IMO 1997 Solutions

7/24/2019 IMO 1997 Solutions

1/10

IMO 1997

Solutions

[Note: From the Post-IMO97 Web Pagein Argentina]

Problem 1

(a) LetABCbe a right-angled triangle whose vertices have integer coordinates and whose

legs lie along edges of the squares with


2/10

(c) Let us co'"utef(%k2 #,%k). /s in (b) we will consider a "ointLonABsuch thatAL=

%k. /sf(%k, %k) = 0 and S#(LBC) = S%(ALC) we have

f(%k2 #,%k) = *S#(LBC) - S%(LBC) *.

he area of the triangleLBCis k. u""ose without loss of generalit! that the diagonalLC

is all blac$ (see figure below). hen the white "art ofLBCconsists of several trianglesBLN%k,M%k-#L%k-#N%k-#, ... ,M#L#N#, each of the' being si'ilar toBAC. heir total area is

.

herefore

and thus

his function ta$es arbitraril! large values.


3/10

Problem 2


4/10

Let us denote the dia'eters which are "er"endicular bisectors ofABandACb! band c

res"ectivel!. Let us fix an orientation for arcs. hen the notation PQ defines a

unique arc on the circle.

LetX, Ybe the second intersections of the circle with the linesBVand CWres"ectivel!.hen the chord CYis the 'irror i'age of the chordAUin c.

hus YC = AU. i'ilarl! XB = AU. hen XB = YCand

hence the line seg'entsBXand YCare s!''etric with res"ect to the dia'eter dwhich"asses through the 'id"oint ofBY, withXbeing the 'irror i'age of Cand Ybeing the

'irror i'age ofB.

/s the "oint Tlies on the line d, the "er"endicular bisector of bothBYand CX, we have

TB= TYand TC = TX.

henAU = BX = BT 2 TX = BT 2CT.

Problem 3

or an! "er'utation = (y#,y%, ... ,yn) of (x#,x%, ... ,xn) denote b! S( ) the value

of the su'y#2 %y%2 ... 2 nyn. Let r= (n2 #)+%. t has to be shown that *S( )* r

for so'e "er'utation .

Let 0be the identit! "er'utation 0= (x#, ... ,xn) and let be the reverse

"er'utation = (xn , ... ,x#). f *S( 0)* ror*S( )* r, the clai' is true.

/ssu'e for the sequel that *S( 0)* 3 rand *S( )* 3 r. 4ote that

S( 0) 2 S( ) = (x#2 %x%2 ... 2 nxn) 2 (xn2 %xn-#2 ... 2 nx#) = (n2 #)(x#2 ... 2xn),

and hence | S( 0) 2 S( ) * = n2 # = %r. ince each one of the nu'bers S( 0)

and S( ) exceeds rin absolute value, the! 'ust have o""osite signs. /ccordingl!, one of

the' is grater than rand the other one is s'aller than -r.


5/10

tarting fro' 0, we are able to obtain an! "er'utation b! successive trans"ositions

of neighbouring ele'ents. n "articular, there exists a chain of "er'utations 0, #,

... , msuch that m= and, for each i 0, ... , m-#1, the "er'utation

i2#arises fro' ib! interchanging two of its neighbouring ter's.

his 'eans that if i= (y#, ... ,yn) and i2#= (#, ... , 5n), then there is an index k

#, ... , n- #1 such that

k= yk2#, k2#= yk6 != y! for! k, k2 #.

/nd since the nu'bersxido not exceed rin absolute value,

| S( i2#) - S( i) * = * kk2 (k2 #)k2#- kyk(k2 #)yk2#*

= *yk-yk2#* *yk* 2 *yk2#* %r.

his shows that the distance between an! two successive nu'bers in the sequence S(

0), S( i) , ... , S( m) is not greater than %r.

7ecall that the nu'bers S( 0) and S( m), regarded as "oints of the real line, lie

"utside the interval 8-r, r, on distinct sides of it. t follows that at least one of the

nu'bers S( i) has to hit an interval. hus we have S( i) rfor a certain i,and the assertion is "roved.

