see the parabolas,take the section of the surface by a plane through l: the locus of points inthat plane that are equidistant from the point where it meets m and fromthe line l is a parabola.

(2) A slight generalization of the problem provides a simple constructionof a spread (which is a collection of skew lines that completely cover the threedimensional space in the sense that every point of space is on exactly one ofthe lines of the spread): the locus of points whose distances from l andm arein the ratio 1 : k is a ruled quadratic. The proof of the original problem canbe modi�ed to a proof of the claim by taking B0 and B00 to be k units fromA0, and using a dilative rotation to take A and B to their primed mates.

To �nish this number of the Corner we give participant or \o�cial" so-lutions to the Canadian Mathematical Olympiad given at the beginning ofthis number. My thanks go to Daryl Tingley, Chair of the Canadian Mathe-matical Olympiad Committee for furnishing the following:

CANADIAN MATHEMATICAL OLYMPIAD 1999SOLUTIONS

Most of the solutions to the problems of the 1999 CMO presented be-low are taken from students' papers. Some minor editing has been done |unnecessary steps have been eliminated and some wording has been changedto make the proofs clearer. But for the most part, the proofs are as submit-ted.

1. Solution| Adrian Chan, Upper Canada College, Toronto, Ontario.

Rearranging the equation we get 4x2 + 51 = 40[x]. It is known that

396

x � [x] > x� 1, so

4x2 + 51 = 40[x] > 40(x� 1) ,

4x2 � 40x+ 91 > 0 ,

(2x� 13)(2x� 7) > 0 .

Hence x > 13=2 or x < 7=2. Also,

4x2 + 51 = 40[x] � 40x ,

4x2 � 40x+ 51 � 0 ,

(2x� 17)(2x� 3) � 0 .

Hence 3=2 � x � 17=2. Combining these inequalities gives 3=2 � x < 7=2or 13=2 < x � 17=2 .

Case 1: 3=2 � x < 7=2.

For this case, the possible values for [x] are 1, 2 and 3.

If [x] = 1 then 4x2 + 51 = 40 � 1 so 4x2 = �11, which has no realsolutions.

If [x] = 2 then 4x2 + 51 = 40 � 2 so 4x2 = 29 and x =p292

. Notice

thatp162

<p292

<p362

so 2 < x < 3 and [x] = 2.

If [x] = 3 then 4x2+51 = 40 �3 and x =p69=2. But

p692>

p642

= 4.So, this solution is rejected.

Case 2: 13=2 < x � 17=2.

For this case, the possible values for [x] are 6, 7 and 8.

If [x] = 6 then 4x2 + 51 = 40 � 6, so that x =p1892

. Notice thatp1442

<p1892

<p1962

, so that 6 < x < 7 and [x] = 6.

If [x] = 7 then 4x2 + 51 = 40 � 7, so that x =p2292

. Notice thatp1962

<p2292

<p2562

, so that 7 < x < 8 and [x] = 7.

If [x] = 8 then 4x2 + 51 = 40 � 8, so that x =p2692

. Notice thatp2562

<p2692

<p3242

, so that 8 < x < 9 and [x] = 8.

The solutions are x =

p29

2,

p189

2,

p229

2,

p269

2.

(Editor: Adrian then checks these four solutions.)

397

2. Solution 1 | Keon Choi, A.Y. Jackson SS, North York, Ontario.

E

Oq

C

A B

D

LetD andE be the intersections ofBC and extendedAC, respectively,with the circle.

Since CO k AB (because both the altitude and the radius are 1)\BCO = 60� and therefore \ECO = 180� � \ACB � \BCO = 60�.

Since a circle is always symmetric about its diameter and line CE is are ection of line CB in CO, line segment CE is a re ection of line segmentCD.

Therefore CE = CD.

Therefore 4CED is an isosceles triangle.

Therefore \CED = \CDE and \CED+ \CDE = \ACB = 60�.

\CED = 30� regardless of the position of centre O. Since \CED isalso the angle subtended from the arc inside the triangle, ifCED is constant,the arc length is also constant.

Editor's Note: This proof has had no editing.

Solution 2 | Jimmy Chui, Earl Haig SS, North York, Ontario.

-x

6y

AB

C

r

O (a; 0)

B0A0

Place C at the origin, point A at�

1p3; 1�and point B at

�� 1p

3; 1�.

Then 4ABC is equilateral with altitude of length 1.

398

Let O be the centre of the circle. Because the circle has radius 1, andsince it touches line AB, the locus of O is on the line through C parallel toAB (since C is length 1 away from AB); that is, the locus of O is on thex{axis.

Let point O be at (a; 0). Then � 1p3� a � 1p

3since we have the

restriction that the circle rolls along AB.

Now, let A0 and B0 be the intersection of the circle with CA and CB,respectively. The equation of CA is y =

p3 x; 0 � x � 1p

3, of CB is

y = �p3 x, � 1p3� x � 0, and of the circle is (x� a)2 + y2 = 1.

We solve for A0 by substituting y =p3 x into (x � a)2 + y2 = 1 to

get x =a�p4� 3a2

4.

Visually, we can see that solutions represent the intersection of ACextended and the circle, but we are only concerned with the greater x{value| this is the solution that is on AC, not on AC extended. Therefore

x =a+

p4� 3a2

4, y =

p3

a+

p4� 3a2

4

!.

Likewise we solve for B0, but we take the lesser x{value to get

x =a�p4� 3a2

4, y = �

p3

a�p4� 3a2

4

!.

Let us �nd the length of A0B0:

jA0B0j2 =

a+

p4� 3a2

4� a�p4� 3a2

4

!2

+

p3a+

p4� 3a2

4

!� �p3a�p4� 3a2

4

!!2

=4� 3a2

4+ 3

a2

4= 1 ,

which is independent of a.

Consider the points O, A0 and B0. 4OA0B0 is an equilateral triangle(because A0B0 = OA0 = OB0 = 1).

Therefore \A0OB0 = �3and arc A0B0 = �

3, a constant.

3. Solution|Masoud Kamgarpour, Carson SS, North Vancouver, BC.Note that n = 1 is a solution. For n > 1 write n in the form

n = P�11 P�2

2 : : : P�mm where the Pi's, 1 � i � m, are distinct prime num-

bers and �i > 0. Since d(n) is an integer, n is a perfect square, so �i = 2�ifor integers �i > 0.

399

Using the formula for the number of divisors of n,

d(n) = (2�1 + 1)(2�2 + 1) : : : (2�m + 1) ,

which is an odd number. Now because d(n) is odd, (d(n))2 is odd, thereforen is odd as well, so Pi � 3; 1 � i � m. We get

P�11 � P�2

2 : : : P�mm = [(�1 + 1)(�2 + 1) : : : (�m + 1)]2

or using �i = 2�i

P�11 P

�22 : : : P �m

m = (2�1 + 1)(2�2 + 1) : : : (2�m + 1) .

Now we prove a lemma:

Lemma: P t � 2t + 1 for positive integers t and P � 3, with equality onlywhen P = 3 and t = 1.

Proof: We use mathematical induction on t. The statement is true for t = 1because P � 3. Now suppose P k � 2k+ 1, k � 1; then we have

P k+1 = P k �P � P k(1+2) > P k+2 � (2k+1)+2 = 2(k+1)+1 .

Thus P t � 2t+ 1 and equality occurs only when P = 3 and t = 1.

Let us say n has a prime factor Pk > 3; then (by the lemma)

P �kk > 2�k + 1 and we have P �1

1 : : : P �mm > (2�1 + 1) : : : (2�m + 1), a

contradiction.

Therefore, the only prime factor ofn isP = 3 and we have 3 = 2 +1.By the lemma = 1.

The only positive integer solutions are 1 and 9.

4. Solution 1 | David Nicholson, Fenelon Falls SS, Fenelon Falls,Ontario.

Without loss of generality let a1 < a2 < a3 : : : < a8.

Assume that there is no such integer k. Let us just look at the sevendi�erences di = ai+1 � ai. Then amongst the di there can be at most two1s, two 2s, and two 3s, which totals to 12.

Now 16 = 17 � 1 � a8 � a1 = d1 + d2 + : : : + d7 so the sevendi�erences must be 1, 1, 2, 2, 3, 3, 4.

Now let us think of arranging the di�erences 1, 1, 2, 2, 3, 3, 4. Notethat the sum of consecutive di�erences is another di�erence. (For example,d1 + d2 = a3 � a1, d1 + d2 + d3 = a4 � a1)

We cannot place the two 1s side by side because that will give us anotherdi�erence of 2. The 1s cannot be beside a 2 because then we have three 3s.They cannot both be beside a 3 because then we have three 4s! So we musthave either 1, 4, �, �, �, 3, 1 or 1, 4, 1, 3, �, �, � (or their re ections).

400

In either case we have a 3, 1 giving a di�erence of 4 so we cannot putthe 2s beside each other. Also we cannot have 2, 3, 2 because then (with the1, 4) we will have three 5s. So all cases give a contradiction.

Therefore there will always be three di�erences equal.

One set of seven numbers satisfying the criteria is f1, 2, 4, 7, 11,16, 17g. [Editor: There are many such sets.]

Solution 2 | The CMO committee.Consider all the consecutive di�erences (that is, di above) as well as

the di�erences bi = ai+2� ai, i = 1, : : : , 6. Then the sum of these thirteendi�erences is 2 � (a8 � a1) + (a7 � a2) � 2(17� 1) + (16� 2) = 46. Nowif no di�erence occurs more than twice, the smallest the sum of the thirteendi�erences can be is 2 �(1+2+3+4+5+6)+7 = 49, giving a contradiction.

5. Solution 1 | The CMO committee.Let f(x; y; z) = x2y + y2z + z2x. We wish to determine where f is

maximal. Since f is cyclic, without loss of generality we may assume thatx � y, z. Since

f(x; y; z)� f(x; z; y) = x2y + y2z + z2x� x2z � z2y � y2x= (y� z)(x� y)(x� z) ,

we may also assume y � z. Then

f(x+ z; y;0)� f(x; y; z) = (x+ z)2y� x2y � y2z � z2x

= z2y+ yz(x� y) + xz(y � z) � 0 ,

so wemay now assume z = 0. The rest follows from the arithmetic-geometricmean inequality:

f(x; y;0) =2x2y

2� 1

2

�x+ x+ 2y

3

�3=

4

27.

Equality occurs when x = 2y, hence at (x; y; z) = (23; 13; 0).

�As well

as (0; 23; 13) and (1

3; 0; 2

3)�.

Solution 2 | The CMO committee.With f as above, and x � y, z

f

�x+

z

2, y +

z

2, 0

�� f(x; y; z) = yz(x� y) + xz

2(x� z) +

z2y

4+z3

8,

so we may assume that z = 0. The rest follows as for solution 1.

That completes this number of the Corner. Sendme your nice solutionsas well as Olympiad materials for use in future issues.

401

BOOK REVIEWS

ALAN LAW

Calculus, The Dynamics of Change edited by A. Wayne Roberts,published by The Mathematical Association of America, 1996.ISBN # 0-88385-098-2, softcover, 166+ pages, $34.95 (U.S.).Reviewed by Jack W. Macki, University of Alberta, Edmonton, Alberta.

This volume is number 39 in the outstanding Mathematical Associationof America series on mathematics education, with the emphasis on calculusand calculus reform. There are four main sections: I. Visions; II. Planning;III. Assessment; IV. Connections, each of which has a number of articles byvarious authors. In addition, there is a prefatory section on how to thinkabout and plan a modern calculus course and three �nal sections on, respec-tively, resources, calculus on the Internet, and a historical and philosophicalsection \Calculus for a New Century". A total of 17 di�erent authors wereinvolved in the various articles.

The prefatory section is a good \how to" guide to e�ective organizationof a course in calculus|every beginning instructor (and all old fogeys) shouldread it. There are ideas you may not accept, but overall any instructor willget a refreshingly succinct and useful guide to preparing an interesting anduseful course.

There are six articles in the Visions section. Sharon Cutler Ross in \Vi-sions of Calculus" gives a historical discussion of the development of reform,with a very balanced discussion of the most important issues. This is a goodway to bring yourself to an understanding of what are the issues, what hasworked, and which ideas are still in question. A short article by ThomasTucker shows, with one simple example, how one can solve calculus prob-lems using numerical or verbal techniques. Mai Gehrke and David Pengelleyreport on their programs at New Mexico State University, and give practi-cal advice for conducting reform courses, and for getting colleagues on side.Deborah HughesHallett contributes a very personal essay on her experienceswith calculus teaching and calculus reform. David A. Smith contributes anessay which emphasizes 10 active verbs which characterize his ideas aboutteaching.

There are four articles in the Planning section. Martin Flashman's arti-cle reports on the full history of the introduction of reform at six institutions.Morton Brown reports on the Michigan program. The last two articles arein fact an outline for change followed by a checklist.

Part III, Assessment, has an introductory article by David Bressoud,followed by a large set of �nal examinations for each of Calculus I, II andIII. Each examination had at least one pleasant surprise for me. Bressoud'sarticle has lots of good thoughts, and begins with an apocryphal quote of

402

Richard Feynman, The biggest problem with being a student is that you'realways too busy getting an education to learn anything, and the rest of thearticle is just as interesting in its insights. Assessment is probably the weak-est area of most mathematics courses, and this section will help most of uslearn to think more deeply about the problem.

Part IV, Connections, is concerned with what in many ways is the mostoverlooked area in Mathematics departments. The degree to which we thinkof our individual courses in isolation, rather than as an integrated whole, isalmost criminal. I am not thinking of the \analysis sequence" or the \algebrasequence", but of how all of our courses �t (or rather, do not) into some kindof integrated system. John Dossey's article on secondary school mathematicsreform is not terribly relevant to Canada|this nation is quite far ahead of theU.S. in designing e�ective high school mathematics programs|whether stu-dents take them and are attracted to the subject by them is another matter!Robert Borrelli and Courtney Coleman of Harvey Mudd report on their ex-periments with modifying the introductory di�erential equations course. Thearticle gives four examples to emphasize how modern ideas can be broughtinto a �rst course. David Carlson and Wayne Roberts give a very brief reporton their experiences with post-calculus linear algebra and analysis. SheldonGordon recorded a round-table discussion between mathematicians, electri-cal engineers, ecologists, physicists, biologists, chemists and chemical engi-neers.

Martin Flashman's article on the Internet gives a couple of key, central,sites. Since this book appeared in 1996, there are by now hordes of other �rstclass sites.

Recent reports from colleagues indicate that at many schools the math-ematics department is more or less completely isolated from its clients, inparticular from the faculty of engineering and the department of physics. Atmany schools these faculties and departments keep their students away fromthe mathematics department as much as possible. The reports indicate anarrogance and lack of respect for other disciplines which is truly amazing.These mathematics departments are recognized as leaders in our profession,yet they have no research in teaching, learning, or assessment, nor wouldthey be valued if they did. This book and others like it show that in our pro-fession there is a large group of committed professionals trying to get us allto think about the teaching and learning side of our work. Thank goodness.And thank the MAA.

403

THE SKOLIAD CORNERNo. 41

R.E. Woodrow

This issue we give the Final Round Parts A and B of the 1998 BritishColumbia Colleges Senior High School Mathematics Contest. My thanks goto Jim Totten, University College of the Cariboo, one of the organizers, forforwarding the materials for use in the Corner.

BRITISH COLUMBIA COLLEGESSENIOR HIGH SCHOOL MATHEMATICS CONTEST

Final Round 1998Part A

1. If�r + 1

r

�2= 3, then r3 + 1

r3=

(a) 0 (b) 1 (c) 2 (d) 2p3 (e) 4

p3

2. Kevin has �ve pairs of socks in his drawer, all of di�erent colours andpatterns and, being a typical teenage boy, they are not folded and have beenthoroughly mixed up. On the �rst day of school Kevin reaches into his sockdrawer without looking and pulls out three socks. What is the probabilitythat two of the socks match?

(a) 310

(b) 35

(c) 13

(d) 124

(e) 115

3. A small circle is drawn within a 16sector of a circle of radius r, as

shown. The small circle is tangent to the two radii and the arc of the sector.The radius of the small circle is:

r

(a) r2

(b) r3

(c) 2p3 r3

(d)p2 r2

(e) none of these

404

4. In the accompanying diagram,the circle has radius one, the centralangleAOB is a right angle andAC andBC are of equal length. The shadedarea is:

A B

O

C

(a) �2

(b)p22

(c) ��p2

2(d)

p2+12

(e) 12

5. The side, front and bottom faces of a rectangular solid have areas2x, y

2, and xy square centimetres, respectively. The volume of the solid is:

(a) xy (b) 2xy (c) x2y2 (d) 4xy (e)impossible to de-termine from thegiven information

6. The numbers from 1 to 25 are each written on separate slips of paperwhich are placed in a pile. You draw slips from the pile without replacingany slip you have chosen. You can continue drawing until the product oftwo numbers on any pair of slips you have chosen is a perfect square. Themaximum number of slips you can choose before you will be forced to quitis:

(a) 13 (b) 14 (c) 15 (d) 16 (e) 17

7. A container is completely �lled from a tap running at a constantrate. The accompanying graph shows the level of the water in the containerat any time while the container is being �lled. The segment PQ is a straightline. The shape of the container which corresponds with the graph is:

Full

WaterLevel

EmptyTime

q

q

Q

P

(a) (b) (c) (d) (e)

405

8. The accompanying diagram is a road plan of a small city. All theroads go east-west or north-south, with the exception of the one short diag-onal road shown. Due to repairs one road is impassable at the point X. Ofall the possible routes from P to Q, there are several shortest routes. Thetotal number of shortest routes is:

X

P

Q

(a) 4 (b) 7 (c) 9 (d) 14 (e) 16

9. Four pieces of timber with the lengths shown are placed in theparallel positions shown. A single cut is made along the line L perpendicularto the lengths of timber so that the total length of timber on each side of Lis the same. The length, in metres, of the longest piece of timber remainingis:

-�

-�

-�

5m

3m 3m

4m 5m

1:5m 4m

L

(a) 4:85 (b) 4:50 (c) 4:75 (d) 3:75 (e) none of the above

10. The positive integers are written in order with one appearing once,two appearing twice, three appearing three times, : : : , ten appearing tentimes, and so on, so that the beginning of the sequence looks like this:

1 , 2 , 2 , 3 , 3 , 3 , 4 , 4 , 4 , 4

The number of 9's appearing in the �rst 1998 digits of the sequence is:

(a) 57 (b) 96 (c) 113 (d) 145 (e) 204

406

Part B

1. A right triangle has an area of 5 and its hypotenuse has length 5.Determine the lengths of the other two sides.

2. Find a set of three consecutive positive integers such that the small-est is a multiple of 5, the second is a multiple of 7 and the largest is a multipleof 9.

3. In the diagram, BD = 2, BC = 8 and

the marked angles are all equal; that is,

\ABC = \BCA = \CDE = \DEC .

Find AB.

A

B C

E

D

2

8

4. The ratio of male to female voters in an election was a : b. If cfewer men and d fewer women had voted, then the ratio would have beene : f . Determine the total number of voters who cast ballots in the electionin terms of a, b, c, d, e and f .

5. Three neighbours named Penny, Quincy and Rosa took part in a localrecycling drive. Each spent a Saturday afternoon collecting all the aluminumcans and glass bottles he or she could. At the end of the afternoon eachperson counted up what he or she had gathered, and they discovered thateven though Penny had collected three times as many cans as Quincy, andQuincy had collected four times as many bottles as Rosa, each had collectedexactly the same number of items, and the three as a group had collectedexactly as many cans as bottles. In total, the three collected fewer than 200items in all. Assuming that each person collected at least one can and onebottle, how many cans and bottles did each person collect?

Last issue we gave the Final Round Parts A and B of the 1998 BritishColumbia Colleges Junior High School Mathematics Contest. My thanks goto Jim Totten, University College of the Cariboo, one of the organizers, forforwarding the \o�cial solutions" which follow.

407

BRITISH COLUMBIA COLLEGESJUNIOR HIGH SCHOOL MATHEMATICS CONTEST

Final Round 1998Part A

1. Each edge of a cube is coloured either red or black. If every face ofthe cube has at least one black edge, the smallest possible number of blackedges is:

(a) 6 (b) 5 (c) 4 (d) 3 (e) 2

Answer: The correct answer is (d).

Suppose that every face of a cube has atleast one black edge. Since every edgebelongs to exactly two faces, and thereare six faces, the cube has at least threeblack edges. On the other hand, threeblack edges su�ce to satisfy the require-ment, as we can see on the diagram.The black edges are represented by thethicker lines.

2. Line AE is divided into four equal parts by the points B, C and D.Semicircles are drawn on segments AC, CE, AD and DE creating semicir-cular regions as shown. The ratio of the area enclosed above the line AE tothe area enclosed below the line is:

r r r r rAB C

DE

(a) 4 : 5 (b) 5 : 4 (c) 1 : 1 (d) 8 : 9 (e) 9 : 8

Answer: The correct answer is (a).

Suppose that AB has length 1. Then both semicircles lying above AEhave radii of 1, while the semicircles below AE have radii of 11

2and 1

2. The

ratio of the enclosed areas is [12�(1)2+ 1

2�(1)2]� [1

2�(11

2)2+ 1

2�(1

2)2] = 4

5.

3. A container is completely �lled from a tap running at a uniform rate.The accompanying graph shows the level of the water in the container at anytime while the container is being �lled. The segment PQ is a straight line.The shape of the container which corresponds with the graph is:

408

Full

WaterLevel

EmptyTime

q

q

Q

P

(a) (b) (c) (d) (e)

Readers may have noticed that this is the same as problem 7 in part Aof the British Columbia Colleges Senior High School Mathematics ContestFinal Round 1998, printed above. For this reason, we delay publishing thiso�cial solution until the next issue.

4. The digits 1, 9, 9, and 8 are placed on four cards. Two of the cardsare selected at random. The probability that the sum of the numbers on thecards selected is a multiple of 3 is:

(a) 14

(b) 13

(c) 12

(d) 23

(e) 34

Answer: The correct answer is (b).

Let a, b, c, d denote the cards with digits 1, 9, 9, and 8, respectively.There are six possible choices of two cards from the set of four: fa; bg, fa; cg,fa; dg, fb; cg, fb; dg, fc; dg. For exactly two of these, fb; cg and fa; dg, thecorresponding sums, 9 + 9 and 1 + 8, are divisible by 3. This gives theprobability of 2

6= 1

3.

5. The surface areas of the six faces of a rectangular solid are 4, 4, 8, 8,18 and 18 square centimetres. The volume of the solid, in cubic centimetresis:

(a) 24 (b) 48 (c) 60 (d) 324 (e) 576

Answer: The correct answer is (a).

If the edges of a rectangular solid have lengthsa, b, and c, then the areasof its nonparallel faces are ab, bc, and ac. Its volume abc =

p(ab)(bc)(ac) .In

our case abc =p4 � 8 � 18 = 24.

409

6. The area of the small triangle in the diagram is 8 square units. Thearea of the large triangle, in square units, is:

(a) 18 (b) 20 (c) 24 (d) 28 (e) 30

Answer: The correct answer is (e).

The length of the base of the larger tri-angle is 5b, while the length of thebase of the smaller triangle is 2b. Thisgives the ratio of 5

2. Similarly, the ra-

tio of the corresponding perpendicularheights, H : h, is 3a : 2a = 3

2.

Hence, the area of the larger triangle is52(32)(8) = 30.

2b 3b

H

h

a

2a

7. At 6:15 the hands of the clock form two positive angles with a sumof 360�. The di�erence of the degree measures of these two angles is:

(a) 165 (b) 170 (c) 175 (d) 180 (e) 185

Answer: The correct answer is (a).

At 6:15 the minute hand points at 3, while the hour hand is 14of the way

from 6 to 7. The smaller angle between the hands is [90+ 14(36012

)]� = 97:5�,while the larger is (360 � 97:5)� = 262:5�. This gives the di�erence of(262:5� 97:5)� = 165�.

8. The last digit of the number 826 is:

(a) 0 (b) 2 (c) 4 (d) 6 (e) 8

Answer: The correct answer is (c).

By inspecting the last digit of the numbers in the sequence 81, 82, 83,84, : : : , we discover a repeating pattern of length four: 8, 4, 2, 6. Since826 = 84(6)+2, we conclude that the last digit of 826 is the same as the lastdigit of 82, that is 4.

9. For the equation Ax+3

+ Bx�3 = �x+9

x2�9 to be true for all values of x

for which the expressions in the equation make sense, the value of AB is:

(a) 2 (b) �1 (c)�2 (d)�3 (e) �6Answer: The correct answer is (c).

The expression makes sense for all values of x, except �3. By mul-tiplying both sides of the equation by the common denominator x2 � 9 =(x � 3)(x+ 3), we get A(x� 3) + B(x+ 3) = �x+ 9. After multiplyingout and collecting the like terms on the left hand side of this equation we get(A + B)x + 3B � 3A = �x + 9. Clearly, the polynomials on both sidesmust be identical; therefore A+B = �1 and 3B� 3A = 9. This system oftwo equations can be solved in any standard way. For example, we can �ndB = 3 + A from the second equation and substitute this for B in the �rstequation. In that way we �nd A = �2 and B = 1.

410

10. A hungry hunter came upon two shepherds, Joe and Frank. Joe hadthree small loaves of bread and Frank �ve loaves of the same size. The loaveswere divided equally among the three people, and the hunter paid $8 for hisshare. If the shepherds divide the money so that each gets an equitable sharebased on the amount of bread given to the hunter, the amount of money thatJoe receives is:

(a) $1 (b) $1:50 (c) $2 (d) $2:50 (e) $3

Answer: The correct answer is (a).Divide each loaf into 3 parts and distribute equally to each of the three

persons. Each person receives 8 parts. The two shepherds start with 9 and15 parts each, so (after removing their own 8 parts) they contribute 1 and7 parts, respectively, to the hunter and should receive compensation fromthe hunter in that ratio. Thus the hunter who originally had 3 loaves shouldreceive $1.

Part B

1. Four positive integers sum to 125. If the �rst of these numbers isincreased by 4, the second is decreased by 4, the third is multiplied by 4 andthe fourth is divided by 4, you produce four equal numbers. What are thefour original numbers?

Solution. The numbers are 16, 24, 5 and 80.If x, y, z, and w are the numbers then x + y + z + w = 125 and

x + 4 = y � 4 = 4z = w4. Hence, y = x + 8, z = x+4

4, w = 4(x + 4).

By substituting these expressions to the �rst equation, we get x+ (x+8)+x+44

+ 4(x + 4) = 125. Thus, x = 16, and consequently, y = 24, z = 5,w = 80.

2. A semi-circular piece of paper of radius 10 cm is formed into a conicalpaper cup as shown (the cup is inverted in the diagram):

V 10 cm

V

10 cm

VN N

Find the height of the paper cup, that is, the depth of water in the cupwhen it is full.

Solution. The height of the paper cup is 5p3 cm.

The base of the conical paper cup is a circle with circumference equalto the length of the given semicircle. Thus, if r is the radius of the base then2�r = 1

2(2�10). Hence, r = 5 cm. The side length of the cone s is the same

as the radius of the semicircle; thus s = 10 cm. Finally, the height of thecone is

h =ps2 � r2 =

p102 � 52 = 5

p3 cm .

411

3. In the diagram a quarter circle is inscribed in a square with sidelength 4, as shown. Find the radius of the small circle that is tangent to thequarter circle and two sides of the square.

Solution. The radius of the small circle is 12� 8p2.

The Pythagorean Theorem implies thatthe diagonal of a square with side ahas length a

p2. Thus, the diagonal

of the larger square has length 4p2.

It is equal to the sum of the radiusof the larger circle, 4, the radius ofthe smaller circle, r, and the diagonalof the smaller square, r

p2. Hence,

4p2 = 4 + r + r

p2. This gives

r =4p2� 4

1 +p2

=4p2� 4

1 +p2

1�p2

1�p2

!= 12� 8

p2 .

4

44

rr

r

4. Using the digits 1, 9, 9 and 8 in that order create expressions equalto 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. Youmay use any of the four basic operations(+, �, �, �), the square root symbol (p) and parentheses, as necessary.For example, valid expressions for 25 and 36 would be

25 = �1 + 9 + 9 + 8 ,

36 = 1 + 9�p9 + 8 .

Note: You may place a negative sign in front of 1 to create �1 if you wish.

Solution. One of several possible solutions is:

1 = �1 +p9� 9 + 8 ,

2 = 1�p9� 9 + 8 ,

3 = �1 +p9 + 9� 8 ,

4 = 1�p9 + 9� 8 ,

5 = 1 +p9 + 9� 8 ,

6 = �1� 9� 9 + 8 ,

7 = �1 + 9� 9 + 8 ,

8 = �1 + 9� 9 + 8 ,

9 = �1 + 9 + 9� 8 ,

10 = 1 + 9� 9 + 8 .

412

5. At 6 am one Saturday, you and a friend begin a recreational climb ofMt. Mystic. Two hours into your climb, you are overtaken by some scouts.As they pass, they inform you that they are attempting to set a record forascending and descending the mountain. At 10 am they pass you again ontheir way down, crowing that they had not stopped once to rest, not even atthe top.

You �nally reach the summit at noon. Assuming that both you and thescouts travelled at a constant vertical rate, both climbing and descending,when did the scouts reach the top of Mt. Mystic?

Solution. The scouts reached the top of Mt. Mystic at 9:20 am.Suppose that during the time period from 8:00 am to 10:00 am you have

travelled from pointA to pointB and you climbed a distance of x kilometres.Then, since you have been climbing at a uniform rate and reached the top atnoon, the distance from B to the top is also x kilometres. During the twohours you climbed x kilometres from A to B, the scouts climbed the distanceof 3x kilometres: x from A to B, x from B to the top, and x on the way backto B from the top. Since their pace was uniform, they needed 2

3of an hour,

that is 40 minutes, to get from the top to point B, where they met you at10:00 am. This implies that they must have reached the top at 9:20 am.

That completes the Corner for this number. Send me contest materialsand suggestions for the evolution of the Skoliad Corner.

413

MATHEMATICAL MAYHEM

Mathematical Mayhem began in 1988 as a Mathematical Journal for and byHigh School and University Students. It continues, with the same emphasis,as an integral part of Crux Mathematicorum with Mathematical Mayhem.

All material intended for inclusion in this section should be sent toMathematical Mayhem, Department of Mathematics, University of Toronto,100 St. George St., Toronto, Ontario, Canada. M5S 3G3. The electronicaddress is still

mayhem@math.toronto.edu

The Assistant Mayhem Editor is Cyrus Hsia (University of Western On-tario). The rest of the sta� consists of Adrian Chan (Upper Canada College),Jimmy Chui (University of Toronto), David Savitt (Harvard University) andWai Ling Yee (University of Waterloo).

Shreds and Slices

A Combinatorial Proof of a Trigonometric Identity

Douglass Grant

A friend who attended university in Germany once told me that hiscourse in �rst year calculus was made memorable by the fact that the basicde�nition of the sine and cosine functions used by his professor were theMaclaurin series for the two functions. At the time, I made the ippantremark that such an approach would make it a challenge even to prove thatsin2 x+cos2 x = 1. The details of that proof, in fact, involve some identitiesmore commonly encountered in discrete mathematics or combinatorics thanin calculus.

Since

sinx =

1Xk=0

(�1)k(2k+ 1)!

x2k+1 ,

squaring the series and shifting the index by one on the inner series yields

414

sin2 x =

1Xk=0

0B@ X

n+m=kn;m�0

(�1)n(�1)m(2n+ 1)!(2m+ 1)!

1CAx2k+2

=1Xk=1

0B@ X

n+m=k�1n;m�0

(�1)k�1(2n+ 1)!(2m+ 1)!

1CA x2k

=

1Xk=1

k�1Xn=0

(�1)k�1(2n+ 1)!(2k� 2n� 1)!

!x2k .

Similarly,

cos2 x =

1Xk=0

kX

n=0

(�1)k(2n)!(2k� 2n)!

!x2k .

Since the constant term in the series for cos2 x is clearly unity, it su�cesto show that the sum of the coe�cients of x2k for sin2 x and cos2 x is zero fork � 1. Note that the integers whose factorials appear in the denominatorsof both inner summations sum to 2k.

For k � 1, let

Sk =

k�1Xn=0

(�1)k�1(2n+ 1)!(2k� 2n� 1)!

+

kXn=0

(�1)k(2n)!(2k� 2n)!

.

Then

(2k)!Sk =

2kXn=0

(�1)n(2k)!(2n)!(2k� 2n)!

=

2kXn=0

(�1)n�2k

n

�.

But by the Binomial Theorem,

(a� b)2k =2kXn=0

(�1)n�2k

n

�a2k�nbn ,

so letting a = b = 1, we obtain

0 =

2kXn=0

(�1)n�2k

n

�= (2k)!Sk ,

whence Sk = 0 for k � 1, as required.

Douglass L. Grant <dlgrant@uccb.ns.ca>University College of Cape Breton

Box 5300, Sydney, Nova ScotiaCanada B1P 6L2

415

Mayhem Problems

The Mayhem Problems editors are:

Adrian Chan Mayhem High School Problems Editor,Donny Cheung Mayhem Advanced Problems Editor,David Savitt Mayhem Challenge Board Problems Editor.

Note that all correspondence should be sent to the appropriate editor |see the relevant section. In this issue, you will �nd only problems | thenext issue will feature only solutions.

We warmly welcome proposals for problems and solutions. With theschedule of eight issues per year, we request that solutions from this issuebe submitted in time for issue 8 of 2000.

High School Problems

Editor: Adrian Chan, 229 Old Yonge Street, Toronto, Ontario, Canada.M2P 1R5 <all238@sprint.com>

We correct problem H253 again, which we tried to correct in Issue 5.

H253. Find all real solutions to the equationp3x2 � 18x+ 52 +

p2x2 � 12x+ 162 =

p�x2 + 6x+ 280 .

H261. Solve for x:�q7�

p48

�x+

�q7 +

p48

�x= 14 .

H262. Proposed by Mohammed Aassila, CRM, Montr �eal, Qu �ebec.Solve the equation

x� xpx2 � 1

=91

60.

H263. Let ABC be an acute-angled triangle such that a = 14,sinB = 12=13, and c, a, b form an arithmetic sequence (in that order).Find tanA+ tanB + tanC.

H264. Find all values of a such that x3� 6x2+11x+ a� 6 = 0 hasexactly three integer solutions.

416

Advanced Problems

Editor: Donny Cheung, c/o Conrad Grebel College, University of Wa-terloo, Waterloo, Ontario, Canada. N2L 3G6 <dccheung@uwaterloo.ca>

A237. Show that for any sequence of decimal digits that does not be-gin with 0, there is a Fibonacci number whose decimal representation beginswith this sequence. (The Fibonacci sequence is the sequence Fn generatedby the initial conditions F0 = 0, F1 = 1 and Fn = Fn�1 + Fn�2 for n � 2.)

A238. Two circles C1 and C2 intersect at P and Q. A line through Pintersects C1 and C2 again at A and B, respectively, and X is the mid-pointof AB. The line through Q and X intersects C1 and C2 again at Y and Z,respectively. Prove that X is the mid-point of Y Z.

(1997 Baltic Way)

A239. Proposed by Mohammed Aassila, CRM, Montr �eal, Qu �ebec.

Let a1, a2, : : : , an be n distinct numbers, n � 3. Prove that

nXi=1

0@ai �Y

j 6=i

1

ai � aj

1A = 0 .

A240. Proposed by Mohammed Aassila, CRM, Montr �eal, Qu �ebec.

Let a, b, and c be integers, not all equal to 0. Show that

1

4a2 + 3b2 + 2c2���� 3p4a+

3p2b+ c

��� .

Challenge Board Problems

Editor: David Savitt, Department of Mathematics, Harvard University,1 Oxford Street, Cambridge, MA, USA 02138 <dsavitt@math.harvard.edu>

C89. Proposed by Tal Kubo, Brown University.

