SOLUTIONS - cms.math.ca · triangle passes through the midpoint of a side of the orthic triangle. This result is This result is an immediate consequence of the \bisection property"

$Page 1: SOLUTIONS - cms.math.ca · triangle passes through the midpoint of a side of the orthic triangle. This result is This result is an immediate consequence of the \bisection property"$
202/ Solutions

SOLUTIONSNo problem is ever permanently closed. The editor is always pleased to consider forpublication new solutions or new insights on past problems.

Statements of the problems in this section originally appear in 2018: 44(4), p. 160–163,and 44(5), p. 214–218.

4336. Proposed by Michel Bataille.

For non-negative integers m and n, evaluate in closed form

n∑k=0

m∑j=0

(j + k + 1)

Çj + k

j

å.

We received 7 submissions, all correct. We present the solution by Paul Bracken,modified and enhanced by the editor.

Denote the given sum by S. Since both summations are finite, they can be inter-changed to yield:

S =m∑j=0

jn∑

k=0

Çj + k

j

å+

n∑k=0

km∑j=0

Çj + k

k

å+

m∑j=0

n∑k=0

Çj + k

j

å(1)

The two formulae below are known for m,n, r, k ∈ N ∪ {0}.

(a)n∑

k=0

Çr + k

r

å=

Çr + k + 1

r + 1

å, (2)

(b)m∑

k=0

Çn+ k + 1

k + 1

å=m(n+ 1)− 1

n+ 2

Çn+m+ 2

n+ 1

å+ 1, (3)

From (1)− (3) we then have

S =m∑j=0

j

Çj + n+ 1

j + 1

å+

n∑k=1

k

Çk +m+ 1

k + 1

å+

m∑j=0

Çj + n+ 1

j + 1

å=m(n+ 1)− 1

n+ 2

Çm+ n+ 2

n+ 1

å+ 1 +

n(m+ 1)− 1

m+ 2

Çm+ n+ 2

m+ 1

å+ 1

+

Çm+ n+ 2

n+ 1

å− 1

=

Åm(n+ 1)− 1

n+ 2+n(m+ 1)− 1

m+ 2+ 1

ãÇm+ n+ 2

n+ 1

å+ 1 (4)

Crux Mathematicorum, Vol. 45(4), April 2019

Solutions /203

Now by straightforward but tedious computations, we have

(m+ 2)(n+ 1)m− (m+ 2) + (m+ 1)(n+ 2)n− (n+ 2) + (m+ 2)(n+ 2)

= mn2 +m2n+m2 + n2 + 5mn+ 3m+ 3n

= mn(m+ n+ 3) +m(m+ n+ 3) + n(m+ n+ 3)

= (m+ n+ 3)(mn+m+ n). (5)

Substitute (5) into (4) we finally obtain

S =(m+ n+ 3)(mn+m+ n)

(m+ 2)(n+ 2)·Çm+ n+ 2

n+ 1

å+ 1

=mn+m+ n

m+ 2·Çm+ n+ 3

n+ 2

å+ 1.

4337. Proposed by Mihaela Berindeanu.

Let ha, hb, hc be the altitudes from vertices A,B,C, respectively, of triangle ABC.Erect externally on its sides three rectangles ABB1A2, BCC1B2, CAA1C2, whosewidths are k times as long as the parallel altitudes; that is,

CC1

ha=AA1

hb=BB1

hc= k.

If X,Y, Z are the respective midpoints of the segments A1A2, B1B2, C1C2, provethat the lines AX,BY,CZ are concurrent.

We received 8 solutions, all of which were correct. We present the solution by IvkoDimitric, who notes that this problem is very similar to the Crux problem 4258by the same proposer, and the published solutions of that problem – in particular

Copyright © Canadian Mathematical Society, 2019

204/ Solutions

solution 2 – are readily adapted to provide the solution of the present problem. Thesolution given here gives an alternative approach.

Let O be the circumcenter of triangle ABC. Then

∠OAC =1

2(180◦ − ∠AOC) = 90◦ − ∠B.

Let E and F be the feet of the altitudes from B and C, respectively, and let EKbe the straight line segment equal in length and parallel to CF that intersects ABperpendicularly. Then the quadrilateral KFCE is a parallelogram and KF ‖ ACso that

∠FKE = ∠KEA = 90◦ − ∠A.

Moreover,∠KEB = ∠EHC = ∠A

since AFHE is cyclic. Since

tanA =hbAE

=hcAF

,

we haveEB

EK=hbhc

=AE

AF,

so 4BEK is similar to 4EAF (and to 4BAC) and consequently

∠BKE = ∠AFE = ∠C.

