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M
ath
Problem
Book
I
compiledby
K
in
Y�Li
DepartmentofMathematics
HongKongUniversityofScienceandTechnology
Copyrightc �����HongKongMathematicalSocietyIMO�HK�Committ
PrintedinHongKong
Preface
Thereareover�ftycountriesintheworldnowadaysthatholdmath
ematicalolympiadsatthesecondaryschoollevelannually�InHungary
RussiaandRomaniamathematicalcompetitionshavealonghistorydat
ingbacktothelate�����sinHungary�scase�Manyprofessionalorama
teurmathematiciansdevelopedtheirinterestinmathbyworkingonthese
olympiadproblemsintheiryouthsandsomeintheiradulthoodsaswell�
Theproblemsinthisbookcamefrommanysources�Forthoseinvolved
ininternationalmathcompetitionstheynodoubtwillrecognizemanyof
theseproblems�Wetriedtoidentifythesourceswheneverpossiblebut
therearestillsomethatescapeusatthemoment�Hopefullyinfuture
editionsofthebookwecan�llinthesemissingsourceswiththehelpofthe
knowledgeablereaders�
Thisbookisforstudentswhohavecreativemindsandareinterestedin
mathematics�Throughproblemsolvingtheywilllearnagreatdealmore
thanschoolcurriculacano erandwillsharpentheiranalyticalskills�We
hopetheproblemscollectedinthisbookwillstimulatethemandseduce
themtodeeperunderstandingofwhatmathematicsisallabout�Wehope
theinternationalmathcommunitiessupportoure ortsforusingthesebril
liantproblemsandsolutionstoattractouryoungstudentstomathematics�
Mostoftheproblemshavebeenusedinpracticesessionsforstudents
participatedintheHongKongIMOtrainingprogram�Weareespecially
pleasedwiththee ortsofthesestudents�Infacttheoriginalmotivation
forwritingthebookwastorewardtheminsomewaysespeciallythosewho
workedsohardtobecomereserveorteammembers�Itisonly�ttingto
listtheirnamesalongwiththeirsolutions�Againthereareunsungheros
iii
whocontributedsolutionsbutwhosenameswecanonlyhopetoident i
infutureeditions�
Asthetitleofthebooksuggestthisisaproblembook�Soverylitt
introductionmaterialscanbefound�Wedopromisetowriteanotherbo
presentingthematerialscoveredintheHongKongIMOtrainingprogram
Thisforcertainwillinvolvethededicationofmorethanoneperson�Als
thisisthe�rstofaseriesofproblembookswehope�Fromtheresults
theHongKongIMOpreliminarycontestswecanseewavesofnewcreati
mindsappearinthetrainingprogramcontinuouslyandtheyareyoung
andyounger�Maybethenextproblembookintheserieswillbewritten
ourstudents�
FinallywewouldliketoexpressdeepgratitudetotheHongKo
QualityEducationFundwhichprovidedthesupportthatmadethisbo
possible�
KinY�
HongKo
April���
iv
Advicesto
the
Readers
Theonlywaytolearnmathematicsistodomathematics�Inthis
bookyouwill�ndmanymathproblemsrangingfromsimpletochallenging
problems�Youmaynotsucceedinsolvingalltheproblems�Veryfew
peoplecansolvethemall�Thepurposesofthebookaretoexposeyouto
manyinterestingandusefulmathematicalideastodevelopyourskillsin
analyzingproblemsandmostimportantofalltounleashyourpotential
ofcreativity�Whilethinkingabouttheproblemsyoumaydiscoverthings
youneverknowbeforeandputtinginyourideasyoucancreatesomething
youcanbeproudof�
Tostartthinkingaboutaproblemveryoftenitishelpfultolookat
theinitialcasessuchaswhenn���������Thesecasesaresimpleenough
toletyougetafeelingofthesituations�Sometimestheideasinthese
casesallowyoutoseeapatternwhichcansolvethewholeproblem�For
geometryproblemsalwaysdrawapictureasaccurateaspossible�rst�
Haveprotractorrulerandcompassreadytomeasureanglesandlengths�
Otherthingsyoucantryintacklingaproblemincludechangingthe
givenconditionsalittleorexperimentingwithsomespecialcases�rst�
Sometimesmaybeyoucanevenguesstheanswersfromsomecasesthen
youcanstudytheformoftheanswersandtracebackward�
Finallywhenyou�gureoutthesolutionsdon�tjuststopthere�You
shouldtrytogeneralizetheproblemseehowthegivenfactsarenecessary
forsolvingtheproblem�Thismayhelpyoutosolverelatedproblemslater
on�Alwaystrytowriteoutyoursolutioninaclearandconcisemanner�
Alongthewayyouwillpolishtheargumentandseethestepsoftheso
lutionsmoreclearly�Thishelpsyoutodevelopstrategiesfordealingwith
otherproblems�
v
Thesolutionspresentedinthebookarebynomeanstheonlywa
todotheproblems�Ifyouhaveaniceelegantsolutiontoaprobleman
wouldliketosharewithothers�infutureeditionsofthisbook�pleasesen
ittousbyemailatmakyli�ust�hk�Alsoifyouhavesomethingyoucann
understandpleasefeelfreetocontactusbyemail�Wehopethisbookw
increaseyourinterestinmath�
Finallywewillo eronelastadvice�Don�tstartwithproblem��Re
thestatementsoftheproblemsandstartwiththeonesthatinterestyout
most�Werecommendinspectingthelistofmiscellaneousproblems�rst�
Haveafuntime�
vi
Table
ofContents
Preface����������������������������������������������������������������iii
AdvicestotheReaders�������������������������������������������������v