Show that the formal power series (in x and y)

1Xn=0

(xy)n cannot be

expressed as a �nite summXi=1

fi(x)gi(y) , where fi(x) and gi(y) are formal

power series in x and y, respectively, 1 � i � m.

417

C90. Proposed by Noam Elkies, Harvard University.

Let S1, S2, andS3 be three spheres inR3 whose centres are not collinear.

Let k � 8 be the number of planes which are tangent to all three spheres. LetAi, Bi, and Ci be the point of tangency between the ith such tangent plane,1 � i � k, and S1, S2, and S3, respectively, and let Oi be the circumcentreof triangle AiBiCi. Prove that all the Oi are collinear. (If k = 0, then thisstatement is vacuously true.)

Problem of the Month

Jimmy Chui, student, University of Toronto

Problem. The point P (a; b) is on the parabola x2 = 4y. The tangentat P meets the line y = �1 at the point A. For the point F (0;1), prove that\AFP = 90� for all positions of P , except (0; 0).

(Descartes 1998, C3)

Solution. This question can be done using calculus; however, here weshow a clever method that makes the problem fall apart quite easily.

-

6

F

O

A

P

P 0

�

�

First we note that the point F is the focus and the line y = �1 is thedirectrix of the given parabola.

418

Drop the perpendicular from P to the line y = �1, and call that pointP 0. Now, let �, �, and be the angles as in the diagram; that is, let � =\APF , let = APP 0, and let � represent the angle opposite \APP 0.

Observe that PF = PP 0, because any point on a parabola is equidis-tant to both the focus and directrix.

From the opposite angle theorem we get � = .

Now, it is known that a ray, parallel to the axis of symmetry in aparabola, will pass through the focus when it re ects o� the interior of theparabola. The line P 0P extended is parallel to the line x = 0 (the axis ofsymmetry) and can be taken as the ray of incidence. The ray PF is then theray of re ection. The line AP extended is the tangent at the point P of theparabola, and the angle of incidence equals the angle of re ection, or � = �.So we have � = .

Hence, from SAS congruency, we have that 4APF � 4APP 0. Thisimplies that \AFP = \AP 0P = 90�, QED.

J.I.R. McKnight Problems Contest 1991

1. (a) The vertices of a right-angled triangle ABC are A(4; 6), B(3;�3)and C(8; y). Find all possible values of y.

(b) Simplify:

53x+1 � 53x�1 + 24

(2:4)53x + 12.

2. (a) A rectangle has been inscribed in a circle and semi-circles have beendrawn on its sides (as shown in the diagram). Determine the ratioof the sum of the area of the four lunes (shaded regions) to the areaof the rectangle.

419

(b) Find the value of (220

+ 1)(221

+ 1)(222

+ 1) � � � (2250 + 1).

3. (a) Find the sum of the �rst one thousand terms for the series:1 � 4 + 2 � 7 + 3 � 10 + � � � .

(b) Solve for x, y 2 R:

x+ y +pxy = 19

x2 + y2 + xy = 133

4. The two equations a2x2 + 192bx + 1991a = 0 and bx2 + 192a2x +1991b = 0 have a common root. If a2 6= b, determine all possiblevalues of the common root.

5. A metal cone of height 12 cmand radius 4 cm just �ts intoa cylinder of radius 4 cm andheight 100 cm which is �lledwith water to a depth of 50cm. The cone is lowered at aconstant rate of 4 cm per sec-ond (with respect to the wallsof the cylinder). At what rateis the water in the cylinder ris-ing when the vertex of the coneis immersed to a depth of 6 cm?

6. A semi-circle has centre C and diameter AB. The point N is on CBand AB is produced to T so that AT : AC = AN : CN . The tangentfrom T meets the semi-circle at P . Prove that \CNP = 90�.

A C N B T

P

7. (a) Prove that the product of 4 consecutive positive integers cannot bea perfect square.

(b) What must be added to the product of 4 consecutive terms of anyarithmetic sequence to produce a perfect square?

(c) What must be added to the product of 4 consecutive terms of anygeometric sequence to produce a perfect square?

420

8. Triangle ABC has sides AB = 6, AC = 4 and BC = 5. Point P ison AB and point Q is on AC such that PQ bisects the area of triangleABC. Prove that the minimum length of PQ is

p42=2.

9. Given b > 0, b 6= 1, �nd the set of values of k for which the equationlogb2(x

2 � b2) = logb(x� bk) has real solutions.

10. Six points are chosen in space such that no three are collinear and nofour are coplanar. The 15 line segments joining the points in pairs aredrawn then painted, some red and some blue. Prove that some trianglehas all its sides the same colour.

IMO Report

Jimmy Chuistudent, University of Toronto

The 1999 Canadian IMO team members commenced their summer withone and a half weeks of training at the University of Waterloo. During thistime, they managed to hike twice, and it was a shame that no one was lostthis year. On the 13th of July, after an exhausting full day of travel, theteam found itself in Bucharest, Romania, ready for the 40th InternationalMathematical Olympiad.

The members of this year's team were David \Pippy" Arthur, Jimmy\The Squeeze" Chui, James \Roadkill" Lee, Jessie \Pyromaniac" Lei, David\Monkey Matrix" Nicholson, and David \23 Across" Pritchard. Team leaderDr. Ed \81" Barbeau gave incessant lectures on continuity while deputy leaderDr. Arthur \Put down that math and deal!" Baragar could be found playingTetris on some particular portable game machine. Meanwhile, the deputyleader observer, Dr. Dorette \Maybe this works..." Pronk, dutifully tookpictures of the other team members while they were not looking. The teamis also grateful to Dr. Ed Wang and Richard Hoshino for their wise words incombinatorics and inequalities, and to Dr. Christopher Small for sharing hisfunctional equations knowledge, as well as for his outstanding hospitality inElora.

This year's contest was immensely challenging, and it continued the lowmedal cut-o� scores the last few IMOs have seen. Considering the di�cultyof the questions, Canada performed respectably and brought home 3 bronzes.The scores were as follows:

CAN 1 David Arthur 18 Bronze MedalCAN 2 Jimmy Chui 16 Bronze MedalCAN 3 James Lee 6CAN 4 Jessie Lei 9CAN 5 David Nicholson 8CAN 6 David Pritchard 17 Bronze Medal

421

Uno�cially, Canada's total score of 74 was enough for 32nd place outof the 83 competing countries. Best of luck to CAN 3 and 5 as they pursuetheir university studies at the University of Waterloo, and to CAN 2 and 4 asthey move on to the University of Toronto. The remaining two members arestill eligible for next year's team. Hopefully there will be shouts of \We liketo party!" after the competition next year!

Special thanks must also go to Dr. Graham Wright of the CanadianMathematical Society for once again supplying the funds for the team, andagain to team leader Dr. Ed Barbeau for his continual e�orts training IMOpotentials through the CMS's correspondence program.

It was the �rst ight to Europe for many of us, and quite an experience itturned out to be. We were surprised at the endless supply of cheese that thecafeteria managed to put on our plates. We found it a great object to ward o�stray dogs. Visits to several museums, including the infamous TransylvaniaCastle, were a real treat to the competitors, but it was a shame that Draculawas nowhere to be found. However, the cheese did �nd its way along withus. With all that said and done, the IMO was once again a success. We wishthe best of luck to all hopefuls for the 2000 Canadian IMO team, bound forthe Republic of Korea for the 41st IMO.

Stan Wagon's e{mail problem of the week

For the bene�t of those readers of CRUX with MAYHEM who makeuse of Stan Wagon's e-mail Problem of the Week, and for the information ofthose who do not know about it, the Math Forum at Swarthmore has justtaken over the handling of the e-list. The instructions for subscribing are:

to subscribe to the Problem of the Week send a message to<<majordomo@forum.swarthmore.edu>>. Body of message shouldread simply SUBSCRIBE MACPOW. Macalester students shouldNOT subscribe to the e{list, but get printed postings instead.

Here is a sample of the type of problem that you can expect:

Problem 887 Square Division

Show how to divide a unit square into two rectangles so that the smallerrectangle can be placed on the larger with every vertex of the smaller onexactly one of the edges of the larger.

Source: 1994 Dutch Mathematical Olympiad; as reported in Crux Mathe-maticorum, Sept 1998, Vol. 24, No. 5, p. 264.

422

An Identity of a Tetrahedron

Murat Aygen

Problem. Let ABCD be a tetrahedron with sides a = BC, b = AC,c = AB, a1 = AD, b1 = BD, and c1 = CD (see Figure 1(a)). Let V and Rdenote the volume and circumradius of the tetrahedron, respectively. Showthat 6V R equals the area of the triangle with sides aa1, bb1, and cc1.

Solution. Consider the plane of triangle ABC and the parallel planethrough D. These intersect the circumsphere of ABCD in two circles; lettheir radii be r1 and r2, respectively (see Figure 1(b)). Let h be the distancebetween the two planes.

Figure 1(a).

D

C

A

B a1b1c1

b

a c

Figure 1(b).

6

?

h=DD0

r2

r1

Let O be the circumcentre of triangle ABC, letD0 be the projection ofD onto the plane of triangle ABC, and let 2� = \D0OC (see Figure 2).

By the Cosine Law on triangle D0OC,

(CD0)2 = r21 + r22 � 2r1r2 cos 2�

= r21 + r22 � 2r1r2(1� 2 sin2 �)

= (r1 � r2)2 + 4r1r2 sin

2 � .

Since DD0 = h, we obtain similarly that

a21 = AD2 = (DD0)2 + (AD0)2

= h2 + (r1 � r2)2 + 4r1r2 sin

2(B � �) ,b21 = BD2 = (DD0)2 + (BD0)2

= h2 + (r1 � r2)2 + 4r1r2 sin

2(A+ �) ,

c21 = CD2 = (DD0)2 + (CD0)2

= h2 + (r1 � r2)2 + 4r1r2 sin

2 � .

Copyright c 1999 Canadian Mathematical Society

423

A

B

C

2C

2A

2(B � �)

2�

c

a

b

D0r2O

r1

Figure 2.

We now present a geometrical derivation of these lengths. Let K andK1 denote the area of the triangle with sides a, b, and c, and the triangle withsides aa1, bb1, and cc1, respectively. We will derive a relationship betweenK andK1.

Let P0 be the plane containing triangle ABC. Consider another planeP1, passing through A, and meeting P0 at an angle of �. Let be the anglebetween AB and the intersection of P0 and P1 (see Figure 3), where � and are angles that will be speci�ed later.

P0

P1C0

C B

B0

A

z

cb

a

�

Figure 3.

424

Let B0 and C0 be the points in P1 that project to B and C in P0, re-spectively. Then by some elementary trigonometry,

(B0C0)2 = a2 + a2 tan2� sin2(B � ) ,

(AC0)2 = b2 + b2 tan2 � sin2(A+ ) ,

(AB0)2 = c2 + c2 tan2 � sin2 .

Now, assign and � such that

= � and tan2 � =4r1r2

h2 + (r1 � r2)2.

Then the equations above become

(B0C0)2 = a2(1 + tan2 � sin2 �)

=a2[h2 + (r1 � r2)

2 + 4r1r2 sin2 �]

h2 + (r1 � r2)2

=a2a21

h2 + (r1 � r2)2,

(AC0)2 = b2[1 + tan2 � sin2(B � �)]

=b2b21

h2 + (r1 � r2)2,

(AB0)2 = c2[1 + tan2 � sin2(A+ �)]

=c2c21

h2 + (r1 � r2)2.

Hence, AB0C0 is proportional to the triangle with sides aa1, bb1, andcc1, with ratio of areas

1

h2 + (r1 � r2)2.

Projecting fromP1 toP0 scales the area by a further factor of 1= cos�. Hence,

K =K1 cos�

h2 + (r1 � r2)2. (1)

We also know that

V =1

3Kh , (2)

h =qR2 � r21 +

qR2 � r22 . (3)

Squaring both sides of (3), we obtain

h2 + r21 + r22 � 2R2 = 2q(R2 � r21)(R2 � r22) .

425

Therefore,

(h2 + r21 + r22)2 � 4R2h2 � 4R2r21 � 4R2r22 + 4R4

= 4R4 � 4R2r21 � 4R2r22 + 4r21r22 ,

yielding

4R2h2 = h4 + r41 + r42 + 2h2r21 + 2h2r22 � 2r21r22

= h4 + r41 + r42 + 2h2r21 + 2h2r22 + 2r21r22 � 4r21r

22

= (h2 + r21 + r22)2 � (2r1r2)

2

= [h2 + (r1 + r2)2][h2 + (r1 � r2)

2] . (4)

Finally, note that

cos2 � =1

1 + tan2 �=

h2 + (r1 � r2)2

h2 + (r1 + r2)2. (5)

Therefore,

(6V R)2 = 4K2R2h2 (from (2))

=4K2

1R2h2 cos2 �

[h2 + (r1 � r2)2]2(from (1))

=K2

1 [h2 + (r1 + r2)

2][h2 + (r1 � r2)2]2

[h2 + (r1 + r2)2][h2 + (r1 � r2)2]2(from (4) and (5))

= K21 ,

so that6V R = K1 .

Murat Aygen <aygenmurat@usa.net>

Cinnah Cad., Alacam Sok., No. 3/406690 Ankara TURKEY

426

A Simple Proof of a Pentagram Theorem

Geo�rey A. Kandall

We will give a short, transparent proof of Eiji Konishi's pentagram the-orem, which was communicated by Hiroshi Kotera [1]. The proof is reallyjust an exercise in the Law of Sines.

Theorem. Let A1A2A3A4A5 be a pentagram with pentagonB1B2B3B4B5 as shown in the �gure. Let Ck be the intersection of linesegment AkBk and side Bk+2Bk+3, for k = 1, 2, : : : , 5. Then

B3C1

C1B4

� B4C2

C2B5

� B5C3

C3B1

� B1C4

C4B2

� B2C5

C5B3

= 1 .

A1

A2

A3

A4

A5

B1

B2

B3B4

B5

C1

��

6

7

�

� 8

9

10

1�

�2

3

�

�

4

5

Copyright c 1999 Canadian Mathematical Society

427

Proof. Let [P ] denote the area of polygon P . Then we have that

B3C1

C1B4

=[A1B3B1]

[A1B4B1]

=A1B3 � B3B1 � sin\A1B3B1

A1B4 � B4B1 � sin\A1B4B1

=sin� � B3B1 � sin �10sin � B4B1 � sin �3

.

Similarly,

B4C2

C2B5

=sin� � B4B2 � sin�2sin� � B5B2 � sin �5

,

B5C3

C3B1

=sin� � B5B3 � sin�4sin � � B1B3 � sin �7

,

B1C4

C4B2

=sin� � B1B4 � sin �6sin� � B2B4 � sin�9

,

B2C5

C5B3

=sin � B2B5 � sin �8sin� � B3B5 � sin�1

.

Multiplying these �ve equations together, we obtain

B3C1

C1B4

� B4C2

C2B5

� B5C3

C3B1

� B1C4

C4B2

� B2C5

C5B3

=sin�10 � sin�2 � sin�4 � sin �6 � sin �8sin�7 � sin �9 � sin �1 � sin�3 � sin�5

=B1B2

B2B3

� B2B3

B3B4

� B3B4

B4B5

� B4B5

B5B1

� B5B1

B1B2

= 1 .

Reference

[1] H. Kotera, The Pentagram Theorem, CRUX with MAYHEM 24:5 (1998),291{295.

Geo�rey A. Kandall230 Hill Street

Hamden, CT 06514{1522 USA

428

PROBLEMSProblem proposals and solutions should be sent to Bruce Shawyer, Department

of Mathematics and Statistics, Memorial University of Newfoundland, St. John's,

Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution,

together with references and other insights which are likely to be of help to the edi-tor. When a submission is submitted without a solution, the proposer must include

su�cient information on why a solution is likely. An asterisk (?) after a number

indicates that a problem was submitted without a solution.

In particular, original problems are solicited. However, other interesting prob-

lems may also be acceptable provided that they are not too well known, and refer-

ences are given as to their provenance. Ordinarily, if the originator of a problem canbe located, it should not be submitted without the originator's permission.

To facilitate their consideration, please send your proposals and solutions

on signed and separate standard 812"�11" or A4 sheets of paper. These may

be typewritten or neatly hand-written, and should be mailed to the Editor-in-Chief, to arrive no later than 1 April 2000. They may also be sent by email to

crux-editors@cms.math.ca. (It would be appreciated if email proposals and solu-

tions were written in LATEX). Graphics �les should be in epic format, or encapsulatedpostscript. Solutions received after the above date will also be considered if there

is su�cient time before the date of publication. Please note that we do not accept

submissions sent by FAX.

Last month's problem section asked, in error, for solutions by 1 January 2000.That was a Y2K bug! It should have read 1 March 2000.

2452. Correction. Proposed by Antal E. Fekete, Memorial Universityof Newfoundland, St. John's, Newfoundland.

Establish the following equalities:

(a)

1Xn=0

(2n+ 1)2

(2n+ 1)!=

1Xn=0

(2n+ 2)2

(2n+ 2)!.

(b) and (c) are correct as originally printed.

2453. Correction. Proposed by Antal E. Fekete, Memorial Univer-sity of Newfoundland, St. John's, Newfoundland.

Establish the following equalities:

(a)1Xn=0

(�1)n (2n+ 1)3

(2n+ 1)!= �3

1Xn=0

(�1)n 1

(2n+ 1)!.

(b)

1Xn=0

(�1)n (2n)3

(2n)!= �3

1Xn=0

(�1)n 1

(2n)!.

(c)

1Xn=0

(�1)n (2n+ 1)2

(2n+ 1)!

!2+

1Xn=0

(�1)n (2n)2

(2n)!

!2

= 2 .

429

2476. Proposed by Mohammed Aassila, CRM, Universit �e deMontr �eal, Montr �eal, Qu �ebec.

Let n be a positive integer and consider the set f1, 2, 3, : : : , 2ng. Givea combinatorial proof that the number of subsets A such that

1. A has exactly n elements, and

2. the sum of all elements in A is divisible by n,

is equal to1

n

Xdjn

(�1)n+d��n

d

� �d

2d

�,

where � is the Euler function.

Note: When n is prime, proving the formula is problem 6 of the 1995IMO. A non-combinatorial proof of the formula is due to Roberto Dvornicichand Nikolay Nikolov.

2477. Proposed by Walther Janous, Ursulinengymnasium, Inns-bruck, Austria.

Given a non-degenerate 4ABC with circumcircle �, let rA be the in-radius of the region bounded by BA, AC and arc(CB) (so that the regionincludes the triangle).

Similarly, de�ne rB and rC. As usual, r and R are the inradius and circum-radius of 4ABC.

Prove that

(a)64

27r3 � rArBrC � 32

27Rr2 ;

(b)16

3r2 � rBrC + rCrA + rArB � 8

3Rr ;

(c) 4r � rA + rB + rC � 4

3(R+ r) ,

with equality occurring in all cases if and only if4ABC is equilateral.

2478. Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,Spain.

For n 2 N, evaluatenX

k=0

n� kk+ 1

�2k

k

��2n� 2k

n� k

�.

430

2479. Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,Spain.

Writing � (n) for the number of divisors of n, and !(n) for the numberof distinct prime factors of n, prove that

nXk=1

(� (k))2=

nXk=1

2!(k)bn=kcXj=1

�bn=kcj

�.

2480. Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,Spain.

Writing �(n) for Euler's totient function, evaluate

Xdjn

dXkjd

�(k)�(d=k)

k.

2481. Proposed by Mih �aly Bencze, Brasov, Romania.Suppose that A, B, C are 2� 2 commutative matrices. Prove that

det�(A+ B +C)(A3 + B3 + C3 � 3ABC)

� � 0 .

2482. Proposed by Mih �aly Bencze, Brasov, Romania.Suppose that p, q, r are complex numbers. Prove that

jp+ qj+ jq + rj+ jr + pj � jpj+ jqj+ jrj+ jp+ q + rj .

2483. Proposed by V�aclav Kone �cn �y, Ferris State University, BigRapids, MI, USA.

Suppose that 0 � A, B, C and A+ B + C � �. Show that

0 � A� sinA� sinB � sinC + sin(A+ B) + sin(A+ C) � � .

There are, of course, similar inequalities with the angles permuted cyclically.

[The proposer notes that this came up during an attempt to generaliseproblem 2383.]

2484. Proposed by Toshio Seimiya, Kawasaki, Japan.Given a square ABCD, suppose that E is a point on AB produced be-

yond B, that F is a point on AD produced beyondD, and that EF = 2AB.Let P and Q be the intersections of EF with BC and CD, respectively.Prove that

(a) 4APQ is acute-angled;

(b) \PAQ � 45�.

431

2485. Proposed by Toshio Seimiya, Kawasaki, Japan.ABCD is a convex quadrilateral with AB = BC = CD. Let P be

the intersection of the diagonals AC and BD. Suppose thatAP : BD = DP : AC.

Prove that either BC k AD or AB ? CD.

2486. Proposed by Joe Howard, New Mexico Highlands University,Las Vegas, NM, USA.

It is well-known that cos(20�) cos(40�) cos(80�) = 18.

Show that sin(20�) sin(40�) sin(80�) =p38.

2487. Proposed by Jos �e Luis D��az, Universitat Polit �ecnicade Catalunya,Terrassa, Spain.

If a, b, c, d are distinct real numbers, prove that

a4 + 1

(a� b)(a� c)(a� d)+

b4 + 1

(b� a)(b� c)(b� d)

+c4 + 1

(c� a)(c� b)(c� d)+

d4 + 1

(d� a)(d� b)(d� c) = a+ b+ c+ d .

2488. Proposed by G. Tsintsifas, Thessaloniki, Greece.Let Sn = A1A2 : : :An+1 be a simplex in En, andM a point in Sn. It

is known that there are real positive numbers �1, �2, : : : , �n+1 such thatn+1Xj=1

�j = 1 and M =

n+1Xj=1

�jAj (here, by a point P, we mean the position

vector�!OP). Suppose also that �1 � �2 � � � � � �n, and let Bk =

1

k

kXj=1

Aj.

Prove that

M 2 convex cover of fB1, B2, : : : , Bn+1g ;

that is, there are real positive numbers �1, �2, : : : , �n+1 such that

M =

n+1Xk=1

�kBk .

432

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.

2370. [1998: 364] Proposed by Walther Janous, Ursulinengymnas-ium, Innsbruck, Austria.

Determine the exact values of the roots of the polynomial equation

x5 � 55x4 + 330x3 � 462x2 + 165x� 11 = 0 .

Solutionby Joe Howard, New MexicoHighlands University, Las Vegas,NM, USA.

Let c = cos �, s = sin �, where � = k�11, k = 1, 3, 5, 7, 9. Then, by

DeMoivre's formula, we have

(c+ is)11 =�ei��11

= e11i� = ek�i = �1 .

Equating the imaginary parts, we get�11

1

�c10s�

�11

3

�c8s3+

�11

5

�c6s5�

�11

7

�c4s7+

�11

9

�c2s9�

�11

11

�s11 = 0 .

Since s 6= 0, we obtain

s10 � 55s8c2 + 330s6c4 � 462s4c6 + 165s2c8 � 11c10 = 0 .

Since c 6= 0, dividing the last equation by c10 yields

tan10 � � 55 tan8 � + 330 tan6 � � 462 tan4 � + 165 tan2 � � 11 = 0 .

Therefore, the roots of the given polynomial equation are given byx = tan2

�k�11

�, k = 1, 3, 5, 7, 9.

Also solved by MANUEL BENITO MU ~NOZ and EMILIO FERN �ANDEZ MORAL,I.B. Sagasta, Logro ~no, Spain; C. FESTRAETS-HAMOIR, Brussels, Belgium; DOUGLASSL. GRANT, University College of Cape Breton, Sydney, Nova Scotia; RICHARD I. HESS,Rancho Palos Verdes, CA, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece(2 solutions); GERRY LEVERSHA, St. Paul's School, London, England; KEE-WAI LAU, HongKong; VEDULA N. MURTY, Visakhapatnam, India; HEINZ-J �URGEN SEIFFERT, Berlin, Ger-many; ARAM TANGBOONDOUANGJIT, Carnegie Mellon University, Pittsburgh, PA, USA;JEREMY YOUNG, student, Nottingham High School, Nottingham, UK; and the proposer.

All the submitted solutions are more or less equivalent to the one given above. Almost

all the solvers gave the answers x = tan2�k�11

�, k = 1, 2, 3, 4, 5. Benito and Fern �andez gave

the answer x = cot2�k�22

�, k = 1, 3, 5, 7, 9. It is easy to see that all these expressions are

the same as the one obtained by Howard.

433

2371. [1998: 364] Proposed by Bill Sands, University of Calgary,Calgary, Alberta.

For n an integer greater than 4, let f(n) be the number of �ve-elementsubsets, S, of f1, 2, : : : , ng which have no isolated points, that is, such thatif s 2 S, then either s� 1 or s+ 1 (not taken modulo n) is in S.

Find a \nice" formula for f(n).

I. Solution byManuel Benito and Emilio Fernandez, I.B. Praxedes Ma-teo Sagasta, Logro ~no, Spain.

Let us represent any such subset S as [a1; a2; a3; a4; a5], wherea1 < a2 < a3 < a4 < a5. It is clear that the number a3 can take oneach of the n� 4 values from 3 to n� 2, but let us ask: how many times canit take on each one of these values?

Well, a1 is not isolated, so that a2 = a1+1, and a5 is not isolated eitherso a4 = a5�1. Then, for each �xed k from 3 to n�2 we can enumerate thesubsets Sk for which a3 = k, in the following manner:

� type [k � 2; k� 1; k; k+ 1; k + 2] ........................... 1 subset;

� from a1 = 1 to a1 = k� 3, type [a1; a2; k; k + 1; k + 2]......................................................... k� 3 subsets;

� from a5 = k + 3 to a5 = n, type [k� 2; k � 1; k; a5 � 1; a5].............................................. n� (k+ 3) + 1 subsets;

which gives a total of n � 4 subsets Sk for each k. Since there are n � 4values of k, this gives

f(n) = (n� 4)2 .

II. Solution by Kathleen E. Lewis, SUNY Oswego, Oswego, NY, USA.The characteristic function for such a subset will contain n � 5 zeros

and �ve ones. To be non-isolated, the ones must either consist of a singleblock or be broken into a block of size two and a block of size three (in eitherorder). Looking at the case of two blocks, we can think of this as a problemof placing two distinct items plus n � 5 indistinguishable items (the zeros)in a sequence, which can be done in (n � 3)(n � 4) ways. But if the twoblocks of ones end up adjacent to each other, we get the case of all �ve onestogether. Since it does not matter in this case which block of ones is �rst, wehave double-counted these arrangements. There are n � 4 ways to put allthe ones together among the n� 5 zeros. Therefore, when we compensatefor double-counting this case earlier, we get a �nal answer of

(n� 3)(n� 4)� (n� 4) = (n� 4)2 .

III. Solutionby Jeremy Young, student, NottinghamHigh School, Not-tingham, UK.

Given f(n), consider f(n+1) and every corresponding subset S. Thereare by de�nition f(n) subsets S with n+ 1 62 S. If n+ 1 2 S, n 2 S too.

434

Consider subsets of this form. Now if n�1 2 S, the remaining two elementsmust be adjacent; that is, k and k + 1 with 1 � k � n � 3, and there aren � 3 such subsets. If n � 1 62 S, the three remaining elements must beadjacent; that is, k; k + 1; k + 2 with 1 � k � n� 4, and there are n � 4such subsets.

We have thus established the recurrence

f(n+ 1) = f(n) + (n� 3) + (n� 4) = f(n)+ 2n� 7 ,

with f(5) = 1 (obviously) giving f(6) = 4, f(7) = 9, f(8) = 16, : : : . Notethat f(k) = (k� 4)2 is true for k = 5 and gives

f(k+ 1) = (k� 4)2 + 2k � 7 = k2 � 6k+ 9 = (k� 3)2 .

Therefore f(n) = (n� 4)2 by induction, for all integers n � 5.

Also solved by SAM BAETHGE, Nordheim, TX, USA; CHRISTOPHER J. BRADLEY,Clifton College, Bristol, UK; C. FESTRAETS-HAMOIR, Brussels, Belgium; RICHARD I. HESS,Rancho Palos Verdes, CA, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece;KEE-WAI LAU, Hong Kong, China; GERRY LEVERSHA, St. Paul's School, London, England;GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; EDWARD T.H. WANG, Wilfrid LaurierUniversity, Waterloo, Ontario; and the proposer. One incorrect solution was received.

Bradley's solution was the same as Lewis's, and Lau, Leversha and Perz all had solutionssimilar to Young's.

2372. [1998: 365] Proposed by Bill Sands, University of Calgary,Calgary, Alberta.

For n and k positive integers, let f(n; k) be the number of k{elementsubsets S of f1, 2, : : : , ng satisfying:(i) 1 2 S and n 2 S; and(ii) whenever s 2 S with s < n� 1, then either s+ 2 2 S or s+ 3 2 S.

Prove that f(n; k) = f(4k� 2� n; k) for all n and k; that is, the sequence

f(k; k) , f(k+ 1; k) , f(k+ 2; k) , : : : , f(3k� 2; k)

of non-zero values of ff(n;k)g1n=1 is a palindrome for every k.

Solution by Gerry Leversha, St. Paul's School, London, England.I shall call sets which satisfy the condition permissible.

In general, it is clear that f(k; k) = 1. Also f(3k� 2; k) = 1, since theonly possible set is f1, 4, 7, : : : , 3k� 5, 3k� 2g, an arithmetic progressionwith common di�erence 3 and k terms. Since this is the \sparsest" possi-ble k{element permissible set, it is clear that f(n; k) = 0 for n � 3k � 1.Equally obviously f(n;k) = 0 for n < k. So we need only consider n suchthat k < n < 3k � 2. For n = 2k � 1, we have 4k � 2 � n = 2k � 1, so

435

the palindromic condition is trivial here. It remains to show that fork < n < 2k� 1, f(n;k) = f(4k� 2� n; k).

Consider a permissible set S � f1, 2, 3, : : : , ng which has k elements.Let T be the sequence of di�erences between successive elements of S; then

� T is an ordered (k�1){tuple each of whose elements is either 1, 2 or 3;

� The sum of the elements of T is n� 1;

� T cannot contain the elements 1, 3 in that order.

Any such T will yield a permissible set S. The restriction on subsequencesof the form 1, 3 ensures that we cannot have a, a+ 1, a+4 appearing in S,which would contravene the condition since neither a + 2 nor a + 3 wouldbe in S. Hence

f(n; k) is equal to the number of such sequences T .

Now from any such sequence T construct a sequence T � as follows:

� Replace every element x of T by 4� x;

� Reverse the sequence so formed.

Then the new sequence T � has the following properties:

� It is an ordered (k�1){tuple formed from the elements 1, 2 and 3;

� The sum of the elements is 4(k� 1)� (n� 1) = 4k� n� 3;

� It cannot contain a subsequence 1, 3 (since there would have to havebeen such a subsequence in T in the �rst place).

Thus there is a one-to-one correspondence between sequences of typeT and those of type T �. But sequences of type T � correspond in turn topermissible k{element sets S � f1, 2, : : : , 4k � n � 2g. Hence we haveshown that

f(n;k) = f(4k� n� 2; k) .

I shall illustrate this process in the case of f(7;5) and f(11;5). Thefour columns below show, respectively,

� a permissible 5{element subset of f1, 2, 3, 4, 5, 6, 7g,

� the corresponding T sequence,

� the corresponding T � sequence,

� the permissible 5{element subset of f1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11g.

436

f1 ,2 ,3 ,5 ,7g (1; 1; 2; 2) (2; 2; 3; 3) f1 ,3 ,5 ,8 ,11gf1 ,2 ,4 ,5 ,7g (1; 2; 1; 2) (2; 3; 2; 3) f1 ,3 ,6 ,8 ,11gf1 ,2 ,4 ,6 ,7g (1; 2; 2; 1) (3; 2; 2; 3) f1 ,4 ,6 ,8 ,11gf1 ,3 ,4 ,5 ,7g (2; 1; 1; 2) (2; 3; 3; 2) f1 ,3 ,6 ,9 ,11gf1 ,3 ,4 ,6 ,7g (2; 1; 2; 1) (3; 2; 3; 2) f1 ,4 ,6 ,9 ,11gf1 ,3 ,5 ,6 ,7g (2; 2; 1; 1) (3; 3; 2; 2) f1 ,4 ,7 ,9 ,11gf1 ,4 ,5 ,6 ,7g (3; 1; 1; 1) (3; 3; 3; 1) f1 ,4 ,7 ,10 ,11g

Also solved by MANUEL BENITO and EMILIO FERNANDEZ, I.B. Praxedes Ma-teo Sagasta, Logro ~no, Spain; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;MICHAEL LAMBROU, University of Crete, Crete, Greece; KATHLEEN E. LEWIS, SUNY Os-wego, Oswego, NY, USA; JEREMY YOUNG, student, Nottingham High School, Nottingham,UK; and the proposer.

Lambrou, Lewis and Young gave \combinatorial" solutions similar to Leversha's.

2375. [1998: 365] Proposed by Toshio Seimiya, Kawasaki, Japan.Let D be a point on side AC of triangle ABC. Let E and F be points

on the segments BD and BC, respectively, such that \BAE = \CAF . LetP and Q be points on BC and BD, respectively, such that EP k DC andFQ k CD.

Prove that \BAP = \CAQ.

Solution by the proposer.A

B C

H

P F

EQ

GD

X

Y

Let PE and FQmeet AB at X andY , respectively.

Since PX k FY and FP , QE andY X concur at B, we have

XE

XP=

Y Q

Y F. (1)

Let the line through F parallel toAB meet AC and AQ at G and H,respectively.

Since AY k FH and QF k AG, we getY Q

Y F=

AQ

AH=

GF

GH. (2)

From (1) and (2), we now have

XE

XP=

GF

GH. (3)

Since XE kAG and AX kGF , we have

\AXP = 180� � \XAG = \AGF . (4)

437

Since \XAE = \BAE = \CAF = \GAF , we get from (4) that

4AXE � 4AGF . (5)

From (3) and (5), we have that

4AXP � 4AGH .

Therefore, we obtain that \XAP = \GAH, and this implies that\BAP = \CAQ.

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain;MANUEL BENITO MU ~NOZ and EMILIO FERN �ANDEZ MORAL, I.B. Sagasta, Logro ~no, Spain;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; NIKOLAOS DERGIADES, Thessa-loniki, Greece; C. FESTRAETS-HAMOIR, Brussels, Belgium; WALTHER JANOUS, Ursulinen-gymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete, Greece; andD.J. SMEENK, Zaltbommel, the Netherlands.

No-one other than the proposer used pure geometric methods. Lambrou made use ofvectors, and all the other solvers made use of trigonometry.

2376. [1998: 424] Proposed by Albert White, St. Bonaventure Uni-versity, St. Bonaventure, NY, USA.

Suppose that ABC is a right-angled triangle with the right angle at C.Let D be a point on hypotenuse AB, and let M be the mid-point of CD.Suppose that \AMD = \BMD. Prove that

1. AC 2MC 2 + 4[ABC] [BCD] = AC 2MB 2;

2. 4AC 2MC 2 � AC 2BD 2 = 4[ACD]2 � 4[BCD]2,

where [XY Z] denotes the area of 4XY Z.(This is a continuation of problem 1812, [1993: 48].)

Solution by Gerry Leversha, St. Paul's School, London, England.

We do not make use of the assumption that \AMD = \BMD. LetAB = c; BC = a; CA = b;BD = �c;\BCD = �, and let X be the footof the perpendicular from D to BC.

MB2 �MC2 = BC2 � 2MC � BC cos �

= a2 � CD � a cos �= a2 � a �CX = a � BX .