Problem 4

(a) Let n3 # be an integer. u""ose that a silver nx n'atrixAexists. Letxbe a fixedele'ent of #, %, ... , %n-#1 which does not a""ear on the diagonal ofA. (uch an

ele'ent exists, since there are onl! nele'ents on the diagonal but %n- # ele'ents in

total.) he union of theith row and the ith colu'n we shall call it ith cross. he ele'entx

a""ears in each cross exactl! once. fxis the (i,!)th entr! ofA, then it belongs to the ithcross and also to the!th cross. n this case we sa! that these crosses arex-lin$ed. (or


6/10

exa'"le, in the 'atrixAbelow the #st and the :th crosses are ;-lin$ed.) his i'"lies that

all ncrosses are "artitioned into "airs ofx-lin$ed ones, and therefore nis even. ut #99

is an odd nu'ber.

(b) or n= %

is a silver 'atrix. or n= : we have alread! several of the', for exa'"le

he construction ofAcan be generali5ed. u""ose that an nxnsilver 'atrix exists. hen

%nx %nsilver 'atrixDcan be constructed as follows>

whereBis the nx n'atrix obtained fro'Ab! adding %nto each entr!, and Cis the

'atrix obtained fro'Ab! adding %nto each entr!, and Cis the 'atrix obtained fro'B

b! re"lacing all its diagonal entries b! %n. his 'atrix will be again a silver 'atrix. o

"rove this let us consider the ith cross ofD. u""ose that i n6 the other case is

si'ilar. his cross is co'"osed of the ith cross ofA, the ith row ofBand the ith colu'nof C. he ith cross ofAcontains nu'bers #, %, ... , %n-#1. he ith row ofBand the ith

colu'n of Ctogether contain the nu'bers %n, %n2 # , ... , :n- #1.

Problem 5

Let ", bbe a solution, let dbe the greatest co''on divisor of "and band let "= d#, b=d$, with integers #, $being co"ri'e, he given equation is equivalent to

(d#)d$% = (d$)#. (#)

o'"aring the ex"onents d$%and #in (#), we distinguish three cases.


7/10

C"%& #. d$%= #. hen (#) i'"lies # =$. ince #, $are co"ri'e, we get # = $ = #6 and d$%

= #!ields d =#. ence " =b = #, which is a solution.

C"%&%. d$%3 #. 7ewriting (#) as

dd$%-u

. #d$%

= $#

, (%)

we see that #d$%divides $#. ince #, $are co"ri'e, # =# and (%) beco'es

dd$%-# = $. (?)

4ow, if d=# then b! (?) $= # and the inequalit! d$% 3 #fails to hold. /nd for d % we

have

dd$%-# %%$%-# %%$-#3 $ for $= #, %, ?, ... 6

he last inequalit! is easil! verified b! induction on $. his contradicts (?) and shows

that there are no solutions in this case.

C"%&?. d$%' #. (ence #3 d.) 7ewriting (#) as

#d$%= d#(d$%. $#, (:)

&e see that $#divides #d$%. ince #, $are co"ri'e, $= # and (:) beco'es

#d= d#(d. (@)

he ex"onentiation bases in (@) satisf! the inequalit! #3 d6 hence the ex"onents satisf!

the o""osite inequalit! d< # ( d.

! (@), an! "ri'e divisor)is of dis also a divisor of #. a$ingy,to be the greatest

ex"onents with)y* #,) | d, we obtain fro' (@)yd = (# ( d), which shows thaty3.

hus ddivides #, so # = kd or a certain integer k. 4ote that k ? because #3%d.ubstituting # = kdinto (@), we get

k= dk(%. (;)

t follows that dcannot be #. herefore d %.


8/10

f k= ?, then b! (;) d= ?, whence #= 9, "= %, b= ?.

f k= :, then b! (;) d%= :, whence d= %, #= A, "= #;, b= %.

f k @, then dk-% %k-%3 k(b! eas! induction) and (;) fails.

hus, the equation has three solutions> (", b) = (#, #), (%, ?) and (#;, %).

Problem 6

f n= %k2 # is an! odd integer greater than #, then ever! re"resentation of nin the given

for' has a B#B as one of its su''ands. Celeting we obtain a re"resentation of the nu'ber%k. /nd conversel!, b! attaching a B#B to a re"resentation of %kwe obtain a re"resentation

of %k2 #. he corres"ondence that results is obviousl! biDective. he recursion for'ula

follows>

f (%k2 #) =f(%k). (#)

urther, if n= %kis an! "ositive even integer, then ever! re"resentation of nin the givenfor' is one of the following two t!"es> either it contains so'e ter's equal to # or it does

not contain such ter's. n the first case we can delete one B#B to obtain a re"resentation