Since BCEF is cyclic we have

∠BKF = ∠BKE − ∠FKE = ∠C − (90◦ − ∠A) = 90◦ − ∠B = ∠OAC,

so BK ‖ OA. Further,

−−→AX =

1

2(−−→AA1 +

−−→AA2) =

k

2(−→hb +

−→hc) =

k

2

−−→BK

and hence AX ‖ BK ‖ OA, meaning that O,A,X are collinear. In the samemanner, each triple O,B, Y and O,C,Z consists of collinear points, which impliesthat the lines AX,BY and CZ are concurrent at the circumcenter O, the isogonalconjugate of the orthocenter H.

4338. Proposed by Daniel Sitaru.

Prove that for any triangle ABC, we have

2∑∣∣∣ cos

A−B2

∣∣∣ ≤ 3 +√

3 + 2∑

cos(A−B).

We received five submissions, all of which are correct. We present two differentsolutions.


Solutions /205

Solution 1, by Shuborno Das, modified and enhanced by the editor.

Note first that the absolute value sign in the statement is redundant since

−π2<A−B

2,B − C

2,C −A

2<π

2.

Since

cos(A−B) = 2 cos2A−B

2− 1,

we have

2 cos2A−B

2= 1 + cos(A−B),

so

2 cosA−B

2=»

2 + 2 cos(A−B).

Similarly,

2 cosB − C

2=»

2 + 2 cos(B − C) (1)

and

2 cosC −A

2=»

2 + 2 cos(C −A).

By the Cauchy-Schwarz Inequality, we have∑cyc

»2 + 2 cos(A−B) ≤

∑cyc

(1 + 1 + 1)(2 + 2 cos(A−B))

=

6∑cyc

(1 + cos(A−B)). (2)

It thus suffices to show that 6∑cyc

(1 + cos(A−B)) ≤ 3 +

3 + 2∑cyc

cos(A−B)

or

6∑cyc

(1 + cos(A−B)) ≤ 12 + 2∑cyc

cos(A−B) + 6

3 + 2∑cyc

cos(A−B)

or

3 + 2∑cyc

cos(A−B) ≤ 3

3 + 2∑cyc

cos(A−B). (3)

Let S =∑

cyc cos(A−B). Then (3) becomes

(2s+ 3)2 ≤ 9(2s+ 3) ⇐⇒ 2s+ 3 ≤ 9 ⇐⇒ s ≤ 3,

which is obviously true. From (1)− (3), the conclusion follows.


206/ Solutions

Solution 2, by Leonard Giugiuc.

Clearly, the absolute value sign in the question is redundant.

Let u = cosA+ i sinA, v = cosB + i sinB, and w = cosC + i sinC.

Then |u| = |v| = |w| and

|u+ v| = |(cosA+ cosB) + i(sinA+ sinB)|

=

∣∣∣∣2 cosA+B

2cos

A−B2

+ 2i sinA+B

2cos

A−B2

∣∣∣∣= 2

∣∣∣∣ÅcosA−B

2

ãÅsin

C

2+ i cos

C

2

ã∣∣∣∣= 2 cos

A−B2

.

Similarly, |v + w| = 2 cosB − C

2and |w + u| = 2 cos

C −A2

.

Also,

|u+ v + w| =»

(cosA+ cosB + cosC)2 + (sinA+ sinB + sinC)2

=

3 + 2∑cyc

cos(A−B).

Since|u+ v|+ |v + w|+ |w + u| ≤ |u|+ |v|+ w|+ |u+ v + w|

by Hlawka’s Inequality, we have

2∑cyc

cosA−B

2≤ 3 +

3 + 2

∑cyc

cos(A−B),

completing the proof.

4339. Proposed by Kadir Altintas and Leonard Giugiuc.

Suppose ABC is an acute-angled triangle, DEF is an orthic triangle of ABC, Sis the symmedian point of ABC, G is the barycenter of DEF . If D is the foot ofthe altitude from A and K is the point of intersection of AS and FE, prove thatD,G and K are collinear.

We received four submissions, all of which were correct, and feature the solutionby Ivko Dimitric.

We use barycentric coordinates based on 4ABC where the coordinates of thevertices are A(1 : 0 : 0), B(0 : 1 : 0) and C(0 : 0 : 1) and a, b, c denote thecorresponding side lengths. The vertices of the orthic triangle are the feet of thealtitudes of 4ABC. Hence, we get

E (SC : 0 : SA) and F (SB : SA : 0 ) ,


Solutions /207

where

SA =b2 + c2 − a2

2, SB =

c2 + a2 − b22

, SC =a2 + b2 − c2

2,

according to Conway’s triangle notation. Since SC + SA = b2 and SB + SA = c2,we get the respective normalized barycentric coordinates as

E

ÅSC

b2, 0,

SA

b2

ãand F

ÅSB

c2,SA

c2, 0

ã.