But by similar triangles BX = �a and so MB2 � MC2 = �a2. Now[ABC] = 1

2ab and [BCD] = 1

2�ab and so

4[ABC][BCD] = �a2b2 = AC2(MB2 �MC2) ,

438

which is part 1 of the problem. In a similar vein

AC2(4MC2 � BD2) = AC2(CD2 � BD2)

= b2(BC2 � 2BC � BD cosB)

= b2(a2 � 2a � �c cosB) = a2b2(1� 2�)

= 4((1� �)2 � �2)[ABC]2

= 4[ACD]2 � 4[BCD]2 ,

as required for part 2.

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain;NIKOLAOS DERGIADES, Thessaloniki, Greece; WALTHER JANOUS, Ursulinengymnasium,Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete, Greece; D.J. SMEENK,Zaltbommel, the Netherlands; PARAGIOU THEOKLITOS, Limassol, Cyprus; and the proposer.

Smeenk's solution is virtually identical to our featured solution, while that of Lambrouis quite similar although he avoids the cosine law by using Cartesian coordinates. All submittedsolutions avoided the unnecessary condition involving the pair of angles at M ; the proposerevidently arrived at his problem while investigating problem 1812 [1993: 48, 1994: 20-22], andhe failed to notice that his result is valid for any choice ofD onAB.

2377. [1998: 425] Proposed by Nikolaos Dergiades, Thessaloniki,Greece.

Let ABC be a triangle and P a point inside it. Let BC = a, CA = b,AB = c, PA = x, PB = y, PC = z, \BPC = �, \CPA = � and\APB = .

Prove that ax = by = cz if and only if ��A = � �B = �C = �3.

Solution by Jun-hua Huang, the Middle School Attached To HunanNormal University, Changsha, China.

Let M , N , L be the feet of the perpendiculars from P to CA, AB,BC, respectively.

Thus ax = 2R sinA � x = 2R �MN , where R is the circumradius of4ABC. Similarly by = 2R �NL and cz = 2R �ML. So ax = by = cz ()MN = NL = ML. Now ��A = \ABP +\ACP = \NLP +\MLP =\NLM and � � B = \LMN: � C = \MNL. So

MN = NL = ML (==) ��A = � � B = �C =�

3,

which completes the proof.

Also solved by WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAELLAMBROU, University of Crete, Crete, Greece; TOSHIO SEIMIYA, Kawasaki, Japan;D.J. SMEENK, Zaltbommel, the Netherlands; G. TSINTSIFAS, Thessaloniki, Greece; JEREMYYOUNG, student, Nottingham High School, Nottingham, UK; and the proposer.

439

2380. [1998: 425] Proposed by Bill Sands, University of Calgary,Calgary, Alberta.

When the price of a certain book in a store is reduced by 1=3 androunded to the nearest cent, the cents and dollars are switched. For ex-ample, if the original price was $43.21, the new price would be $21.43 (thisdoes not satisfy the \reduced by 1=3" condition, of course). What was theoriginal price of the book? [For the bene�t of readers unfamiliar with NorthAmerican currency, there are 100 cents in one dollar.]

Solution by Gerry Leversha, St. Paul's School, London, England.Let the price be a dollars and b cents. Then

2

3(100a+ b) = 100b+ a+

x

3,

where x 2 f�1; 0; 1g is included to deal with any possible rounding. Thissimpli�es to

197a = 298b+ x .

Now 197 and 298 are coprime, so the smallest solution in the case x = 0is a = 298, b = 197, which is impossible since 0 � b � 99. The usualEuclidean algorithm procedure yields

59� 197 � 39� 298 = 1 ,

and this shows that we should take a = 59, b = 39 and x = 1. Hence theprice of the book was $59.39.

Also solved by CHARLES ASHBACHER, Cedar Rapids, IA, USA; MIGUELCARRI �ON �ALVAREZ, Universidad Complutense de Madrid, Spain; NIKOLAOS DERGIADES,Thessaloniki, Greece; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX,USA; SHAWN GODIN, Cairine Wilson S.S., Orleans, Ontario; RICHARD I. HESS, RanchoPalos Verdes, CA, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; V �

ACLAV KONE �CN �Y, Ferris State University, Big Rapids, MI, USA; MICHAEL LAMBROU,University of Crete, Crete, Greece; KATHLEEN E. LEWIS, SUNY Oswego, Oswego, NY, USA;J.A. McCALLUM, Medicine Hat, Alberta; GOTTFRIED PERZ, Pestalozzigymnasium, Graz,Austria; D.J. SMEENK, Zaltbommel, the Netherlands; DIGBY SMITH, Mount Royal College,Calgary, Alberta; JEREMY YOUNG, student, Nottingham High School, Nottingham, UK;and the proposer. Two incorrect solutions were received, at least one of which was due tomisunderstanding the problem.

Many solvers noted that the answer is unique, as can be seen from the above solution.

Diminnie opines that \in the U.S. there would be no nontrivial solution to this problem,since U.S. stores seem to insist on rounding up in all circumstances!".

2381. [1998: 425] Proposed by Angel Dorito, Geld, Ontario.Solve the equation log2 x = log4(x+ 1).

Solution by Chris Cappadocia, student, St. Joseph Scollard Hall SS,North Bay, Ontario.

Let both of them equal y. Then

2y = x and 4y = 22y = x+ 1 .

440

And so

2y =22y

2y=

x+ 1

x.

Comparing, we get x = (x+ 1)=x, and solving for x and throwing away thenegative value, we get a �nal answer of

x =1+

p5

2.

Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia andHerzegovina; MICHEL BATAILLE, Rouen, France; FRANK P. BATTLES, Massachusetts Mar-itime Academy, Buzzards Bay, Massachusetts, USA; FRANCISCO BELLOT ROSADO, I.B.Emilio Ferrari, Valladolid, Spain; BOOKERY PROBLEM GROUP, Walla Walla, Washington,USA; MIGUEL CARRI �ON �ALVAREZ, Universidad Complutense de Madrid, Spain; NIKOLAOSDERGIADES, Thessaloniki, Greece; RUSSELL EULER and JAWAD SADEK, NW Missouri StateUniversity, Maryville, MO, USA; C. FESTRAETS-HAMOIR, Brussels, Belgium; SHAWN GODIN,CairineWilson S. S., Orleans, Ontario; RICHARD I. HESS, RanchoPalos Verdes, CA, USA; JOHNG. HEUVER, Grande Prairie Composite High School, Grande Prairie, Alberta; JOE HOWARD,NewMexico Highlands University, Las Vegas, NM, USA; JUN-HUAHUANG, theMiddle SchoolAttached To Hunan Normal University, Changsha, China; WALTHER JANOUS, Ursulinengym-nasium, Innsbruck, Austria; ANGEL JOVAL ROQUET, Instituto Espa ~nol de Andorra, Andorra;V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, MI, USA; MICHAEL LAMBROU,University of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul's School, London, England;KATHLEEN E. LEWIS, SUNY Oswego, Oswego, NY, USA; J.A. McCALLUM, Medicine Hat, Al-berta; JOHN GRANT McLOUGHLIN, Faculty of Education, Memorial University, St. John's,Newfoundland; HENRY J. RICARDO, Medger Evers College (CUNY), Brooklyn, New York,USA; JUAN-BOSCOROMEROM �ARQUEZ, Universidad de Valladolid, Valladolid, Spain; HEINZ-J �URGEN SEIFFERT, Berlin, Germany; TOSHIO SEIMIYA, Kawasaki, Japan; MAX SHVARAYEV,Tucson, Arizona, USA; SKIDMORE COLLEGE PROBLEM GROUP, Saratoga Springs, New York,USA; D.J. SMEENK, Zaltbommel, the Netherlands; DIGBY SMITH, Mount Royal College, Cal-gary, Alberta; ARAM TANGBOONDOUANGJIT, Carnegie Mellon University, Pittsburgh, PA,USA; PARAGIOU THEOKLITOS, Limassol, Cyprus; PANOS E. TSAOUSSOGLOU, Athens, Greece;UNIVERSITY OF ARIZONA PROBLEM SOLVING GROUP, Tucson, Arizona, USA; JEREMYYOUNG, student, Nottingham High School, Nottingham, UK; and the proposer.

Joval's solution must be the �rst one ever received by this journal from Andorra!Welcome | and can we now hear from Liechtenstein and Monaco?

2382. [1998: 425] Proposed by Mohammed Aassila, Universit �eLouis Pasteur, Strasbourg, France.

If4ABC has inradius r and circumradius R, show that

cos2�B � C

2

�� 2r

R.

Solution by Vedula N. Murty, Dover, PA, USA.

We have �cos

B � C

2� 2 sin

A

2

�2� 0 ,

441

so

cos2�B �C

2

�� 4 cos

B � C

2sin

A

2� 4 sin2

A

2

= 4 sinA

2

�cos

B �C2

� cosB + C

2

�

= 8 sinA

2sin

B

2sin

C

2=

2r

R,

since r = 4R sin A2sin B

2sin C

2[see for example, Roger A. Johnson, Modern

Geometry (1929) 298a].

Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosniaand Herzegovina; FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain;NIKOLAOS DERGIADES, Thessaloniki, Greece; JUN-HUA HUANG, the Middle School At-tached To Hunan Normal University, Changsha, China; WALTHER JANOUS, Ursulinengym-nasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, MI,USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul'sSchool, London, England; ISAO NAOI and HIDETOSHI FUKAGAWA, Gifu, Japan; ISTV �ANREIMAN, Budapest, Hungary; JUAN-BOSCO ROMERO M �ARQUEZ, Universidad de Valladolid,Valladolid, Spain; HEINZ-J �URGEN SEIFFERT, Berlin, Germany; TOSHIO SEIMIYA, Kawa-saki, Japan; D.J. SMEENK, Zaltbommel, the Netherlands; ARAM TANGBOONDOUANGJIT,Carnegie Mellon University, Pittsburgh, PA, USA; PANOS E. TSAOUSSOGLOU, Athens, Greece;G. TSINTSIFAS, Thessaloniki, Greece; PAUL YIU, Florida Atlantic University, Boca Raton, FL,USA; and the proposer.

Arslanagi�c, Bellot Rosado, Janous, Naoi and Fukagawa, and Sei�ert all note that wehave equality precisely when 2a = b + c. Kone �cn �y, Smeenk and Yiu provide the equivalentcondition, tan B

2tan C

2= 1

3.

Iftimie Simion (Math Teacher, Stuyvesant HS, New York, NY, USA) points out that thisproblem appears in a 10th grade textbook [Matematic �a: Geometrie si trigonometrie, A. Cotaet al., p. 106] used in Romania.

Bellot Rosado refers us to two notes on this inequality by Dan Plaesu (Iasi) and GheorgeMarchidan (Suceava) in the Romanian journal Gazeta matematica (1991) nos. 6 and 7. In thesecond note the authors prove the related inequality

cos2B � C

2� a2bc

R2(b+ c)2.

Bellot Rosado also shows thata2bc

R2(b + c)2>

2r

R

precisely when as2 (3�p5; 1).

2383. [1998: 425] Proposed by Mohammed Aassila, Universit �eLouis Pasteur, Strasbourg, France.

Suppose that three circles, each of radius 1, pass through the same pointin the plane. Let A be the set of points which lie inside at least two of thecircles. What is the least area that A can have?

Solution by Kathleen E. Lewis, SUNY Oswego, Oswego, NY, USA.Let the point of intersection of the circles be labelledO and the centres

be C1, C2 and C3. Consider �rst the area of intersection of the �rst two

442

circles as a function of the angleC1OC2, whichwe will call �. IfD is the otherpoint of intersection of these circles, then the area of intersection is twice thedi�erence between the area of the sector created by the angleOC1D and thearea of the triangleOC1D. The angleOC1D has measure ���, so the sectorhas area (���)=2 and the triangle has area sin(�=2) cos(�=2) = (1=2) sin�.Therefore the area of overlap of the two circles is ���� sin �, which we callf(�). Since f 0(�) = �1� cos �, which is never positive, the area decreasesas � increases from 0 to �.

Suppose that the circles with centres at C1 andC2 are in �xed position,with an angle of � � � between them, and we wish to place the third circlein a way that minimizes the total area of overlap. Since we want large anglesbetween the centres of the circles, we want to put the third circle so that thenew angles formed at O divide the angle 2� � � rather than �. If we callthese angles � and �, then the total area of overlap is

(� � � � sin�) + (� � �� sin�) + (� � �� sin�) .

Since � is �xed and � is constant, we need to minimize

��� sin�� �� sin� = ��� sin�� (2� � � � �)� sin(2� � � � �)

= � � 2� � sin�� sin(2� � � � �)

as a function of �. Taking the derivative, we get � cos�+ cos(2�� ���),which is zero if and only if cos� = cos(2� � � � �). Since the sum ofthese two angles is strictly less than 2�, the cosines can only be equal if theangles are. Thus, the minimum area occurs when � = �. This argumentcould equally well be used to argue that if any two of the angles are unequal,then the area could be reduced by moving the circle between those anglesto equalize them (leaving the other two circles �xed). The minimum areatherefore occurs when all three angles are equal to 2�=3, giving an overlaparea of 3[�=3� sin(2�=3)] = � � 3

p3=2.

Also solved by WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAVKONE �CN �Y, Ferris State University, Big Rapids, MI, USA; MICHAEL LAMBROU, University ofCrete, Crete, Greece; MAX SHKARAYEV, Tuscon, AZ, USA; THE UNIVERSITY OF ARIZONAPROBLEM SOLVING GROUP, Tuscon, AZ, USA; JEREMY YOUNG, student, Nottingham HighSchool, Nottingham, UK; and the proposer. There were one incomplete and three incorrectsolutions.

2384. [1998: 425] Proposed by Paul Bracken, CRM, Universit �e deMontr �eal, Qu �ebec.

Prove that 2(3n� 1)n � (3n+ 1)n for all n 2 N.Solution by Michel Bataille, Rouen, France.The inequality is obvious for n = 0, so we may assume n � 1. We

have to prove �3n� 1

3n+ 1

�n� 1

2,

443

or, equivalently,

n ln3n� 1

3n+ 1� ln

1

2. (1)

To this aim, we introduce the function

f(x) = x ln3x� 1

3x+ 1

de�ned on [1;1). We compute

f 0(x) = ln3x� 1

3x+ 1+

6x

9x2 � 1and f 00(x) =

�12(9x2 � 1)2

.

Since f 00(x) < 0, f 0(x) is strictly decreasing on [1;1). Moreover,

limx!1

f 0(x) = 0 ,

so f 0(x) > 0 for all x 2 [1;1). Hence f is increasing on [1;1) and, sincef(1) = ln 1

2, the inequality (1) follows.

Also solved by MIGUEL CARRI �ON �ALVAREZ, Universidad Complutense de Madrid,Spain; �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia andHerzegovina; JAMEST. BRUENING, Southeast Missouri State University, Cape Girardeau, MO, USA; NIKOLAOSDERGIADES, Thessaloniki, Greece; CHARLES R. DIMINNIE, Angelo State University, San An-gelo, TX, USA; KEITH EKBLAW, Walla Walla, WA, USA; RUSSELL EULER and JAWAD SADEK,NW Missouri State University, Maryville, MO, USA; C. FESTRAETS-HAMOIR, Brussels, Bel-gium; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; JUN-HUA HUANG, the MiddleSchool Attached To Hunan Normal University, Changsha, China; WALTHER JANOUS, Ursu-linengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete, Greece(three solutions); NORVALD MIDTTUN, Royal Norwegian Navy Academy, Norway; VEDULAN.MURTY, Visakhapatnam, India; VICTOROXMAN, University of Haifa, Haifa, Israel; HEINZ-J �URGEN SEIFFERT, Berlin, Germany; MAX SHKARAYEV, Tucson, AZ, USA; DIGBY SMITH,Mount Royal College, Calgary, Alberta; PANOS E. TSAOUSSOGLOU, Athens, Greece; JEREMYYOUNG, student, Nottingham High School, Nottingham, UK; and the proposer. There werealso three incorrect solutions submitted.

Janous and Sei�ert have proved the more general inequality

a + b

a � b(ax� b)x � (ax+ b)x

for all real x 2 [1;1) and real a and b such that a > b > 0.

2385. [1998: 426] Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel,Alcorc �on, Spain.

A die is thrown n � 3 consecutive times. Find the probability that thesum of its n outcomes is greater than or equal to n+6 and less than or equalto 6n� 6.

Solutionby Jeremy Young, student, NottinghamHigh School, Notting-ham, UK, modi�ed by the editor.

444

Consider the number of ways of scoring a sum of n+k for some k suchthat 0 � k � 5. This is the number of ways of expressing n+ k as a sum ofn positive integers, where the order of the summands matters. This numberis well-known to be

�n+k�1n�1

�. [Ed.: see Theorem 1.5.3 on p. 142 of [1].]

Furthermore, the number of ways of scoring a sum of 6n�k for some k suchthat 0 � k � 5 is the same as that of scoring a sum of n + k. This canbe seen by replacing each outcome �j by 7 � �j for j = 1, 2, : : : , n, since

n �nXj=1

�j � n+ 5 if and only if 6n� 5 �nXj=1

(7� �j) � 6n.

The total number of possible scoring sequences is 6n. Hence, the re-quired probability is

pn =1

6n

6n � 2

5Xk=0

�n+ k � 1

n� 1

�!

= 1� 2

6n

5Xk=0

�n+ k� 1

n� 1

�= 1� 2

6n

�n+ 5

n

�

by the well-known combinatorial identity

nXm=r

�m

r

�=

�n+ 1

r+ 1

�, where

0 � r � n. [Ed.: See Theorem 1.6.4 on p. 156 of [1].]

Reference:

[1]. H. Joseph Straight, Combinatorics, An Invitation, Brooks/Cole, 1993.

Also solved by RICHARD I. HESS, Rancho Palos Verdes, CA, USA; WALTHER JANOUS,Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete,Greece; GERRY LEVERSHA, St. Paul's School, London, England; NORVALD MIDTTUN, RoyalNorwegian Navy Academy, Norway; HEINZ-J �URGENSEIFFERT, Berlin, Germany; and the pro-poser. There were two partially incorrect solutions.

Janous, Leversha and Midttun all obtained the answer 1 � 2

6n

5Xk=0

� 5

5� k

��nk

��rst,

and then stated or showed that5X

k=0

� 5

5� k

��nk

�=

�n+ 5

5

�, which, of course, is just a

special case of the well-known and easy-to-prove Vandermonde's Identity:

lXk=0

�mk

�� n

l� k

�=�m+ n

l

�for integers l,m, n such that 0 � l �m, n.

[Ed.: See Ex. 26 on p. 167 of [1].]

Janous remarked that, in general, for t 2 f0, 1, 2, : : : , b5n=2cg, we have

p(n + t � X � 6n � t) = 1� 2

6n

�n + t+ 1

t� 1

�,

where X is the random variable denoting the sum of the n outcomes.

Leversha based his solution on the fact that the probability generating function forX isG(t) = 1

6n

�t+ t2 + t3 + t4 + t5 + t6

�n.

445

2388 [1998, 503; Correction 1999, 171]. Proposedby Daniel Kupper,B �ullingen, Belgium.

Suppose that n � 1 2 N is given and that, for each integerk 2 f0, 1, : : : , n� 1g, the numbers ak, bk, zk 2 C are given, with the z2kdistinct. Suppose that the polynomials

An(z) = zn +

n�1Xk=0

ak zk and Bn(z) = zn +

n�1Xk=0

bk zk

satisfy An(zj) = Bn(z2j ) = 0 for all j 2 f0, 1, : : : , n� 1g.

Find an expression for b0, b1, : : : , bn�1 in terms of a0, a1, : : : , an�1.

Solution by Kee-Wai Lau, Hong Kong.

We have An(z) �nY

k=1

(z � zk) and Bn(z) =

nYk=1

�z � z2k

�.

Hence, Bn

�z2�=

nYk=1

(z � zk) (z + zk) � (�1)nAn(z)An(�z).

Thus,

nXk=0

bkz2k � (�1)n

nX

k=0

akzk

! nX

k=0

(�1)kakzk!.

It follows that, for k = 0, 1, 2, : : : , n,

bk =

minfn;2kgXj=maxf0;2k�ng

(�1)n�jaja2k�j .

Also solved by MICHAEL LAMBROU, University of Crete, Crete, Greece; GERRYLEVERSHA, St. Paul's School, London, England; JOS �E LUIS DIAZ, Universidad Polit �ecnica deCatalu ~na, Terrassa, Spain; HEINZ-J �URGEN SEIFFERT, Berlin, Germany; and the proposer.There was one incorrect solution.

We thank Michael Lambrou for the correction to the statement of this problem.

The answer obtained by Sei�ert is the same as the one given above, while those givenby Lambrou, Leversha and the proposer are minor variations thereof. On the other hand, theanswer obtained by Luis has a di�erent \appearance":

bk = (�1)n�k

0@a2k + 2

n�k+1Xj=1

(�1)jak�jak+j

1A , k = 0, 1, : : : , n

with aj = 0 if j < 0 or j > n.

446

2392. [1998: 504] Proposed by G. Tsintsifas, Thessaloniki, Greece.Suppose that xi, yi, (1 � i � n) are positive real numbers. Let

An =

nXi=1

xiyi

xi + yi, Bn =

�Pni=1 xi

� �Pni=1 yi

�Pn

i=1 (xi + yi),

Cn =

�Pni=1 xi

�2+�Pn

i=1 yi�2Pn

i=1 (xi + yi), Dn =

nXi=1

x2i + y2ixi + yi

.

Prove that

1. An � Cn,

2. Bn � Dn,

3. 2An � 2Bn � Cn � Dn.`

Solution by Heinz-J �urgen Sei�ert, Berlin, Germany.It is su�cient to prove the inequalities in the third part.

Let

Xn =

nXj=1

xj , Yn =

nXj=1

yj , and Zn =

nXj=1

(xj � yj)2

xj + yj.

The Cauchy-Schwarz Inequality gives

0@ nXj=1

(xj � yj)

1A2

=

0@ nXj=1

pxj + yj

xj � yjpxj + yj

1A2

�0@ nXj=1

(xj + yj)

1A0@ nXj=1

(xj � yj)2

xj + yj

1A ,

or �Xn � Yn

�2 � �Xn + Yn

�Zn . (1)

Using the easily veri�ed identities

4An = Xn + Yn � Zn , Bn =XnYn

Xn + Yn,

Cn =X2n + Y 2

n

Xn + Yn, and 2Dn = Xn + Yn + Zn ,

we see that the inequalities An � Bn and Cn � Dn both follow from (1),while the inequality 2Bn � Cn is an immediate consequence of theAM{GM Inequality.

447

Also solved by MICHEL BATAILLE, Rouen, France; NIKOLAOS DERGIADES, Thessa-loniki, Greece; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAM-BROU, University of Crete, Crete, Greece; PANOS E. TSAOUSSOGLOU, Athens, Greece; JEREMYYOUNG, student, Nottingham High School, Nottingham, UK; and the proposer.

All the solvers, except Tsaoussoglou and the proposer, had solutions similar to the onegiven above. Tsaoussoglou proved stronger inequalities than the proposed 1. and 2., to obtainhis solution.

2393. [1998: 504] Proposed by G. Tsintsifas, Thessaloniki, Greece.Suppose that a, b, c and d are positive real numbers. Prove that

1.�(a+ b)(b+ c)(c+ d)(d+ a)

�3=2� 4abcd(a+ b+ c+ d)2;

2.�(a+ b)(b+ c)(c+ d)(d+ a)

�3� 16(abcd)2

Ya; b; c; dcyclic

(2a+ b+ c).

Solution by Jun-hua Huang, the Middle School Attached To Hunan NormalUniversity, Changsha, China.

1. The inequality is equivalent to

[(a+ b)(b+ c)(c+ d)(d+ a)]3 � 16(abcd)2(a+ b+ c+ d)4 ,

or, dividing by (abcd)6,��1

a+

1

b

��1

b+

1

c

��1

c+

1

d

��1

d+

1

a

��3

� 16

�1

bcd+

1

acd+

1

abd+

1

abc

�4.

Let x = a�1, y = b�1, z = c�1, and w = d�1. The inequality becomes

[(x+ y)(y+ z)(z +w)(w+ x)]3 � 16(xyz+ yzw+ zwx+wxy)4 .

We prove it as follows. Applying the Geometric Mean{Arithmetic Mean In-equality, we obtain

4(xyz+ yzw+ zwx+ wxy)2

= 4 [xw(y + z) + yz(x+ w)]2

= 4�pxwpxw(y+ z) +

pyzpyz(x+ w)

�2� �p

xw(x+ w)(y+ z) +pyz(y+ z)(x+ w)

�2= (x+ w)2(y+ z)2(

pxw +

pyz)2

= (x+ w)2(y+ z)2(xw+ yz+ 2pxwyz)

� (x+ w)2(y+ z)2(xw+ yz+ xy + wz)

= (x+ w)2(y+ z)2(x+ z)(w+ y) .

448

Hence

4(xyz+ yzw+ zwx+ wxy)2 � (x+ w)2(y+ z)2(x+ z)(w+ y) .

Similarly,

4(xyz+ yzw+ zwx+ wxy)2 � (x+ w)(y+ z)(x+ z)2(w+ y)2 .

The result follows by multiplying the last two inequalities.

2. The inequality follows easily from the �rst one, because

Ycyclic

(2a+ b+ c) �0@1

4

Xcyclic

(2a+ b+ c)

1A4

= (a+ b+ c+ d)4 ,

by the Geometric Mean{Arithmetic Mean Inequality.

Also solved by WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAELLAMBROU, University of Crete, Crete, Greece; PANOS E. TSAOUSSOGLOU, Athens, Greece; andthe proposer.

Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell

Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil

Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,

J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

449

THE ACADEMY CORNERNo. 29

Bruce ShawyerAll communications about this column should be sent to BruceShawyer, Department of Mathematics and Statistics, Memorial Universityof Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

Here are the hints and answers to the 1999 Bernoulli Trials [1999: 321].Many thanks to Christopher Small for sending them to us.

The Bernoulli Trials 1999

Christopher G. Small & Ravindra MaharajUniversity of Waterloo

Hints and Answers:

1. TRUE. We have

1

1999=

1

x+

1

y> max

�1

x,1

y

�.

Therefore x, y � 2000. Moreover, if

x > 3998000 = 2000� 1999

then 1=x < 1=3998000, so that

1

1999<

1

3998000+

1

y.

This reduces to y < 2000, which is a contradiction.

2. TRUE. The equation implies that

cosx = �=2� sinx (mod 2�)

As j cosxj � 1 we must have cosx = �=2� sinx, or

cos2 x� 2 cosx sinx+ sin2 x = �2=4 .

Therefore j sin2xj = �2=4� 1 > 1. But this cannot hold.

450

3. FALSE. As n3�n = n(n�1)(n+1), where n is odd, it follows that theright hand side is divisible by 9 and 16. As the sum of the digits mustbe 0 (mod 9) it follows that A+B = 7 (mod 9). In addition, B mustbe even. The possibilities for AB are 70, 52, 34, 16 and 88. However,48AB must be divisible by 16.

So the answer is AB = 16. In fact, CD = 76, although this is notneeded.

4. FALSE.

A

O C

B

We have AO = 12� 8 = 4.

Let x denote one-half of the side length of the square. Then AC = 2x.Therefore OC = AC � AO = 2x� 4. Thus:

OC = 2x� 4, BC = x, and OB = 12.

Therefore x is a solution to the quadratic equation (2x�4)2+x2 = 122.

Solving this equation gives 2x =16 +

p2816

5.

5. TRUE. The colouring of the squares in the �rst row and �rst column isarbitrary. There are 215 choices for colouring these squares. Once thecolours of these squares are determined, the rest of the board can onlybe coloured in one way to ful�l the condition.

6. TRUE. Consider polynomials. Let n be the degree of f . The equationimplies that 2n = n2. This requires that f be constant or a quadratic.It can be checked that f(x) = x2 works.

7. FALSE. Formally di�erentiating term by term gives us back the sameseries. Thus �0(x) = �(x). This implies that �(x) = Aex for someconstant A. We note that f(x) = A works. Since �(0) = 1999, we�nd that �(x) = 1999 ex. Therefore �(�1999) > 0.

8. TRUE. WriteZ 1

0

x�x dx =

Z 1

0

ex ln (1=x) dx

=

1Xn=0

1

n!

Z 1

0

xn [ln (1=x)]ndx

451

Substituting xn+1 = e�y reduces the right-hand side to

1Xn=1

n�n. The

result follows easily from this.

9. FALSE. First note that the event that a triangle is acute is independentof the choice of labels for the vertices.

Shrink the circle around 4ABX until one of the 3 points hits theboundary. Now relabel the triangle so that the two points in the inte-rior of the new circle are A and B in random order, and the point onthe boundary of the new circle is called X0.

This shows that with respect to events that are symmetric in the labels,4ABX and4ABX0 have the same distribution. Thus� = �0.

10. TRUE. Since it takes 4 counters to block the lines in any 4 � 4 slice,any solution must have exactly 4 counters in each such horizontal slice.There are two basic patterns, A and B as shown, which will block thelines with 4 counters within a slice. Other con�gurations of four coun-ters are obtained as rotations or re ections of these two basic cases.

XX

A

XX

BX

X

XX

XX

XX

IX

X

XX

II

X

X

XX

III

X

X

XX

IV

Stacking I, II, III, and IV in order can be checked to block all lines par-allel to a face or edge.

11. FALSE. Try

Day 1: M1W1 v M2W2 and M3W3 v M4W4

Day 2: M1W3 v M3W1 and M2W4 v M4W2

Day 3: M1W2 v M4W3 and M2W1 v M3W4

452

We now present the questions of the 1999 Atlantic Provinces Council on theSciences Annual Mathematics Competition, held this year at Memorial Uni-versity, St. John's, Newfoundland. The winning team consisted of Ian Cainesand Jacky Pak Ki Li from Dalhousie University. The runners-up were theteam of Shannon Sullivan and Jerome Terry from Memorial University ofNewfoundland. You may view their pictures at

www.math.mun.ca/~apics/picture/shots.html

Send me your nice solutions!

APICS 1999 Mathematics Competition

1. Find the volume of the solid formed by one complete revolution aboutthe x{axis of the area in common to the circles with equationsx2 + y2� 4y + 3 = 0 and x2 + y2 = 3.

2. The Memorial University Philosophers' Jockey Club has just receivedthe bronze busts of the ten members of their hall of fame. Each will beplaced in its designated place on a single shelf, above the gold plaquebearing the name of the member. The ten busts are drawn at randomfrom the crate. What is the probability that at no time will there be anempty space between two busts already placed on the shelf?

3. Prove that sin2(x+�)+sin2(x+�)�2 cos(���) sin(x+�) sin(x+�)is a constant function of x.

4. In Scottish Dancing, there are three types of dances, two of which arefast rhythms, Jigs and Reels, and one is a slow rhythm, Strathspey.

A Scottish Dance program always starts with a Jig. The following dancesare selected (by type) according to the following rules:

(i) the next dance is always of a di�erent type from the previous one,

(ii) no more than two fast dances can be consecutive.

Find how many di�erent arrangements of Jigs, Reels and Strathspeysare possible in a Scottish Dance list which has (a) seven dances, (b)�fteen dances.

5. Find all di�erentiable functions f(x)which satisfy the integral equation

�f(x)

�2000=

Z x

1

�f(t)

�1999dt .

453

6. Inside a square of side r, four quarter circles are drawn, with radius rand centres at the corners of the square.

Find the area of the shaded region.

7. Pat has a method for solving quadratic equations. For example, Patsolves 6x2 + x� 2 = 0 as follows:

Step 1. Pat multiplies the leading coe�cient by the constant coe�cient andsolved the simpler equation x2+x�12 = 0 to get (x+4)(x�3) =0.

Step 2. Pat then replaces each x by 6x (x times the leading coe�cient) toget (6x+ 4)(6x� 3) = 0.

Step 3. Pat then simpli�es this equation to get (3x+2)(2x�1) = 0, whichsolves the original equation.

Prove or disprove that Pat's method always works.

8. An arbelos consists of three semicircular arcs as shown:

A circle is placed inside the arbelos so that it is tangent to all threesemicircles.

Suppose that the radii of the two smaller semicircles are a and b, andthat the radius of the circle is r.

Assuming that a > b > r and that a, b and r are in arithmetic progres-sion, calculate a=b.

454

THE OLYMPIAD CORNERNo. 202

R.E. Woodrow

All communications about this column should be sent to Professor R.E.Woodrow, Department of Mathematics and Statistics, University of Calgary,Calgary, Alberta, Canada. T2N 1N4.

In this issue, we give the problems of the 13 th Iranian Mathemati-cal Olympiad, 1996 Second Round and the two exams of the Final Round.My thanks go to J.P. Grossman, Team Leader of the Canadian MathematicalOlympiad Team at Mumbai, India for collecting the problems for our use.

13th IRANIANMATHEMATICAL OLYMPIAD 1996Second Round

Time: 2� 4 hours

Problems are of equal value.

1. Prove that for every natural number n � 3 there exist two setsA = fx1, x2, : : : , xng and B = fy1, y2, : : : , yng such that

(a) A \ B = ;;(b) x1 + x2 + � � �+ xn = y1 + y2 + � � �+ yn;

(c) x21 + x22 + � � �+ x2n = y21 + y22 + � � �+ y2n.

2. Let L be a line in the plane of an acute triangle ABC. Let the linessymmetric to L with respect to the sides of ABC intersect each other in thepoints A0, B0 and C0. Prove that the incentre of triangle A0B0C0 lies on thecircumcircle of triangle ABC.

3. 12k persons have been invited to a party. Each person shakes handswith 3k + 6 persons. We know also that the number of persons who shakehands with any two persons is constant. Find the number of the personsinvited.

4. Let n be a natural number. Prove that n can be written as a sum ofsome distinct numbers of the form 2p3q such that none of them divides anyother. For example, 19 = 4 + 6 + 9.

5. Prove that for any natural number nlpn+

pn+ 1 +

pn+ 2

m=lp

9n+ 8m.

455

6. In tetrahedron ABCD let A0, B0, C0, and D0 be the circumcen-tres of faces BCD, ACD, ABD and ABC. We mean by S(X;Y Z), theplane perpendicular from point X to the line Y Z. Prove that the planesS(A;C0D0), S(B;D0A0), S(C;A0B0), and S(D;B0C0) are concurrent.

Final Round { First Exam

Time: 2 � 4 hours

Each problem is worth seven points.

1. Prove the following inequality

(xy+ xz + yz)

�1

(x+ y)2+

1

(y+ z)2+

1

(x+ z)2

�� 9

4

for positive real numbers x, y, z.

2. Prove that for every pair m; k of natural numbers, m can be ex-pressed uniquely as

m =

�ak

k

�+

�ak�1k � 1

�+ � � �+

�at

t

�,

where

ak > ak�1 > � � � > at � t � 1 .

3. In triangle ABC we have \A = 60�. Let O, H, I, and I0 be thecircumcentre, orthocentre, incentre, and the excentre with respect to A ofthe triangle ABC. Consider points B0 and C0 on AC and AB such thatAB = AB0 and AC = AC0. Prove that

(a) eight points B, C, H, O, I, I0, B0, and C0 are concyclic;

(b) ifOH intersects AB andAC inE and F respectively, then triangleAEFhas a perimeter equal to AC +AB;

(c) OH = jAB � ACj.4. Let k be a positive integer. Prove that there are in�nitely many

perfect squares in the arithmetic progression fn� 2k � 7gn�1.5. Let ABC be a non-isosceles triangle. Medians of the triangleABC

intersect the circumcircle in points L, M , N . If L lies on the median of BCand LM = LN , prove that 2a2 = b2 + c2.

6. We have attached the up and down, and left and right sides of ann � n chessboard (forming a torus) in order that a \tour" be constructed.Find the maximum number of knights which can be placed on this tour insuch a way that none of them controls the square on which another lies.

456

Final Round { Second Exam

Time: 2 � 4 hours

Each problem is worth seven points.

1. Find all real numbers a1 � a2 � � � � � an satisfying

nXi=1

ai = 96 ,

nXi=1

a2i = 144 ,

nXi=1

a3i = 216 .