of %k- #> as before, we see that there is a biDection between the Bfirst-t!"eB

re"resentations of %kand all re"resentations of %k- #. n the case of a Bsecond-t!"eBre"resentation (with no ter's equal to #), we can divide all its ter's b! % and obtain a

re"resentation of k6 and also this corre"ondence is biDective. he second recursion

for'ula follows>

f (%k) =f(%k- #) 2f(k). (%)

oth of these for'ulae hold for an! integer k #. Ebviousl!,f(#) = #. Let us

additionall! definef(0) = #6 then the for'ula (#) holds for k= 0 as well. t follows fro'(#) and (%) that the functionfis non-decreasing.

/ccording to (#), the nu'berf(%k- #) in the for'ula (%) can be re"laced b!f(%k- %),

!ielding the equation

f (%k) -f(%k- %) =f(k) for k= #, %, ?, ... .

a$ing an! integer n # and su''ing these equations u" over k= # , ... , n, we obtain

the following for'ula>


9/10

f (%n) =f(0) 2f(#) 2 ... 2f(n) for n= #, %, ?, ... . (?)

&e "ass the two-sided esti'ation of the sequence (f(%n)> n= #, %, ?, ... ). he u""er

esti'ate is eas!> in the for'ula (?) all the su''ands are not greater than the last one.

/nd since % =f(%) f(n) for n %, we get

f (%n) = % 2 (f(%) 2 ... 2f(n) ) % 2 (n - #)f(n)

f(n) 2 (n- #)f(n) = nf(n) for n= %, ?, :, ... .

onsequentl!,

f(%n

) %n-#

.f(%n-#

) %n-#

. %n-%

.f(%n-%

) %n-#

. %n-%

. %n-?

.f(%n-?

)

... %(n-#) 2 (n-%) 2 ... 2 # .f(%) = %n(n-#)+%. %.

/nd since %n(n-#)+%. % < %n%+%for n ?, the u""er esti'ate results.

o wor$ out the lower esti'ate, we first show that

f(b2 #) -f(b) f("2 #) -f(")

(:)

whenever b " 0 are integers of the sa'e "art!. ndeed> if "and bare both even,

then in view of (#) we have 5eros on both sides of (:)6 and if "and bare odd, then b! (%)

we getf(b2 #) -f(b) =f( (b2 #)+% ),f("2 #) -f(") =f( ("2 #)+% ) and the inequalit! in (:)holds because the functionfis non-decreasing.

Let us ta$e an! integers r k #, reven, and substitute subsequentl! in (:) thenu'bers " = r ( !, b = r 2!for!= 0, ... , k ( #. hen add the inequalities that result, to

obtain

f(r2 k) -f(r) f(r2 #) -f(r ( k 2 #).

ince ris even, we havef(r2 #) =f(r), and hence


10/10

f(r2 k) 2f(r ( k 2 #) * f(r) for k= #, ... , r..

u''ing these inequalities u" over k= # , ... , rwe get

f(#) 2f(%) 2 ... 2f(%r) %rf(r).

n view of (?) the su' on the left-hand side is equal to f(:r) - #. hus

f(:r) %r f(r) 2 # 3 %rf(r) for ever! integer r %.

a$ing r= %m-%we hence obtain

f(%m) 3 %m-#f(%m-%). (@)

o ensure that r= %m-%is even, mought to be an integer greater than %6 note, however, that(@) is also true for m= %.

inall!, letnbe an! integer greater than #. f +is a "ositive integer such that %+ n,

then a""l!ing the inequalit! (@) for m= n, n-#, ... , n- %+ , % we get

f(%n) 3%n-#.f(%n-%) 3 %n-#. %n-?.f(%n-:) 3 %n-#. %n-?.%[email protected](%n-;)

3 ... 3 %(n-#) 2 (n-?) 2 ... 2 (n- %+2 #) .f(%n-%+) = %+(n ( +).f(%n- %+).

4ow, if n is even, ta$e += n+%6 and if nis odd, ta$e += (n- #)+%. he resulting inequalities

are>

f(%n) 3 %n%+:.f(%0) = %n%+: for neven,

f(%n) 3 %(n% - #)+:.f(%#) = %(n% - #)+:. % 3 %n%+: for nodd.

&e have obtained the required lower esti'ate for ever! integer n %. (t is also validfor n= #, as can be verified directl!.)

Documents

IMO 1997 Solutions