Contributors����������������������������������������������������������ix
AlgebraProblems�������������������������������������������������������
GeometryProblems�����������������������������������������������������
NumberTheoryProblems�����������������������������������������������
CombinatoricsProblems�������������������������������������������������
MiscellaneousProblems�������������������������������������������������
SolutionstoAlgebraProblems�������������������������������������������
SolutionstoGeometryProblems�����������������������������������������
SolutionstoNumberTheoryProblems����������������������������������
SolutionstoCombinatoricsProblems������������������������������������
SolutionstoMiscellaneousProblems�������������������������������������
Contributors
ChanKinHang����������������HongKongteammember
ChanMingChiu����HongKongteamreservemember
ChaoKhekLun����HongKongteammember
ChengKeiTsi����HongKongteammember
CheungPokMan��������HongKongteammember
FanWaiTong����HongKongteammember
FungHoYin����HongKongteamreservemember
HoWingYip������������HongKongteammember
KeeWingTao����HongKongteamreservemember
LamPoLeung����HongKongteamreservemember
LamPeiFung����HongKongteammember
LauLapMing��������HongKongteammember
LawKaHo������������HongKongteammember
LawSiuLung����HongKongteammember
LeeTakWing����HongKongteamreservemember
LeungWaiYing����HongKongteammember
LeungWingChung��������HongKongteammember
MokTzeTao������������HongKongteammember
NgKaMan����HongKongteamreservemember
NgKaWing��������HongKongteammember
PoonWaiHoi������������HongKongteammember
PoonWingChi����HongKongteamreservemember
TamSiuLung����HongKongteamreservemember
ToKarKeung��������HongKongteammember
WongChunWai��������HongKongteammember
WongHimTing��������HongKongteammember
YuKaChun����HongKongteammember
YungFai����HongKongteammember
ix
Problem
s
AlgebraProblems
Polynomials
���CruxMathematicorumProblem
��Find�withoutcalculus�a�fth
degreepolynomialp�x�suchthatp�x���isdivisibleby�x����and
p�x���isdivisibleby�x�����
��ApolynomialP�x�ofthenthdegreesatis�esP�k��
�kfork�
����������n�FindthevalueofP�n����
�������PutnamExam�LetP�x�beapolynomialwithrealcoe�cients
suchthatP�x���foreveryrealx�Provethat
P�x��f ��x���f ��x�������f n�x��
forsomepolynomialsf ��x��f ��x������f n�x�withrealcoe�cients�
�������RussianMathOlympiad�Isitpossibleto�ndthreequadratic
polynomialsf�x��g�x��h�x�suchthattheequationf�g�h�x�����has
theeightroots����������������
�������PutnamExam�Determineallpolynomialswhosecoe�cientsare
all��thathaveonlyrealroots�
�������Putnam
Exam�Isthereanin�nitesequencea��a��a�����of
nonzerorealnumberssuchthatforn�
����������thepolynomial
Pn�x��a��a�x�a�x������anxnhasexactlyndistinctrealroots�
�������AustrianPolishMathCompetition�LetP�x�beapolynomial
withrealcoe�cientssuchthatP�x���for��x���Showthat
therearepolynomialsA�x��B�x��C�x�withrealcoe�cientssuchthat
�a�A�x����B�x����C�x���forallrealxand
�b�P�x��A�x��xB�x�����x�C�x�forallrealx�
�ForexampleifP�x��x���x��thenP�x����x���x������x�x���
�
�������IMO�Letf�x��xn��xn�����wheren��isaninteg e
Provethatf�x�cannotbeexpressedasaproductoftwopolynomia
eachhasintegercoe�cientsanddegreeatleast��
��Provethatiftheintegeraisnotdivisibleby�thenf�x��x��x�
cannotbefactoredastheproductoftwononconstantpolynomialswi
integercoe�cients�
��������SovietMathOlympiad�Given�ndistinctnumbersa��a������a
b ��b������bn�ann�ntableis�lledasfollows�intothecellinthei
rowandjthcolumniswrittenthenumberai�b j�Provethatift
productofeachcolumnisthesamethenalsotheproductofeachro
isthesame�
���Leta��a������anandb ��b������bnbetwodistinctcollectionsofnpo
itiveintegerswhereeachcollectionmaycontainrepetitions�Ifthetw
collectionsofintegersai�aj���i�j�n�andb i�b j���i�j�
arethesamethenshowthatnisapowerof��
RecurrenceRelations
���Thesequencexn
isde�nedby
x����
xn���
��xn
���xn
�
n�����������
Provethatxn��
� �
or�forallnandthetermsofthesequenceare
distinct�
��������NanchangCityMathCompetition�De�nea����a���an
an���
a� n����
an
forpositiveintegern�Provethat�anan����i s
perfectsquareforeverypositiveintegern�
����ProposedbyBulgariafor����IMO�De�nea����a���andan
�an���an��forn���Showthatforpositiveintegerk�anisdivisib
by�kifandonlyifnisdivisibleby�k�
�
����AmericanMathematicalMonthlyProblem
E�����Letxandybe
distinctcomplexnumberssuchthat
xn�yn
x�y
isanintegerforsome
fourconsecutivepositiveintegersn�Showthat
xn�yn
x�y
isaninteger
forallpositiveintegersn�
Inequalities
���Forrealnumbersa��a��a������ifan���an����an
forn���������
thenprovethat
An���An����An
forn���������
whereAn
istheaverageofa��a������an�
���Leta�b�c��andabc���Provethat
a c�
b a�
c b�a�b�c�
��������MoscowMathOlympiad�Usetheidentity����������n��
n��n����
�
toprovethatfordistinctpositiveintegersa��a������an�
�a� ��a
� ������a
� n���a
� ��a
� ������a
� n����a
� ��a
� ������a
� n���
Canequalityoccur�
��������IMOshortlistedproblem�Leta������an
�an����bea
sequenceofrealnumbers�Provethat
v u u tn X k��
ak�
n X k��
p k�pak�p ak����
�
��������ChineseTeamSelectionTest�For��a�b�c�d�ea n
a�b�c�d�e���showthat
ad�dc�cb�be�ea�
� ��
��������WuhuCityMathCompetition�Letx�y�zberealnumberssu
thatx�y�z���Showthat
��x
��y
��z
�����x
��y
��z
����
��������IMO�Letnbea�xedintegerwithn���
�a�DeterminetheleastconstantCsuchthattheinequality
X��i�j�n
xixj�x
� i�x
� j��C
� X ��i�n
xi�
holdsforallnonnegativerealnumbersx��x������xn�
�b�ForthisconstantC�determinewhenequalityholds�
��������BulgarianMathCompetition�Letn��and��xi��f
i���������n�Provethat
�x��x������xn���x�x��x�x������xn��xn�xnx���
h n �i
where�x�isthegreatestintegerlessthanorequaltox�
���Foreverytripletoffunctionsf�g�h������R�provethattherea
numbersx�y�zin�����suchthat
jf�x��g�y��h�z��xyzj�
� ��
����ProposedbyGreatBritainfor����IMO�Ifx�y�zarerealnumbe
suchthatx��y��z����thenshowthatx�y�z�xyz���
�
����ProposedbyUSAfor����IMO�Provethatforpositiverealnumbers
a�b�c�d�
a
b��c��d
�
b
c��d��a
�
c
d��a��b
�
d
a��b��c
�� ��
���Leta��a������an
andb ��b������bn
be�npositiverealnumberssuch
that
�a�a��a������an
and
�b�b �b ����bk�a�a����akforallk���k�n�
Showthatb ��b ������b n�a��a������an�
����ProposedbyGreecefor����IMO�Leta�b�c��andmbeapositive
integerprovethat
am
b�c
�
bmc�a
�
cma�b
�� �� a�b�c
�
� m���
���Leta��a������an
bedistinctpositiveintegersshowthat
a� ��
a� ������
ann�n
���
� �n�
��������WestGermanMathOlympiad�Ifa��a������an
��anda�
a��a������an�thenshowthat
n X i��
ai
�a�ai
�
n�n��
�
���Provethatifa�b�c���then
a�
b�c
�
b�c�a
�
c�a�b
�a��b��c�
�
�
���Leta�b�c�d��and �
��a�
���b
�
���c
�
���d���
�
Provethatabcd���
����DuetoPaulErd�os�Eachofthepositiveintegersa������anislessth
�����Theleastcommonmultipleofanytwooftheseisgreaterth
�����Showthat
� a�
�����
� an
���
n����
�
���Asequence�Pn�ofpolynomialsisde�nedrecursivelyasfollows�
P��x���
andforn���
Pn���x��Pn�x��
x�Pn�x��
�
�
Provethat
��
p x�Pn�x��
�n��
foreverynonnegativeintegernandallxin������
��������IMOshortlistedproblem�LetP�x�betherealpolynomialfun
tionP�x��ax��bx��cx�d�ProvethatifjP�x�j��forallxsu
thatjxj���then
jaj�jbj�jcj�jdj���
����AmericanMathematicalMonthlyProblem�����LetP�z��az�
bz��cz�d�wherea�b�c�darecomplexnumberswithjaj�jbj�jcj
jdj���ShowthatjP�z�j�
p �foratleastonecomplexnumber
satisfyingjzj���
��������HungarianIsraeliMathCompetition�Findallrealnumbers
withthefollowingproperty�foranypositiveintegern�thereexists
integermsuchthat
� � ���m n� � ��� �n�
��������BritishMathOlympiad�Ifnisapositiveintegerdenotebyp�
thenumberofwaysofexpressingnasthesumofoneormorepositi
integers�Thusp������asthereare�vedi erentwaysofexpressi
�intermsofpositiveintegers�namely
����������������������and��
�
Provethatp�n�����p�n��p�n�����foreachn���
FunctionalEquations
���Findallpolynomialsfsatisfyingf�x���f�x�f�x������
��������GreekMathOlympiad�Letf�����Rbeafunctionsuch
that
�a�fisstrictlyincreasing
�b�f�x���
� x
forallx��and
�c�f�x�f�f�x��� x���forallx���
Findf����
��������E�otv�osK�ursch�akMathCompetition�Thefunctionfisde�ned
forallrealnumbersandsatis�esf�x��xandf�x�y��f�x��f�y�
forallrealx�y�Provethatf�x��xforeveryrealnumberx�
����ProposedbyIrelandfor����IMO�Supposef�R
Rsatis�es
f������f�a�b��f�a��f�b�foralla�b�Randf�x�f�
� x���for
x����Showthatf�x��xforallx�
��������PolishMathOlympiad�LetQ�
bethepositiverationalnumbers�
Determineallfunctionsf�Q�
Q�
suchthatf�x����f�x���
andf�x���f�x��foreveryx�Q��
��������IMOshortlistedproblem�LetRdenotetherealnumbersand
f�R������satisfy
f� x
��� ��� �
f�x��f
� x�
� �� �
f� x
�� ��
foreveryx�R�Showthatfisaperiodicfunctioni�e�thereisa
nonzerorealnumberTsuchthatf�x�T��f�x�foreveryx�R�
���LetNdenotethepositiveintegers�Supposes�NNisanincreasing
functionsuchthats�s�n����nforalln�N�Findallpossiblevalues
ofs�������
�
���LetNbethepositiveintegers�Isthereafunctionf�NNsucht h
f���� �n���nforalln�N�wheref� �x��f�x�andfk�� �x�
f�fk �x���
����AmericanMathematicalMonthlyProblemE����LetRdenotet
realnumbers�Findallfunctionsf�RRsuchthatf�f�x���x��
orshownosuchfunctioncanexist�
���LetRbetherealnumbers�Findallfunctionsf�RRsuchthatf
allrealnumbersxandy�
f� xf�y��x
� �xy�f�x��
��������IMO�Determineallfunctionsf�RRsuchthat
f�x�f�y���f�f�y���xf�y��f�x���
forallx�yinR�
��������ByelorussianMathOlympiad�LetRbetherealnumbers�Fin
allfunctionsf�RRsuchthat
f�f�x�y���f�x�y��f�x�f�y��xy
forallx�y�R�
��������CzechoslovakMathOlympiad�LetZbetheintegers�Find
functionsf�ZZsuchthat
f�����f���andf�x��f�y��f�x��xy��f�y��xy�
forallintegersx�y�
��������SouthKoreanMathOlympiad�LetAbethesetofnonnegati
integers�Findallfunctionsf�AAsatisfyingthefollowingtw
conditions�
�a�Foranym�n�A��f�m��n����f�m�����f�n����
�
�b�Foranym�n�Awithm�n�f�m���f�n���
����AmericanMathematicalMonthlyProblemE�����LetQdenotethe
rationalnumbers�Findallfunctionsf�QQsuchthat
f�����
and
f� x�y
x�y
� �f�x��f�y�
f�x��f�y�
forx��y�
����MathematicsMagazineProblem�����Findallfunctionsf�RR
suchthat
f�x�yf�x���f�x��xf�y�
forallx�yinR�
Maximum�Minimum
��������AustrianMathOlympiad�Forpositiveintegersn�de�ne
f�n���n��n����n��������n���
���n���
��n�
Whatistheminimumoff�n����f�n��
��������PutnamExam�Giventhatfx��x������xng�f��������ng��nd
thelargestpossiblevalueofx�x��x�x������xn��xn�xnx�interms
ofn�withn����
�
GeometryProblems
��������BritishMathOlympiad�TriangleABChasarightangleat
TheinternalbisectorsofanglesBACandABCmeetBCandC
atPandQrespectively�ThepointsM
andN
arethefeetoft
perpendicularsfromPandQtoAB�FindangleMCN�
��������LeningradMathOlympiad�SquaresABDE
andBCFG
a
drawnoutsideoftriangleABC�ProvethattriangleABCisisosceles
DGisparalleltoAC�
���ABisachordofacirclewhichisnotadiameter�ChordsA�B�an
A�B�intersectatthemidpointPofAB�Letthetangentstothecirc
atA�andB�intersectatC��Similarlyletthetangentstothecirc
atA�andB�intersectatC��ProvethatC�C�isparalleltoAB�
��������HunanProvinceMathCompetition�TwocircleswithcentersO
andO�intersectatpointsAandB�AlinethroughAintersectst
circleswithcentersO�
andO�
atpointsY�Z�respectively�Lett
tangentsatYandZintersectatX
andlinesYO�andZO�interse
atP�Letthecircumcircleof�O�O�BhavecenteratOandinterse
lineXBatBandQ�ProvethatPQisadiameterofthecircumcirc
of�O�O�B�
��������BeijingCityMathCompetition�InadiskwithcenterO�the
arefourpointssuchthatthedistancebetweeneverypairofthem
greaterthantheradiusofthedisk�Provethatthereisapairofpe
pendiculardiameterssuchthatexactlyoneofthefourpointsliesinsi
eachofthefourquarterdisksformedbythediameters�
���Thelengthsofthesidesofaquadrilateralarepositiveintegers�T
lengthofeachsidedividesthesumoftheotherthreelengths�Pro
thattwoofthesideshavethesamelength�
��������SichuanProvinceMathCompetition�Supposethelengthsoft
threesidesof�ABCareintegersandtheinradiusofthetriangleis
Provethatthetriangleisarighttriangle�
��
GeometricEquations
��������IMO�AcirclehascenteronthesideABofthecyclicquadri
lateralABCD�Theotherthreesidesaretangenttothecircle�Prove
thatAD�BC�AB�
��������RussianMathOlympiad�CirclesS�andS�withcentersO��O�
respectivelyintersecteachotheratpointsAandB�RayO�Bintersects
S�atpointFandrayO�BintersectsS�atpointE�Thelineparallel
toEFandpassingthroughBintersectsS�andS�atpointsM
and
N�respectively�Provethat�Bistheincenterof�EAFand�MN�
AE�AF�
���PointCliesontheminorarcABofthecirclecenteredatO�Suppose
thetangentlineatCcutstheperpendicularstochordABthroughA
atEandthroughBatF�LetDbetheintersectionofchordABand
radiusOC�ProvethatCE�CF�AD�BDandCD��AE�BF�
���QuadrilateralsABCPandA� B� C� P�areinscribedintwoconcentric
circles�IftrianglesABCandA� B� C�areequilateralprovethat
P� A��P� B��P� C��PA���PB���PC���
���LettheinscribedcircleoftriangleABCtouchssideBCatDsideCA
atEandsideABatF�LetGbethefootofperpendicularfromDto
EF�Showthat
FG
EG
�BF
CE
�
��������IMOshortlistedproblem�LetABCDEFbeaconvexhexagon
suchthat
�B��D��F�����and
AB
BC
�CD
DE
�EF
FA
���
Provethat
BC
CA
�AE
EF
�FD
DB
���
��
SimilarTriangles
��������BritishMathOlympiad�P�Q�andRarearbitrarypointsont
sidesBC�CA�andABrespectivelyoftriangleABC�Provethatt
threecircumcentresoftrianglesAQR�BRP�andCPQformatriang
similartotriangleABC�
���HexagonABCDEFisinscribedinacirclesothatAB�CD�E
LetP�Q�RbethepointsofintersectionofACandBD�CEandD
EAandFB
respectively�ProvethattrianglesPQRandBDFa
similar�
��������IMOshortlistedproblem�LetABCDbeacyclicquadrilater
LetEandFbevariablepointsonthesidesABandCD�respective
suchthatAE�EB�CF�FD�LetPbethepointonthesegme
EFsuchthatPE�PF�AB�CD�Provethattheratiobetweent
areasoftrianglesAPDandBPCdoesnotdependonthechoiceof
andF�
TangentLines
���TwocirclesintersectatpointsAandB�Anarbitrarylinethrough
intersectsthe�rstcircleagainatCandthesecondcircleagainatD
Thetangentstothe�rstcircleatCandtothesecondcircleat
intersectatM�TheparalleltoCM
whichpassesthroughthepoi
ofintersectionofAM
andCDintersectsACatK�ProvethatBK
tangenttothesecondcircle�
��������IMO�Twocircles��and��arecontainedinsidethecircle
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whosedecimalrepresentationinvolvesalltendigits�
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valueofthenaturalnumbern��withthepropertythatwhenever
thesetSn
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positiveintegerkshowthatthereexistsapositiveintegermsuchthat