Then the midpoint M of EF is

M =1

2(E + F ) =

Åb2SB + c2SC

2b2c2,SA

2c2,SA

2b2

ã.

Since the symmedian point has homogeneous coordinates S = (a2 : b2 : c2), theline through A and S has an equation c2y− b2z = 0. The coordinates of M clearlysatisfy this equation, so M belongs to AS and, therefore, M ≡ K. Since thecentroid G of 4DEF belongs to the median from D, namely DM = DK, thepoints D,G, and K are collinear.

Editor’s Comments. In other words Dimitric has shown that each symmedian of atriangle passes through the midpoint of a side of the orthic triangle. This result isan immediate consequence of the “bisection property” discussed by Michel Batailleat the start of his Crux article, “Characterizing a Symmedian” [Vol. 43(4), April2017, 145-150], namely that the A-symmedian is the locus of the midpoints of theantiparallels to BC bounded by the lines AB and AC. A very simple coordinate-free proof is given there.

4340. Proposed by Digby Smith.

Let a, b, c and d be positive real numbers such that

a+ b+ c+ d =1

a+

1

b+

1

c+

1

d.

Show that

a+ b+ c+ d ≥ maxß

4√abcd,

4√abcd

™.

We received 11 submissions of which all but two were correct and complete. Wepresent a solution followed by a generalization.

Solution by Roy Barbara.

First, it suffices to prove that for any a > 0, b > 0, c > 0, and d > 0 that satisfy

the hypothesis, a+ b+ c+ d =1

a+

1

b+

1

c+

1

d, the following inequality holds

a+ b+ c+ d ≥ 4√abcd. (1)


208/ Solutions

Indeed, positive numbers x, y, z, and t defined by

x =1

a, y =

1

b, z =

1

cand t =

1

d

satisfy the hypothesis and consequently the inequality (1). However, inequality(1) for x, y, z, and t is

1

a+

1

b+

1

c+

1

d≥ 4√

abcd, (2)

which proves the statement inequality.

Second, we establish inequality (1). Let

S = a+ b+ c+ d and T = abc+ bcd+ cda+ dab.

The following identity relates S and T :

S3 − 16T = S(a+ b− c− d) + 4(a+ b)(c− d)2 + 4(c+ d)(a− b)2.

Therefore S3 − 16T ≥ 0. (Ed.:This inequality is known as Maclaurin’s inequality,as several solvers indicated.) The hypothesis, is equivalent to abcd = T/S. Hence

16abcd =16T

S≤ S3

S= S2

and (1) follows.

Generalization by Leonard Giugiuc, Ardak Mirzakhmedov, and Roy Barbara. Leta1, a2, . . . , and an be n strictly positive numbers that satisfy

a1 + · · ·+ an =1

a1+ · · ·+ 1

an.

Thena1 + · · ·+ an ≥ max

{n n−2√a1 . . . an,

nn−2√a1 . . . an

}. (3)

Solution for generalized inequality.

Maclaurin’s inequality is a refinement of the AM-GM inequality, which you canfind at https://en.wikipedia.org/wiki/Maclaurin%27s_inequality

Due to Maclaurin’s inequality for n positive numbers,

(a1 + · · ·+ an)n−1 ≥ nn−2(a1 . . . an)( 1

a1+ · · ·+ 1

an

).

However, a1 + · · ·+ an = 1a1

+ · · ·+ 1an

, leading to

(a1 + · · ·+ an)n−1 ≥ nn−2(a1 . . . an)(a1 + · · ·+ an),

(a1 + · · ·+ an)n−2 ≥ nn−2(a1 . . . an),

a1 + · · ·+ an ≥ n n−2√a1 . . . an. (4)


https://en.wikipedia.org/wiki/Maclaurin%27s_inequality

Solutions /209

Denote by xi = 1/ai for i = 1, . . . , n. Then

x1 + · · ·+ xn =1

x1+ · · ·+ 1

xn

and according to (4)

x1 + · · ·+ xn ≥ n n−2√x1 . . . xn, or

a1 + · · ·+ an ≥ n1

n−2√a1 . . . an

. (5)

Based on (4) and (5), generalized inequality (3) follows.

4341. Proposed by Daniel Sitaru and Leonard Giugiuc.

Let ABC be an arbitrary triangle. Show that∑cyc

sinA(| cosA| − | cosB cosC|) = sinA sinB sinC.

Eleven correct solutions were received. Most of the solvers used the approach inthe solution given here.