2. Points D and E are situated on the sides AB and AC of triangleABC in such a way that DEkBC. Let P be an arbitrary point inside thetriangleABC. Lines PB and PC intersect DE at F andG respectively. LetO1 be the circumcentre of triangle PDG and let O2 be that of PFE. Showthat AP ? O1O2.

3. Let P (x) be a polynomial with rational coe�cients such thatP�1(Q) � Q. Show that P is linear.

4. Let S = fx1, x2, : : : , xng be an n{element subset offx 2 R j x � 1g. Find the maximum number of the elements of the form

nXi=1

"ixi , "i = 0, 1

which belong to I, where I varies over all open intervals of length one andS over all n{element subsets.

5. Does there exist a function f : R! R that ful�ls all of the follow-ing conditions:

(a) f(1) = 1;

(b) there existsM > 0 such that �M < f(x) < M ;

(c) if x 6= 0 then

f

�x+

1

x2

�= f(x) +

�f

�1

x

��2.

6. Let a, b, c be positive real numbers. Find all real numbers x, y, zsuch that

x+ y + z = a+ b+ c ,

4xyz � (a2x+ b2y+ c2z) = abc .

457

Next we turn to readers' comments and solutions for problems andsolutions previously given in the Corner. First, a comment about a commentto a solution given in the October number of the Corner.

2. [1998: 196{197; 1999: 333] 44 th LithuanianMathematical Olympiad.What is the least number of positive integers such that the sum of their

squares is 1995?

Comment by R.K. Guy, University of Calgary.In fact 1992 = 402 + 142 + 142.

Here is a simpler solution to a problem discussed in the October Corner.

2. [1998: 133; 1999: 328] VIII Nordic Mathematical Contest.A �nite set S of points in the plane with integer coordinates is called a

two{neighbour set, if for each (p; q) in S exactly two of the points (p+1; q),(p; q + 1), (p � 1; q), (p; q � 1) are in S. For which n does there exist atwo-neighbour set which contains exactly n points?

Solution by Carl Johan Ragnarsson, Lund, Sweden.Put the points in the coordinate plane. Let (x; y) be even if

x + y = 0 (mod 2) and odd otherwise. Note that the even (odd) pointshave only odd (even) neighbours. So, if we have k even points, we will count2k odd neighbours. But each odd point has two even neighbours, so eachpoint was counted twice and there are k odd points as well. This means thenumber of points is even (it is equal to 2k). 4 works by taking a 2 by 2 square2k, k > 3 works by starting with a 3 by 3 square with centre removed andthen stretching it appropriately. Finally, it is evident that a set of six pointsdoes not work.

We now turn to solutions from our readers to the First Round problemsof the 8 th Korean Mathematical Olympiad [1998: 197{198].

8th KOREAN MATHEMATICAL OLYMPIADFirst Round

1. Consider �nitely many points in a plane such that, if we choose anythree points A, B, C among them, the area of4ABC is always less than 1.Show that all of these �nitely many points lie within the interior or on theboundary of a triangle with area less than 4.

Comments and solutions by Mohammed Aassila, Strasbourg, France;

and by Pierre Bornsztein, Courdimanche, France. We give Aassila's solution.Let ABC be the triangle having the maximal area [ABC]. Then

[ABC] � 1. Let A0B0C0 be the triangle whose medial triangle is ABC.

458

Then [A0B0C0] � 4[ABC] � 4. We will show that A0B0C0 contains all n ofthe points.

Assume, for a contradiction, that there is a point, D, outside A0B0C0.Then ABC and D lie on di�erent sides of at least one of the lines A0B0,B0C0, C0A0.

A

BC

D

A0

B0 C0

Then [ABD] � [ABC], contradicting the maximality of [ABC].

Comment by P. Bornsztein: This problem and a solution appeared ear-lier [1993: 163].

2. For a given positive integer m, �nd all pairs (n; x; y) of positiveintegers such thatm, n are relatively prime and (x2+y2)m = (xy)n, wheren, x, y can be represented by functions ofm.

Solutions by Mohammed Aassila, Strasbourg, France; and by PierreBornsztein, Courdimanche, France. We give Aassila's solution.

Let p be a common prime divisor of x and y, and let � and � be thelargest integers such that p� j x and p� j y.

Now p2� j x2, p2� j y2 and

p(�+�)n j (xy)n = (x2 + y2)m .

We claim that � = � (and x = y). Indeed, if � < �, then p2� j x2 + y2 andp2�m j (x2 + y2)m; that is

2�m = (�+ �)n > 2�n ,

and thenm > n. But this is impossible since

(xy)m < (2xy)m � (x2 + y2)m = (xy)n .

Similarly if� > � we obtain a contradiction. Hence x = y and (x2+y2)m =(xy)n reduces to 2mx2m = x2n. The solutions are of the form (n;x; y) =(m+ 1; 2m=2; 2m=2).

Comment by P. Bornsztein. This problemwas posed on the 1992WilliamLowell Putnam contest, and a solution was published in the American Math.Monthly (1993, pbA3, p. 760).

Ifm is odd there is no solution. For m even n = m+1, x = y = 2m.

459

Moreover, if we do not suppose (n;m) = 1 the solutions are thosenumbers of the form m = 2a�, n = m+ �, x = y = 2a.

3. Let A, B, C be three points lying on a circle, and let P , Q, R bethe mid-points of arcs BC, CA, AB, respectively. AP , BQ, CR intersectBC, CA, AB at L, M , N , respectively. Show that

AL

PL+BM

QM+CN

RN� 9 .

For which triangle ABC does equality hold?

Solutionsand comments byMiguel Amengual Covas, Cala Figuera,Mal-

lorca, Spain; Mohammed Aassila, Strasbourg, France; by Pierre Bornsztein,Courdimanche, France; and by Murray S. Klamkin, University of Alberta,Edmonton, Alberta. We give Amengual's solution.

To be de�nite, we have taken P to lie on the arc BC that does notcontain A.

Since P is the mid-point of BC, \BAP = \PAC and AL is the in-ternal bisector of the angle A of triangle ABC formed by joining the pointsA, B, C.

Let AA0 = ha be the altitude from A and let P0 be the foot of theperpendicular from P to the side BC.

The right triangles AA0L and PP0L are similar, so

AL

PL=

AA0

PP0or

AL

PL=

ha

PP0. (1)

A

B C

P

P0

A0

L

q

If the lengths of the sides of4ABC areBC = a,CA = b andAB = c,then the area of 4ABC is 1

2a � ha and also 1

2bc � sinA; hence

ha =bc sinA

a.

460

In 4BP0P , we have \P0BP = \CBP = 12\A (since both \CBP

and \A=2 are inscribed in the circular arc PC). Hence

PP0 = BP0 � tanA

2=

a

2� tan A

2.

Substituting these expressions for ha and PP0 in (1) gives

AL

PL=

4bc cos2 A2

a2.

Since cos2A

2=

s(s� a)

bcwhere s =

a+ b+ c

2, this is equivalent

to

AL

PL=

2s

a� 2(s� a)

a

=

�1 +

b

a+c

a

���1 + b

a+c

a

�

=

�b

a+c

a

�2� 1 .

We have conducted our discussions with respect to the side BC of4ABC. Applying the same reasoning to either of the other sides instead,we obtain

AL

PL+

BM

QM+

CN

RN

=

"�b

a+c

a

�2� 1

#+

"�c

b+a

b

�2� 1

#+

"�a

c+b

c

�2� 1

#

=

�a2

b2+b2

a2

�+

�b2

c2+c2

b2

�+

�c2

a2+a2

c2

�

+2

�ab

c2+bc

a2+ca

b2

�� 3 .

According to the arithmetic mean{geometric mean inequality,

a2

b2+b2

a2� 2;

b2

c2+c2

b2� 2 ,

c2

a2+a2

c2� 2;

ab

c2+bc

a2+ca

b2� 3 .

Therefore,

AL

PL+BM

QM+CN

RN� 2 + 2 + 2 + 2 � 3� 3 = 9 .

461

Equality occurs when a = b = c; that is, when4ABC is equilateral.

Aassila comments that this problem and a solution appeared in Crux[1989: 74; 1990: 158; 1991:48]. It was proposed by Mihaly Bencze, Brasov,Romania.

4. A partition of a positive integer n is a sequence (�1, �2, : : : , �k) ofpositive integers such that �1 + �2 + � � � + �k = n and �1 � �2 � � � � ��k � 1. Each �i is called a summand. For example, (4; 3; 1) is a partitionof 8 whose summands are distinct. Show that, for a positive integer m withn > 1

2m(m+1), the number of all partitions of n into distinctm summands

is equal to the number of all partitions of n� 12m(m+1) into r summands

(r � m).

Solution by Mohammed Aassila, Strasbourg, France.

By the de�nition of a summand we have that �m � 1, �m�1 � 2, : : : ,�1 � m.

De�ne �k =: �k � (m� k+ 1). Then �k � 0 for all k and �1 + �2 +� � �+ �m = n� 1

2m(m+ 1).

5. If we select at random three points on a given circle, �nd the prob-ability that these three points lie on a semicircle.

Solutions by Mohammed Aassila, Strasbourg, France; and by MurrayS. Klamkin, University of Alberta, Edmonton, Alberta. We give Klamkin'ssolution and extensions.

It should have been stated that the random selection is one with a uni-form distribution with respect to arc length; otherwise one could get varyinganswers. Also, this is a fairly widely known problem.

If �1, �2, �3 are measures of the three arcs between successive points,then their sum is 2�. So equivalently, we want the probability that �1, �2,�3 do not satisfy the triangle inequality; for example, each angle is less than�. Now consider an equilateral triangle4 with altitude 2�. For any randompoint within or on the triangle the sum of the perpendiculars to the sides= 2�, so these perpendiculars can be taken as �1, �2, �3 (we can assume thatthe random point is uniformly distributed with respect to area). Then �1, �2,�3 will form a triangle if the point lies within the medial triangle, the onewhose vertices are the mid-points of the sides of 4. Since the area of thistriangle is 1

4of4, the desired probability is 3

4.

Comments: It may be of interest to give the analogous 3{dimensionalproblem from the Educational Times back in the 19th century:

\2621. (proposed by Rev. M.M.U. Wilkinson, M.A.) If four points betaken at random on the surface of a sphere, show that the chance of them alllying on some one hemisphere is 7=8.

Solution by Stephan Watson. If great circles of the sphere be describedthrough each pair of the three points, they will divide the surface of the

462

sphere in eight portions, in seven of which the fourth point may lie so that thefour points may all be in some one hemisphere. Moreover, when the threepoints take all positions on the surface of the sphere, the eight portions willall pass through the same magnitudes; hence the required chance is 7=8."

As before, it is tacitly assumed that the random points are distributeduniformly with respect to surface area.

In \A problem in geometric probability", Math. Scand. 11 (1962), 109{111, J.G. Wendell gives an elegant proof that if one hasN points scattered atrandom on the surface of a sphere in En, the probability that all the pointslie in the same hemisphere is given by

pn;N = 2�N+1n�1Xk=0

�N � 1

k

�,

and in particular, pn;n+1 = 1 � 2�n. He also notes that pn;N equals theprobability that in tossing a fair coin repeatedly, the nth \head" occurs on orafter theN th toss and that it does not seem possible to �nd an isomorphismbetween coin-tossing and the given problem that would make the result im-mediate.

6. Show that any positive integer n > 1 can be expressed by a �nitesum of numbers satisfying the following conditions:

(i) they do not have factors except 2 or 3;

(ii) any two of them are neither a factor nor a multiple of each other.

That is,

n =

NXi=1

2�i3�i ,

where �i, �i (i = 1, 2, : : : , N ) are non-negative integers and(�i � �j)(�i � �j) < 0 whenever i 6= j.

Solutions by Mohammed Aassila, Strasbourg, France; and by PierreBornsztein, Courdimanche, France. We give Aassila's solution.

We use induction on n. Assume that all integers� n have the requiredrepresentation, and let us prove that n + 1 has the same representation aswell.

Case 1: n is odd. By the induction hypothesis, n+12

has the representation

and hence n + 1 = 2 � n+12

has the representation also. (Multiply the

representation of n+12

by 2).

Case 2: n is even.

Case 2a: n+ 1 is a power of 3.Nothing to prove. The result is true.

Case 2b: n+ 1 is not a power of 3.

463

In this case, there exists an integer m such that 3m < n+ 1 < 3m+1.The integer (n+1� 3m) is even, and (n+1� 3m)=2 is less than n. By theinduction hypothesis it has a representation, whence n+ 1� 3m does, and�nally n+ 1 is the sum of 3m and the representation of n+ 1� 3m.

7. Find all real-valued functions f de�ned on real numbers except 0such that

1

xf(�x) + f

�1

x

�= x , x 6= 0 .

Solutions by Mohammed Aassila, Strasbourg, France; and by PierreBornsztein, Courdimanche, France. We give Bornsztein's solution.

L'identit �e s' �ecrit : pour t 6= 0

f(�t) + tf

�1

t

�= t2 .

Pour t = 1x, l'identit �e s' �eerit

f(t) + xf

��1

t

�=

1

t(i)

be meme, pour t = �x 6= 0, on obtient

f(t)� tf

��1

t

�=

1

t. (ii)

On additionnant (i) et (ii), pour x 6= 0

2f(x) = x2 +1

x;

c. �a.d.

f(x) =x3 + 1

2x.

R �eciproquement, si x 6= 0 et f(x) = x3+12x

,

alors1

xf(�x) + f

�1

x

�=

x3 � 1

2x2+x3 + 1

2x2= x .

Finalement, f :�x 7! x3+1

2x

�est l'unique solution.

8. Two circles O1, O2 of radii r1, r2 (r1 < r2), respectively, intersectat two pointsA andB. P is any point on circle O1. Lines PA, PB and circleO2 intersect at Q and R, respectively.

(i) Express y = QR in terms of r1, r2, and � = \APB.

(ii) Show that y = 2r2 is a necessary and su�cient condition that circle O1

be orthogonal to circle O2.

464

Solution by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain.(i) In �gure 1, \PBA = \PQR, both being supplements of \ABR.

Let W1 and W2 be the centres of O1 and O2. In Figure 2, the lineW1W2, joining the centres of the intersecting circles, is perpendicular to theircommon chord AB which subtends at the centre W1 (respectively W2) twicethe angle it subtends at P (respectively Q) on the circumference O1 (respec-tively O2), implying

\QPB = \W2W1B and \BQP = \BW2W1 .

W1q qW2

P

Q

B

A

R

W1q qW2

P

Q

B

A

R

Figure 1. Figure 2.

The similar triangles 4PAB � 4PRQ and 4PBQ � 4W1BW2

yieldQR

AB=

PQ

PB,

PQ

PB=

W1W2

W1B,

whence

y = QR =AB

W1B�W1W2 (QR is constant) .

Since sin� =AB=2

W1B, we have

AB

W1B= 2 sin �, and hence

y = 2 �W1W2 � sin � .

By the Law of Sines, applied to 4W1W2B with \BW2W1 = �,

W2B

sin �=

W1B

sin�; that is, sin� =

r1

r2sin � .

Hence

W1W2 = r1 cos � + r2 cos�

= r1 cos � + r2

s1�

�r1

r2sin �

�2

= r1 cos � +

qr22 � r21 sin

2 � .

465

We conclude that

y = 2

�r1 cos � +

qr22 � r21 sin

2 �

�sin � .

(ii) In the equation y = 2r2, we replace y by

2 ��r1 cos � +

qr22 � r21 sin

2 �

�sin�

and write it in the form�qr22 � r21 sin

2 �

�sin � = r2 � r1 sin � cos � ,

which is equivalent to the one obtained by squaring both sides; that is,

(r22 � r21 sin2 �) sin2 � = r22 � 2r1r2 sin � cos � + r21 sin

2 � cos2 � ,

which is equivalent to

(r2 cos � � r1 sin �)2 = 0 .

This, in turn, is equivalent to

tan � =r2

r1,

or �nally\W1BW2 = 90� .

(Of course, this angle is the same at both intersections A and B).

Since the tangent line to a circle is perpendicular to the radius at thepoint of contact, the angle between the tangents at both intersections A andB is 90� and O1, O2 cut orthogonally.

This completes the Olympiad Corner for this issue. Please send meyour nice solutions and Olympiad contest sets.

466

BOOK REVIEWS

ALAN LAW

Which way did the Bicycle Go? : : : and Other IntriguingMathematical Mys-teries by Joseph D.E. Konhauser, Dan Velleman, Stan Wagon,publishedby theMathematical Association of America, 1996 (DolcianiMath-ematical Expositions, No. 18)ISBN # 0-88385-325-6, softcover, 245+ pages.Reviewed by Einar Rodland, University of Oslo.

\Which way did the bicycle go?" Such is the problem facing SherlockHolmes having found the tracks in the snow. \It was undoubtably headingaway from the school," he observes, presenting an incorrect argument aboutthe hind wheel \obliterating the more shallow mark of the front one." Oh,no, my dear mister Holmes; this time, you should have turned to mathemat-ics.

This problem sets the tone of the book: problems are formulated so asto be understandable and attractive without relying on a mathematical back-ground. This was Joe Konhauser's style when he posted the \Problem of theWeek" at Macalester College, a column later taken over by Stan Wagon. Thisreviewer believes them successful in doing so; though most of the problemsmay demand some mathematical interest from the students, there are somethat may even appeal to those normally not attracted by the subject. Themain target group of the problems is advanced high school pupils and be-ginning college students. However, there are plenty of problems that maybe given to both younger and to older. As such, it may in particular be avaluable source for teachers trying to engage their students.

The problems, counting 191, are ordered by type|plane geometry,number theory, algebra, combinatorics and graph theory, three-dimensionalgeometry, andmiscellaneous|which is handywhen trying to recover a previ-ously read problem. However, as the problems generally do not match anycurriculum particularly well, I would expect a teacher seeking an inspiringproblem to largely ignore this ordering, picking problems from throughoutthe book: it would not be easy to �nd problems to illustrate a particulartechnique.

Well over two thirds of the book is devoted to giving complete solutionsto all the problems, and occasionally there are several alternative solutions.The solutions are clearly written and, though occasionally knowledge of somenon-trivial results is needed, should be su�ciently detailed for even the lessadvanced students to read. Of course, the question \How did they think ofthat?" will occasionally arise: some solutions rely on �nding the right trick,an art which is not easily explained. On the other hand, some of the problems

The complete set is ISBN 0-88385-300-0.

467

lead naturally to more traditional mathematical theory, or refer to researchdone by others, which helps indicate that mathematics goes beyond merepuzzle-solving.

It was said that the book contains 191 problems. This, however, isnot fully true. In the solution of several problems, new problems are statedwhich allows one to test one's understanding of the solution: to try it outon a similar problem or proceed with a more advanced problem, perhaps ageneralization. This may help coach students (and teachers) into not onlytrying out methods on new problems, but trying to pose new problems, askill which is also useful: as useful as being able to solve the problems, manywould claim.

The overall conclusion is that the book would have its greatest strengthin the hands of a teacher; with this book at hand, a high-school or undergrad-uate teacher would have an ample supply of problems, and could, withouttoo much of an e�ort (�nding good problems can be quite hard), regularlychallenge his or her students with problems of a di�erent kind from whatthey traditionally meet. As such, it is highly recommended.

The Mathematical Olympiad Handbook; an Introduction to Problem Solvingby Tony Gardiner,published by Oxford University Press, 1997,ISBN # 0-19-850105-6, softcover, 229+ pages, US$29.95.Reviewed by Catherine Shevlin, Wallsend upon Tyne, England.

Mathematical Olympiads began in Hungary in the nineteenth century.They are now held for high school students throughout the world. Theyfeature problems which, though they require only high school mathemat-ics, seem very di�cult because they are unpredictable and have no obviousstarting point. This book introduces readers to these delightful and challeng-ing problems and aims to convince them that Olympiads are not just for aselect minority. The book contains problems from the British MathematicalOlympiad (BMO) competitions between 1965 and 1996. It includes hints andsolutions for each problem from 1975 on, a review of the basic mathematicalskills needed, and a list of recommended reading, making it an ideal sourcefor enriching one's experience in mathematics.

The World Championship Mathematical Olympiad, the InternationalMathematical Olympiad, began in Romania in 1959, with the participationof a few countries from eastern Europe. It has now grown to a worldwidecompetition with participation from all continents. Tony Gardiner has beenthe Leader of the United Kingdom team to the International MathematicalOlympiad, and so is well quali�ed to write a book such as this.

The contents are:

� Problems and Problem Solving� How to Use this Book

468

� A Little Useful MathematicsIntroduction

1. Numbers 2. Algebra3. Proof 4. Elementary number theory5. Geometry 6. Trigonometric formulae

� Some Books for Your Bookshelf� The Problems� Hints and Outline Solutions� Appendix: The International Mathematical Olympiad: UK teams and results1967 - 1996

The problems consist of all the BMO's from the �rst (1965) up to the32nd (1996), together with hints and outline solutions for the 11th (1975) upto the 32nd (1996). These are given in su�cient detail to enable the inter-ested reader to work out full solutions. For example, for question 5 of the32nd BMO, he states:

If you have a clever idea about how to solve both parts at once, you will

probably not need this outline solution. So I shall begin as though youhave not yet solved the problem.

He ends the solution to question 3 of the 31st BMO with the admonition:

There are many other correct ways, But beware: most arguments are awed!

The author states that the book is \unashamedly" for beginners. Heaims to convince many people that solvingMathematical Olympiad problemsis for them, and \not just for some bunch of freaks". There is this commonmisconception around that those students who are good at mathematics areabnormal. From my personal experience in coaching students for Mathemat-ical Olympiads, nothing can be further from the truth. They are as normal apopulation as any other segment of the population.

What is sad is that they are not as highly regarded amongst their peersas are, say, those students who excel in games.

Gardiner's book covers the main topics required for Olympiad problemsin a compact and readable form. As he says, those who want more mustread more. His section entitled \Some Books for Your Bookshelf" is verycomprehensive. It ends up with referring to this journal as \the ProblemSolver's Bible"!

Tony Gardiner is well quali�ed to write such a book. He has beeninvolved in the British Mathematical Olympiad for several years. He wasLeader of the UK International Mathematical Olympiad team from 1990 to1995, and has been Vice-President of the World Federation of NationalMath-ematics Competitions. The Federation, in recognition of his work in Mathe-matical Olympiads, awarded him the 1995 Paul Erd }os National Award.

This book is a must for any high school student desirous of a challengein mathematical problem solving. It is also a must for anyone involved inthe encouragement and training of potential Mathematical Olympiad partic-ipants. Problem solving is, after all, the essence of mathematics.

469

Mathematical Competitions in Croatia edited by �Zeljko Hanj�s,published by the Croatian Mathematical Society, Zagreb, 1998.Softcover, 31 pages.Reviewed by Richard Hoshino, University of Waterloo.

There are many mathematical competitions administered in Croatia,for students in Grades 5 through 12. There are municipal contests in March,which are followed by regional contests in April, and the process culminatesin a national olympiad contest in May, which only the very top students areinvited to write. The Croatian contest system is quite similar to the Americanthree-contest system, where students write the AHSME to qualify for theAIME, and outstanding performance in these contests quali�es a student forthe USAMO.

In Croatia, students write di�erent contests according to their grade.Thus a student in the I class (Grade 9) writes a di�erent contest than does afellow student in the II class (Grade 10). Each of the three contests (munici-pal, regional, and national) consists of four questions.

This book contains the 48 problems from the 1998 Croatian math con-tests, and solutions are given for each of the problems. Many of these prob-lems are quite demanding for even the very best students, especially theproblems intended for the IV class (Grade 12). To illustrate this, one of thequestions on the IV class municipal contest is identical to Question 2 of the1967 IMO!

The problems are both engaging and challenging, and will certainly beof bene�t to anyone who is interested in problem-solving. We close o� withone of the problems that appeared on the III class (Grade 11) regional contest.

Prove that between every 79 successive natural numbers thereexists at least one whose sum of digits is divisible by 13. Find asequence of 78 successive natural numbers with the property thatthe sum of digits of any of its members is not divisible by 13.

Women in Mathematics: Scaling the Heights edited by Deborah Nolan,published by the Mathematical Association of America, 1997,ISBN 0-88385-156-3, softcover, 121+ pages, $29.95 (U.S.)Reviewed by Julia Johnson, University of Regina, Regina, Saskatchewan.

This book will engender emotion in any woman who has been drivenby the joy of mathematics and met with obstacles to the ful�lment of thatjoy. It is#46 in the MAA Notes series which comprises a breadth of works inmathematics education. This particular volume is an augmented version ofa report based on an NSF funded conference aimed at developing programsto advance women in mathematics.

Several introductory articles provide information about why \Some-where between the beginning of graduate school and the tenure decision, a

470

lot of things can and do go wrong for women, things that do not seem to gowrong as often for men." (Carol Wood, page 14). One of the factors is thatwomen in mathematics are very often isolated. A special program for womenis advocated, not because women have to learn mathematics in a special way,but because most successful male mathematicians have had such supportiveenvironments. The strategies presented are therefore not gender speci�c.They can be used to create an exciting and challenging environment for bothwomen and men to learn mathematics.

The conference centres on the program of the Mills College SummerMathematics Institute (SMI). This augmented version of the conference re-port aims at giving insight into the type of women who attend the summerinstitute in terms of their math backgrounds and future career plans. An ar-ticle entitled \A View of Mathematics from an Undergraduate Perspective"provides data which indicate that women are less con�dent in their ability tosucceed in the �eld of mathematics than are men. This e�ect snowballs tomake it less likely that women will succeed.

A number of modules complete with exercises and solutions for teach-ing various topics in upper level courses in undergraduate mathematics areprovided. This section provides a terri�c resource. The course designs weresupplied by faculty of the SMI, who include advice on how to incorporatetheir techniques into the traditional classroom. A seminar that stands out isentitled \What are numbers", by Svetlana Katok, which uses Kirillov's gen-eral philosophy of consequent extensions to reveal the concept of number inmodern mathematics.

The �nal section describes a variety of summer mathematics programsincluding the Carleton College and the St. Olaf College programs, the Pro-gram for Women in Mathematics at George Washington University and theMills College SMI which has since moved to UC Berkeley and is now calledthe Summer Institute for the Mathematical Sciences (SIMS). Subjects rangefrom \Where are the students now?" to \What do students and faculty thinkabout the program?" At SIMS all instructors are women, not because menwould not be e�ective teachers of women, but because women instructorsserve as role models for students to break the stereotypical image of a math-ematician being a man. Women generally come back from such a programwith con�dence in their ability to do mathematics and inspired to pursuegraduate work in mathematics.

One is left with the realization that speci�c courses are critical to gener-ating student interest in studying advanced mathematics. It is also apparentfrom this reading that an environment for learning mathematics which doesnot encourage working on math problems in teams provides a greater obsta-cle to those who are passionate about mathematics than to those who areless so.

471

After Math, Puzzles and Brainteasers by Ed Barbeau, published by Wall &Emerson, Inc., Toronto, Ontario/Dayton, Ohio. 1995.ISBN # 0-921332-42-4, softcover, 198 pages.Reviewed byMogens Esrom Larsen, University of Copenhagen, Copen-hagen, the Netherlands.

I love the book and heartily recommend it to all puzzle addicts.

It is a collection of challenging problems in elementary mathematicspartly from the journal Alumni Magazine of the University of Toronto, fromwhere it got the title.

Most of the problems are small gems with solutions varied from \Aha"to a systematic analysis of the inspired generalization. The problems are frommost mathematical disciplines, but numbers and geometry are the favourites.

As an example, make a 9{digit number of the digits 1, : : : , 9 such thatthe number created by the �rst n digits is divisible by n. The solution isunique (381654729). Or, �nd the locus of the mid-point of a ladder slidingalong the oor while leaning against a wall (a circular arc). Or, having 10numbers less than 100, is it always possible to select two disjoint subsetshaving the same sum? Well, the total number of subsets is 1023, so, as thesums are always smaller than 1000, two of them must be equal.

Besides the solutions, there are hints before and explanations after toplace the problems in the proper context. Perhaps too pedagogical to somebut an understandable temptation for a university teacher, I agree.

472

THE SKOLIAD CORNERNo. 42

R.E. Woodrow

This number we give the thirty problems of Maxi �eliminatoire 1996 ofthe 21i �eme Olympiade Belge organized by the Belgian Mathematics Teach-ers' Association. Twenty-six of the questions are multiple choice. For theremaining four the answer is an integer in the interval [0; 999]. Correct an-swers score 5 points, 2 points are given for no response and 0 for an incorrectanswer. No calculators allowed! Time allowed, 90 minutes. My thanks go toRavi Vakil for collecting this set when he was Canadian Team Deputy Leaderto the International Mathematical Olympiad at Mumbai, India.

21i �eme OLYMPIADE MATH �EMATIQUE BELGEMaxi �eliminatoire 1996

Mercredi 17 janvier 1996

1. Dans un article de journal au sujet de la construction de logements,les sch �emas suivants sont suppos �es repr �esenter le nombre d'habitationsbaties durant deux ann �ees de r �ef �erence.

1975 1985

Quel est, selon ce graphique, le rapport du nombre de constructions de 1985�a celui de 1975?

(a) 2 (b) 4 (c) 8 (d) 12 (e) 16

2. Sans r �eponse pr �eformul �ee | Dans une pi �ece obscure, un tiroircontient des bougies de meme forme et de meme taille : 7 blanches, 5 bleues,5 jaunes, 3 vertes et 3 rouges. Combien faut-il en prendre pour etre sur d'enavoir au moins deux de la meme couleur ?

3. Laquelle des formules suivants traduitl'organigrame ci-contre ?

(a) x = (p� q � r)� (s+ t)

(b) x = (p� (q� r))� (s+ t)

(c) x = (p� (q � r))� s+ t

(d) x = (p� q � r)� s+ t

(e) x = p� q� r � s+ t

� +

�

�

p q r s t

x

473

4. Voici une partie d'und �eveloppement d'une py-ramide �a base carr �ee dontles faces lat �erales sont destriangles �equilat �eraux. La-quelle des �gures ci-dessousen donne un d �eveloppementcomplet ?

(a) (b) (c) (d) (e)

5. Sans r �eponse pr �eformul �ee | Combien existe-t-il de nombres de 4chi�res constitu �es de deux paires, distinctes de chi�res identiques, commepar exemple 1661, 1122 on 1414 (mais non 3333) ?

6. Si p = 500 0005, q = 200 000 0004, r = 1025 et s = ( 110)1000,

laquelle des chaines d'in �egalit �es suivantes est exacte ?

(a) p < q < r < s (b) q < p < r < s (c) s < r < p < q(d) s < r < q < p (e) Aucune des pr �ec �edentes

7. Pour tout nombre r �eel a di� �erent de 0, de 1, de 2 et de 3, on poseb = a� 1, c = b� 1 et d = c� 1. La somme

a� 1

(a� 2)(a� 3)+

a� 2

(a� 3)(a� 1)+

a� 3

(a� 1)(a� 2)

est toujours �egale �a l'une des expressions suivantes. Laquelle ?

(a) 3bcd

(b) 1bcd

(c) b+c+dbcd

(d) b2+c2+d2

bcd(e) 1

a+ 1

b+ 1

c

8. Quel est le reste de la division par 5 de l'entier n si 3n+45

2 Z ?

(a) 0 (b) 1 (c) 2 (d) 3 (e) 4

9. Sans r �eponse pr �eformul �ee |Deux lignes de chemin de fer mesurentl'une 3672 m et l'autre 5472 m; elles ont �et �e construites avec des rails tousidentiques, sans qu'il faille en recouper. Quelle est, en m�etres, la longueur deceux-ci, sachant que c'est un nombre entier et qu'il a �et �e choisi aussi grandque possible?

10. Quelle valeur faut-il donner au r �eel k pour que, quelle que soit lavaleur du r �eel a, le polynome

2x4 � 5ax3 + 2a2x2 � 5a3x+ 2k

soit divisible par x+ 2a ?

(a) �90a4 (b) 90a2 (c)�45a4 (d) 5a4 (e) 45

474

11. Un vol a �et �e commis �a l' �ecole par une seule des quatre �el �eves ci-dessous, qui font successivement �a la directrice les d �eclarations suivantes :

ALICE : \Barbara est coupable."

BARBARA : \Charlotte est coupable."

CHARLOTTE : \Barbara a menti."

DANI �ELE : \Je ne suis pas coupable."

Si une et une seule de ces a�rmations est vraie, qui a commis le vol ?

(a) Alice (b) Barbara (c) Charlotte (d) Dani �ele

(e) Les donn �ees ne permettent pas de la d �eterminer

12. Laquelle des transformations suivantes de Z est une bijection ?

(a) f : Z! Z : x 7! 3x (b) f : Z! Z : x 7! 12x(x+ 1)

(c) f : Z! Z : x 7! 5� x (d) f : Z! Z : x 7! (3x� 1)2

(e) f : Z! Z : x 7! x� jxj + jx� 1j13. Voici le graphe d'une fonction f : R! R, p �eriodiquede p �eriode�.

1

0�2� �� 2��

Sachant que f(x) est donn �e par une des cinq expressions suivantes, delaquelle s'agit-il?

(a) sin2 x (b) cos2 x (c) j sinxj (d) sin jxj (e) 1 + sinx

14. Par lequel des nombres suivants 21996 � 1 n'est-il pas divisible?

(a) 1 (b) 3 (c) 5 (d) 7 (e) 15

15. Une fonction f : R ! R est partout d �erivable et sa d �eriv �ee s'an-nule en 0 ainsi qu'en 1. Lequel des graphes suivants est, demani �ere plausible,celui de f sur l'intervalle [0; 1] ?

(a) (b) (c) (d) (e)

1q

1q

1q

1q

1q

16. Une seule des in �egalit �es ci-dessous est vraie pour tout, r �eel stric-tement positif a et tout r �eel x. Laquelle ?

(a) ax � a�x � 0 (b) ax � a�x � 0 (c) ax + a�x � 2(d) ax + a�x � 2 (e) ax + a�x � 1

475

17. �Etant donn �e f : R! R : x 7! �x et g : R! R : x 7! jxj, lequeldes graphes suivants est celui de f + g ?

(a) (b) (c) (d) (e)

1

1

1

1

1

1

1

1

1

1

18. L'Alg �ebristan est constitu �e de quatre provinces qui ont des den-sit �es de population de 20, 24, 36 et 84 habitants/km2. Il r �esulte de ces in-formations que la densit �e de la population de l'Alg �ebristan

(a) est de 30 habitants/km2 ;(b) est de 41 habitants/km2 ;(c) est inf �erieure �a 11 habitants/km2 ;(d) est sup �erieure �a 20 habitants/km2 ;(e) est comprise entre 22 et 60 habitants/km2.

19. Si h est la compos �ee f � g des deux fonctions g : R ! R : x 7!x+ 2x3 et f : R! R : x 7! 1 + x2, alors, pour tout r �eel x, h(x) =

(a) 1 + x+ x2 + 2x3 (b) x+ 3x3 + 2x5 (c) 1 + x2 + 4x6

(d) 1 + x2 + 4x4 + 4x6 (e) 3 + 7x2 + 6x4 + 2x6

20. Soit a et b deux naturels, avec a > b. Soit x le reste de la divisionde a1996 par a � b et y le reste de la division de b1996 par a � b. Alors,n �ecessairement,

(a) x+ y > 1995 (b) x� y = 0 (c) ax� by = 0

(d) x1996 � y1996 < 0 (e) x1996 � y1996 > 0

21.

1 2

����

3 4

Claude a dessin �e sur chacune des quatre �ches ci-dessus un disque d'un cot �eet un carr �e de l'autre, et m'a�rme que, si le disque est noir, le carr �e l'est�egalement. Pour m'assurer que cela est exact,

(a) je dois retourner les quatre �ches ;

(b) il me su�t de retourner les �ches 1 et 2 ;

(c) il me su�t de retourner les �ches 1 et 3 ;

(d) il me su�t de retourner les �ches 1 et 4 ;

(e) il me su�t de retourner la �che 1.