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Representations
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k����������thesumofthe�rstktermsisdivisiblebyk�
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divisibleby�n�
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therearein�nitelymanyperfectsquaresoftheformn�k���wheren
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lettersa�b�c��ndthenumberofwordsofnletterswhichcontain
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containasubsetAconsistingof����numberssuchthatx�Aimpl
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asquareofunitareainsuchawaythatthecenterofthesquareisn
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than��
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thereexistatleast��rectangleswithverticesofthesamecoloran
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testpaperineachofn�n���subjects�Itisknownthatforany
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conditionsimplythatexactlyonestudentgetsthebestscoreinevery
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placedatdi erentsquaresofa���squareboard�Ineachmoveevery
beetlecreepstoahorizontalorverticaladjacentsquare�Ifnobeetle
makeseithertwohorizontalmovesortwoverticalmovesinsuccession
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samesquare�
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areonaplanesuch
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thenumbers
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enoughtoguaranteeawinningticketingeneral�
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Aboytakesnrides�Betweeneachridehemovesclockwiseacerta
number�lessthann�ofplacestoanewhorse�Eachtimehemoves
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eachhorse�
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boardthatconsistsof���squares�PlayerIchoosesasquarean
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onitswayaroundthetrackintheclockwisedirection�
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eightpairwisedistinctnaturalnumbersandoneachofitsedgesis
writtenthegreatestcommondivisorofthenumbersattheendpoints
oftheedge�Canthesumofthenumberswrittenattheverticesbethe
sameasthesumofthenumberswrittenattheedges�
����Canthepositiveintegersbepartitionedintoin�nitelymanysubsets
suchthateachsubsetisobtainedfromanyothersubsetbyaddingthe
sameintegertoeachelementoftheothersubset�
���������RussianMathOlympiad�Isitpossibleto�llinthecellsofa
���tablewithpositiveintegersrangingfrom�to��insuchaway
thatthesumoftheelementsofevery���squareisthesame�
���������GermanMathematicalOlympiad�Showthatforeverypositive
integern���thereexistsapermutationp ��p������pn
of��������n
suchthatp k��dividesp ��p ������p kfork���������n���
����Eachlatticepointoftheplaneislabeledbyapositiveinteger�Each
ofthesenumbersisthearithmeticmeanofitsfourneighbors�above
belowleftright��Showthatallthenumbersareequal�
���������TournamentoftheTowns�Inapartynboysandngirlsare
paired�Itisobservedthatineachpairthedi erenceinheightisless
than��cm�Showthatthedi erenceinheightofthekthtallestboy
andthekthtallestgirlisalsolessthan��cmfork���������n�
���������LeningradMathOlympiad�Onemayperformthefollowingtwo
operationsonapositiveinteger�
�a�multiplyitbyanypositiveintegerand
�b�deletezerosinitsdecimalrepresentation�
ProvethatforeverypositiveintegerX�onecanperformasequenceof
theseoperationsthatwilltransformXtoaonedigitnumber�
��
���������IMOshortlistedproblem�Fourintegersaremarkedonacir c
Oneachstepwesimultaneouslyreplaceeachnumberbythedi eren
betweenthisnumberandnextnumberonthecircleinagivendirecti
�thatisthenumbersa�b�c�darereplacedbya�b�b�c�c�d�d�a
Isitpossibleafter����suchstepstohavenumbersa�b�c�dsuchth
thenumbersjbc�adj�jac�bdj�jab�cdjareprimes�
���������NanchangCityMathCompetition�Thereare����coinson
table�Someareplacedwiththeheadsidesupandsomethetailsid
up�Agroupof����personswillperform
thefollowingoperation
the�rstpersonisallowedturnoveranyonecointhesecondperson
allowedturnoveranytwocoins����thekthpersonisallowedtu
overanykcoins����the����thpersonisallowedtoturnovereve
coin�Provethat
���nomatterwhichsidesofthecoinsareupinitiallythe����perso
cancomeupwithaprocedureturningallcoinsthesamesidesu
attheendoftheoperations
���intheaboveprocedurewhethertheheadorthetailsidesturn
upattheendwilldependontheinitialplacementofthecoins�
�����ProposedbyIndiafor����IMO�Showthatthereexistsaconv
polygonof����sidessatisfyingthefollowingconditions�
�a�itssidesare��������������insomeorder�
�b�thepolygoniscircumscribableaboutacircle�
����Thereare��white��black��redchipsonatable�Inonestepy