From the expansion of

0 = tanπ = tan(A+B + C),

we obtaintanA tanB tanC = tanA+ tanB + tanC,

whence

sinA sinB sinC = sinA cosB cosC + sinB cosC cosA+ sinC cosA cosB.

Also,

2 sinA cosA+ 2 sinB cosB + 2 sinC cosC

= sin 2B + sin 2C + 2 sinA cosA

= 2 sin(B + C) cos(B − C) + 2 sinA cosA

= 2 sinA(cos(B − C)− cos(B + C))

= 4 sinA sinB sinC

= 2(sinA sinB sinC + sinA cosB cosC + sinB cosC cosA+ sinC cosA cosB).

This yields the result for acute and right triangles. If, say, ∠A > π/2, then cosineof A is negative and all the other trigonometric ratios are positive. Note that

cosA+ cosB cosC = − cos(B + C) + cosB cosC = sinB sinC,

with analogous identities for other arrangements of A,B,C.


210/ Solutions

The left side of the equation is then

− sinA cosA− sinA cosB cosC + sinB cosB

+ sinB cosC cosA+ sinC cosC + sinC cosA cosB

= − sinA(cosA+ cosB cosC) + sinB(cosB + cosC cosA)

+ sinC(sinC + cosA cosB)

= − sinA sinB sinC + 2 sinA sinB sinC

= sinA sinB sinC.

4342. Proposed by Oai Thanh Dao and Leonard Giugiuc.

In a convex quadrilateral ABA1C, construct four similar triangles ABC1, A1BC2,ACB1 and A1CB2 as shown in the figure.

Show that C1C2 = B1B2 and that the directed angles satisfy

∠(C1C2, B1B2) = 2∠C1BA.

We received 6 submissions, all of which were correct, including fixing the typo-graphical error in the problem’s statement (which had one unwanted subscript).We present a composite of the similar solutions by AN-anduud Problem SolvingGroup and Michel Bataille.

Let us denote the pairs of equal oriented angles at B and at C by ψ; more precisely,

ψ = ∠C1BA = ∠C2BA1 = ∠ACB1 = ∠A1CB2.

Denote by k the common ratio of the corresponding adjacent sides of the givensimilar triangles:

k =BA

BC1=BA1

BC2=

CA

CB1=CA1

CB2.


Solutions /211

The spiral similarity about center B with angle ψ and ratio k takes C1 to A andC2 to A1, while the spiral similarity about center C with angle −ψ and ratio ktakes B1 to A and B2 to A1.

Consequently, C1C2 = B1B2 (because they equal 1kAA1). Moreover,

ψ = ∠(−−−→C1C2,

−−→AA1) = ∠(

−−→AA1,

−−−→B1B2),

Therefore

∠(−−−→C1C2,

−−−→B1B2) = ∠(

−−−→C1C2,

−−→AA1) + ∠(

−−→AA1,

−−−→B1B2)

= 2ψ

= 2∠C1BA.

4343. Proposed by Mihaela Berindeanu.

Let ABC be an acute triangle and E be the center of the excircle tangent at Fand G to the extended sides AB and AC, respectively. If GF ∩ BE = {B1},FG ∩CE = {C1} and B′ and C ′ are feet of the altitudes from B, respectively C,show that B1C1B

′C ′ is a cyclic quadrilateral.

All of the 11 submissions we received were correct. Seven were essentially thesame as our featured solution by Shuborno Das, and most solvers observed that therestriction to acute triangles is not required.

Note that

∠CEB1 = ∠CEB

= 180◦ −Å

90◦ − B

2+ 90◦ − C

2

ã= 90◦ − A

2.

Moreover, because AF = AG,

∠CGB1 =180◦ −A

2= 90◦ − A

2.

Thus CGEB1 is cyclic.

Since EG ⊥ CG it follows that ∠EB1C = 90◦. Similarly we get ∠EC1B = 90◦.

Hence, the points B1 and C1 lie on the circle whose diameter is BC. But B′ and C ′

are the feet of the altitudes from B and C (that is, BB′ ⊥ B′C and CC ′ ⊥ C ′B),so they also must lie on the same circle. In other words, for any given triangleABC, acute or not, the quadrilateral B1C1B

′C ′ is inscribed in the circle whosediameter is BC, and we are done.


212/ Solutions

4344. Proposed by Michel Bataille.

Let n be a positive integer. Find all polynomials p(x) with complex coefficientsand degree less than n such that x2n + xn + p(x) has no simple root.

We received 4 correct solutions. We present 3 of them.

Solution 1, by the proposer.

The only polynomial is the constant p(x) = 14 .