476

22. Du cube ci-contre, de vo-lume V , sont enlev �ees les huit py-ramides AILM , BIJN , etc., donttoutes les faces sont des triangles

isoc �eles. Si�!AI= 1

3

�!AB, quel est le

volume du solide restant ?

AL

DK

C

O

G

RF

QE

M

I

N

BJ

(a) 12V (b) 2

3V (c) 3

4V (d) 5

6V (e) 7

9V

23. Un trap �eze isoc �ele a sa petite base et les deux cot �es adjacents delongueur �x �ee L. Pour quelle valeur (en radians) de l'angle entre la grandebase et un cot �e adjacent l'aire de ce trap �eze est-elle maximale ?

(a) �2

(b) �3

(c) �4

(d) �5

(e) �6

24. La suite (a0; a1; a2; : : : ) est d �e�nie par a0 = a1 = 1 et (8n 2 N)an+2 = an+1+an. Sachant que limn!1

an+1an

existe, que vaut cette limite ?

(a) 0 (b) 1 (c) 2 (d) 12(p5� 1) (e) 1

2(p5 + 1)

25. Quel est le reste de la division de 683 + 883 par 49 ?

(a) 0 (b) 2 (c) 28 (d) 35 (e) 42

26. Lequel des graphes ci-dessous est celui de la fonction

f : R! R : x 7! j sinxj � jxj?

(a) (b) (c) (d) (e)

2�

2�

2�

2�

2�

2�

2�

2�

2�

2�

27. Sans r �eponse pr �eformul �ee | Voici deux suites arithm �etiques :

3 ,7 ,11 ,15 , : : : ,4072 ,9 ,16 ,23 , : : : ,709

de raisons respectives 4 et 7. Combien ont-elles de termes en commun ?

477

28. Un octa �edre r �egulier est inscritdans un cube, chacun de ses sommets �etantle centre de l'une des faces du cube. Quelest le rapport de l'aire du cube �a celle del'octa �edre ?

(a)p36

(b) 34

(c)p3 (d) 2

p3 (e) 6

29. Quels que soient les nombres r �eels a, b et x tels que 0 < a < b etx 62 f�

2+ k� j k 2 Zg, l'expression

1pb� a

�

qb�aa

sinxs1 +

�qb�aa

sinx

�2 �pa+ b tan2 x

se simpli�e en l'une des suivantes ; laquelle ?

(a) 1 (b) tanx (c) tanxa

(d) j tanxj (e) sinxj cosxj

30. Quel est le nombre de solutions de l' �equation sin(100x) = x ?

(a) 59 (b) 60 (c) 61 (d) 62 (e) 63

Last number we gave the problems of the British Columbia Senior HighSchool Mathematics Contest, Final Round, Parts A and B. Next we give the\o�cial" solutions. My thanks go to Jim Totten, one of the contest organiz-ers, for furnishing them for our use.

BRITISH COLUMBIA COLLEGES SENIOR HIGHSCHOOL MATHEMATICS CONTEST

Final Round 1998

Part A

1. If�r + 1

r

�2= 3, then r3 + 1

r3=

(a) 0 (b) 1 (c) 2 (d) 2p3 (e) 4

p3

Answer: The correct answer is (a).If (r+ 1

r)2 = 3 then r + 1

r= �p3, and

r3 +1

r3=

�r +

1

r

�3� 3r2

1

r� 3r

1

r2

=

�r +

1

r

�3� 3r

1

r

�r +

1

r

�= (�

p3)3 � 3

p3 = 0:

478

2. Kevin has �ve pairs of socks in his drawer, all of di�erent colours andpatterns and, being a typical teenage boy, they are not folded and have beenthoroughly mixed up. On the �rst day of school Kevin reaches into his sockdrawer without looking and pulls out three socks. What is the probabilitythat two of the socks match?

(a) 310

(b) 35

(c) 13

(d) 124

(e) 115

Answer: The correct answer is (c).

There are 120 =�103

�= 10�9�8

1�2�3 di�erent three socks subsets in the set

of ten socks. On the other hand, every three sock subset with two matchingsocks contains one of the �ve matching pairs accompanied by one of the eightremaining socks. Therefore, there are 40 = 5 � 8 such subsets. This givesthe probability of 40

120= 1

3:

3. A small circle is drawn within a 16sector of a circle of radius r, as

shown. The small circle is tangent to the two radii and the arc of the sector.The radius of the small circle is:

(a) r2

(b) r3

(c) 2p3 r3

(d)p2 r2

(e) none of these

Answer: The correct answer is (b).

If � denotes the radius of the small circle then AO = 2�, since triangleAOB is a half of an equilateral triangle AOD. Furthermore, OC = �, andr = AO + OC = 2�+ � = 3�. Thus, � = r

3.

A

C

B

D

O

4. In the accompanying diagram the circle has radius 1, the centralangle AOB is a right angle and AC and BC are of equal length. The shadedarea is:

(a) �2

(b)p22

(c) ��p2

2(d)

p2+12

(e) 12

Answer: The correct answer is (b).

479

A B

O

C

D

The shaded area can be evaluated by subtracting the area of trian-gle AOB from the area of triangle ACB. The area of triangle AOB is12(AO)(OB) = 1

2, since AOB is a right angle. For the same reason AB =p

2, and consequently, DO = 12AB =

p22

. The area of triangle ABC is12(AB)(DC) = 1

2(AB)(DO + OC) = 1

2

p2(p22

+ 1) =p2+12

. Conse-

quently, the shaded area isp2+12

� 12=

p22:

5. The side, front and bottom faces of a rectangular solid have areas2x, y

2, and xy square centimetres, respectively. The volume of the solid is:

(a) xy (b) 2xy (c) x2y2 (d) 4xy (e)

impossible todetermine fromthe given infor-mation

Answer: The correct answer is (a).The volume of a rectangular solid is equal to the square root of the

product of the areas of its nonparallel faces. For, if a, b, and c are the edgesof the solid then ab, bc, and ca are the corresponding areas of its faces, andits volume is abc =

p(ab)(bc)(ca) : Consequently, the volume of our solid

isq(2x)(y

2)(xy) =

p(xy)2. Obviously, x and y are non-negative, since 2x

and y2represent areas. Therefore,

p(xy)2 = xy.

6. The numbers from 1 to 25 are each written on separate slips of paperwhich are placed in a pile. You draw slips from the pile without replacingany slip you have chosen. You can continue drawing until the product oftwo numbers on any pair of slips you have chosen is a perfect square. Themaximum number of slips you can choose before you will be forced to quitis:

(a) 13 (b) 14 (c) 15 (d) 16 (e) 17

Answer: The correct answer is (d).Let a, b, and c be three numbers from our set. If ab and bc are perfect

squares then ac is a perfect square too. For, if ab = m2 and bc = n2 then

ac = (ab)(bc)b2

= (mnb)2 and, obviously, ac is an integer. This implies that

the set f1, 2, 3, : : : , 25g can be partitioned into disjoint subsets with thefollowing properties:

480

1. A product of any two elements from the same subset is a perfect square.

2. A product of any two elements selected from di�erent subsets is not aperfect square.

In force of these properties we can easily �nd the partition, because in orderto form a subset containing a number a, all we need is to collect all numbersb, such that ab is a perfect square. We start by �nding the subset containing1, then �nd the other subsets by successively completing the smallest numbernot yet selected to a subset. The partition is: f1, 4, 9, 16, 26g, f2, 8, 18g,f3, 12g, f5, 20g, f6, 24g, f7g, f10g, f11g, f13g, f14g, f15g, f17g, f19g,f21g, f22g, f23g. Clearly, we can pick at most one number slip from eachsubset. This gives a maximum of 16 slips.

Note: The reader familiar with the notion of an equivalence relationwill notice that the relation a � b if and only if ab is a perfect square isan equivalence relation and that the optimal choice of slips corresponds to aselector from the disjoint equivalence classes of this relation.

7. A container is completely �lled from a tap running at a constantrate. The accompanying graph shows the level of the water in the containerat any time while the container is being �lled. The segment PQ is a straightline. The shape of the container which corresponds with the graph is:

Full

WaterLevel

EmptyTime

q

q

Q

P

(a) (b) (c) (d) (e)

Answer: The correct answer is (b).It is clear that wider parts of a container will take more time to be �lled

with water than its narrower parts. More precisely, the rate of increase ofthe level of water is inversely proportional to the cross-sectional area of thecontainer at that level. Thus, the constant slope of the last part of the graph,PQ, implies the constant width of the top of the container. This leaves uswith two possible shapes: (b) or (c). However, the smaller slope of the �rstpart of the graph indicates the wide bottom of the container. This agreeswith the shape shown in (b).

481

Comment. As noted last issue, this problem is identical to problem 3in part A of the British Columbia Colleges Junior High School Mathemat-ics Contest Final Round 1998 [1999:341{343]. This solution completes theJunior Contest solutions which were given last issue.

8. The accompanying diagram is a road plan of a small city. All theroads go east-west or north-south, with the exception of the one short diag-onal road shown. Due to repairs one road is impassable at the point X. Ofall the possible routes from P to Q, there are several shortest routes. Thetotal number of shortest routes is:

X

P

Q

(a) 4 (b) 7 (c) 9 (d) 14 (e) 16

Answer: The correct answer is (d).

Method I. It is clear that every shortest route must consist of stepsdue west or north only, and that the diagonal road must be included. Inthe following we will consider only the shortest routes or their parts. Thus,the diagonal road D can be reached from P in two di�erent ways, WN orNW , where W denotes a step due west and N a step due north. If all theroads were passable then in order to reach Q from the end of D we wouldneed a �ve step route consisting of two Ns and three W s performed in anyorder. Since the two Ns can be chosen in

�52

�= 10 ways in that sequence,

there are 10 such roads. However, in order to count only passable routes,we need to remove all the routes with point X from the set of ten routes.The impassable routes begin byWW and then continue from the end of theimpassable road to Q. There are three such routes, corresponding to threechoices of W on the way to Q from the end of the impassable road. Thisgives 2� (10� 3) = 14 possibilities.

Method II. The number of shortest routes can be found by counting thenumber of last generation branches in a tree diagram:

482

PW

N

N

W

D

D

N

W

W

NW

N

N

W

N

N

W

W

W

W

W

N

N

N

N

W W

W

W

W

W

W

W

W

W

W

N

N

N

N

W

W

W

W

N

W

W

W

W

N

N

N

N

N

Again, we �nd 2� 7 = 14 last generation branches in the tree.

9. Four pieces of timber with the lengths shown are placed in theparallel positions shown. A single cut is made along the line L perpendicularto the lengths of timber so that the total length of timber on each side of Lis the same. The length, in metres, of the longest piece of timber remainingis:

-�

-�

-�

5m

3m 3m

4m 5m

1:5m 4m

L

(a) 4:85 (b) 4:50 (c) 4:75 (d) 3:75 (e) none of the above

Answer: The correct answer is (c).Let x be the distance from the left-hand side to the line L. It is clear

that the total length of timber on each side of L is the same for exactly onevalue of x. The expressions for the total length of timber on either sidedepend, however, on the value of x. If 4 < x < 5 the cut passes through

483

each of the four logs and x+(x� 3)+ (x� 4)+ (x� 1:5) = (5�x)+ (3+3�x)+(4+5�x)+(1:5+4�x). The solution of this equation, x = 4:25,is indeed between 4 and 5. Since the value of x is unique, we do not needto examine equations corresponding to x � 4 or x � 5. The longest pieceof timber left to the cut has length x, while the longest piece right to the cuthas length 9� x. The latter is the longest and its length is 4:75 m.

10. The positive integers are written in order with one appearing once,two appearing twice, three appearing three times, : : : , ten appearing tentimes, and so on, so that the beginning of the sequence looks like this:

1 , 2 , 2 , 3 , 3 , 3 , 4 , 4 , 4 , 4

The number of 9's appearing in the �rst 1998 digits of the sequence is:

(a) 57 (b) 96 (c) 113 (d) 145 (e) 204

Answer: The correct answer is (b).At �rst, we need to �nd the number in which the 1998th digit occurs.

This is a two-digit number, since there are (1+99)(99)2

numbers with one ortwo digits that appear in our sequence, which is greater than 1998. If x is thenumber with 1998th digit, then x is the smallest positive integer for which9(1+9)

2(1) + (x�10+1)(10+x)

2(2) � 1998. By solving this inequality, we get

x � �1+p81722

. Hence x = 45, since x is a positive integer. Now, we cancount the 9s in the �rst 1998 digits: 9 + 19 + 29 + 39 = 96.

Part B

1. A right triangle has an area of 5 and its hypotenuse has length 5.Determine the lengths of the other two sides.

Solution. The lengths of the other two sides arep5 and 2

p5.

Let a and b be the lengths of the sides forming the right angle of thetriangle. Then 1

2ab = 5, since the area of the triangle is 5, and a2+ b2 = 52,

by the Pythagorean Theorem. By �nding b = 10a, from the �rst equation,

substituting this to the second and simplifying we get a4 � 25a2 + 100 = 0.This gives a2 = 5 or a2 = 20. Consequently, a =

p5 and b = 2

p5, or

a = 2p5 and b =

p5.

2. Find a set of three consecutive positive integers such that the small-est is a multiple of 5, the second is a multiple of 7 and the largest is a multipleof 9.

Solution. The set is of the form f160 + 315m, 161 + 315m,162 + 315mg, where m is a non-negative integer.

Let fx; x + 1; x + 2g be such a set. Then x = 5k, x + 1 = 7s, andx + 2 = 9t, where k, s, and t are positive integers. The �rst two condi-tions give x + 1 = 5k + 1 = 7s. The integer k can be expressed in theform k = 7l + r, where l is a non-negative integer and 0 � r < 7. Hence,

484

5k + 1 = 5(7l + r) + 1 = 5l(7) + 5r + 1 = 7s. From the last equationwe �nd r = 4. The last condition, x + 2 = 9t, now impliesx + 2 = 5k + 2 = 35l + 22 = 9t. The last part of this equation can bealso written as 36l + 18� l + 4 = 9t. This is satis�ed whenever �l + 4 isa multiple of 9; that is when l is of the form 9m+ 4, with m � 0. Finally,x = 5k = 5(7l + 4) = 5(7(9m + 4) + 4) = 160 + 315m. Thus, our setis f160 + 315m, 161 + 315m, 162 + 315mg, where m is a non-negativeinteger.

Note: The reader interested in this kind of problem may �nd it usefulto familiarize oneself with the so called Chinese Remainder Theorem whichcan be found in any textbook of Elementary Number Theory.

3. In the diagram, BD = 2, BC = 8 andthe marked angles are all equal; that is,

\ABC = \BCA = \CDE = \DEC .

Find AB.

A

B C

E

D

2

8

Solution. AB = 16p3

3.

We have: BE = BC = 8, since triangle BCE is isosceles;DE = BE � BD = 8 � 2 = 6; CE : 6 = 8 : CE, by similarity of trian-gles CBE and DCE. From the last equation we �nd (CE)2 = 48. Hence,CE = 4

p3. Finally, AB : 8 = 4

p3 : 6, by similarity of triangles ABC and

CDE. This gives AB = 16p3

3.

4. The ratio of male to female voters in an election was a : b. If cfewer men and d fewer women had voted, then the ratio would have beene : f . Determine the total number of voters who cast ballots in the electionin terms of a, b, c, d, e and f .

Solution. The total number of voters is (a+b)(cf�de)af�be .

Let M and F denote the number of male and female voters, respec-tively. Then M

F= a

band M�c

F�d = ef. These equations can be transformed to

the following system of linear equations:�bM � aF = 0 ,fM � eF = cf � de .

The system has no solution if be = af , that is a : b = e : f , butc : d 6= e : f . If the three ratios, a : b, e : f , and c : d are equal thesystem has in�nitely many solutions. Finally, if e : f 6= a : b the system hasexactly one solution (M;F ). The solution can be found by �rst �nding M

485

or F from one of the equations and substituting this to the second equation,or by eliminating one of the unknowns in another way. After some algebrawe get M = acf�ade

af�be and F = bcf�bdeaf�be . Thus, the total number of voters is

M + F = acf�ade+bcf�bdeaf�be = (a+b)(cf�de)

af�be .

5. Three neighbours named Penny, Quincy and Rosa took part in a localrecycling drive. Each spent a Saturday afternoon collecting all the aluminumcans and glass bottles he or she could. At the end of the afternoon eachperson counted up what he or she had gathered, and they discovered thateven though Penny had collected three times as many cans as Quincy, andQuincy had collected four times as many bottles as Rosa, each had collectedexactly the same number of items, and the three as a group had collectedexactly as many cans as bottles. In total, the three collected fewer than 200items in all. Assuming that each person collected at least one can and onebottle, how many cans and bottles did each person collect?

Solution. Penny collected 18 cans and 16 bottles, Quincy collected6 cans and 28 bottles, Rosa collected 27 cans and 7 bottles.

Let Cp, Cq and Cr denote the number of cans collected by Penny,Quincy and Rosa, respectively. Similarly, let Bp, Bq, and Br denote thecorresponding numbers of bottles. Now, the given conditions can be trans-lated to:

Cp = 3Cq , (1)

Bq = 4Br , (2)

Cp + Bp = Cq+ Bq = Cr + Br , (3)

Cp + Cq + Cr = Bp + Bq + Br , (4)

Cp + Cq + Cr + Bp + Bq + Br < 200 . (5)

The third equation represents a pair of single equations:

Cp + Bp = Cq + Bq (3a)

andCq + Bq = Cr + Br . (3b)

Now, we use the equations (1), (2), (3a), (3b), in the given order, to succes-sively eliminate the unknowns. We get: Cp = 3Cq, Bq = 4Br,Bp = 4Br � 2Cq, Cr = Cq+ 3Br. By substituting this into the equation(4) and simplifying, we get 7Cq = 6Br. Hence, Cq = 6k, Br = 7k. andconsequently, Cp = 18k, Cr = 27k, Bp = 16k, Bq = 28k, where k is apositive integer. This gives a total of 100k items. Thus k = 1, since the totalis less than 200.

That completes the Skoliad Corner for this issue. My bank of contestsat a suitable level is getting low. Please send me your contest materials,together with comments and suggestions about the Skoliad Corner.

486

MATHEMATICAL MAYHEM

Mathematical Mayhem began in 1988 as a Mathematical Journal for and byHigh School and University Students. It continues, with the same emphasis,as an integral part of Crux Mathematicorum with Mathematical Mayhem.

All material intended for inclusion in this section should be sent toMathematical Mayhem, Department of Mathematics, University of Toronto,100 St. George St., Toronto, Ontario, Canada. M5S 3G3. The electronicaddress is still

mayhem@math.toronto.edu

The Assistant Mayhem Editor is Cyrus Hsia (University of Western On-tario). The rest of the sta� consists of Adrian Chan (Upper Canada College),Jimmy Chui (University of Toronto), David Savitt (Harvard University) andWai Ling Yee (University of Waterloo).

Mayhem Problems

The Mayhem Problems editors are:

Adrian Chan Mayhem High School Problems Editor,Donny Cheung Mayhem Advanced Problems Editor,David Savitt Mayhem Challenge Board Problems Editor.

Note that all correspondence should be sent to the appropriate editor |see the relevant section. In this issue, you will �nd only solutions | thenext issue will feature only problems.

We warmly welcome proposals for problems and solutions. With theschedule of eight issues per year, we request that solutions from the previousissue be submitted in time for issue 8 of 2000.

High School Solutions

Editor: Adrian Chan, 229 Old Yonge Street, Toronto, Ontario, Canada.M2P 1R5 <all238@sprint.com>

H245. Determine how many distinct integers there are in the set

��12

1998

�,

�22

1998

�,

�32

1998

�, : : : ,

�19982

1998

��.

487

Solution. If a � 44, then a2 � 1936 < 1998, and ba2=1998c = 0. Ifa = 45, then ba2=1998c = 1, and if a = 999, then ba2=1998c = b999=2c =499. Now for k with 45 � k � 999, we have that

k2

1998� (k� 1)2

1998=

2k� 1

1998< 1 ,

so �k2

1998

���(k� 1)2

1998

�= 0 or 1 ,

so that we do not skip any integers from 1 to 499 when we list out the valuesof bk2=1998c, from k = 45 to k = 999. Thus, each of the 500 integers from0 to 499 appears in our list.

Now for 1000 � k � 1998,

k2

1998� (k� 1)2

1998=

2k� 1

1998> 1 ,

so �k2

1998

���(k� 1)2

1998

�� 1 .

In other words, no integer in the set fb10002=1998c, b10012=1998c, : : : ,b19982=1998cg appears more than once, since consecutive terms di�er by atleast one. So there are 999 integers in this set.

Hence, in total, there are 500 + 999 = 1499 integers in the set.

H246. Let S(n) denote the sum of the �rst n positive integers. Wesay that an integer n is fantastic if both n and S(n) are perfect squares. Forexample, 49 is fantastic, because 49 = 72 and S(49) = 1+2+3+ � � �+49 =1225 = 352 are both perfect squares. Find another integer n > 49 that isfantastic.

Solution. Let n = k2. We wish to �nd a positive integer k such thatS(n) = n(n+1)=2 = k2(k2+1)=2 is a perfect square. Since k2 is a perfectsquare, we need (k2 + 1)=2 to be a perfect square.

So, let (k2+1)=2 = m2, for some positive integerm. Then k2�2m2 =(k + m

p2)(k � m

p2) = �1. Note that (k;m) = (7; 5) satis�es this

equation. Now, (3+2p2)(3�2

p2) = 1, so multiplying these two equations

together, we obtain

(7 + 5p2)(7� 5

p2)(3 + 2

p2)(3� 2

p2)

= [(7 + 5p2)(3 + 2

p2)][(7� 5

p2)(3� 2

p2)]

= (41 + 29p2)(41� 29

p2)

= 412 � 2 � 292 = �1 .The last equation above tells us that (k;m) = (41;29) is a solution to

k2 � 2m2 = �1. Therefore, n = 412 = 1681 is another fantastic number.

488

H247. Say that the integers a, b, c, d, p, and r form a cyclic sextuple(a; b; c; d; p; r) if there exists a cyclic quadrilateral with circumradius r, sidesa, b, c, and d, and diagonals p and 2r.

(a) Show that if r < 25, then no cyclic sextuple exists.

(b) Find a cyclic sextuple (a; b; c; d; p; r) for r = 25.

Solution. Since one of the diagonals of the cyclic quadrilateral is 2r, itmust be a diameter of the circle. Suppose that the sides of the quadrilateralare a, b, c, and d (in that order). Then a2 + b2 = c2 + d2 = 4r2, since anangle subtended by a diameter is a right angle.

(a) Let us attempt to �nd two distinct Pythagorean triples (a; b; t) and (c; d; t),where t = 2r < 50. Listing all Pythagorean triples with hypotenuse lessthan 50, we �nd that only one value of t, t = 25, appears in two di�erentPythagorean triples as the hypotenuse. These triples are (15; 20;25) and(7; 24; 25). But then 2r = 25, for which r is not an integer. Thus, we con-clude that there is no cyclic sextuple for r < 25.

(b) If r = 25, we can have a = 14, b = 48, c = 40, and d = 30, since(14; 48; 50) and (40; 30;50) are both Pythagorean triples with t = 2r = 50.Then by Ptolemy's Theorem, 2pr = ac+bd, or 50p = 2000, so p = 40, whichis an integer. So a cyclical sextuple with r = 25 is (14;48; 40; 30; 40; 25).

H248. Consider a tetrahedral die that has the four integers 1, 2, 3,and 4 written on its faces. Roll the die 2000 times. For each i, 1 � i � 4,let f(i) represent the number of times that i turned up. (So, f(1) + f(2)+f(3) + f(4) = 2000). Also, let S denote the total sum of the 2000 rolls.

If S4 = 6144 � f(1)f(2)f(3)f(4), determine the values of f(1), f(2),f(3), and f(4).

Solution. We have f(1)+ f(2)+ f(3)+ f(4) = 2000 and S = f(1)+2f(2)+3f(3)+4f(4). Also, f(i) � 0 for i = 1, 2, 3, and 4. By the AM-GMInequality,

S

4=

f(1) + 2f(2)+ 3f(3) + 4f(4)

4� 4p24f(1)f(2)f(3)f(4)

===) S � 4 4p24f(1)f(2)f(3)f(4)

===) S4 � 44 � 24f(1)f(2)f(3)f(4)= 6144f(1)f(2)f(3)f(4).

Equality occurs if and only if f(1) = 2f(2) = 3f(3) = 4f(4), and sincef(1)+ f(2)+ f(3)+ f(4) = 2000, we can quickly solve to get f(1) = 960,f(2) = 480, f(3) = 320, and f(4) = 240.

489

Advanced Solutions

Editor: Donny Cheung, c/o Conrad Grebel College, University of Wa-terloo, Waterloo, Ontario, Canada. N2L 3G6 <dccheung@uwaterloo.ca>

A221. Construct, using ruler and compass only, the common tangentsof two non-intersecting circles.

Solution. First, we show how to construct a tangent from a given pointP to a given circle C1. Find the centre of C1 by taking any two non-parallelchords in the circle and drawing their perpendicular bisectors. These bisec-tors intersect at O1, the centre of circle of C1, as shown (Why?).

C1

O1qP

Next, construct the mid-point of PO1 and label it O2. Construct thecircle with radius PO2 centred at O2. Call this circle C2. Let one of theintersection points of C1 and C2 be R. See the following diagram.

PO2

O1

R

C1

q

Line PR is the tangent from the point P to the given circle C1. Thisfollows from the fact that PO1 is the diameter of circle C2 so \O1RP = 90�

and that O1R is the radius of C1.

Now we are ready to construct the common tangents of two non-inter-secting circles. We give the construction for the outer common tangents oftwo circles with radii of di�erent lengths. The cases of inner common tan-gents and common tangents of circles with equal radii are left to the reader.

Let the circles C1 and C2 have centres O1 and O2 respectively. Con-struct two points P1 and P2 as follows. Draw the line ` connecting the twocentres O1 and O2. Draw the lines perpendicular to ` through O1 and O2.Let P1 andP2 be the intersection points of these perpendicular lines with thecircles C1 and C2 respectively on the same side of the line as shown. Extendthe line of centres and the line P1P2 to intersect at P .

490

qPP1

P2

O1 O2

C2C1

Now, we use the result shown at the very beginning to construct thetangent from point P to circle C1. Let the point of tangency be X. ExtendPX. We claim that PX is tangent to circle C2, say at Y . Then XY is thecommon outer tangent to C1 and C2.

qPO1 O2

C2C1

qqX

Y

The point P is a centre of homothety for circles C1 and C2; that is, wecan dilate circle C1 with respect to P to obtain circle C2. In particular, pointP1 is taken to point P2 under this dilatation | this is how we found point P .Therefore, the point of tangency X gets taken to the point of tangency Y .

A222. Does there exist a set of n consecutive positive integers suchthat for every positive integer k < n, it is possible to pick k of these numberswhose mean is still in the set?

Solution. We claim that the answer is yes. Consider the set ofn integersf0, 1, : : : , n� 1g.

If k is odd, then take the �rst k integers 0, 1, : : : , k � 1. The mean is[(k � 1)k=2]=k = (k� 1)=2, which is an integer in the set.

If k is even, then take the k integers 0, 1, : : : , k=2� 1, k=2 + 1, : : : ,k. The sum of these numbers is k(k+1)=2� k=2 = k2=2. The mean is thenk=2.

Now to obtain a set of n consecutive positive integers, just add anypositive constant, say c, to each number. The means will also be increasedby c.

A223. Proposed by Mohammed Aassila, Strasbourg, France.

Suppose p is a prime with p � 3 (mod 4). Show that for any set ofp � 1 consecutive integers, the set cannot be divided into two subsets sothat the product of the members of the one set is equal to the product of themembers of the other set.

(Generalization of Question 4, IMO 1970)

491

Solution by the Proposer.We will show that the two products will not even be congruent to each

other modulo p. Henceforth, we will be taking congruence modulo p. If oneof the numbers is congruent to 0 modulo p, then the product in which thatterm appears is congruent to 0 as well, whereas the other is not. Thus wemay assume that the numbers we are given are congruent to 1, 2, : : : , p� 1modulo p. Suppose that the two products are � and �. Then � and � arecongruent modulo p and hence so are �� and �2. But �� � (p� 1)! � �1modulo p by Wilson's Theorem. Hence so is �2.

To complete the proof, it su�ces to show that the congruence �2 � �1has no solution. Suppose on the contrary that it does. Let p = 4n+3. Then�p�1 � �4n+2 � �2(2n+1) � (�1)2n+1 � �1. This contradicts Fermat'sLittle Theorem.

A224. Proposed by Waldemar Pompe, student, University of War-saw, Poland.

LetP be an interior point of triangleABC such that \PBA = \PCA =(\ABC + \ACB)=3. Prove that

AC

AB + PC=

AB

AC + PB.

SolutionLet D be the point on the opposite side of BC to A such that BACD

is a parallelogram. Let � = (B + C)=3 so A+ 3� = 180�.

Let Q be the point in triangle BDC such that BPCQ is a parallelo-gram. Extend CQ to meet AB extended at R and likewise extend BQ tomeet AC extended at S as shown.

A

D

CB

R

S

P

Q

�

�

�

�

�

�

�

�

Now \ARC = \ABP = � and \ASB = \ACP = �. Then\PCR = 180�� (A+ 2�) = �. Similarly, \PBS = 180�� (A+2�) = �.Further, \BQR and \CQS both equal �, since BQkPC and CQkPB.

492

Thus AR = AB + BR = AB + BQ = AB + PC. SimilarlyAS = AC+PB. TrianglesARC andASB are similar with anglesA, �, and2�. Therefore, AC=AR = AB=AS, orAC=(AB+PC) = AB=(AC+PB).

Also solved by Andrei Simion, Brooklyn Technical High School, NewYork, NY, USA.

Challenge Board Solutions

Editor: David Savitt, Department of Mathematics, Harvard University,1 Oxford Street, Cambridge, MA, USA 02138 <dsavitt@math.harvard.edu>

C81. Let fang be the sequence de�ned as follows: a0 = 0, a1 = 1,and an+1 = 4an � an�1 for n = 1, 2, 3, : : : .

(a) Prove that a2n � an�1an+1 = 1 for all n � 1.

(b) Evaluate1Xk=1

arctan

�1

4a2k

�.

Solution. (a) We will proceed by induction. The statement holds forn = 1. Assume it holds for some n = k � 1, so a2k�1 � ak�2ak = 1. Then

a2k � ak�1ak+1 = ak(4ak�1 � ak�2)� ak�1(4ak � ak�1)

= a2k�1 � ak�2ak= 1 .

Hence, the statement holds for n = k, and by induction, for all n � 1.

(b) Note that the �rst few terms of the sequence are a0 = 0, a1 = 1, a2 = 4,a3 = 15, a4 = 56, etc. Let Sn denote the nth partial sum

nXk=1

arctan

�1

4a2k

�.

We will compute the �rst few Sn using the following result: let a and b bepositive reals such that ab < 1. Then

arctana+ arctan b = arctan

�a+ b

1� ab

�,

where the arctangents are chosen to lie in the interval (0; �=2). To see this,let � = arctan a and � = arctan b. Then

tan(arctana+ arctan b) = tan(�+ �)

=tan�+ tan�

1� tan� tan�

=a+ b

1� ab> 0 .

493

Hence, 0 < arctana+ arctan b < �=2, and the result follows.

The �rst few Sn are then

S1 = arctan

�1

4

�,

S2 = arctan

�1

4

�+ arctan

�1

64

�= arctan

14+ 1

64

1� 14� 164

!

= arctan

�4

15

�,

S3 = arctan

�4

15

�+ arctan

�1

900

�= arctan

415

+ 1900

1� 415� 1900

!

= arctan

�15

56

�; etc:

By induction, we will prove that Sn = arctan(an=an+1). The state-ment holds for n = 1. Assume that it holds for some n = k � 1, soSk�1 = arctan(ak�1=ak). Then

Sk = arctan

�ak�1ak

�+ arctan

�1

4a2k

�

= arctan

ak�1ak

+ 14a2

k

1� ak�1ak

� 14a2

k

!

= arctan

�ak(4ak�1ak + 1)

4a3k � ak�1

�.

Now

4a3k � ak�1 = (4ak)(a2k)� ak�1 = 4ak(ak�1ak+1 + 1)� ak�1

= 4ak�1akak+1 + 4ak � ak�1 = 4ak�1akak+1 + ak+1

= ak+1(4ak�1ak + 1) ,

so

Sk = arctan

�ak(4ak�1ak + 1)

ak+1(4ak�1ak + 1)

�= arctan

�ak

ak+1

�.

Hence, the statement holds for n = k, and by induction, for all n � 1.

Finally, solving for fang using its characteristic equation, we obtain

an =(2 +

p3)n � (2�p3)n

2p3

.

494

Therefore,

1Xk=1

arctan

�1

4a2k

�= lim

n!1arctan

�an

an+1

�

= arctan

�1

2 +p3

�

= arctan(2�p3) =

�

12.

C82. Find the smallest multiple of 1998 which appears as a partialsum of the increasing sequence

1 , 1 , 2 , 2 , 2 , 4 , 4 , 4 , 4 , 8 , : : : ,

in which the number 2k appears k+ 2 times (for k a non-negative integer).

Solution by Ivailo Dimov, freshman, M.I.T., MA, USA.

We will prove the following lemma: the partial sums of this sequenceare the integers of the form m � 2n, where n is a non-negative integer andm runs from n + 1 to 2(n+ 1). To solve the problem using the lemma, ifan integer m � 2n is divisible by 1998, thenm must be divisible by 999. So,we are looking for the smallest positive integer n such that a multiple of 999lies between n+ 1 and 2(n+ 1). Evidently the smallest such n is 499, andso the smallest partial sum divisible by 1998 is 999 � 2499 = 1998 � 2498.

To prove the lemma, we proceed by induction on n. To begin with, the�rst two partial sums are 1�20 and 2�20. Next, assume that the lemma is truefor n = k�1, which entails that 1+1+2+ � � �+2k�1 = 2(k�1+1) �2k�1,where the left-hand side contains every term up to and including the last ofthe k + 1 terms which are equal to 2k�1. Then the next k + 2 partial sumsare 2k � 2k�1 + i � 2k = (k + i) � 2k with i running from 1 to k + 2, exactlyas desired, and so the lemma is true for n = k.

495

Problem of the Month

Jimmy Chui, student, University of Toronto

Problem. If p1 and p2 are distinct odd primes andA = (p1p2+1)4�1,show that A has at least 4 distinct prime divisors.

(1999 Descartes, problem D1)

Solution. Let p3 = p1p2 + 2. Observe that

A = (p21p22 + 2p1p2 + 1)2 � 12

= (p21p22 + 2p1p2 + 1� 1)(p21p

22 + 2p1p2 + 1 + 1)

= p1p2(p1p2 + 2)(p21p22 + 2p1p2 + 2)

= p1p2p3(p1p2p3 + 2) .

Now, p3 � 2 (mod p1) and p3 � 2 (mod p2). Since p1, p2 � 3, weknow that p1 - p3 and p2 - p3. Furthermore, note that p3 > p1, p2.

Thus, two cases exist. If p3 is composite, then A has at least fourdistinct prime factors, and we are done. However, if p3 is prime, then letp4 = p1p2p3 + 2.

Then, p4 � 2 (mod p1), p4 � 2 (mod p2), and p4 � 2 (mod p3). Thismeans that p1 - p4, p2 - p4, and p3 - p4, which leads us to the conclusionthat A contains at least four distinct prime factors, QED.

J.I.R. McKnight Problems Contest 1992

1. (a) If p and q are the roots of 2x2 � 5x+ 1 = 0, what is the value oflog2 p+ log2 q?

(b) Solve for x:

x� 7

x� 9� x� 9

x� 11=x� 13

x� 15� x� 15

x� 17.