maychoose�chipsofdi erentcolorsandreplaceeachonebyachip
thethirdcolor�Canallchipsbecomethesamecoloraftersomestep
����Thefollowingoperationsarepermittedwiththequadraticpolynom
ax��bx�c�
�a�switchaandc
�b�replacexbyx�t�wheretisarealnumber�
Byrepeatingtheseoperationscanyoutransformx��x��intox�
x���
��
����Fivenumbers���������arewrittenonablackboard�Astudentmay
eraseanytwoofthenumbersaandbontheboardandwritethe
numbersa�bandabreplacingthem�Ifthisoperationisperformedre
peatedlycanthenumbers����������������everappearontheboard�
����Nine���cellsofa�����squareareinfected�Inoneunittimethe
cellswithatleast�infectedneighbors�havingacommonside�become
infected�Cantheinfectionspreadtothewholesquare�Whatifnine
isreplacedbyten�
���������ColombianMathOlympiad�Weplaythefollowinggamewith
anequilateraltriangleofn�n�����dollarcoins�withncoinsoneach
side��Initiallyallofthecoinsareturnedheadsup�Oneachturnwe
mayturnoverthreecoinswhicharemutuallyadjacent�thegoalisto
makeallofthecoinsturnedtailsup�Forwhichvaluesofncanthisbe
done�
���������ChineseTeamSelectionTest�Everyintegeriscoloredwithone
of���colorsandall���colorsareused�Forintervals�a�b���c�d�having
integersendpointsandsamelengthsifa�chavethesamecolorand
b�dhavethesamecolorthentheintervalsarecoloredthesameway
whichmeansa�xandc�xhavethesamecolorforx���������b�a�
Provethat�����and����havedi erentcolors�
��
Solutions
SolutionstoAlgebraProblems
Polynomials
���CruxMathematicorumProblem
��Find�withoutcalculus�a�fth
degreepolynomialp�x�suchthatp�x���isdivisibleby�x����and
p�x���isdivisibleby�x�����
Solution��DuetoLawKaHoNgKaWingTamSiuLung�Note
�x����dividesp�x���andp��x����so�x����dividestheirsum
p�x��p��x��Also�x����dividesp�x���andp��x����so�x����
dividesp�x��p��x��Then�x�����x����dividesp�x��p��x��whichis
ofdegreeatmost��Sop�x��p��x���forallx�Thentheevendegree
termcoe�cientsofp�x�arezero�Nowp�x�����x�����Ax��Bx����
Comparingthedegree�and�coe�cientswegetB��A��and
���B�A���whichimpliesA�����andB������Thisyields
p�x����x�����x������x���
��ApolynomialP�x�ofthenthdegreesatis�esP�k��
�kfork�
����������n�FindthevalueofP�n����
Solution�For��r�n�thepolynomial� x r
� �x�x�������x�r���
r�
isofdegreer�Considerthedegreenpolynomial
Q�x��
� x �� �� x �� �
����
� x n� �
BythebinomialtheoremQ�k�������k��kfork�����������n�
SoP�x��Q�x�forallx�Then
P�n����Q�n����
� n��
�
� �� n��
�
� �����
� n��
n
� ��n
�����
�������PutnamExam�LetP�x�beapolynomialwithrealcoe�cients
suchthatP�x���foreveryrealx�Provethat
P�x��f ��x���f ��x�������f n�x��
��
forsomepolynomialsf ��x��f ��x������f n�x�withrealcoe�cients�
Solution��DuetoCheungPokMan�WriteP�x��aR�x�C�x��whe
aisthecoe�cientofthehighestdegreetermR�x�istheproductof
realrootfactors�x�r�repeatedaccordingtomultiplicitiesandC�x�
theproductofallconjugatepairsofnonrealrootfactors�x�zk��x�zk
Thena���SinceP�x���foreveryrealxandafactor�x�r��n
wouldchangesignneararealrootrofoddmultiplicityeachrealro
ofPmusthaveevenmultiplicity�SoR�x��f�x��forsomepolynom
f�x�withrealcoe�cients�
Nextpickonefactorfromeachconjugatepairofnonrealfacto
andlettheproductofthesefactors�x�z k�beequaltoU�x��iV�x
whereU�x��V�x�arepolynomialswithrealcoe�cients�Wehave
P�x��af�x���U�x��iV�x���U�x��iV�x��
��paf�x�U�x��
���paf�x�V�x��
��
�������RussianMathOlympiad�Isitpossibleto�ndthreequadrat
polynomialsf�x��g�x��h�x�suchthattheequationf�g�h�x�����h
theeightroots����������������
Solution�Supposetherearesuchf�g�h�Thenh����h��������h���w
betherootsofthe�thdegreepolynomialf�g�x���Sinceh�a�
h�b��a��bifandonlyifa�baresymmetricwithrespecttotheax
oftheparabolaitfollowsthath����
h����h����
h����h���
h����h����h���andtheparabolay�h�x�issymmetricwithr
specttox�����Alsowehaveeitherh����h����h����h���
h����h����h����h����
Nowg�h�����g�h�����g�h�����g�h����aretherootsofthequadra
polynomialf�x��sog�h�����g�h����andg�h�����g�h�����whi
impliesh����h����h����h����Forh�x��Ax��Bx�C�thiswou
forceA���acontradiction�
�������PutnamExam�Determineallpolynomialswhosecoe�cientsa
all��thathaveonlyrealroots� ��
Solution�Ifapolynomiala�xn�a�xn�������anissuchapolynomial
thensoisitsnegative�Hencewemayassumea����Letr ������rnbe
theroots�Thenr� ������r� n�a� ���a�andr� ����r� n�a� n�Iftheroots
areallrealthenbytheAMGMinequalityweget�a� ���a���n�a
��nn
�
Sincea��a�
�
���wemusthavea�
�
��andn���Bysimple
checkingwegetthelist
��x����
��x����
��x
��x����
��x
��x����
��x
��x
��x����
��x
��x
��x����
�������Putnam
Exam�Isthereanin�nitesequencea��a��a�����of
nonzerorealnumberssuchthatforn�
����������thepolynomial
Pn�x��a��a�x�a�x������anxnhasexactlyndistinctrealroots�
Solution�Yes�Takea�
�
��a�
�
��andproceedbyinduction�
Supposea������anhavebeenchosensothatPn�x�hasndistinctreal
rootsandPn�x�or�asxdependinguponwhethern