Let q(x) = x2n + xn + p(x) have s distinct roots r1, r2, . . . , rs with respectivemultiplicities k1, k2, . . . , ks, all exceeding 1. Then

2s ≤ k1 + k2 + · · ·+ ks = 2n,

whence s ≤ n. Suppose that

p(x) = an−1xn−1 + an−2x

n−2 + · · ·+ a1x+ a0

and that f(x) = 2nq(x)− xq′(x). Then f(x) is a polynomial of degree n equal to

nxn + (n+ 1)an−1 + (n+ 2)an−2 + · · ·+ (2n− 1)a1x+ 2na0.

Since each of the roots ri has multiplicity exceeding 1, it is a root of q′(x) withmultiplicity ki − 1. Thus, each of the roots ri is a root of f(x) with multiplicityat least ki − 1, so that

2n− s = (k1 + k2 + · · ·+ ks)− s = (k1 − 1) + (k2 − 1) + · · ·+ (ks − 1) ≤ n,

whence s ≥ n. Therefore n = s, k1 = k2 = · · · = kn = 2, and

q(x) =n∏

i=1

(x− ri)2 = (xn + un−1xn−1 + · · ·+ u1x+ u0)2,

for some complex coefficients ui. Plugging this into the equation defining q andnoting that all the coefficients of powers of x between xn and x2n vanish, we findthat

2un−1 = 0;

2un−2 + u2n−1 = 0;

2un−3 + 2un−1un−2 = 0;

...

2u1 + 2u2un−1 + · · · = 0;

...

2u0 + 2u1un−1 + · · · = 1.

Thus q(x) = (xn + 12 )2 and p(x) = 1

4 is the unique solution to the problem.


Solutions /213

Solution 2, AN-anduud Problems Solving Group.

Using the notation and argument of the first solution, we have that

q(x) = x2n + xn + p(x) = (u(x))2

for some monic polynomial u(x) of degree n. Then

p(x)− 1

4= u(x)2 −

Åxn +

1

2

ã2=

Åu(x)− xn − 1

2

ãÅu(x) + xn +

1

2

ã.

Either p(x) = 1/4 and u(x) = xn + 1/2 or the polynomials on both sides of theequation are nontrivial. But the latter possibility is precluded by the fact that thedegree of p(x)− 1/4 is less than n while that of u(x) + xn + 1/2 is equal to n.

Solution 3, by Leonard Giugiuc and Ramanujan Srihari (done independently).

With the notation of the previous solutions and from the fact that s ≤ n, we notethat the coefficients of the powers x2n−s+1 up to x2n−1 in q(x) vanish, along withthe corresponding symmetric functions of the roots. From Newton’s formulae forthe powers of the roots, we obtain the equation

k1 + k2 + · · ·+ ks = 2n

as well as the equations in the system

S = {k1rt1 + k2rt2 + · · ·+ ksr

ts = 0 : 1 ≤ t ≤ s− 1 ≤ n− 1 < 2n− n},

for integers k1, k2, . . . , ks in terms of distinct complex numbers r1, r2, . . . , rs.

If, say, rs = 0, then S becomes a system of s − 1 equations in s − 1 unknownswith a nonzero Vandermonde determinant of its coefficients, so it has the uniquesolution k1 = k2 = · · · = ks−1 = 0. But this forces ks = 2n and q(x) = x2n, apalpable falsehood.

Thus, all the ri are nonzero. If s < n, we can append to S the equation

k1rs1 + k2r

s2 + · · ·+ ksr

ss = 0

and obtain the solution k1 = k2 = · · · = ks = 0, which is again false. Therefores = n, k1 = k2 = · · · = kn = 2, and x2n + xn + p(x) is the square of a polynomialof degree n. Since the roots of this polynomial satisfy

rt1 + rt2 + · · ·+ rtn = 0

for 1 ≤ t ≤ n− 1, we must have that

x2n + xn + p(x) = (xn + c)2

for some constant c. A check of the coefficient of xn reveals that c = 12 .


214/ Solutions

4345. Proposed by Mihai Miculita and Titu Zvonaru.

Let ABC be a triangle with AB < BC and incenter I. Let F be the midpoint ofAC. Suppose that the C-excircle is tangent to AB at E. Prove that the pointsE, I and F are collinear if and only if ∠BAC = 90◦.

We received 16 submissions, all of which were correct, and will feature two of them.

Solution 1, by Ivko Dimitric (typical of the five submissions that used barycentriccoordinates).

We will see that that the assumption AB < BC is superfluous. We use barycentriccoordinates based on 4ABC where A(1 : 0 : 0), B(0 : 1 : 0) and C(0 : 0 : 1) arethe coordinates of the vertices and a, b, c denote the corresponding side lengths.As usual, s = 1

2 (a+ b+ c) is the semi-perimeter. Then F (1 : 0 : 1) and I(a : b : c).Since BE = s− a and AE = s− b, it follows that E(s− a : s− b : 0). Hence, F, Iand E are collinear if and only if∣∣∣∣∣∣

1 0 1a b c

s− a s− b 0

∣∣∣∣∣∣ = 0.