2. (a) Given x2 � x+ 1 = 0, evaluate x9 � 3x6 + 4x3.

(b) The L-shaped room has a oor made with 231 identical square tiles,each of side 1 m. Find the least perimeter of the room.

a

a

b

b

496

3. Determine the rational values of x, y, and k that satisfy the system:

x+ y = 2k ,

x2 + y2 = 5k ,

x3 + y3 = 14k .

4. Two identical cones of radius 8 cm and height 8 cm are shown. At thestart of the problem the upper is full of water and the lower one isempty. Water drains from the upper cone into the lower one. At thetime the depth of the water in the upper cone is 4 cm and falling at0:1 cm per second, how deep is the water in the lower cone, and howquickly is it rising?

5. Show that 2n � 32n � 1 is always divisible by 17 for n 2 N.

6. (a) Three urns are arranged in a row. The left urn contains two redballs and one white ball. The centre urn contains one red ball andone white ball. The right urn contains one red ball and two whiteballs. A ball is randomly selected from the appropriate urn andthen discarded. If the chosen ball is white, then one moves one urnto the left; however, if the chosen ball is red, then one moves oneurn to the right. [Ed. It is the person selecting the ball that movesone urn over and proceeds to select another ball if it is possible.]The game continues until either an empty urn is reached or a moveis made past an end urn. Assuming that the game commences witha selection from the centre urn, what is the probability that thegame ends because an empty urn was reached?

(b) Every time John Olerud gets a hit, his con�dence increases. Thenext time he bats, he has a 42 per cent probability of getting a hit.But when he does not get a hit, the probability of his getting a hitin his next at bat drops to 23 per cent. Determine John Olerud'sbatting average for the season (that is, in the long term).

497

7. The parabola y = 4x2�24x+31 crosses the x{axis at (y;0) and (g;0),both to the right of the origin. A circle also passes through these twopoints. Find the length of the tangent from the origin to any such circle.

8. Triangle ABC has sides 8, 15 and 17. A point P is inside the triangle.Find the minimum value of PA2 + PB2 + PC2.

9. (a) Consider the system of equations:

x1 + x2 = 2 ,

x2 + x3 = 4 ,

x3 + x4 = 8 ,

.

.

.

xk�1 + xk = 2k�1 ,

.

.

.

x2000 + x2001 = 22000 ,

x2001 + x1 = 22001 .

Evaluate x2001.

(b) In the original system, replace 2001 by n. Find the solution forxn, for all values of n.

10. Two circles are tangent at P as shown. Lines AC and BD are drawnthrough P . If points A, B, C, D are concyclic, prove AC = BD.

A

B

D

C

P

498

Three Gems in Geometry

Naoki Sato

Geometry may well be the most elegant branch of mathematics, andunfortunately the most under-appreciated, with respect to the high schoolcurriculum. Here we present three pretty results, which come in the formof do-it-yourself exercises. Get ready to roll up your sleeves, it will be wellworth it!

The QM-AM-GM-HM Inequality

First, we introduce some de�nitions. For non-negative reals a and b,let s

a2 + b2

2,

a+ b

2,

pab , and

21a+ 1

b

=2ab

a+ b

denote the quadratic mean (QM), arithmetic mean (AM), geometric mean(GM), and harmonic mean (HM) of a and b, respectively.

Now, let A1A2B2B1 be a trapezoid as shown, with a = A1A2 andb = B1B2. Let P andQ be points onA1B1 and A2B2 respectively such thatPQ is parallel to the bases.

A1 A2

Q

B2B1

P

(a) Show that the line PQ dividing the trapezoid into two trapezoids ofequal area has length QM(a; b).

(b) Show that the line PQ at equal distance to both bases has lengthAM(a; b).

(c) Show that the linePQ dividing the trapezoid into two similar trapezoidshas length GM(a; b).

(d) Show that the line PQ passing through the intersection of A1B2 andA2B1 has length HM(a; b). Furthermore, show that the intersectionis the mid-point of PQ.

Copyright c 1999 Canadian Mathematical Society

499

(e) Show that for all a, b > 0,

HM(a; b) � GM(a; b) � AM(a; b) � QM(a; b) ,

with equality occurring in any of the above inequalities if and only ifa = b.

The Area of a Pedal Triangle

Let ABC be a triangle and P a point in the plane, and let P1, P2,and P3 be the feet of the perpendiculars from P to sides BC, AC, and ABrespectively. Then triangle P1P2P3 is called the pedal triangle of P . We willderive a simple formula for K0, the area of this triangle.

A

P3

B P1 C

P2

P

(a) We have K0 =1

2P1P2 � P1P3 sin(\P2P1P3) .

(b) Show that P1P2 = PC sinC and P1P3 = PB sinB.

(c) Extend BP to Q on the circumcircle of triangle ABC. Show that\QCP = \P1P2P3.

A

P3

B P1 C

P2

P

Q

500

(d) Show thatsin\QCP

sin\BQC=

PQ

PC.

(e) Show that K0 =1

2� PB � PQ sinA sinB sinC .

(f) Recall by power of a point that PB � PQ = R2 � OP 2, where R andO are the circumradius and circumcentre of triangle ABC respectively,and that

sinA sinB sinC =K

2R2.

Conclude that K0 =R2 � OP 2

4R2�K .

Note that the formula depends only on OP , namely the distance from O toP . There are two interesting cases:

(i) First, K0 is maximized when P = O (for P in triangle ABC), in whichcase P1P2P3 is the medial triangle of ABC, and K0 = K=4.

(ii) Second, we see that K0 = 0 when OP = R; that is, when P lies on thecircumcircle of triangleABC. In this case, triangleP1P2P3 degeneratesinto a line, called the Simson line.

Ptolemy's Theorem

Let ABCD be a cyclic quadrilateral.

D

AB

CP

(a) Let P be the point on BD such that \PAB = \DAC. Show that tri-angles PAB andDAC are similar, and that triangles PAD and BACare similar.

(b) Show thatBP

CD=

AB

ACand

DP

BC=

AD

AC.

(c) Prove Ptolemy's Theorem: AC � BD = AB � CD +BC � AD .

501

Hints

The QM-AM-GM-HM Inequality

(a) Let h1 and h2 be the heights of the upper and lower trapezoids respectively.

Show that, in general,h1h2

=PQ� A1A2

B1B2 � PQ.

(c) Show that the ratios a=PQ and PQ=b must be equal.

(d) Let M be the intersection of A1B2 and A2B1. Show that triangles A1A2Mand B2B1M are similar, so A1M :MB2 = a=b. Compute PM .

(e) Assume without loss of generality that a � b. Let A be the area of the wholetrapezoid. The length of PQ determines its position, and hence the area of theupper trapezoid, say U . Let f(PQ) = U=A. Thus, f(a) = 0, f(b) = 1, andby part (a), f(QM(a; b)) = 1=2. It is clear that f is an increasing function.The idea of this problem is to compare f at di�erent values. For example,f(AM(a; b)) � 1=2 = f(QM(a; b)), so AM(a; b) � QM(a; b).

To prove thatGM(a; b) � AM(a; b), let g(PQ) = h1�h2, where h1 and h2are the heights of the upper and lower trapezoid respectively. Is g increasingor decreasing? Compute g(GM(a; b)) and g(GM(a; b)), and compare.

The Area of a Pedal Triangle

(b) Quadrilateral PP1CP2 is cyclic, and furthermore, PC is a diameter of the cir-cumcircle of PP1CP2. Use the Sine Law.

(c) Show that \PP1P3 = \PBP3 and \PP1P2 = \PCP2. The rest is an anglechase; use \P2P1P3 = \PP1P3 + \PP1P2 = \PBP3 + \PCP2, etc.

(d) Use the Sine Law in triangle QCP .

(e) Put (a), (b), (c), and (d) together.

Ptolemy's Theorem

(a) Since ABCD is cyclic, \ABP = \ABD = \ACD. Similarly, \ADP =\ADB = \ACB.

(b) Use similar triangles.

(c) Expand BD = BP + PD.

Naoki Sato403 Renforth Drive

Etobicoke, ONM9C 2M3

mayhem@math.toronto.edu

502

A Do-It-Yourself Proof of the n = 4 case of

Fermat's Last Theorem

Ravi Vakil

IntroductionThe impact of Fermat's Last theorem on the development of mathemat-

ics is immeasurable. This is not simply because it resisted humankind's beste�orts for centuries | there are many other such problems that are long-forgotten. But the mathematics that has come up in uncounted attemptsto solve it has proved to be fundamental in number theory (and in math ingeneral).

Here, for the record, is Fermat's Last Theorem.

Theorem (Wiles, Wiles-Taylor). Ifn, x, y, and z are integers with n � 3and xn + yn = zn, then xyz = 0.

We will refer to Fermat's Last Theorem for a particular n by \FLTn".

Even individual cases are di�cult, and proofs of special cases shed lighton advanced mathematical ideas. In this article, you the reader will provethe n = 4 case of Fermat's Last Theorem, and in the sequel (to appearnext issue), you will prove the n = 3 case. (FLT3 is often referred to in theliterature as the easiest case, for example [D] p. 96{104, but this is not true.)

Although Fermat's Last Theorem turned out to follow from Wiles'sproof of the more important Taniyama-Shimura-Weil conjecture, problems inthe vicinity of FLT remain on the cutting edge of current research. For exam-ple, a powerful generalization of Fermat, the ABC Conjecture (due to Masserand Oesterl �e) is considered an important and fundamental open problem.

Warning: This is a very interactive article! You will really have to try all of the

problems. If you get stuck, then skip to the next one. Even getting stuck is a good

thing (much better than not trying at all), because the seeds of ideas that come up in

one problem invariably turn up again in a later one. You will see that no advanced

background is required; ambitious high school students should be able to tackle it.

Warming Up: Primitive Pythagorean Triples

The proof of FLT4 will be similar in character to one method for gen-erating primitive Pythagorean triples, and these triples will come up in theproof, so we will start there.

A primitive Pythagorean triple is an ordered triple of positive integers(a; b; c), pairwise relatively prime, that are the sides of a right-angled tri-angle, that is a2 + b2 = c2. Familiar examples are (3;4; 5), (5; 12; 13),(7; 24; 25), and (8; 15; 17).

Copyright c 1999 Canadian Mathematical Society

503

1. Show that if (a; b; c) is a primitive Pythagorean triple, then exactly oneof a, b is odd.

(Hint: Check modulo 4.) Without loss of generality, say a is odd.

2. Then a2 = c2 � b2 = (c� b)(c+ b). Show that c� b and c + b haveno common factor.

3. Two relatively prime odd numbers multiplying to a perfect square mustboth be odd perfect squares. In the previous problem, show that c� band c+b can be taken to be (m�n)2 and (m+n)2 respectively, wherem and n are positive integers, m > n, and exactly one of m and n iseven. (Remember that we are assuming that a is odd!)

4. In the previous problem, show that m and n are relatively prime.

(Hint: Show that if they have a common factor d, then d is a factor ofboth b and c, which are relatively prime.)

5. Solve for b and c to get b = 2mn, c = m2 + n2. Then show thata = m2 � n2.

In conclusion, any primitive Pythagorean triple is of the form (a; b; c) =(m2�n2; 2mn;m2+n2) or (2mn;m2�n2; m2+n2), wherem and n arerelatively prime positive integers, one of which is even, with m > n. Withthis result, you can now do lots of interesting things involving Pythagoreantriples. (You will not need to do them to tackle FLT4.) For example:

6. Plug in some large values ofm andn to get ridiculously hugePythagoreantriples.

7. Show that any Pythagorean triple (a; b; c) can be written as a multiplek of a primitive Pythagorean triple.

8. 3192 + 4592 = 5552. Which k, m, and n give this triple?

9. How many Pythagorean triangles are there with hypotenuse 60?

10. Suppose that (a; b; c) is a primitive Pythagorean triple, and a is odd.Show that (c�a)=2, (c+a)=2, c+ b, and c� b are all perfect squares.

11. If you know some trigonometry, try \breaking the rules" and substi-tuting m = cos �, n = cos � in the formula for primitive Pythagoreantriples. What formula do you get? (Remember the double angle formu-las: cos 2� = cos2 � � sin2 � and sin2� = 2 sin � cos �.)

(Many of these examples appear on [V] p. 130.)

Are you all warmed up? Then let's get to it : : :

The proof of FLT4

12. Show that if there is a solution of FLT4, then there is a solution wherex, y, and z are pairwise relatively prime.

504

13. Show that if there is a solution of FLT4, then there is a solution of

x4 + y4 = z2 , (1)

where x, y, and z are pairwise relatively prime positive integers.

14. (This is the big one!) Assume that there is a solution (x; y; z) = (a; b; c)to equation (1). Then (a2; b2; c) is a primitive Pythagorean triple, soyou can use what you know about such triples. Play around with thealgebra. (Another primitive Pythagorean triple may come up.) You willhopefully end up with another solution to equation (1) that is in somesense smaller than the solution (x; y; z) = (a; b; c). (Make that pre-cise.) Then the argument by contradiction will go as follows: Suppose(x; y; z) is the \smallest" solution of equation (1). Then this methodproduces a smaller solution | contradiction.

If you prove FLT4, then please send it in | it will be problem A241 nextissue. And congratulations | reach over your shoulder and pat yourself onthe back! If you are able to tackle other advanced-level problems but thinkthis one must be too hard, you're wrong | just be ambitious and try it; youmight surprise yourself!

Acknowledgements. Suggestions by David Savitt of Harvard University on anearlier version of this article have greatly improved the exposition. He also suggesteda couple of fun-related problems.

1. Show that if a, b, and c are integers that are the sides of a right-angledtriangle, then 60 divides abc.

2. Find a right-angled triangle with rational sides and area 5. (Hint: Try to scalewell-known Pythagorean triples.) One such triangle was discovered by Fibonacci,among others. In fact, 5 is the smallest integer which is the area of a right-angledtriangle with rational sides.

It is a classical unsolved problem to determine all of the integers which areareas of right triangles with rational sides. In 1983, it was shown that a solution toone of the most important conjectures in number theory, the Birch-Swinnerton-Dyerconjecture, would give a solution to this problem as well. For more on this fascinatingconnection between classical diophantine equations and the frontier of mathematics,see an upcoming CRUX with MAYHEM article by David Savitt.

References

[D] H. D�orrie, 100 Great Problems of Elementary Mathematics: Their History and

Solution, Dover: New York, 1965.

[V] R. Vakil, A Mathematical Mosaic: Patterns and Problem Solving, Brendan Kelly

Publ.: Toronto, 1996.

Ravi VakilMIT Dept. of Mathematics

77 Massachusetts Ave. Rm. 2-248Cambridge MA USA 02139

vakil@math.mit.edu

505

PROBLEMSProblem proposals and solutions should be sent to Bruce Shawyer, Department

of Mathematics and Statistics, Memorial University of Newfoundland, St. John's,

Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution,

together with references and other insights which are likely to be of help to the edi-

tor. When a submission is submitted without a solution, the proposer must includesu�cient information on why a solution is likely. An asterisk (?) after a number

indicates that a problem was submitted without a solution.

In particular, original problems are solicited. However, other interesting prob-

lems may also be acceptable provided that they are not too well known, and refer-

ences are given as to their provenance. Ordinarily, if the originator of a problem can

be located, it should not be submitted without the originator's permission.

To facilitate their consideration, please send your proposals and solutionson signed and separate standard 81

2"�11" or A4 sheets of paper. These may

be typewritten or neatly hand-written, and should be mailed to the Editor-in-

Chief, to arrive no later than 1 May 2000. They may also be sent by email tocrux-editors@cms.math.ca. (It would be appreciated if email proposals and solu-

tions were written in LATEX). Graphics �les should be in epic format, or encapsulated

postscript. Solutions received after the above date will also be considered if thereis su�cient time before the date of publication. Please note that we do not accept

submissions sent by FAX.

2489 Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,Spain.

The set of twelve vertices of a regular icosahedron can be partitionedinto three sets of four vertices, each being such that none of the sets havetheir four vertices forming a golden rectangle. In how many di�erent wayscan this be done?

2490 Proposed by Mih �aly Bencze, Brasov, Romania.Let � > 1. Denote by xn the only positive root of the equation:

(x+ n2)(2x+ n2)(3x+ n2) : : : (nx+ n2) = �n2n .

Find limn!1

xn.

2491 Proposed by Mih �aly Bencze, Brasov, Romania.Suppose that f : R! R is a continuous function and that fakgnk=1 and

fbkgnk=1 are two geometric sequences for which

nXk=1

f(ak) < 0 <

nXk=1

f(bk) .

Prove that there exists a geometric sequence fckgnk=1 for which

nXk=1

f(ck) = 0 .

506

2492 Proposed by Toshio Seimiya, Kawasaki, Japan.

In4ABC, suppose that \BAC is a right angle. Let I be the incentreof 4ABC, and that D and E are the intersections of BI and CI with ACand AB respectively. Let points P and Q be on BC such that IPkAB andIQkAC.

Prove that BE + CD = 2PQ.

2493 Proposed by Toshio Seimiya, Kawasaki, Japan.

Suppose that ABCD is a convex cyclic quadrilateral, that\ACB = 2\CAD, and that \ACD = 2\BAC.

Prove that BC + CD = AC.

2494 Proposed by Toshio Seimiya, Kawasaki, Japan.

Given 4ABC with AB < AC, let I be the incentre and M be themid-point of BC. The lineMI meets AB and AC at P and Q respectively.A tangent to the incircle meets sides AB and AC at D and E respectively.

Prove thatAP

BD+AQ

CE=

PQ

2MI.

2495 Proposed by G. Tsintsifas, Thessaloniki, Greece.

Let P be the interior isodynamic point of 4ABC; that is,AP

bc=BP

ca=CP

ab(a, b, c are the side lengths, BC, CA, AB, of4ABC).

Prove that the pedal triangle of P has area

p3

d2F , where F is the area

of 4ABC and d =a2 + b2 + c2

2+ 2

p3F .

2496 Proposed by Paul Yiu, Florida Atlantic University, Boca Raton,FL, USA.

Given a triangle ABC, let CA be the circle tangent to the sides AB,AC, and to the circumcircle internally. De�ne CB and CC analogously. Findthe triangle, unique up to similarity, for which the inradius and the radii ofthe three circles CA, CB , and CC are in arithmetic progression.

2497 Proposed by Nikolaos Dergiades, Thessaloniki, Greece.

Given4ABC and a point D on AC, let \ABD = � and \DBC = .Find all values of \BAC for which �

> AD

DC.

507

2498 Proposed by K.R.S. Sastry, Dodballapur, India.A Gergonne cevian is the line segment from a vertex of a triangle to the

point of contact, on the opposite side, of the incircle. The Gergonne point isthe point of concurrency of the Gergonne cevians.

In an integer triangle ABC, prove that the Gergonne point � bisectsthe Gergonne cevian AD if and only if b, c, 1

2j3a � b � cj form a triangle

where the measure of the angle between b and c is �3.

2499 Proposed by K.R.S. Sastry, Dodballapur, India.A Gergonne cevian is the line segment from a vertex of a triangle to the

point of contact, on the opposite side, of the incircle. The Gergonne point isthe point of concurrency of the Gergonne cevians.

Prove or disprove:

two Gergonne cevians may be perpendicular to each other.

2500 Proposed by G. Tsintsifas, Thessaloniki, Greece.In the lattice plane, the unit circle is the incircle of4ABC.

Determine all possible triangles ABC.

508

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.

1492. [1989: 297; 1991: 50] Proposed by G. Tsintsifas, Thessaloniki,Greece.

LetA0B0C0 be a triangle inscribed in a triangleABC, so thatA0 2 BC,B0 2 CA, C0 2 AB. Suppose also that BA0 = CB0 = AC0.

1. If either the centroids G,G0 or the circumcentres O, O0 of the trianglescoincide, prove that 4ABC is equilateral.

2.? If either the incentres I, I0 or the orthocentres H, H0 of the trianglescoincide, characterize 4ABC.

III. Solution to part 2? by C.R. Pranesachar and B.J. Venkatachala,Department of Mathematics, Indian Institute of Science, Bangalore560 012 India.

1. The case in which triangles ABC and A0B0C0 have the same ortho-centre.

We �rst prove that when A0, B0, C0 lie in the interior of the sidesBC; CA;AB, andH = H0, then4ABC is equilateral.

Let triangles ABC and A0B0C0 have H as their common orthocentre.TakingH as the origin, we have

��!HA � ��!HB =

��!HB � ��!HC =

��!HC � ��!HA = T .

Since HA = 2R cosA, HB = 2R cosB and \AHB = � � C, it followsthat T = �4R2 cosA cosB cosC (for all triangles ABC). Further, asBA0 = CB0 = AC0 = x (say), we obtain

��!HA0 =

x��!HC + (a� x)

��!HB

a,

��!HB0 =

x��!HA+ (b� x)

��!HC

b,

and��!HC0 =

x��!HB + (c� x)

��!HA

c.

We also have

��!HA0 � ��!HB0 =

��!HB0 � ��!HC0 =

��!HC0 � ��!HA0 .

509

Here the �rst term��!HA0 � ��!HB0 is equal to

1

ab

�x��!HC + (a� x)

��!HB

���x��!HA+ (b� x)

��!HC

�=

1

ab

�x2T + x(b� x)4R2 cos2C + x(a� x)T + (a� x)(b� x)T

�=

1

ab

�(x2 � bx+ ab)T + x(b� x)4R2 cos2C

�=

1

abx(b� x)(4R2 cos2C + 4R2 cosA cosB cosC) + T

=1

abx(b� x)4R2 cosC � sinA sinB + T

= x(b� x) cosC + T .

Similarly��!HB0 � ��!HC0 = x(c � x) cosA + T , and

��!HC0 � ��!HA0 =

x(a� x) cosB + T . Hence we have

(c� x) cosA = (a� x) cosB = (b� x) cosC . (1)

Because 0 < x < a; 0 < x < b and 0 < x < c, we see thatcosA; cosB and cosC are all of the same sign and hence are all positive.Thus ABC is an acute triangle. If some two sides were equal, say a = b,then we get cosB = cosC from (1) and so b = c. Thus triangle ABC isequilateral. Next, suppose to the contrary that a < b < c. Then

0 < a� x < b� x < c� x

and

0 < cosC < cosB < cosA ,

giving

(a� x) cosB < (c� x) cosA ,

a contradiction.

Finally, suppose a < c < b. Then

0 < a� x < c� x < b� x

and

0 < cosB < cosC < cosA ,

giving

(a� x) cosB < (c� x) cosA ,

once again a contradiction.

By the cyclic symmetry of the relations (1), it is enough to consider thesecases. Hence we conclude that triangle ABC is necessarily equilateral.

510

Remark: If we allow x to exceed one of the sides, say x > a, then it ispossible that triangle ABC be non-equilateral and still triangles ABC andA0B0C0 have a common orthocentre. We proceed as follows:

Eliminating x from the relations (1), we get

(a� b) cosB cosC + (b� c) cosC cosA+ (c� a) cosA cosB = 0 .

Multiplying this by a2b2c2, [setting cosA = (b2+ c2�a2)=2bc, etc.], andremoving the factor (a+ b+ c), we obtain

0 = a5b+ b5c+ c5a+ a2b4 + b2c4 + c2a4

�a4b2 � b4c2 � c4a2 � a3b3 � b3c3 � c3a3

�2abc(a3 + b3 + c3) + 2abc(a2b+ b2c+ c2a)

+abc(ab2 + bc2 + ca2)� 3a2b2c2 .

Setting a = 8, b = 27 and solving the last equation for c (usingMAPLE), we get c = 24:159993. Hence from (1), x = 21:248900. If wetake B = (0; 0) and C = (8; 0) in the xy{plane, we obtain

A = (�5:0809203; 23:619685) ,A0 = (21:248900; 0) ,

B0 = (�2:2946361; 18:588605) ,C0 = (�0:61221170; 2:8459897) ,

and the common orthocentre

H = (�5:0809203; �2:8138865) .

2. The case in which triangles ABC and A0B0C0 have the same ortho-centre.

Editor's comment. The authors prove that again in this case, when A0,B0,C0 lie in the interior of the sidesBC,CA, AB, and I = I0 then4ABCis equilateral. They further provide an example of nonequilateral trianglesfor which the incentre of 4ABC coincides with an excentre of 4A0B0C0.Their treatment relies heavily on the use of MAPLE for their algebraic ma-nipulations. While their proof of this interesting result is certainly valid (andrather clever), computer calculations seem out of place here. Readers whowould like to see the 5-page proof can apply to the solvers.

511

2206. [1997: 46; 1998: 61, 311] Proposed by Heinz-J �urgen Sei�ert,Berlin, Germany.

Let a and b denote distinct positive real numbers.

(a) Show that if 0 < p < 1, p 6= 12, then

1

2

�apb1�p + a1�pbp

�< 4p(1� p)

pab+

�1� 4p(1� p)

� a+ b

2.

(b) Use (a) to deduce P �olya's Inequality:

a� b

log a� log b<

1

3

�2pab+

a+ b

2

�.

Note: \log" is, of course, the natural logarithm.

Comment on part (a) by the proposer, slightly adapted by the editor.

The proposer writes that, in the displayed equation on page 312, theterm x2k+1 should be x2k+2.

2373. [1998: 365] Proposed by Toshio Seimiya, Kawasaki, Japan.

Given triangle ABC with AB > AC. Let M be the mid-point of BC.Suppose thatD is the re ection ofM across the bisector of \BAC, and thatA, B, C and D are concyclic.

Determine the value ofAB � AC

BC.

Solution by Michel Bataille, Rouen, France.

The answer is :AB � AC

BC=

1p2.

Suppose the bisector of \BAC meets the segment BC at K. SinceKC=AC = KB=AB and AB > AC, we have BK > KC and M belongsto segment KB. Hence \KAM < \KAB, so that \KAD < \KAC(by re ection across AK). From this, we see that B and D are on thesame side of the line AC, and, since A, B, C and D are concyclic, we have\CDA = \CBA.

Furthermore \CAD = \BAM , so that4ACD and4AMB are sim-ilar. Therefore we have: AB=AD = AM=AC and, since AD = AM , weobtain

AM2 = AB � AC . (1)

But it is well known that the median AM of4ABC sati�es

4AM2 +BC2 = 2AB2 + 2AC2 . (2)

512

From (1) and (2), we immediately get 2(AB �AC)2 = BC2, and the resultfollows (since AB > AC).

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain;MANUEL BENITO and EMILIO FERNANDEZ, I.B. Praxedes Mateo Sagasta, Logro ~no, Spain;NIKOLAOS DERGIADES, Thessaloniki, Greece; C. FESTRAETS-HAMOIR, Brussels, Belgium;WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAVKONE �CN �Y, Ferris StateUniversity, Big Rapids, MI, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece (2solutions); D.J. SMEENK, Zaltbommel, the Netherlands; PARAYIOU THEOKLITOS, Limassol,Crete; and the proposer.

2374. [1998: 365, 424] Proposed by Toshio Seimiya, Kawasaki,Japan.

Given triangle ABC with \BAC > 60�. Let M be the mid-point ofBC. Let P be any point in the plane of4ABC.

Prove that AP + BP + CP � 2AM .

Solution by Michael Lambrou, University of Crete, Crete, Greece.It is su�cient to prove the stated inequality for the point P for which

PA + PB + PC is a minimum. It is well known that if \A < 120� thenP is the intersection of the concurrent lines AD, BE, CF , where D, E,F are the three vertices of the external equilateral triangles on sides BC,CA, AB, respectively, but if \A � 120�, then P coincides with A (see forexample Courant & Robbins,What isMathematics, the discussionof Steiner'sproblem).

If\A � 120� thenmin(PA+PB+PC) = AB+AC. ButAB+AC >2AM , as is easily seen by completing the parallelogram ofBAC (the medianAM is half the diagonal of this parallelogram and the triangle inequalityapplies). So this case is dealt with and we may assume \A < 120�.

In this case, it is well known and easy to see that AD = BE = CF =(the stated minimum). Thus we are reduced to showing CF � 2AM ; thatis, CF 2 � 4AM2. By the Cosine Rule on4AFC we have

CF 2 = AF 2 + AC2 � 2AF � AC cos(\A+ 60�)

= c2 + b2 � 2bc cos(\A+ 60�) .

But AM is a median of ABC so that

4AM2 = 2b2 + 2c2 � a2 = b2 + c2 + (b2 + c2 � a2)

= b2 + c2 + 2bc cos\A .

In other words we are to show � cos(\A+ 60�) � cos\A. Recall now theproblem has the restriction \A > 60�; that is, \A+ 30� > 90�, and so

cos\A+ cos(\A+ 60�) = 2 cos(\A+ 30�) cos 30�

=p3 cos(\A+ 30�) < 0 ,

513

showing that cos\A < � cos(\A + 60�), as required. This completes theproof. In fact, we have shown the strict inequalityPA+PB+PC > 2AM .

Also solved by MICHEL BATAILLE, Rouen, France; MANUEL BENITO and EMILIOFERNANDEZ, I.B. Praxedes Mateo Sagasta, Logro ~no, Spain; NIKOLAOS DERGIADES, Thes-saloniki, Greece; JUN-HUA HUANG, the Middle School Attached To Hunan Normal Univer-sity, Changsha, China; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAVKONE �CN �Y, Ferris State University, Big Rapids, MI, USA; VEDULAN.MURTY, Dover, PA, USA;VICTOR OXMAN, University of Haifa, Haifa, Israel; D.J. SMEENK, Zaltbommel, the Nether-lands; PANOS E. TSAOUSSOGLOU, Athens, Greece; G. TSINTSIFAS, Thessaloniki, Greece; andthe proposer. There was one partial solution.

Kone �cn �y suggests that readers might be interested in the article, \An Advanced CalculusApproach to Finding the Fermat Point" by Mowa�aq Hajja inMathMagazine 67, no. 1 (1994).

2378. [1998: 425] Proposed by David Doster, Choate RosemaryHall, Wallingford, CT, USA.

Find the exact value of: cot

��

22

�� 4 cos

�3�

22

�.

I. Solution by Nikolaos Dergiades, Thessaloniki, Greece.Let x = �

22and z = cosx+ i sinx; then 11x = �

2and z2 +1 6= 0. We

shall prove that

cotx� 4 cos(3x) =p11 . (1)

We have

z22 = cos(22x) + i sin(22x) = �1 . (De Moivre)

This is equivalent to each of the following:

z22 + 1 = 0 ,

(z2 + 1)(z20 � z18 + z16 � � � � � z2 + 1) = 0 ,

z10 +1

z10�

�z8 +

1

z8

�+

�z6 +

1

z6

��

�z4 +

1

z4

�+

�z2 +

1

z2

�� 1 = 0 ,

2 cos(10x)�2 cos(8x) + 2 cos(6x)�2 cos(4x)+2 cos(2x)�1 = 0 . (2)

Squaring both sides of (1), we get�cosx� 4 sinx � cos(3x)�2 = 11 sin2 x .

This is equivalent to each of the following:

cos2 x� 8 cos x � sinx � cos(3x) + 16 sin2 x � cos2(3x) = 11 sin2 x ,

1+cos(2x)�8 sin(2x) � cos(3x)+8[1�cos(2x)][1+cos(6x)] = 11[1�cos(2x)] ,

�2 + 4 cos(2x)� 4[sin(5x) � sinx] + 8 cos(6x) � 8 cos(2x) � cos(6x) = 0 ,

�2 + 4 cos(2x)� 4[cos(6x)� cos(10x)] + 8 cos(6x)� 4[cos(4x) + cos(8x)] = 0 ,

2 cos(10x) � 2 cos(8x) + 2 cos(6x) � 2 cos(4x) + 2 cos(2x) � 1 = 0 ,

which is true by (2).

514

Editor's Remark. We now have that cotx � 4 cos(3x) = �p11. Theexpression is clearly positive (by calculator, for example). Diminnie and

White make it clearer: since 0 < sin

��

22

�<

�

22<

1

4, it follows that

cot��22

�� 4 cos�3�22

�> 4 cos

��22

�� 4 cos�3�22

�> 0.

II. Solution by Allen Herman, Regina, Canada.

The quadratic Gauss sum for p prime and p � 3 (mod 4) has the value

p�1Xk=0

�k

p

��k = i

pp ,

where � = e2�ip is a primitive pth root of unity and

�kp

�is the Legendre

symbol (equal to 1 when k is a square modulo p, and to �1 otherwise).See, for example, x6.3 of Kenneth Ireland and Michael Rosen, A ClassicalIntroduction to Modern Number Theory, 2nd ed., Springer 1990, p. 75. Inparticular, when p = 11 then � = e

2�i11 and 1, 3, 4, 5, 9 are the squares

modulo 11. Therefore,

� � �2 + �3 + �4 + �5 � �6 � �7 � �8 + �9 � �10 = ip11 .

As in solution I let z = e�i22 . Then z = �i�3, and so (making ample use of

�11 = 1)

cos�

22=

z + z�1

2=

i

2(�8 � �3) , and

sin�

22=

z � z�1

2i= �1

2(�3 + �8) .

Thus

cot�

22= i

��3 � �8

�3 + �8

�= i

��6 � 1

�6 + 1

�= i

�1� 2

1 + �6

�.

Since (�6)11 = 1, we have

(1 + �6)(1� (�6) + (�6)2 � (�6)3 + � � �+ (�6)10) = 2 ,

so that

cot�

22= i(1� (1� �6 + � � �7 + �2 � �8 + �3 � �9 + �4 � �10 + �5))

= i(�� � �2 � �3 � �4 � �5 + �6 + �7 + �8 + �9 + �10) .

515

Finally, cos3�

22=

z3 + z�3

2=

i

2(�9 � �2), and thus

cot�

22� 4 cos

3�

22

= �i((� + �2 + �3 + �4 + �5 � �6 � �7 � �8 � �9 � �10) + 2(�9 � �2))

= �i(� � �2 + �3 + �4 + �5 � �6 � �7 � �8 + �9 � �10)

= �i(ip11)

=p11 .

Also solved by MICHEL BATAILLE, Rouen, France; CHARLES DIMINNIE and LARRYWHITE, San Angelo, TX, USA; C. FESTRAETS-HAMOIR, Brussels, Belgium; FLORIAN HERZIG,student, Cambridge, UK; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; JUN-HUAHUANG, theMiddle School Attached ToHunanNormalUniversity, Changsha, China;MICHAELLAMBROU, University of Crete, Crete, Greece; M. PERISASTRY, Vizianagaram, and VEDULAN. MURTY, Visakhapatnam, India; ARAM TANGBOONDOUANGJIT, Carnegie Mellon Univer-sity, Pittsburgh, PA, USA; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer.

Perisastry and Murty deduce the result as the special casem = 2 of the identity

tan3m�

11+ 4 sin

2m�

11= �p11 ,

where \+" is used form = 1, 3, 4, 5, 9 | the perfect squares modulo 11 as in solution II |and \�" for the non-squares. (Compare the recent proposal 2463* [1999: 366].) An instanceof this identity is problem 218 ( posed by Murty!) in the College Mathematics Journal 14:4(Sept. 1983) 358-359. References there trace it back to the Math. Tripos of 1895. Murty addsthat Murray Klamkin found it as problem #29 in Hobson's Treatise on Plane Trigonometry, 7thed. p. 123. Most solvers used trigonometric identities as in solution I; Doster and Herzig bothnoticed a connection with the squares modulo 11, with Doster using the Gauss sum much as insolution II.

2379. [1998: 425] Proposed by D.J. Smeenk, Zaltbommel, the Neth-erlands.

Suppose thatM1,M2 andM3 are the mid-points of the altitudes fromA to BC, from B to CA and from C to AB in4ABC. Suppose that T1, T2and T3 are the points where the excircles to 4ABC opposite A, B and C,touch BC, CA and AB.

Prove that M1T1, M2T2 andM3T3 are concurrent.

Determine the point of concurrency.

Nearly identical solutions by J.F. Rigby, Cardi�, Wales, and Peter Y.Woo, Biola University, La Miranda, CA, USA.