isevenorodd�SupposetherootsofPn�x�isintheinterval��T�T��
Letan�������n���MwhereM
ischosentobeverylargesothat
Tn���M
isverysmall�ThenPn���x��Pn�x����x�n���M
isvery
closetoPn�x�on��T�T�becausejPn���x��Pn�x�j�Tn���M
for
everyxon��T�T��SoPn���x�hasasignchangeveryclosetoevery
rootofPn�x�andhasthesamesignasPn�x�atT�SincePn�x�and
Pn���x�takeondi erentsignwhenxtheremustbeanother
signchangebeyondT�SoPn���x�musthaven��realroots�
�������AustrianPolishMathCompetition�LetP�x�beapolynomial
withrealcoe�cientssuchthatP�x���for��x���Showthat
therearepolynomialsA�x��B�x��C�x�withrealcoe�cientssuchthat
�a�A�x����B�x����C�x���forallrealxand
�b�P�x��A�x��xB�x�����x�C�x�forallrealx�
�ForexampleifP�x��x���x��thenP�x����x���x������x�x���
Solution��Belowallpolynomialshaverealcoe�cients��Weinduct
onthedegreeofP�x��IfP�x�isaconstantpolynomialc�thenc��
��
andwecantakeA�x��c�B�x��C�x����Nextsupposethedegr
ncaseistrue�ForthecaseP�x�isofdegreen���IfP�x���f
allrealx�thensimplyletA�x��P�x��B�x��C�x����Otherwis
P�x�hasarootx�in�����or������
Casex�in������ThenP�x���x�x��Q�x�andQ�x�isofdegree
withQ�x���forallxn������SoQ�x��A��x��xB��x�����x�C��x
whereA��x��B��x��C��x���forallxin������Usingx���x�
x���x������x�x��wehave
P�x���x�x���A��x��xB��x�����x�C��x��
���x�A��x��x
�B��x��
z
�
Ax
�x�A��x��x�B��x�����x��C��x
zBx
����x���x�C��x��x
�B��x��
z
�
Cx
�
wherethepolynomialsA�x��B�x��C�x���forallxin������
Casex�in������ConsiderQ�x��P���x��Thisreducestot
previouscase�WehaveQ�x��A��x��xB��x�����x�C��x��whe
thepolynomialsA��x��B��x��C��x���forallxin������Then
P�x��Q���x��A����x�
z�
Ax
�xC����x�
z�
Bx
����x�B����x�
z�
Cx
�
wherethepolynomialsA�x��B�x��C�x���forallxin������
�������IMO�Letf�x��xn��xn�����wheren��isanintege
Provethatf�x�cannotbeexpressedasaproductoftwopolynomia
eachhasintegercoe�cientsanddegreeatleast��
Solution�Supposef�x��b�x�c�x�fornonconstantpolynomialsb�
andc�x�withintegercoe�cients�Sincef������wemayassum
b����
��andb�x��
xr�������Sincef��������r�
��L
z ������zrbetherootsofb�x��Thenjz ����zrj�jb���j��and
jb����j�j����z ���������z r�j�
r Y i��
jzn��
i
�zi���j��r���
��
Howeverb����alsodividesf�������acontradiction�
��Provethatiftheintegeraisnotdivisibleby�thenf�x��x��x�a
cannotbefactoredastheproductoftwononconstantpolynomialswith
integercoe�cients�
Solution�Supposefcanbefactoredthenf�x��
�x�b�g�x�or
f�x���x��bx�c�g�x��Inthefomercaseb��b�a�f�b����Now
b��b�mod��byFermat�slittletheoremorsimplycheckingthecases
b�����������mod���Then�dividesb�b��a�acontradiction�In
thelattercasedivdingf�x��x��x�abyx��bx�c�wegetthe
remainder�b��b�c�c����x��b�c��bc��a��Sincex��bx�cis
afactoroff�x��bothcoe�cientsequal��Finally
��b�b��b
�c�c�������b�c��bc��a��b��b��bc���a
implies�a�b��b��bc�isdivisibleby��Thenawouldbedivisible
by�acontradiction�
��������SovietMathOlympiad�Given�ndistinctnumbersa��a������an�
b ��b������bn�ann�ntableis�lledasfollows�intothecellintheith
rowandjthcolumniswrittenthenumberai�b j�Provethatifthe
productofeachcolumnisthesamethenalsotheproductofeachrow
isthesame�
Solution�Let
P�x���x�a���x�a������x�an���x�b ���x�b ������x�b n��
thendegP�n�NowP�bj���bj�a���b j�a������bj�an��c�some
constantforj���������n�SoP�x��chasdistinctrootsb ��b������bn�
ThereforeP�x��cforallxandso
c�P��ai������n���ai�b ���ai�b ������ai�b n�
fori���������n�Thentheproductofeachrowis����n��c�
���Leta��a������anandb ��b������bnbetwodistinctcollectionsofnpos
itiveintegerswhereeachcollectionmaycontainrepetitions�Ifthetwo
��
collectionsofintegersai�aj���i�j�n�andb i�b j���i�j�
arethesamethenshowthatnisapowerof��
Solution��DuetoLawSiuLung�Considerthefunctionsf�x�
n X i��
xai
andg�x��
n X i��
xbi�Sincetheai�sandb i�saredistinctfand
aredistinctpolynomials�Now
f�x���
n X i��
x�ai
��
X��i�j�n
xai�aj
�f�x
����
X��i�j�n
xai�aj�
Sincetheai�aj�sandtheb i�b j�sarethesamesof�x���f�x��
g�x���g�x���Sincef����g����n�n���sof�x��g�x���x
��kQ�x�forsomek��andpolynomialQsuchthatQ�������Then
f�x��g�x��
f�x���g�x��
f�x��g�x�
��x����kQ�x��
�x���kQ�x�
��x���kQ�x��
Q�x�
�
Settingx���wehaven��k���
RecurrenceRelations
���Thesequencexn
isde�nedby
x����
xn���
��xn
���xn
�
n�����������
Provethatxn��
� �
or�forallnandthetermsofthesequenceare
distinct�
Solution��DuetoWongChunWai�Thetermsxn�sareclearlyration
byinduction�Writexn�pn�q n�wherepn�qnarerelativelyprimeint
gersandq n���Thenq ���andpn���q n�����q n�pn���qn��pn
Soq n��dividesq n��pn�whichimplieseveryq nisoddbyinductio
Henceeveryxn��
� ��
Nexttoshoweveryxn
����let��arctan��thenxn
�tann
byinduction�Supposexn
��andnistheleastsuchindex�Ifn
��
�miseventhen��x�m
�tan�m���xm����x� m�wouldimply
xm
���acontradictiontonbeingleast�Ifn��m��isoddthen
��x�m��
�tan����m������x�m������x�m�wouldimply