Expanding this determinant gives

−c(s− b) + a(s− b)− b(s− a) = 0

orbc− (−a+ b+ c) s = 0.

This, in turn, is equivalent to

0 = (−a+ b+ c)(a+ b+ c)− 2bc

= (b+ c)2 − a2 − 2bc

= b2 + c2 − a2,

which reduces to the Pythagorean condition b2 + c2 = a2. In other words, startingwith an arbitrary triangle ABC, we have that F, I and E are collinear if and onlyif 4ABC is a right-angled triangle with ∠BAC = 90◦.

Solution 2, by AN-anduud Problem Solving Group, with some details supplied bythe editor.

Let σ be the dilatation with center C that takes the C-excircle to the incircle.Side AB of ∆ABC, which is tangent to the excircle at E, is taken by σ to a linesegment A′B′ (with A′ on AC and B′ on BC) that is tangent to the incircle ata point E′. If D is the point where the incircle is tangent to AB, then DE′ is adiameter of the incircle.

Assume now that ∠BAC = 90◦. Then ∆EDE′ and ∆EAC are homothetic righttriangles that share the angle at E; because I and F are the midpoints of the


Solutions /215

parallel sides opposite E, namely DE′ and AC, we deduce that E, I, and F arecollinear.

Conversely, we assume that E, I, and F are collinear. Consider the dilatation withcenter E that takes I to F . It takes E′ to a point E′′ on the line EE′ and D toa point D′ on AB such that E′′D′ ⊥ BD′ and F is the midpoint of D′E′′. Thisforces D′ to coincide with A and E′′ with C as follows: as a moving point X slidesalong the line AB, the locus of points Y for which F is the midpoint of XY is theline through C that is parallel to AB, and that line intersects EE′ in the uniquepoint C. We conclude that ∠BAC = ∠BD′E′′ = 90◦, as desired.

4346. Proposed by Daniel Sitaru.

Find all x, y, z ∈ (0,∞) such that®64(x+ y + z)2 = 27(x2 + 1)(y2 + 1)(z2 + 1),

x+ y + z = xyz.

We received 17 correct solutions, along with 1 incorrect solution. Eight solutionsproceeded as in Solution 1; they all followed the same strategy, some depending onthe Hermite-Hadamard inequality. We present 3 solutions in total.

Solution 1, by Paul Bracken.

Let

(x, y, z) = (tanA, tanB, tanC),

where 0 < A,B,C < π/2. Then the two equations become

tanA+ tanB + tanC = tanA tanB tanC


216/ Solutions

and64(tanA tanB tanC)2 = 27(sec2A)(sec2B)(sec2 C).

These are equivalent to A+B + C = π (expand tan(A+B + C)) and

sin2A sin2B sin2 C =27

64.

Since 2 ln sin t is a strictly concave function of t on (0, π/2), by Jensen’s inequalitywe get

ln sin2A+ ln sin2B + ln sin2 C ≤ 3 ln sin2

ÅA+B + C

3

ã= 3 ln sin2

(π3

).

Hence

sin2A sin2B sin2 C ≤Å

3

4

ã3=

27

64,

with equality if and only if A = B = C = π/3.

Therefore the equations are satisfied if and only if (x, y, z) = (√

3,√

3,√

3).

Solution 2, by Nghia Doan.

Letp = x+ y + z = xyz and q = xy + yz + zx.

Sincex2 + y2 + z2 = p2 − 2q

andx2y2 + y2z2 + z2x2 = q2 − 2p2,

the first equation becomes

64p2 = 27[p2 + (q2 − 2p2) + (p2 − 2q) + 1] = 27(q − 1)2.

Since

q =

Å1

xy+

1

yz+

1

zx

ã(xy + yz + zx) ≥ 9,

then 8p = 3√

3(q − 1).

By the AM-GM inequality,

p3 = (x+ y + z)3 ≥ 27xyz = p,

so that p ≥ 3√

3, with equality if and only if x = y = z =√

3.

On the other hand,

(xy + yz + zx)2 ≥ 3((xy)(yz) + (yz)(zx) + (zx)(xy))

= 3xyz(x+ y + z)

= 3p2,


Solutions /217

so that q ≥√

3p with equality if and only if xy = yz = zx. Hence

8p ≥ 3√

3(√

3p− 1) = 9p− 3√

3,

so that p ≤ 3√

3.

It follows that the only solution of the given system of equations is x = y = z =√

3.