Let D be the foot of the altitude from A to BC, P be the point wherethe incircle touches BC, and Q be the point of the incircle diametricallyopposite P . There is a dilatation with centre A mapping the incircle and itscentre I to the excircle opposite A and its centre I1. This dilatation mapsQ to T1, so that A;Q, and T1 are collinear. Because M1 is the mid-pointof AD while I is the mid-point of the parallel segment PQ, it follows that

516

M1; I, and T1 are also collinear. Similarly I lies onM2T2 andM3T3, so thethree lines are concurrent at the incentre I.

Also solved by MICHEL BATAILLE, Rouen, France; FRANCISCO BELLOT ROSADO, I.B.Emilio Ferrari, Valladolid, Spain; NIKOLAOS DERGIADES, Thessaloniki, Greece; HIDETOSHIFUKAGAWA, Gifu, Japan; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;MICHAEL LAMBROU, University of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul's School,London, England; G. TSINTSIFAS, Thessaloniki, Greece; PAUL YIU, Florida Atlantic University,Boca Raton, FL, USA; and the proposer.

2386?. [1998: 426] Proposed by Clark Kimberling, University ofEvansville, Evansville, IN, USA.

Write

1 ! 11! 3

1! 4 1

1 3! 6 2 1

1 3 4! 8 1 3 2 1

1 2 3 4 6!

(The last ten numbers shown indicate that up to this point, eight 1's,one 2, three 3's, two 4's and one 6 have been written.)

(a) If this is continued inde�nitely, will 5 eventually appear?

(b) Will every positive integer eventually be written?

Note: 11 is a number and not two 1's.

Solution.All solvers pointed out that 5 appears in the very next iteration. So the

answer to part (a) is trivially \yes". No solver was able to solve part (b), butall seemed to believe the answer here was also \yes". So part (b) remainsopen.

Solved by CHARLES ASHBACHER, Cedar Rapids, IA, USA; RICHARD I. HESS, Ran-cho Palos Verdes, CA, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; andJ.A. McCALLUM, Medicine Hat, Alberta.

2387. [1998: 426] Proposed by Walther Janous, Ursulinengymnas-ium, Innsbruck, Austria.

For �xed p 2 N, consider the power sums

Sp(n) :=

nXk=1

(2k� 1)p; where n � 1 ;

so that Sp(n) is a polynomial in n of degree p+ 1 with rational coe�cients.

Prove that

(a)? If all coe�cients of Sp(n) are integers, then p = 2m � 1 for somem 2 N.

517

(b)? The only values of p yielding such polynomials are p = 1 and p = 3(with S1(n) = n2 and S3(n) = 2n4 � n2).

Solution by Florian Herzig, student, Cambridge, UK.

(a) LetRp(n) =Pn

k=1 kp. To explicitly �nd the coe�cients in the poly-

nomialRp(n), we can use the following approach using generating functions:

R0(n)� R1(n)x+R2(n)x

2

2!� R3(n)x

3

3!+� � � �

=

1Xp=0

(�1)pxpp!

nXk=1

kp =

nXk=1

1Xp=0

(�kx)pp!

= e�x + e�2x + � � �+ e�nx = e�x1� e�nx

1� e�x

=x

ex � 1� 1� e�nx

x. (1)

Now the �rst factor in the last expression has power seriesPi�0Bix

i=i!

where the Bi are the Bernoulli numbers (by de�nition) [see equation 6.81 of[1]].

So the above expression becomes 1Xi=0

Bi

i!xi

! 1Xk=1

(�1)k�1nkxk�1

k!

!.

Comparing the coe�cient of xp on both sides of (1) we get

(�1)pRp(n)

p!=

p+1Xk=1

(�1)k�1nkk!

Bp+1�k(p+ 1� k)!

,

and so

Rp(n) = (�1)pp!p+1Xk=1

1

k!(p+ 1� k)!nk(�1)1�kBp+1�k

=1

p+ 1

p+1Xk=1

�p+ 1

k

�nk(�1)p+1�kBp+1�k . (2)

Therefore from Sp(n) = Rp(2n)� 2pRp(n),

Sp(n) =1

p+ 1

p+1Xk=1

�p+ 1

k

�(2k � 2p)nk(�1)p+1�kBp+1�k . (3)

518

Note that we get a recursion for theBi from (ex�1)P

i�0Bixi=i! = x

by equating coe�cients, namely B0 = 1 andPn�1

k=1

�nk

�Bk = 0 for all n � 2

[see equation 6.79 of [1]]. From this we easily �nd that B1 = �1=2,B2 = 1=6, etc. Since

x

ex � 1+x

2

is an even function of x, it also follows that Bn = 0 for odd n > 1. [See6.84 of [1].]

If we look back at equation 3 we see that the coe�cient of np+1 equals2p=(p+1). Thus if we want Sp(n) to have integer coe�cients, it is necessarythat p+ 1 = 2m for some integer m, as required.

(b) I will show that 1 and 3 are the only positive integers p such thatSp(n) has only integer coe�cients. Assume for the proof that p > 3 is aninteger with this property (p = 2 is clearly not possible, by (a)). The strat-egy is to show that p has to be divisible by arbitrarily large powers of 3. Tosimplify the notation introduce the function v from the non-zero rationalsinto the integers, de�ned by r = 3v(r)q where q is a rational number withnumerator and denominator not divisible by 3 (that is, \v(r) is the exponentof the power of 3 contained in r"). This is clearly well-de�ned and a homo-morphism of the multiplicative group of non-zero rationals into the additivegroup of integers; that is, v(rs) = v(r)+v(s) for all non-zero rationals r, s.We will need a few lemmas now.

Lemma 1 v(B2n) = �1 for all n � 1.

Proof. The proof is by induction on n.

First it is true for n = 1. Assume then that v(B2k) = �1 for all1 � k < n. Then R2n(3) = 1 + 22n + 32n � 2 (mod 3) and

R2n(3) = 3B2n +1

2n+ 1

2n+1Xk=2

�2n+ 1

k

�3k(�1)2n+1�kB2n+1�k

= 3B2n +

2n+1Xk=2

�2n

k� 1

�1

k3k(�1)2n+1�kB2n+1�k

from equation (2). Note now that 3k�1 = (1 + 2)k�1 � 1 + 2(k � 1) > kfor all k > 1, and so v(3k�1=k) � 0 for k > 1 (as the power of 3 in k iscompletely cancelled). Also v(3B2n+1�k) � 0 when k > 1 is odd, by theinduction hypothesis. Thus (using also that B2n+1�k = 0 for even k > 0)the whole sum in the above displayed equation has denominator not divisibleby 3. Thus

1

3(R2n(3) + 1) = B2n +

1

3+ s ,

519

where s is a rational number with denominator not divisible by 3. Since theleft-hand side is an integer and v(1

3+ s) = �1, it follows that v(B2n) = �1

as claimed.

Actually a much more general result is true: the denominator of B2n

is the product of all primes p such that (p � 1)j2n; compare [1], p. 315,problem 54.

Lemma 2 v(2k� 2p) = 0 whenever k � p is odd.

Proof. First v(2k � 2p) = v(1� 2jp�kj). The claim follows because

2n � 1 (mod 3) (==) n � 0 (mod 2) .

Lemma 3 p does not equal a power of 3 (if p is as assumed).

Proof. By (a) we know that p = 2m � 1 for some natural number m.But p > 3 and 3n 6� �1 (mod 8) for all natural numbersn so that the lemmafollows.

We can �nish the proof now. By induction we show that v(p) � afor all natural numbers a. For the induction basis consider the coe�cient ofnp�1 in Sp(n). From (3) it follows that it equals

� =1

p+ 1

�p+ 1

2

�B2(�2p�1) =

p

12(�2p�1) .

So v(�) = v(p)�1 � 0 since we assumed that Sp(n) has integer coe�cients.Hence v(p)� 1 as we wanted to show.

For the induction step suppose that v(p) � a for some a � 1. ByLemma 3 we know that p > 3a. So we can consider the coe�cient of np�3

a

in Sp(n), which is the integer

� =1

p+ 1

�p+ 1

3a + 1

�B3a+1(2

p�3a � 2p)

=p

3a(3a + 1)� (p� 1)(p� 2) � � � (p� 3a + 1)

(3a � 1)(3a � 2) � � � 1 B3a+1(2p�3a � 2p) .

As p is divisible by 3a, the big fraction has v{value 0 (the powers of 3cancel in corresponding terms of the numerator and the denominator). Thus,using Lemmas 1 and 2, v(�) � v(p)�a+0�1+0� 0, whence v(p) � a+1.By induction this completes the proof of (b).

Reference

[1] R.L. Graham, D.E. Knuth, O. Patashnik, Concrete Mathematics, 2nd Ed.,Addison-Wesley, 1994.

Both parts also solved by RICHARD I. HESS, Rancho Palos Verdes, CA, USA; andMICHAEL LAMBROU, University of Crete, Crete, Greece. Part (a) only solved by NIKOLAOSDERGIADES, Thessaloniki, Greece; and HEINZ-J �URGEN SEIFFERT, Berlin, Germany.

520

\Late" solutions to 2388 [1998: 503; 1999: 171; 1999: 445] were re-ceived from MICHEL BATAILLE, Rouen, France and WALTHER JANOUS, Ur-sulinengymnasium, Innsbruck, Austria.

2389. [1998: 503; 1999, 171] Proposedby NikolaosDergiades, Thes-saloniki, Greece.

Suppose that f is continuous on Rn and satis�es the condition thatwhen any two of its variables are replaced by their arithmetic mean, thevalue of the function increases; for example:

f(a1; a2; a3; : : : ; an) � f

�a1 + a3

2; a2;

a1 + a3

2; a4; : : : ; an

�.

Let m =a1 + a2 + : : :+ an

n. Prove that

f(a1; a2; a3; : : : ; an) � f (m;m;m; : : : ;m) .

Solution by Michael Lambrou, University of Crete, Crete, Greece.We show how to construct a sequence of closed nested intervals

J1 � J2 � : : : and a sequence of n{tuples (aN;1; aN;2 : : : ; aN;n) 2 Rn

for N = 0, 1, 2, : : : with (a0;1; a0;2; : : : ; a0;n) = (a1; a2; : : : ; an) such that

(i)

nXi=1

aN;i =

nXi=1

ai,

(ii) aN;i 2 JN for i = 1, 2, : : : , n,

(iii) limN!1

jJN j = 0, where jJ j denotes the length of the interval J , and

(iv) the sequence tN = f(aN;1; aN;2; : : : ; aN;n) is increasing.

Assuming this for the moment, we complete the proof as follows: By the

nested interval theorem the intersection1\

N=1

JN is non-empty, consisting of

a unique number (because of (iii)) which we call m. By (ii) and (iii) we have

limN!1

aN;i ! m for each i = 1, 2, : : : , n and so by (i) we have

nXi=1

ai = nm.

Finally by the continuity of f at (m;m; : : : ;m) 2 Rn we have

tN = f(aN;1; aN;2; : : : ; aN;n) ! f(m;m; : : : ;m) .

Then by (iv) we have

f(a1; a2; : : : ; an) = f(a0;1; a0;2; : : : ; a0;n) � f(m;m; : : : ; m) ,

521

as required.

Now for the details. Set J1 = [a; b] where a = minfa1; a2; : : : ; angand b = maxfa1, a2, : : : , ang. Consider the intervals:

I1 =

�a;

2a+ b

3

�, I2 =

�2a+ b

3;a+ 2b

3

�, I3 =

�a+ 2b

3; b

�.

As long as each of I1, I3 contains at least one of the ak's, say ai 2 I1 andaj 2 I3, repeat the following procedure: Replace ai, aj by their averagex = 1

2(ai + aj). Notice that by assumption the value of f increases since

f(a1; a2; : : : ; an) � f(a1; : : : ; x; : : : ; x; : : : ; an) ,

and that the sum

a1 + a2 + � � �+ an

= a1 + � � �+ ai�1 + x+ ai+1 + � � �+ aj�1 + x+ aj+1 + � � �+ an (1)

does not change value. Also, as a � ai <2a+ b

3,a+ 2b

3< aj � b we have

2a+ b

3<

ai + aj

2<

a+ 2b

3.

In other words x 2 I2. After a �nite number of repetitions (at most 12n) of

this replacing procedure, at least one of I1, I3 will not contain any of the ak'sor their replacements (they move to I2). Thus we end up with numbers a1;1,a1;2, : : : , a1;n, none of which is in I1 or none of which is in I3. That is, theyare all either in I2[I3 or in I1[I2. Call J2 that one of I2[I3 or I1[I2 thatcontains all the ai;k's. Observe that (i) holds (because of (1)), that J1 � J2and that jJ2j = 2

3(b� a) = 2

3jJ1j.

Repeating the whole process with J2 in place of J1, etc. we �nd grad-

ually J1 � J2 � J3 � : : : such that jJN+1j = 23jJN j =

�23

�N(b� a) ! 0

such that conditions (i), (ii), (iii), and (iv) are satis�ed. This completes theconstruction and the proof.

Also solved by WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MARKLYON and MAX SHKARAYEV, students, University of Arizona, Tuscon, AZ, USA; JEREMYYOUNG, student, Nottingham High School, Nottingham, UK; and the proposer.

2390. [1998: 504] Proposed by Walther Janous, Ursulinengymnas-ium, Innsbruck, Austria.

For � � 0 and p; q � 1, let Sn(�; p; q) :=

n��Xi=1

nXj=i+�

ip jq, where

n > �.

Given the statement: \Sn(�; p; q), understood as a polynomial in Q[n],is always divisible by (n� �)(n� �+ 1)(n� �+ 2)",

522

(a) give examples for � = 0, 1, 2, 3, 4;

(b)? prove the statement in general.

Solution to both parts by G.P. Henderson, Garden Hill, Campbellcroft,Ontario.

We will show that the statement is true for 0 � � < n, p � 1, andq � 0.

First consider the case � = 0. We have

Sn(0; p; q) =

nXi=1

nXj=i

ipjq =

nXj=1

jXi=1

ipjq .

Using Newton's interpolation formula we can write

ip =

pXr=1

apr

�i

r

�, p � 1 ,

where the a's are independent of i. Then

Sn(0; p; q) =

nXj=1

jqpX

r=1

apr

jXi=1

�i

r

�.

The inner sum is

jXi=1

��i+ 1

r + 1

���

i

r + 1

��=

�j + 1

r+ 1

�,

and so

Sn(0; p; q) =

nXj=1

pXr=1

aprjq

�j + 1

r + 1

�. (1)

The following is easily proved by induction on q.

Lemma. If P (j) is a polynomial in j with 0 � deg(P ) � q then P canbe written as a linear combination of the polynomials�

j + s+ 1

s

�, s = 0, 1, : : : , q.

[Or just note that these polynomials are linearly independent and thus mustform a basis of the vector space Q[j]. {Ed.] In particular, there existconstants bqs, s = 0, 1, : : : , q, such that

jq =

qXs=0

bqs

�j + s+ 1

s

�, q � 0 .

523

Using this in (1),

Sn(0; p; q) =

pXr=1

qXs=0

aprbqs

nXj=1

�j + s+ 1

s

��j + 1

r+ 1

�.

The inner sum is

nXj=1

�j + s+ 1

r + s+ 1

��r + s+ 1

s

�

=

�r + s+ 1

s

� nXj=1

��j + s+ 2

r + s+ 2

���j + s+ 1

r + s+ 2

��

=

�r + s+ 1

s

��n+ s+ 2

r + s+ 2

�,

and so

Sn(0; p; q) =

pXr=1

qXs=0

aprbqs

�r + s+ 1

s

��n+ s+ 2

r+ s+ 2

�.

The polynomial�n+s+2r+s+2

�is

(n+ s+ 2)(n+ s+ 1) : : : (n� r + 1)

(r + s+ 2)!.

Since s � 0 and r � 1, Sn(0; p; q) is divisible by n(n + 1)(n + 2). This�nishes the case � = 0.

In the general case, if � < n, then

Sn(�; p; q) =

n��Xi=1

nXj=i+�

ipjq =

n��Xi=1

n��Xk=i

ip(k+ �)q

where j = k + �. Thus

Sn(�; p; q) =

n��Xi=1

n��Xk=i

qXt=0

�q

t

��q�tipkt =

qXt=0

�q

t

��q�tSn��(0; p; t) .

Since Sn��(0; p; t) is divisible by (n��)(n��+1)(n��+2), this completesthe proof.

Also solved by MANUEL BENITO and EMILIO FERNANDEZ, I.B. Praxedes MateoSagasta, Logro ~no, Spain; andMICHAEL LAMBROU, University of Crete, Crete, Greece. Part (a)only solved by RICHARD I. HESS, Rancho Palos Verdes, CA, USA; and the proposer.

For part (a), here is Sn(�; 1; 1), listed by almost all solvers:

Sn(�; 1; 1) =1

24(n� �)(n� �+ 1)(n� �+ 2)(3n + �+ 1) .

524

2391. [1998: 504] Proposed by G. Tsintsifas, Thessaloniki, Greece.Consider d + 1 points, B1, B2, : : : , Bd+1 in the unit sphere in Rd,

so that the simplex Sd(B) = B1B2 : : : Bd+1 includes the origin O.Let P = fx j Bi � x � 1g for all i between 1 and d+ 1.

Prove that there is a point y 2 P such that jyj � d.

Solution by the proposer.Let �i be the tangent plane to the sphere (0,1) at the point Bi and we

denote Ai = \d+1j=1;j 6=i�j . Therefore (0,1) is the inscribed sphere in the

simplex Sd(A) = A1A2 : : : Ad+1 and denoting����!0Ai

��� = xi we will have:

x1 + 1 � h1 ,

x1a1 + a1 � h1a1 ,

where h1, a1 we denote the altitude from A1 and the volume of the facetopposite A1. But h1 � a1 = d � Vol(Sd(A)) or

x1a1 + a1 � a1 + a2 + : : :+ ad+1 or

x1 � a2

a1+a3

a1+ : : :+

ad+1

a1. (1)

From the remaining vertices, we have d inequalities like (1). Adding, we take

d+1Xi=1

xi �1;d+1Xi<j

�ai

aj+aj

ai

�� 2d(d+ 1)

2.

Therefore maxxi = jyj � d.

There were no other solutions submitted.

2394. [1998: 505] Proposed by Vedula N. Murty, Visakhapatnam,India.

The inequality aabb ��a+ b

2

�a+b, where a, b > 0, is usually proved

using Calculus. Give a proof without the aid of Calculus.

I. Solution by Joe Howard, New Mexico Highlands University, Las Ve-gas, NM, USA.By the weighted Geometric-Harmonic Mean Inequality we have

a+bpaabb � a+ b

aa+ b

b

=a+ b

2.

Therefore,

aabb ��a+ b

2

�a+b.

525

II. Solutionby John G. Heuver, Grande Prairie Composite High School,Grande Prairie, Alberta.By the Arithmetic-Geometric Mean Inequality we have�1 +

b

a

�a �1 +

a

b

�b��

a

a+ b

�1 +

b

a

�+

b

a+ b

�1 +

a

b

��a+b= 2a+b .

This implies that �a+ b

2

�a+b� aabb .

Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia andHerzegovina; NIELS BEJLEGAARD, Stavanger, Norway; NIKOLAOS DERGIADES, Thessaloniki,Greece; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y,Ferris State University, Big Rapids, MI, USA; MICHAEL LAMBROU, University of Crete,Crete, Greece; EFSTRATIOS RAPPOS, University of Cambridge, Cambridge, UK; PANOS E.TSAOUSSOGLOU, Athens, Greece; JEREMY YOUNG, student, Nottingham High School,Nottingham, UK; and the proposer. There were also two solutions which used Calculus (Taylorseries).

The proposer also notes that the generalization is true:

nYi=1

aaii �

�a1 + a2 + � � �+ an

n

�a1+a2+���+an

,

where ai > 0, i = 1, 2, : : : , n. Indeed, Kone �cn �y actually proves this result; his proof is basedon a theorem on p. 141 of the Czech book,Metody �re�sen�imatematick �ych �uloh I, by Ji�r�i Herman,Radan Ku�cera, Jarom�ir �Sim�sa, and published by Masaryk University in 1996. (As an applicationof this theorem they actually prove the above generalization of our result for n = 3.)

2395. [1998: 505] Proposed by Witold Janicki, Jagiellonian Univer-sity, Krakow, Poland,Michael Sheard, St. Lawrence University, Canton, NY,USA, Dan Velleman, Amherst College, Amherst, MA, USA, and Stan Wagon,Macalester College, St. Paul, MN, USA.

Let P be such that

(A) P (0) is true, and

(B) P (n) =) P (n+ 1).

Find an integer n > 106 such that P (n) can be proved without usinginduction, but rather using

(L) the Law of Implication (that is, X and (X =) Y ) yield Y )

ten times only.

Solution by E.B. Davies, King's College, London, and MichaelLambrou, University of Crete, Crete, Greece.

We can considerably improve the outcome.

We show that with ten uses of the Law of Implication (L) we can showthat P (n) is true for all n � 265536 � 2 � 1019728. This will follow by

526

showing the validity of S(n) for n = 4 and k = m = n = 0, where S(n),n 2 N, is the statement

S(n) : 8k�8m

�8n�P (0)^ P (1)^ � � � ^ P (n)

=) P (0)^ P (1)^ � � � ^ P�2k � 22m � 222

N

+ n

����. (1)

For this purpose de�ne Q(m),m 2 N, to be the statement

Q(m) : 8n �P (0)^ P (1)^ � � � ^ P (n)=) P (0)^ P (1)^ � � � ^ P (2m + n)

�. (2)

We show Q(0) with one use of (L) by arguing

P (0)^ � � � ^ P (n)P (n) =) P (n+ 1)

P (0)^ � � � ^ P (n) =) P (0)^ � � � ^ P (n+ 1)

By a second use of (L) we show that 8m, (Q(m) =) Q(m+ 1)). Indeed, ifQ(m) is true, we have for all n:

P (0)^ � � � ^ P (n) =) P (0) ^ � � � ^ P (2m + n) (this is (2))

P (0)^ � � � ^ P (2m + n) =) P (0)^ � � � ^ P (2m+1 + n)

( (2) with 2m + n in place of n)

P (0)^ � � � ^ P (n) =) P (0) ^ � � � ^ P (2m+1 + n)

as claimed. In particular, we have shown the theorem (call it Theorem 1)that if P satis�es P (0) and 8n (P (n) =) P (n+ 1)), then Q given by (2)satis�es Q(0) and 8n (Q(n) =) Q(n+ 1)).

De�ne R(k), k 2 N and S(N), N 2 N by

R(k) : 8n �Q(0) ^ � � � ^ Q(n) =) Q(0) ^ � � � ^ Q �2k + n

��(3)

S(N) : 8n �R(0) ^ � � � ^R(n) =) R(0) ^ � � � ^R �2N + n

��. (4)

Note that by one use of (L) (see Remarks), Theorem 1 with Q, R inplace of P , Q gives R(0) and 8k (R(k) =) R(k+ 1)). Similarly, one moreuse of Theorem 1 gives S(0) and 8N (S(N) =) S(N + 1)). Two uses of(L) give us consecutively, 8m,

S(m) =) S(m+ 1) S(m) =) S(m+ 2)

S(m+ 1) =) S(m+ 2) and so S(m+ 2) =) S(m+ n)

S(m) =) S(m+ 2) S(m) =) S(m+ 4) .

By a seventh use of (L), we have

527

S(0)

S(m) =) S(m+ 4) (applied to m = 0)

S(4)

Going back, we have, by the de�nition of S(4) and an eighth use of (L),that

R(0)

R(0) =) R(0) ^ � � � ^R(24)R(24)

Thus R(16) is true. By the de�nition of R(16) we have

Q(0)

Q(0) =) Q(0) ^ � � � ^ Q(216)Q(216)

That is, Q(65536) is true. Finally, a tenth use of (L) gives

P (0)

P (0) =) P (0) ^ � � � ^ P (265536)P (0) ^ � � � ^ P (265536)

showing that P (n) is true for all n � 265536.

Remarks:

1. It is easy to see that S(n) given in (4) can be re-written as the one givenin (1).

2. We deduced R(0) and 8k (R(k) =) R(k + 1)) by using Theorem 1 andcounting one use of (L). By invoking a proof schema called \law of generaliza-tion" we could just as legitimately count this as \zero" uses of (L). However,some people might �nd this unsatisfactory and argue that since Theorem 1used the Law of Implication twice, each use of Theorem 1 must also countas two uses of (L). If such is the case, here is how we modify our argument:After proving Q(0) and 8n (Q(n) =) Q(n+ 1)), two uses of (L) give R(0)and 8k (R(k) =) R(k + 1)). Three more uses give R(k) =) (R(k + 2),R(k) =) R(k + 4), R(k) =) R(k + 8) (k 2 N). The eighth use is

R(0)

R(k) =) R(k + 8)

R(8)

from which, as before, we get Q(28) = Q(512), and �nally P (0), P (1), : : : ,P (2512) are true.

528

Also solved by NIKOLAOS DERGIADES, Thessaloniki, Greece; and the proposers.

Dergiades showed that P (n) can be proved for n = 1953125. The proposers gaveshorter and longer versions of the proof of P (134217728).

2396. [1998: 505] Proposed by Jose Luis Diaz, Universitat Politec-nica de Catalunya, Colum, Terrassa, Spain.

Suppose that A(z) =

nXk=0

akzk is a complex polynomial with an = 1,

and let r = max0�k�n�1

njakj1=(n�k)

o. Prove that all the zeros of A lie in the

disk C =

�z 2 C : jzj � r

21=n � 1

�.

I. Solution by Michael Lambrou, University of Crete, Crete, Greece.Let z0 denote a zero of A(z): Note that for 0 � k � n � 1 we have

jakj � rn�k so that

2jz0jn = jz0jn + jz0jn= j � a0 � a1z0 � a2z

20 � � � � an�1z

n�10 j+ jz0jn

� ja0j+ ja1j jz0j+ ja2j jz0j2 + � � � + jan�1j jz0jn�1 + jz0jn

� rn + rn�1jz0j+ rn�2jz0j2 + � � � + rjz0jn�1 + jz0jn

� rn +�n1

�rn�1jz0j+

�n2

�rn�2jz0j2 + � � �+ � n

n�1�rjz0jn�1 + jz0jn

=�r + jz0j

�n.

Extracting nth roots, we have 21=njz0j � r + jz0j, from which the requiredresult follows.

II. Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Aus-tria (slightly modi�ed by the editor.)

We shall prove the stronger result that, in fact, all the zeros of A(z) liein the disk D = fz 2 C : jzj � �n rg where �n denotes the uniquepositive solution of xn+1 � 2xn + 1 = 0 which lies in the interval (1; 1)when n � 2 and �1 = 1. The case n = 1 is trivial since A(z) = a0 + z,the only root of which is z = �a0; and r = ja0j: If r = 0; then A(z) = zn,which has z = 0 as the only root (with multiplicity n) and the conclusionclearly holds. Hence we assume that n � 2 and r > 0:

We �rst show that the function f(x) = xn+1 � 2xn + 1 has a uniquezero �n in (1; 1): Since f 0(x) = xn�1

�(n+ 1)x� 2n

�; f 0(x) = 0 if and

only if x = x0 =2n

n+ 1= 2 � 2

n+ 12 (1; 2). Since f is decreasing on

(1; x0) and increasing on (x0; 1) we see that f has a relative and absoluteminimum over (1; 1) at x = x0: Since f(1) = 0 and f(2) = 1 we concludethat there is a unique �n in (1; 1) such that f(�n) = 0:

529

Now let z0 be a zero of A(z): If jz0j � r; then z0 is clearly inside D.Thus we assume that jz0j > r and show that jz0j � �n r.

Using the triangle inequality, we have

0 = jA(z0)j =

�����zn0 +

n�1Xk=0

ak zk0

������ jz0jn �

n�1Xk=0

jakj jz0jk � jz0jn �n�1Xk=0

rn�kjz0jk

= jz0jn � r

� jz0jn � rn

jz0j � r

�.

Multiplying both sides by jz0j�r, we get jz0jn+1�2rjz0jn+rn+1 � 0.

Thus

� jz0jr

�n+1

� 2

� jz0jr

�n+ 1 � 0; that is, f

� jz0jr

�� 0.

Hencejz0jr

� �n from which jz0j � �nr follows.

Also solved by NIKOLAOS DERGIADES, Thessaloniki, Greece; KEITH EKBLAW, WallaWalla, WA, USA; JUN-HUAHUANG, the Middle School Attached To Hunan Normal University,Changsha, China; KEE-WAI LAU, Hong Kong; GERRY LEVERSHA, St. Paul's School, London,England; HEINZ-J �URGEN SEIFFERT, Berlin, Germany and the proposer.

Using similar arguments, Lambrou also obtained the stronger result given by Janous.He pointed out that as a consequence, all the zeros of A(z) lie in the disk D0 = fz 2 C :

jzj � 2rg and commented that the upper bound 2r is considerably better thanr

21=n � 1for

large values of n since1

21=n � 1! 1 as n ! 1. He also gave an example to show that

the bound �nr is the best possible. A similar comment about the \crudeness" of the boundr

21=n � 1was also given by Leversha.

2397. [1998: 505] Proposed by Toshio Seimiya, Kawasaki, Japan.Given a right-angled triangle ABC with \BAC = 90�. Let I be the

incentre, and let D and E be the intersections of BI and CI with AC andAB respectively.

Prove thatBI2 + ID2

CI2 + IE2=

AB2

AC2.

Nearly identical solutions by Miguel Amengual Covas, Cala Figuera,Mallorca, Spain; Michel Bataille, Rouen, France; Gottfried Perz, Pestalozzi-gymnasium, Graz, Austria; Iftimie Simion, Stuyvesant HS, New York, NY,USA; and by D.J. Smeenk, Zaltbommel, the Netherlands.

Let r be the inradius of4ABC. By the relevant de�nitions,

sinB

2=

r

BI, cos

B

2=

r

ID, sin

C

2=

r

CI, and cos

C

2=

r

IE.

530

Hence,

BI2 + ID2

CI2 + IE2=

�r

sin(B2 )

�2+

�r

cos(B2 )

�2�

r

sin(C2 )

�2+

�r

cos(C2 )

�2 =sin2 C

2� cos2 C

2

sin2 B2� cos2 B

2

=sin2C

sin2B=AB2

AC2,

as desired.

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain;NIKOLAOS DERGIADES, Thessaloniki, Greece; RICHARD I. HESS, Rancho Palos Verdes, CA,USA; JUN-HUA HUANG, the Middle School Attached To Hunan Normal University, Chang-sha, China; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; GEOFFREY A.KANDALL, Hamden, CT, USA; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, MI,USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul'sSchool, London, England; ANDREI SIMION, student, Brooklyn Tech. HS, NY, USA; HEINZ-J �URGEN SEIFFERT, Berlin, Germany; ECKARD SPECHT, Magdeburg, Germany; PANOS E.TSAOUSSOGLOU, Athens, Greece; G. TSINTSIFAS, Thessaloniki, Greece; PETER Y. WOO, BiolaUniversity, La Miranda, CA, USA; PAUL YIU, Florida Atlantic University, Boca Raton, FL, USA;JEREMY YOUNG, student, Nottingham High School, Nottingham, UK; and the proposer.

2398. [1998: 505] Proposed by Toshio Seimiya, Kawasaki, Japan.Given a square ABCD with points E and F on sides BC and CD

respectively, let P and Q be the feet of the perpendiculars from C to AE

and AF respectively. Suppose thatCP

AE+CQ

AF= 1.

Prove that \EAF = 45�.

The solution is a combination of the solutions by Miguel AmengualCovas, Cala Figuera, Mallorca, Spain and Paul Yiu, Florida Atlantic Univer-sity, Boca Raton, FL, USA.

Suppose that the square has unit side length. Let \CAP = � and\CAQ = �. Then 0 � �, � � 45� and

CP

AE=p2 sin� cos(45� � �) =

1

2+

1p2sin(2� � 45�) ,

since sin� cos� = 12(sin(�+ �) + sin(�� �)): Similarly,

CQ

AF=

1

2+

1p2sin(2�� 45�) .

It follows that CPAE

+ CQAF

= 1 if and only if sin(2��45�)+sin(2��45�) = 0.This is possible only when 2� � 45� = 45� � 2�; that is, � + � = 45�.

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain;NIKOLAOS DERGIADES, Thessaloniki, Greece; RICHARD I. HESS, Rancho Palos Verdes, CA,

531

USA; JUN-HUA HUANG, the Middle School Attached To Hunan Normal University, Chang-sha, China; �ANGEL JOVAL ROQUET, Instituto Espa ~nol de Andorra, Andorra; GEOFFREY A.KANDALL, Hamden, CT, USA; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, MI,USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul'sSchool, London, England; HEINZ-J �URGEN SEIFFERT, Berlin, Germany; D.J. SMEENK, Zalt-bommel, the Netherlands; PANOS E. TSAOUSSOGLOU, Athens, Greece; G. TSINTSIFAS, Thes-saloniki, Greece; PETER Y. WOO, Biola University, La Miranda, CA, USA; JEREMY YOUNG,student, Nottingham High School, Nottingham, UK; and the proposer.

2399?. [1998: 506] Proposed by David Singmaster, South BankUniversity, London, England.

In James Dodson's The Mathematical Repository, 2nd ed., J. Nourse,London, 1775, pp 19 and 31, are two variations on the classic \Ass andMule"problem:

\What fraction is that, to the numerator of which 1 be added,the value will be 1=3; but if 1 be added to the denominator, itsvalue is 1=4?"

This is easily done and it is easy to generalize to �nding x=y such that(x + 1)=y = a=b and x=(y + 1) = c=d, giving x = c(a + b)=(ad � bc)and y = b(c+ d)=(ad � bc). We would normally take a=b > c=d, so thatad � bc > 0, and we can also assume a=b and c=d are in lowest terms.

\A butcher being asked, what number of calves and sheep hehad bought, replied, `If I had bought four more of each, I shouldhave four sheep for every three calves; and if I had bought fourless of each, I should have had three sheep for every two calves'.How many of each did he buy?"

That is, �nd x=y such that (x + 4)=(y + 4) = 4=3 and (x � 4)=(y � 4) =3=2. Again, this is easily done and it is easy to solve the generalisation,(x + A)=(y + A) = a=b and (x � A)=(y � A) = c=d, gettingx = A(2ac � bc � ad)=(bc � ad) and y = A(ad + bc � 2bd)=(bc� ad).We would normally take a=b < c=d so that bc � ad > 0, and we can alsoassume a=b and c=d are in lowest terms.

In either problem, given that a, b, c and d are integers, is there a con-dition (simpler than computing x and y) to ensure that x and y are integers?

Alternatively, is there a way to generate all the integer quadruples a,b, c, d, which produce integers x and y?

To date, no solutions have been received.

532

2400. [1998: 506] Proposed by V�aclav Kone �cn �y, Ferris State Uni-versity, Big Rapids, MI, USA.

(a) Show that 1 + (� � 2)x <cos(�x)

1� 2x< 1 + 2x for 0 < x < 1=2.

[Proposed by Bruce Shawyer, Editor-in-Chief.]

(b)? Show thatcos(�x)

1� 2x<

�

2� 2(�� 2)

�x� 1

2

�2for 0 < x < 1=2.

Solution by Kee-Wai Lau, Hong Kong (modi�ed and expanded by theeditor).

(a) [Ed: We present only the proof of the left inequality since, as pointedout by Sei�ert, the right inequality was already established by Herzig in hissolution to #2296 [1998: 533].]

It su�ces to prove that for 0 < x < 12,

1 + (� � 4)x� 2(� � 2)x2 < cos(�x) (1)

Let f(x) = cos(�x)+2(��2)x2� (��4)x�1, for 0 � x � 12. Then

f 0(x) = �� sin(�x)+4(��2)x+4�� and f 00(x) = ��2 cos(�x)+4(��2).Setting f 00(x) = 0, we �nd that x = x0 =

1

�cos�1

�4(� � 2)

�2

�= 0:346 � � �.