Solution 3, by Madhav Modak.

From the previous solution, we have that

p2 = (27/64)(q − 1)2, p2 ≥ 3q and q2 ≥ 3p2.

Substituting for p2 in these two inequalities yield respectively

0 ≤ 9q2 − 82q + 9 = (9q − 1)(q − 9)

and0 ≤ −(17q2 − 162q + 81) = (9− q)(17q − 9).

The only value of the pair (p, q) that allows both inequalities to hold is (√

3, 9)and this in turn forces x = y = z = 3

√3 as the unique solution of the system.

Editor’s comment. Some solvers tackled the problem by examining when thepolynomial t3 − pt2 + qt− p has three real roots with 64p2 = 27(q − 1)2. This ledto some straightforward technical gymnastics that are not sufficiently edifying toreproduce here.

4347. Proposed by J. Chris Fisher.

Given a cyclic quadrilateral ABCD with diameter AC (and, therefore, right anglesat B and D), let P be an arbitrary point on the line BD and Q a point on AP .Let the line perpendicular to AP at P intersect CB at R and CD at S. Finally,let E be the point where the line from R perpendicular to SQ meets AP . Provethat P is the midpoint of AE if and only if CQ is parallel to BD.


218/ Solutions

Comment by the proposer: This is a slightly generalized restatement of OC266[2017:137-138], Problem 5 on the 2014 India National Olympiad. The solutionfeatured recently in Crux was a long and uninformative algebraic verification; inparticular, it failed to explain why the triangle had to be acute (it probably didn’t),and it hid the true nature of the problem.

One of the five submissions we received was incomplete; the other four were correctand complete. We feature the solution by Michel Bataille.

Let γ be the circle with diameter AS. Since AP ⊥ PS and AD ⊥ DS, the pointsP and D are on γ and it follows that

∠(DP,DA) = ∠(SP, SA);

that is,

∠(DB,DA) = ∠(SR, SA).

But

∠(DB,DA) = ∠(CB,CA)

(since A,B,C,D are concyclic), hence

∠(CB,CA) = ∠(SR, SA);

that is,

∠(CR,CA) = ∠(SR, SA)

and so C,R,A, S are concyclic on a circle that we denote by γ′. (They cannotbe collinear since otherwise B would be on CA.) Assuming that Q 6= P , we alsoobserve that E is the orthocentre of ∆QRS (since QP ⊥ RS and RE ⊥ QS).


Solutions /219

Now, suppose that P is the midpoint of AE. Then A is the reflection of E inRS, which implies that A is on the circumcircle of ∆QRS, and therefore thiscircumcircle coincides with γ′. We deduce that

∠(CD,CQ) = ∠(CS,CQ) = ∠(AS,AQ) = ∠(AS,AP ).

Since

∠(AS,AP ) = ∠(DS,DP )

(A,D, S, P being concyclic) and

∠(DS,DP ) = ∠(CD,BD),

we finally obtain

∠(CD,CQ) = ∠(CD,BD),

which proves that CQ ‖ BD.

Conversely, suppose that CQ is parallel to BD. Then, we have

∠(CS,CQ) = ∠(DS,DP ) = ∠(AS,AP ) = ∠(AS,AQ),

so that Q is on the circumcircle γ′ of ∆CSA. Since R is also on this circle, we seethat γ′ is the circumcircle of ∆QRS. Since A is on γ′ and on the altitude QP , thepoint A is the reflection of the orthocentre E of ∆QRS and so P is the midpointof AE.

Editor’s comments. In terms of the notation of the foregoing problem, ProblemOC266 (referred to by the proposer) states that

For any point P on the side BD of an arbitrary triangle ABD, letO1, O2 denote the circumcentres of triangles ABP and APD, respec-tively. Prove that the line joining the circumcentre of triangle ABDto the orthocentre H ′ of triangle O1O2P is parallel to BC.

To see that this is a special case of the foregoing problem, define Q to be theintersection of AP with the line through C parallel to BD. Then, according toProblem 4347, the dilatation with center A and ratio 2 will take ∆O1O2P to∆RSE, the orthocentre H ′ (of ∆O1O2P ) to the orthocentre Q (of ∆RSE), andthe circumcentre of ∆ABD to C. Consequently the line joining the circumcentreof triangle ABD to the orthocentre H ′ is parallel to CQ, which is parallel to BC.

4348. Proposed by Marius Dragan.

Let p ∈ [0, 1]. Then for each n > 1, prove that

(1− p)n + pn ≥ (2p2 − 2p+ 1)n + (2p− 2p2)n.

We received 11 correct solutions. We present the solution by Digby Smith.


220/ Solutions

Note that for p = 0 and p = 1, there is equality. Let f(t) = tn for t ∈ [0, 1]. Sincen > 1, it follows that f is convex. Suppose that 0 < t < 1. Applying convexity, itfollows that

f(p) + f(1− p) = [pf(p) + (1− p)f(1− p)] + [(1− p)f(p) + pf(1− p)]≥ f [p(p) + (1− p)(1− p)] + f [(1− p)p+ p(1− p)]= f [p2 + (1− p)2] + f [p(1− p) + p(1− p)]= f(2p2 − 2p+ 1) + f(2p− 2p2),

with equality if and only if p = 1 − p. That is, there is equality if and only ifp = 1/2.

It follows that

(1− p)n + pn ≥ (2p2 − 2p+ 1)n + (2p− 2p2)n,

with equality if and only if p ∈ {0, 1/2, 1}.

4349. Proposed by Hoang Le Nhat Tung.

Let x, y and z be positive real numbers such that x+y+z = 3. Find the minimumvalue of

x3

y√x3 + 8

+y3

z√y3 + 8

+z3

x√z3 + 8

.

We received 7 correct solutions. We present the solution by Leonard Giugiuc.

Due to Cauchy’s inequalityÇx3

y ·√x3 + 8

+y3

z ·√y3 + 8

+z3

x ·√z3 + 8

ã(xy + yz + zx)

≥Ç

x2

4√x3 + 8

+y2

4√y3 + 8

+z2

4√z3 + 8

å2

,

implying that

x3

y ·√x3 + 8

+y3

z ·√y3 + 8

+z3

x ·√z3 + 8

≥

Çx2

4√x3 + 8

+y2

4√y3 + 8

+z2

4√z3 + 8

å2

3.

But3(xy + yz + ax) ≤ (x+ y + z)2 = 9,

soÇx2

4√x3 + 8

+y2

4√y3 + 8

+z2

4√z3 + 8

å2

xy + yz + zx≥

Çx2

4√x3 + 8

+y2

4√y3 + 8

+z2

4√z3 + 8

å2

3.


Solutions /221

Thus

x3

y ·√x3 + 8

+y3

z ·√y3 + 8

+z3

x ·√z3 + 8

≥

Çx2

4√x3 + 8

+y2

4√y3 + 8

+z2

4√z3 + 8

å2

3.

We will show that Çx2

4√x3 + 8

+y2

4√y3 + 8

+z2

4√z3 + 8

å2

3≥ 1,

or equivalently

x2

4√x3 + 8

+y2

4√y3 + 8

+z2

4√z3 + 8

≥√

3.

Consider the function f : (0, 3)→ R defined by

f(t) =t2

4√t3 + 8

.

For all t ∈ (0, 3), we have

f ′′(t) = k(5t6 − 64t3 + 2048

) (t3 + 8

)−9/4,

where k is a positive constant, so that f ′′(t) > 0. Hence f is convex.

By Jensen’s inequality,

f(x) + f(y) + f(z) ≥ 3f(1) =√

3;

that is,x2

4√x3 + 8

+y2

4√y3 + 8

+z2

4√z3 + 8

≥√

3.

Thusx3

y ·√x3 + 8

+y3

z ·√y3 + 8

+z3

x ·√z3 + 8

≥ 1.

Note that if x = y = z = 1, then the left-hand side of the last inequality equals 1.In conclusion,

min

Çx3

y ·√x3 + 8

+y3

z ·√y3 + 8

+z3

x ·√z3 + 8

å= 1.


222/ Solutions

4350. Proposed by Leonard Giugiuc.

Let f : [0, 1] 7→ R be a decreasing, differentiable and concave function. Prove that

f(a) + f(b) + f(c) + f(d) ≤ 3f(0) + f(d− c+ b− a),

for any real numbers a, b, c, d such that 0 ≤ a ≤ b ≤ c ≤ d ≤ 1.

We received 5 correct solutions. We present the solution by AN-anduud ProblemSolving Group.

Since f(x) is decreasing, we have

f(a) ≤ f(0), f(b) ≤ f(0), f(c) ≤ f(0), (1)

and0 ≤ (d− c) + (b− a) = d− (c− b)− a ≤ d ≤ 1,

so thatf(d) ≤ f(d− c+ b− a). (2)

From (1) and (2), we get

f(a) + f(b) + f(c) + f(d) ≤ 3f(0) + f(d− c+ b− a).

Editor’s comments. Note that all of the solvers, other than the proposer, explicitlyor implicitly noted that neither concavity nor differentiability were needed.


Documents

SOLUTIONS - cms.math.ca · triangle passes through the midpoint of a side of the orthic triangle. This result is This result is an immediate consequence of the \bisection property"