Since f 00(x) < 0 for 0 < x < x0, the graph of f is concave down on [0; x0].Since f(0) = 0 and f(x0) = 0:035 � � � > 0, we conclude that f(x) > 0on (0; x0]. On the other hand, since f 00(x) > 0 for x0 < x < 1

2, f 0(x)

is increasing on [x0;12]. This, together with the fact that f 0(1

2) = 0, imply

that f 0(x) < 0 for x0 < x < 12. Hence f(x) is decreasing on [x0;

12]. Since

f(12) = 0 we conclude that f(x) > 0 on (x0;

12) as well and (1) follows.

(b) Let y = �2(1 � 2x). Then 0 < y < �

2and the given inequality

becomes� cos(�

2� y)

2y<

�

2� 2(� � 2)

4

�2y

�

�2or

siny < y � 4(� � 2)

�3y3 . (2)

Let g(y) = siny� y+4(� � 2)

�3y3, for 0 � y � �

2. We show that g(y)< 0

for 0 < y < �2. Note �rst that

g(y) <

�y � y3

6+

y5

120

�� y +

4(� � 2)

�3y3

=y3

120�3(�3y3 + 480� � 20�3 � 960):

533

Straightforward computations show that �3y3 + 480� � 20�3 � 960 < 0when y = 1:52. Hence g(y)< 0 for 0 < y � 1:52. Next,

g0(y) = cos y� 1 +12(� � 2)

�3y2

= 2y2

6(� � 2)

�3� sin2(y

2)

y2

!:

Note thatsiny

yis decreasing for 0 < y < �

2.

[Ed:d

dy

�siny

y

�= y�2(y cosy � siny) and y cosy < siny as it is well

known that cos y <siny

y].

If 1:52 � y < �2, then 0:76 � y

2< �

4and so

sin2(y2)

y2=

1

4

sin(y

2)

y2

!2

<1

4

�sin(0:76)

0:76

�2

<(0:907)

2

4= 0:2057 <

6(� � 2)

�3.

Hence g0(y) > 0 for 1:52 � y < �2which implies that g(y) is increasing on

[1:52; �2]. Since g(�

2) = 0 we conclude that g(y) < 0 for 1:52 � y < �

2as

well. Therefore, g(y)< 0 on (0; �2) and (2) follows.

Also solved by OSCAR CIAURRI, Logro ~no, Spain; WALTHER JANOUS, Ursulinengym-nasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete, Greece; PHILMCCARTNEY, Northern Kentucky University, Highland Heights, KY, USA; HEINZ-J �URGENSEIFFERT, Berlin, Germany; and the proposer who apparently found a proof of (b)� after theproblem had appeared. There was also one incorrect and one incomplete solution.

Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell

Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil

Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,

J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

534

YEAR END FINALE

Again, a year has own by! It is di�cult to realize that I have nowdone this job for four years. My term has one year to go, but I have agreedto a two year extension until the end of 2002, when you will have a newEditor-in-Chief.

There are many people that I wish to thank most sincerely for par-ticular contributions. Again, �rst and foremost is BILL SANDS. Bill is ofsuch value to me and to the continuance of CRUX with MAYHEM. Aswell, I thank most sincerely, CATHY BAKER, ILIYA BLUSKOV, ROLANDEDDY, CHRIS FISHER, BILL SANDS, JIM TOTTEN, and EDWARD WANG,for their regular yeoman service in assessing the solutions; DENIS HANSON,DOUG FARENICK, CHRIS FISHER, CHRIS LEIER, TOMMacDONALD, JUDIMcDONALD, RICHARD McINTOSH, DIETER RUOFF, NABIL SHALABY,JASON STEIN, MICHAEL TSATSOMEROS, BRUCE WATSON, HARLEYWESTON, for ensuring that we have quality articles; ALAN LAW, RICHARDCHARRON, MURRAY S. KLAMKIN, T.W. LEUNG, ANDY LIU, JACK W.MACKI, CHRISTOPHER SMALL, IAN VANDERBURGH, for ensuring that wehave quality book reviews, ROBERT WOODROW, who carries the heavy loadof two corners, and RICHARD GUY for sage advice whenever necessary.

The editors of the MAYHEM section, NAOKI SATO, CYRUS HSIA,ADRIAN CHAN, DONNY CHEUNG, JIMMY CHUI, DAVID SAVITT and WAILING YEE, all do a sterling job.

I also thank those who assist with proofreading. The quality of all thesepeople is a vital part of what makes CRUX with MAYHEM what it is. Thankyou one and all.

As well, I would like to give special thanks to our Associate Editor,CLAYTON HALFYARD, for continuous sage advice, and for keeping me fromprinting too many typographical and mathematical errors; and to my col-leagues, YURI BAHTURIN, RICHARD CHARRON, ROLAND EDDY, EDGARGOODAIRE, MIKE PARMENTER, DONALD RIDEOUT, NABIL SHALABY, inthe Department of Mathematics and Statistics at Memorial University, andto JOHN GRANT McLOUGHLIN, Faculty of Education, Memorial University,for their occasional sage advice. I have also been helped by some Memo-rial University students, KARELYN DAVIS, PAUL MARSHALL, SHANNONSULLIVAN, TREVOR RODGERS, as well as WISE Summer students, JANINERYDER, DENISE VATCHER and REBECCA WHITE.

The staff of the Department of Mathematics and Statistics at Memor-ial University deserve special mention for their excellent work and support:ROS ENGLISH, MENIE KAVANAGH, WANDA HEATH, and LEONCEMORRISSEY; as well as the computer and networking expertise of STEVEPOPE and CRAIG SQUIRES.

535

Also the LATEX expertise of JOANNE LONGWORTH at the University ofCalgary, ELLEN WILSON at Mount Allison University, the MAYHEM sta�,and all others who produce material, is much appreciated.

GRAHAM WRIGHT, the Managing Editor for the �rst four issues, wasa tower of strength and support. Graham kept so much on the right track.He has been a pleasure to work with. We welcome BOB QUACKENBUSH asthe new Managing Editor. The CMS's TEX Editor, MICHAEL DOOB has beenvery helpful in ensuring that the printed master copies are up to the standardrequired for the U of T Press, who continue to print a �ne product.

The online version of CRUX with MAYHEM continues to attract atten-tion. We recommend it highly to you. Thanks are due to LOKI JORGENSON,JEN CHANG, CONG LY, FREDERIC TESSIER, PAUL WOLSTENHOLME, andthe rest of the team at SFU who are responsible for this.

Finally, I would like to express real and heartfelt thanks to the Head ofmy Department, HERBERT GASKILL, to the former Acting Dean of Science ofMemorial University, WILLIE DAVIDSON, and to the new Dean of Science,BOB LUCAS. Without their support and understanding, I would not be ableto do the job of Editor-in-Chief.

Last but not least, I send my thanks to you, the readers of CRUX with

MAYHEM. Without you, CRUX with MAYHEM would not be what it is.Keep those contributions and letters coming in. We need your ARTICLES,PROPOSALS and SOLUTIONS to keep CRUX with MAYHEM alive and well.I do enjoy knowing you all.

CRUX MAYHEM

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536

INDEX TO VOLUME 25, 1999

Crux Articles

Generalisations of a Four-Square Theorem

Hiroshi Okumura and John F. Rigby .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 20

Euler's Triangle Theorem

G.C. Shephard . .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. 148

Apropos Bell and Stirling Numbers

Antal E. Fekete . .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... . 274

The Academy Corner Bruce Shawyer

February No. 22 APICS Mathematics Competition 1998 .... .. ... .. .. ... 1

March No. 23 Canadian Undergraduate Mathematics

Conference 1998 | Part 1 .... .. ... .. .. .. ... .. .. ... . 65

April No. 24 Canadian Undergraduate Mathematics

Conference 1998 | Part 2 .... .. ... .. .. ... .. .. .. ... 129

May No. 25 Canadian Undergraduate Mathematics

Conference 1998 | Part 3 .... .. ... .. .. ... .. .. .. ... 193

September No. 26 Canadian Undergraduate Mathematics

Conference 1998 | Part 4 .... .. ... .. .. ... .. .. .. ... 257

October No. 26 The Bernoulli Trials 1998 Questions

Christopher Small and Ravindra Maharaj .. ... .. .. 321

November No. 27 A Trial Balloon

Vedula N. Murty .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. . 385

December No. 28 The Bernoulli Trials 1998 Answers

Christopher Small and Ravindra Maharaj .. ... .. .. 449

The APICS 1999 Mathematics Competition ...... .. .. ... .. .. .. ... .. .. ... .. . 452

The Olympiad Corner R.E. Woodrow

February No. 195 ...... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. 3

March No. 196 ...... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. .. 67

April No. 197 ...... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. . 132

May No. 198 .... ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. 196

September No. 199 ..... .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. . 261

October No. 200 ..... .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... . 325

November No. 201 .... .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. 387

December No. 202 ..... .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. . 454

The Skoliad Corner R.E. Woodrow

February No. 35 ..... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. . 25

March No. 36 ..... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. . 88

April No. 37 .... ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... 154

May No. 38 ..... ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 214

September No. 39 .... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 282

October No. 40 .... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. . 341

November No. 41 ..... .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. 403

December No. 42 .... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 472

537

Book Reviews Alan Law

Over and Over Again

by Genghe Chang and Thomas Sederberg

Reviewed by Murray S. Klamkin . .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. . 18

A Primer of Real Functions

by Ralph Boas Jr.

Reviewed by Murray Klamkin .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 86

Mathematically Speaking: A Dictionary of Quotations

selected and arranged by Carl C. Gaither and Alma E. Cavazos-Gaither

Reviewed by Bruce Shawyer ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. . 86

Interdisciplinary Lively Application Projects

edited by David C. Amey

Reviewed by T.W. Leung .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... . 146

ElementaryMathematicalModels

by Dan Kalman

Reviewed by Richard Charron .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. . 212

Principles of Mathematical Problem Solving

by Martin J. Erickson and Joe Flowers

Reviewed by Christopher Small . .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... 213

In Polya's Footsteps

by Ross Honsberger

Reviewed by Murray S. Klamkin .. .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. . 270

Problem-Solving Strategies for E�cient and Elegant Solutions

by Alfred S. Posamentier and Stephen Krulik

Reviewed by Ian VanderBurgh . .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... . 396

The art of problem solving, Volume 2

by S. Lehoczky and R. Rusczyk

Reviewed by Andy Liu .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. 475

Calculus, The Dynamics of Change

by A. Wayne Roberts

Reviewed by Jack W. Macki . .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. 401

Which way did the Bicycle Go? : : :

and Other Intriguing MathematicalMysteries

by Joseph D.E. Konhauser, Dan Velleman, Stan Wagon

Reviewed by Einar Rodland .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... 466

The Mathematical Olympiad Handbook; an Introduction to Problem Solving

by Tony Gardiner

Reviewed by Catherine Shevlin . .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. 467

Mathematical Competitions in Croatia

edited by �Zeljko Hanj�s

Reviewed by Richard Hoshino ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. .. 469

Women in Mathematics: Scaling the Heights

edited by Deborah Nolan

Reviewed by Julia Johnson . ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. . 469

After Math, Puzzles and Brainteasers

by Edward Barbeau

Reviewed by Mogens Esrom Larsen .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. 471

538

Miscellaneous

Congratulations, Andy Liu! ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. .. 65

Challenge ..... ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. . 32

Note of Thanks to Ken Williams .... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. 112

Challenge Answer ..... .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 211

Announcement: ATOM Vol. II .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... . 288

Stan Wagon's e{mail problem of the week ..... ... .. .. .. ... .. .. ... .. .. .. ... 421

Year End Finale .... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... 534

Problems

February 2401{2413,2324 .... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. 47

March 2414{2425 ..... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 110

April 2388{2389, 2426{2438 .... .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. . 171

May 2439{2450 ..... ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. . 238

September 2446, 2451{2462 ..... ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. . 306

October 2463{2475 .... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... 366

November 2452{2453, 2476{2488 ..... .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. 428

December 2489{2400..... .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... . 505

Solutions

February 2299{2308, 2310{2311 ..... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... . 51

March 2255, 2309, 2312{2320, 2323 ..... ... .. .. ... .. .. .. ... .. .. ... . 113

April 2321{2322, 2325{2329, 2330{2336 .... .. ... .. .. .. ... .. .. ... . 174

May 2329, 2337{2338, 2340{2345, 2347, 2349{2350 ..... .. .. ... .. 241

September 2324, 2339, 2346, 2351{2355 .... ... .. .. ... .. .. .. ... .. .. ... .. 309

October 2356{2369 .... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... 369

November 2370{2372, 2375{2377, 2380{2385, 2388,

2392{2393 .... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. 432

December 1492, 2206, 2373{2374, 2378{2379, 2386{2387,

2389{2391, 2394-2400 .... .. ... .. .. .. ... .. .. .. ... .. .. .. 508

Mathematical MayhemFebruary .... .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. .. 33

March .... .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 94

April . .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... 158

May .... ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 221

September ..... .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 289

October ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 349

November .... ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. 413

December ..... .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 486

Shreds and Slices

March .... .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 94

Non-mathematical Problem

Awaiting a Combinatorial Proof

May .... ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 221

A Combinatorial Proof

September ..... .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 289

Another Combinatorial Proof

Miscellaneous

Contest Dates .... .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. . 38

IMO Report Jimmy Chui . .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. 420

539

Mayhem Articles

Counting Snakes, Di�erentiating the Tangent Function,

and Investigating the Bernoulli-Euler Triangle

Harold Reiter . ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... . 39

Discovering the Human Calculator in You

Richard Hoshino . ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. . 95

Discovering the Human Calculator in You

Richard Hoshino . ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. . 95

Self-Centred Triangles

Cyrus Hsia .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. 162

Four Ways to Count

Jimmy Chui .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. .. 235

Derangements and Stirling Numbers

Naoki Sato . .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 299

Waiting in Wonderland

Cyrus Hsia .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. 361

An Identity of a Tetrahedron

Murat Aygen ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. 422

A Simple Proof of a Pentagram Theorem

Geo�rey A. Kandall .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. 426

Three Gems in Geometry

Naoki Sato . .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 498

A Do-It-Yourself Proof of the n = 4 case of Fermat's Last Theorem

Ravi Vakil . ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. 502

Mayhem High School Problems

February H249{H252 ..... ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. 34

April H253{H256 ..... .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. .. 158

September H253, H257{H260 ..... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. 291

November H253, H261{H264 .... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... 415

Mayhem High School Solutions

March H225{H226 ...... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 100

May H228, H237{H240 .... .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. 223

October H241{H244 .... ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. 349

December H245{H248.... .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... . 486

Mayhem Advanced Problems

February A225{A228 .... .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. 35

April A229{232 .... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... 159

September A233{A236 ...... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. 292

November A237{A240 .... ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. . 416

Mayhem Advanced Solutions

March A209{A211 .... ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. . 104

May A212{A216 ..... .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... 226

October A217{A220 ..... ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. 352

December A221{A224..... .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... . 489

540

Mayhem Challenge Board Problems

February C83{C84 ...... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. .. 36April C85{C86 ..... ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. . 160September C87{C88 ..... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... . 293November C89{C90 .... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. 416

Mayhem Challenge Board Solutions

May C77 .... .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. 230October C79{C80 ..... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. .. 355December C81{C82 ..... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... . 492

Problem of the Month: Jimmy Chui

February .... .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. .. 36March .... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. 106April . .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... 161May .... ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 233September ..... .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 294October ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 358November .... ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. 417December ..... .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 495

Contests

J.I.R. McKnight Problems Contest 1986 ..... .. .. .. ... .. .. ... .. .. .. ... .. .. ... 37J.I.R. McKnight Problems Contest 1987 .... ... .. .. .. ... .. .. ... .. .. .. ... .. . 107Swedish Mathematics Olympiad 1985 .... .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. 109J.I.R. McKnight Problems Contest 1988 .... ... .. .. .. ... .. .. ... .. .. .. ... .. . 167Swedish Mathematics Olympiad 1990 Qualifying Round ...... .. .. ... .. .. . 168J.I.R. McKnight Problems Contest 1986 Solutions .... .. .. .. ... .. .. ... .. .. . 295J.I.R. McKnight Problems Contest 1989 .... ... .. .. .. ... .. .. ... .. .. .. ... .. . 297J.I.R. McKnight Problems Contest 1990 .... ... .. .. .. ... .. .. ... .. .. .. ... .. . 360J.I.R. McKnight Problems Contest 1991 .... ... .. .. .. ... .. .. ... .. .. .. ... .. . 418J.I.R. McKnight Problems Contest 1992 .... ... .. .. .. ... .. .. ... .. .. .. ... .. . 495

541

Proposers, solvers and commentators in the PROBLEMS and SOLUTIONSsections for 1998 are:

Names of problem proposers

Mohammed Aassila 2362, 2382, 2383, 2426, 2474, 2476, 2461

Doru Popescu Anastasiu 2343

Miguel Amengual Covas 2473

Federico Arboleda 2369

�Sefket Arslanagi �c 2347

Iliya Bluskov 2337, 2368

Mansur Boase 2408

Paul Bracken 2384

Christopher J. Bradley 2301, 2309, 2327, 2350, 2352, 2353,

2418, 2407, 2417, 2403, 2402

Nikolaos Dergiades 2377, 2389, 2458

Jos �e Luis D��az 2396, 2487

Angel Dorito 2381

David Doster 2321, 2378

Antal E. Fekete 2453, 2452

Joaqu��n G �omez Rey 2307, 2308, 2324, 2385, 2479, 2480, 2478,

2324, 2475

Herbert G�ulicher 2354

G.P. Henderson 2355, 2405

Florian Herzig 2319, 2330, 2344

Richard I. Hess 2305, 2317

Joe Howard 2486

Peter Hurthig 2438

Witold Janicki 2395Walther Janous 2325, 2326, 2328, 2329, 2340, 2341, 2363, 2364,

2370, 2387, 2390, 2422, 2477, 2423, 2467, 2468

Clark Kimberling 2386

V�aclav Kone�cn �y 2315, 2318, 2349, 2400, 2472, 2436, 2460, 2435,

2463, 2416, 2410, 2483, 2404

Daniel Kupper 2388

Michael Lambrou 2442, 2443, 2464, 2451, 2444, 2445

Gerry Leversha 2357, 2358, 2449, 2450, 2448, 2447, 2455, 2454,

2457, 2456

Mih�aly Bencze 2482, 2481

Vedula N. Murty 2306, 2345, 2359, 2394, 2459, 2462, 2471

Victor Oxman 2356, 2365, 2466

Ice B. Risteski 2421Juan-Bosco RomeroM�arquez 2346

Bill Sands 2371, 2372, 2380, 2413

K.R.S. Sastry 2310, 2311, 2312, 2322, 2323, 2360, 2361, 2367,

2419, 2432, 2425, 2424, 2433, 2434

Heinz-J�urgen Sei�ert 2313

Toshio Seimiya 2302, 2303, 2304, 2314, 2316, 2334, 2335, 2336,

2338, 2339, 2373, 2374, 2375, 2397, 2398, 2427, 2485, 2440, 2428, 2439,

2484

Zun Shan 2411

Bruce Shawyer 2400

Michael Sheard 2395

Catherine Shevlin 2366, 2446

David Singmaster 2399

D.J. Smeenk 2320, 2332, 2333, 2342, 2348, 2379, 2401, 2409, 2429,

2430, 2420

Jill Taylor 2431

G. Tsintsifas 2391, 2392, 2393, 2488

Murali Vajapeyam 2344

Darko Veljan 2412, 2395

Stan Wagon 2395

Edward T.H. Wang 2411, 2414

Albert White 2376, 2465

Wu Wei Chao 2414

Paul Yiu 2331, 2351, 2437, 2406, 2441, 2469, 2415, 2470

Names of featured solversMiguel Amengual Covas 2397

Sam Baethge 2361

Edward J. Barbeau 2327

Michel Bataille 2332, 2384, 2397

Francisco Bellot Rosado 2318, 2336, 2359

Manuel Benito 2364, 2365, 2371

Mansur Boase 2310, 2344

Chris Cappadocia 2381

Theodore Chronis 2299

Con Amore Problem Group 2330

E.B. Davies 2395

Nikolaos Dergiades 2302, 2303, 2322, 2328, 2333, 2341, 2346,

2348, 2352, 2363, 2378

Charles R. Diminnie 2324

David Doster 2316

Diane and Roy Dowling 2319

Emilio Fern�andez Moral 2364, 2365, 2371G.P. Henderson 2325, 2355, 2390

John G. Heuver 2394

Allen Herman 2378

Florian Herzig 2301, 2309, 2312, 2314, 2320, 2321, 2327, 2335,

2341, 2349, 2350, 2387

Richard I. Hess 2347, 2348

Joe Howard 2370, 2394

Jun-hua Huang 2362, 2377, 2393

Walther Janous 2305, 2306, 2308, 2324, 2357, 2359, 2366, 2396

V�aclav Kone �cn �y 2302

Vjekoslav Kova�c 2350

Michael Lambrou 2304, 2307, 2313, 2317, 2318, 2319, 2326,

2329, 2333, 2337, 2339, 2340, 2351, 2359, 2369, 2374, 2389, 2395, 2396

Kee-Wai Lau 2299, 2345, 2388, 2400

Gerry Leversha 2318, 2334, 2353, 2372, 2376, 2380

Kathleen E. Lewis 2371, 2383

Keivan Mallahi 2319Vedula N. Murty 2382

Gottfried Perz 2397

G�unter Pickert 2354

C.R. Pranesachar 1492

J.F. Rigby 2379

Christo Saragiotis 2343

Heinz-J�urgen Sei�ert 2315, 2364, 2392

Toshio Seimiya 2318, 2319, 2367, 2373, 2375

Iftimie Simion 2397

Shanxi Ankang 1137

D.J. Smeenk 2397

Digby Smith 2323

Jill S. Taylor 2331

G. Tsintsifas 2300

B.J. Venkatachala 1492

Edward T.H. Wang 2324

Kenneth M. Wilke 2311

Peter Y. Woo 2338, 2379

Paul Yiu 2331, 2360, 2361

Jeremy Young 2340, 2342, 2348, 2356, 2368, 2371, 2385

Zhao Linlong 1137

542

Names of other solvers

Hayo Ahlburg 2326, 2332

M. Paros Alexandros 2362

Miguel Carri �on �Alvarez 2380, 2381, 2384

Miguel Amengual Covas 2302, 2303, 2316, 2319, 2332, 2361,

2367

Zavosh Amir-Khosravi 2321, 2327, 2332�Sefket Arslanagi �c 2362, 2381, 2382, 2384, 2394

Charles Ashbacher 2347, 2380, 2386

Sam Baethge 2303, 2323, 2332, 2348, 2359, 2367, 2371

Chetan Balwe 2358

Edward J. Barbeau 2332, 2340

Michel Bataille 2308, 2309, 2319, 2321, 2327, 2352, 2353, 2358,

2367, 2374, 2378, 2379, 2381, 2392

Frank P. Battles 2381

Niels Bejlegaard 2300, 2309, 2338, 2342, 2347, 2350, 2394

Francisco Bellot Rosado 2303, 2304, 2309, 2314, 2316, 2319,

2320, 2322, 2334, 2335, 2338, 2342, 2346, 2348, 2350, 2358, 2360, 2367,

2373, 2375, 2376, 2379, 2381, 2382, 2397, 2398

Manuel Benito 2347, 2348, 2366, 2367, 2368, 2369, 2370, 2372,

2373, 2374, 2375, 2390

Mansur Boase 2301, 2303, 2309, 2312, 2314, 2316, 2319, 2321,

2323, 2328, 2347

Paul Bracken 2302, 2303, 2304, 2305, 2308, 2308, 2311, 2312,

2314, 2315, 2316, 2317, 2318, 2319, 2320, 2321, 2323, 2328, 2331, 2332,

2333, 2334, 2337, 2338, 2339, 2340, 2342, 2346, 2347, 2348, 2349, 2357,

2358, 2359, 2360, 2361, 2363, 2364, 2365, 2366, 2367, 2371, 2375

Duane Broline 2351

James T. Bruening 2327, 2384

Theodore Chronis 2300, 2305, 2315, 2321, 2322, 2323, 2328,

2332, 2340, 2343, 2343, 2345, 2347, 2349, 2351, 2352, 2353, 2364

Oscar Ciaurri 2345, 2343, 2400Goran Conar 2301, 2302, 2303, 2304, 2306, 2309, 2310, 2311, 2314,

2316, 2318, 2319, 2320, 2332, 2346, 2347, 2349

Nikolaos Dergiades 2299, 2300, 2301, 2304, 2309, 2310, 2311,

2314, 2315, 2316, 2318, 2319, 2320, 2326, 2327, 2329, 2332, 2334, 2335,

2336, 2338, 2340, 2342, 2345, 2347, 2349, 2350, 2353, 2358, 2359, 2360,

2361, 2362, 2364, 2365, 2367, 2373, 2374, 2375, 2376, 2379, 2380, 2381,

2382, 2384, 2387, 2392, 2394, 2395, 2396, 2397, 2398

Jos �e Luis D��az 2388

Charles R. Diminnie 2308, 2327, 2332, 2347, 2348, 2378, 2380,

2384

David Doster 2302, 2303, 2308, 2314, 2319, 2327, 2332, 2337, 2348

Diane and Roy Dowling 2305, 2348

Richard Eden 2345, 2348

Keith Ekblaw 2347, 2384, 2396

Russell Euler 2381, 2384

Emilio Fern�andez Moral 2347, 2348, 2366, 2367, 2368, 2369,2370, 2372, 2373, 2374, 2375, 2390

C. Festraets-Hamoir 2352, 2353, 2358, 2361, 2366, 2367, 2370,

2371, 2373, 2375, 2378, 2381, 2384

Anthony Fong 2327, 2332

Hidetoshi Fukagawa 2365, 2367, 2379, 2382

Ian June L. Garces 2302, 2332

Shawn Godin 2305, 2380, 2381

Douglass L. Grant 2332, 2370

Aissa Guesmia 2306

Stergiou Harafapos 2319

Florian Herzig 2299, 2300, 2302, 2303, 2304, 2305, 2308, 2310,

2311, 2316, 2323, 2331, 2332, 2333, 2334, 2337, 2340, 2345, 2378

Richard I. Hess 2299, 2306, 2310, 2311, 2312, 2315, 2319, 2322,

2323, 2327, 2328, 2331, 2332, 2334, 2337, 2340, 2343, 2343, 2345, 2349,

2351, 2356, 2357, 2361, 2364, 2365, 2367, 2368, 2370, 2371, 2378, 2380,

2381, 2384, 2385, 2386, 2387, 2390, 2397, 2398

John G. Heuver 2381

Joe Howard 2381

Jun-hua Huang 2352, 2353, 2354, 2374, 2378, 2381, 2382, 2384,

2396, 2397, 2398

Peter Hurthig 2353

Walther Janous 2300, 2301, 2302, 2303, 2309, 2310, 2311, 2312,

2314, 2315, 2316, 2318, 2319, 2320, 2321, 2322, 2323, 2327, 2330, 2331,

2332, 2333, 2335, 2336, 2338, 2342, 2343, 2343, 2344, 2345, 2346, 2347,

2348, 2349, 2350, 2352, 2353, 2358, 2361, 2362, 2367, 2372, 2373, 2374,

2375, 2376, 2377, 2379, 2380, 2381, 2382, 2383, 2384, 2385, 2388, 2389,

2392, 2393, 2394, 2397, 2400

�Angel Joval Roquet 2348, 2381, 2398

Masoud Kamgarpour 2319, 2359

Geo�rey A. Kandall 2347, 2348, 2359Geo�rey A. Kandall 2361, 2367, 2397, 2398Yeo Keng Hee 2332

Murray S. Klamkin 2299, 2300, 2305

V�aclav Kone �cn �y 2300, 2303, 2306, 2316, 2326, 2329, 2331, 2332,

2334, 2340, 2341, 2342, 2345, 2347, 2348, 2353, 2363, 2365, 2373, 2374,

2380, 2381, 2382, 2383, 2394, 2397, 2398

Vjekoslav Kova �c 2338, 2340, 2342, 2343, 2343, 2346, 2348

Mitko Kunchev 2359

Michael Lambrou 2299, 2300, 2301, 2302, 2303, 2305, 2306,

2308, 2309, 2310, 2311, 2312, 2315, 2316, 2320, 2321, 2322, 2323, 2327,

2330, 2331, 2332, 2334, 2335, 2336, 2341, 2342, 2343, 2343, 2345, 2346,

2347, 2348, 2349, 2350, 2352, 2353, 2356, 2358, 2360, 2361, 2362, 2363,

2364, 2365, 2366, 2367, 2368, 2370, 2371, 2372, 2373, 2375, 2376, 2377,

2378, 2379, 2380, 2381, 2382, 2383, 2384, 2385, 2386, 2387, 2388, 2390,

2392, 2393, 2394, 2397, 2398, 2400

Kee-Wai Lau 2300, 2305, 2308, 2310, 2313, 2314, 2321, 2323, 2327,

2329, 2330, 2332, 2334, 2340, 2344, 2347, 2348, 2349, 2352, 2353, 2362,

2365, 2366, 2370, 2371, 2396

Jessie Lei 2332, 2340Laurent Lessard 2343, 2345

Gerry Leversha 2300, 2301, 2302, 2303, 2304, 2305, 2309, 2310,

2311, 2314, 2316, 2320, 2321, 2322, 2323, 2331, 2332, 2333, 2342, 2343,

2343, 2345, 2346, 2347, 2348, 2349, 2352, 2356, 2359, 2360, 2361, 2363,

2364, 2365, 2367, 2370, 2371, 2379, 2381, 2382, 2385, 2388, 2396, 2397,

2398

Kathleen E. Lewis 2319, 2322, 2357, 2368, 2372, 2380, 2381

Mar��a Ascensi �on L �opez Chamorro 2304, 2346, 2348

Max Lyon 2389

David E. Manes 2311, 2323, 2332

Pavlos Maragoudakis 2305

Giovanni Mazzarello 2302, 2332

J.A. McCallum 2380, 2381

Phil McCartney 2306, 2345, 2362, 2400

John Grant McLoughlin 2381

Norvald Midttun 2384, 2385

Vedula N.Murty 2299, 2340, 2348, 2352, 2353, 2362, 2370, 2374,

2378, 2384

Isao Naoi 2382Jos �e H. Nieto 2319

Victor Oxman 2367, 2374, 2384

Michael Parmenter 2310, 2327

M. Perisastry 2378

Gottfried Perz 2316, 2319, 2347, 2348, 2358, 2359, 2361, 2365,

2367, 2371, 2380

Bob Prielipp 2332, 2345, 2347, 2348

Efstratios Rappos 2394

Istv �an Reiman 2382

Henry J. Ricardo 2381

Juan-Bosco RomeroM�arquez 2381, 2382

Jawad Sadek 2381, 2384

Hadi Salmasian 2327

Christos Saragiotis 2332, 2347

Joel Schlosberg 2301, 2303, 2305, 2309, 2311, 2316, 2327, 2331,

2337

543

Robert P. Sealy 2310

Heinz-J�urgen Sei�ert 2299, 2306, 2308, 2310, 2311, 2326, 2330,

2331, 2332, 2333, 2337, 2340, 2341, 2345, 2349, 2352, 2353, 2359, 2361,

2367, 2370, 2381, 2382, 2384, 2385, 2387, 2388, 2396, 2397, 2398, 2400

Toshio Seimiya 2301, 2309, 2320, 2333, 2342, 2346, 2348, 2349,

2350, 2352, 2353, 2358, 2359, 2360, 2361, 2365, 2377, 2381, 2382

Ron Shepler 2316

Max Shkarayev 2359, 2361, 2381, 2383, 2384, 2389

Andrei Simion 2397

D.J. Smeenk 2300, 2301, 2302, 2303, 2304, 2309, 2314, 2316, 2318,

2319, 2322, 2334, 2335, 2336, 2338, 2352, 2358, 2359, 2360, 2361, 2365,

2367, 2373, 2374, 2375, 2376, 2377, 2380, 2381, 2382, 2398

Digby Smith 2332, 2345, 2347, 2380, 2381, 2384

Eckard Specht 2302, 2303, 2304, 2365, 2397

David R. Stone 2310, 2311

J. Suck 2348, 2359

Aram Tangboondouangjit 2352, 2353, 2370, 2378, 2381, 2382

Nicolas Th �eriault 2322, 2327

Parayiou Theoklitos 2302, 2303, 2304, 2309, 2314, 2316, 2318,

2319, 2320, 2321, 2352, 2353, 2359, 2362, 2367, 2373, 2376, 2381

Todd Thompson 2327

George Tsapakidis 2319

Panos E. Tsaoussoglou 2302, 2315, 2316, 2323, 2332, 2340,

2345, 2347, 2348, 2349, 2352, 2361, 2362, 2367, 2374, 2378, 2381,2382,

2384, 2392, 2393, 2394, 2397, 2398

G. Tsintsifas 2301, 2309, 2374, 2377, 2379, 2382, 2397, 2398

Eric Umega�

rd 2332

John Vlachakis 2302, 2303, 2309

Stan Wagon 2347, 2363

Edward T.H. Wang 2371

Hoe Teck Wee 2320

Larry White 2378

Kenneth M. Wilke 2310, 2317, 2321, 2331, 2332

Peter Y. Woo 2397, 2398

Paul Yiu 2303, 2316, 2319, 2348, 2349, 2352, 2353, 2358, 2379, 2382,

2397

Jeremy Young 2338, 2345, 2346, 2347, 2349, 2351, 2352, 2353,

2357, 2358, 2359, 2360, 2361, 2362, 2363, 2364, 2367, 2370, 2372, 2377,

2380, 2381, 2383, 2384, 2389, 2392, 2394, 2397, 2398

Bookery Problem Group 2381

Skidmore College Problem Group 2381

University of Arizona Problem Solving Group 2359,

2381, 2383

Proposers, solvers and commentators in the MAYHEM PROBLEMS andSOLUTIONS sections for 1998 are:

Proposers

High School

Mohammed Aassila H260 H262

Alexandre Trichtchenko H240 H254 H259

The EditorsH223 H224 H225 H226 H228 H237 H238 H239 H241

H242 H243 H244 H245 H246 H247 H248 H249 H250 H251 H252

H253 H255 H256 H257 H258 H261 H263 H264

Advanced

Mohammed Aassila A218 A219 A223 A231 A235 A236 A239

A240 A227

Waldemar Pompe A220 A224

Naoki Sato A210 A226 A230 A233

Alexandre Trichtchenko A217

1997 Baltic Way A228 A232 A234 A238

International Competition in Mathematics for Uni-versity Students A212

Polish Mathematical Olympiad A211 A215 A216 A229

The Editors A209 A213 A214 A221 A222 A225 A237

Challenge Board

Mohammed Aassila C79

Dima Arinkin C83

Noam Elkies C90

Mark Krusemeyer C87

Tal Kubo C89

Christopher Long C84 C85

The Editors C77 C80 C81 C82 C86 C88

Solvers

High School

Mara Apostol H244

Keon Choi H244

Shawn Godin H241

Nick Harland H244

Kenneth Ho H244

Masoud Kamgarpour H241 H242 H243 H244

Murray S. Klamkin H228

Condon Lau H244

Penny Nom H241

D.J. Smeenk H241

Edward T.H. Wang H237 H238 H239 H240 H241 H242 H244

Dale Whitmore H244

Wendy Yu H244

The Editors H223 H224 H225 H226 H242 H245 H246 H247 H248

Advanced

Mohammed Aassila A218 A223

Andrei Simion A224

D.J. Smeenk A219

Edward T.H. Wang A213 A215

The Editors A209 A210 A211 A212 A213 A214 A215 A216 A217

A220 A221 A222

Challenge Board

Mohammed Aassila C79

Ivailo Dimov C82

The Editors C77 C80 C81

544

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