x�m
����Then����xm����x� m�wouldimplyxm
����
p ����is
irrationalacontradiction�Finallyifxm
�xn
forsomem�n�then
xm�n�tan�m��n����xm
�xn�����xmxn����acontradiction�
Thereforethetermsarenonzeroanddistinct�
��������NanchangCityMathCompetition�De�nea����a���and
an���
a� n����
an
forpositiveintegern�Provethat�anan����isa
perfectsquareforeverypositiveintegern�
Solution��DuetoChanKinHang��Sincean��dependsonan��and
an�itisplausiblethatthesequencesatis�esalinearrecurrencerelation
an���can���c�an�Ifthisissothenusingthe�rst�termswe�nd
c���c������De�neb ��a��b��a��bn����bn���b nforn���
Thenb �����a��Supposeak�b kfork�n���then
an���
b� n����
b n
���b n�b n������
b n
���b n���b n���b n��
��bn���b n�b n���
Soak�b kforallk�
Nextwritingoutthe�rstfewtermsof�anan����willsuggest
that�anan������an�an�����Thecasen��istrueas������
�������Supposethisistrueforn�k�Usingtherecurrencerelations
and� ��a� k������akak�����akak����a� k�wegetthecasen�k��
asfollow�
�ak��ak������ak����ak���ak���
���a
� k����ak�ak������
���a
� k����akak���a
� k��
���a
� k�����ak��ak�a
� k
by� �
���ak���ak����ak���ak�����
��
����ProposedbyBulgariafor����IMO�De�nea����a���andan
�an���an��forn���Showthatforpositiveintegerk�anisdivisib
by�kifandonlyifnisdivisibleby�k�
Solution�Bythebinomialtheoremif���
p ��n�An�Bn
p ��th
���
p ��n�An�Bn
p ��Multiplyingthese�equationswegetA� n
�B� n
�
����n�ThisimpliesAn
isalwaysodd�Usingcharacterist
equationmethodtosolvethegivenrecurrencerelationsonan�we�n
thatan
�Bn�Nowwriten��km�wheremisodd�Wehavek�
�i�e�nisodd�ifandonlyif�B� n
�A� n�����mod����i�e�Bn
odd��Nextsupposecasekistrue�Since���
p ���n��An�Bn
p ���
A�n�B�n
p ��soB�n
��AnBn�Thenitfollowscasekimpliesca
k���
����AmericanMathematicalMonthlyProblem
E�����Letxandy
distinctcomplexnumberssuchthat
xn�yn
x�y
isanintegerforsom
fourconsecutivepositiveintegersn�Showthat
xn�yn
x�y
isaninteg
forallpositiveintegersn�
Solution�Fornonnegativeintegern�lett n��xn�yn���x�y��
t ����t ���andwehavearecurrencerelation
t n���btn���ctn���
whereb���x�y��c�xy�
Supposet nisanintegerform�m���m���m���Sincecn��xy�n
t� n���t nt n��isanintegerforn�m�m���socisrational�Sin
cm��isintegercmustinfactbeaninteger�Next
b�
t mt m���t m��t m��
cm
�
Sobisrational�Fromtherecurrencerelationitfollowsbyinducti
thatt n�f n���b�forsomepolynomialfn��ofdegreen��withinteg
coe�cients�Notethecoe�cientofxn��inf n��is�i�e�f n��ismon
Sincebisarootoftheintegercoe�cientpolynomialfm�z��t m���
bmustbeaninteger�Sotherecurrencerelationimpliesallt n�sa
integers�
��
Inequalities
���Forrealnumbersa��a��a������ifan���an����an
forn���������
thenprovethat
An���An����An
forn���������
whereAn
istheaverageofa��a������an�
Solution�Expressinginak�therequiredinequalityisequivalentto
a������an���
n��n��
�
an�
n�n���
�
an�����
�Fromthecasesn�����weeasilyseethepattern��Wehave
a������an���
n��n��
�
an�
n�n���
�
an��
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n X k��
k�k���
�
�ak����ak�ak������
���Leta�b�c��andabc���Provethat
a c�
b a�
c b�a�b�c�
Solution��DuetoLeungWaiYing�Sinceabc���weget���bc��a�
���ac��band���ab��c�BytheAMGMinequality
�a c�
c b�
a c�
a c�
c b��
�r a� bc��a�
Similarly�b�a�a�c��band�c�b�b�a��c�Addingtheseand
dividingby�wegetthedesiredinequality�
��
Alternativelyletx�
�p ab�c��y�
�p ca�b�andz�
�p bc�a
Wehavea�x�y�b�z�x�c�y�zandxyz�
�p abc���Usingth
andtherearrangementinequalityweget
a c�
b a�
c b�
x�
yz
�z�
xy
�y�
zx
�xyz
� x� yz
�z�
xy
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zx
� �x
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�z�z
�x�a�b�c�
��������MoscowMathOlympiad�Usetheidentity����������n�
n��n����
�
toprovethatfordistinctpositiveintegersa��a������an
�a� ��a
� ������a
� n���a
� ��a
� ������a
� n����a
� ��a
� ������a
� n��
Canequalityoccur�
Solution�Forn���a� ��a� ����a� ����a� ��a�������andsoca
n��istrue�Supposethecasen�kistrue�Forthecasen�k�
withoutlossofgeneralitywemayassumea��a������ak���No
��a
� ������a
� k�������a
� ������a
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��
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So�a� ������a� k�����a� ������a� k������a� ������a� k����follow
Equalityoccursifandonlyifa��a������anare��������n�
��������IMOshortlistedproblem�Leta������an
�an����be
sequenceofrealnumbers�Provethat
v u u tn X k��
ak�
n X k��
p k�pak�p ak����
��
Solution��DuetoLeeTakWing�Letxk�p ak�p ak���Thenak�
�xk�xk�������xn���So
n X k��
ak�
n X k��
�xk�xk�������xn���
n X k��
kx
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ixixj
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kx
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�
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�BOC����������������and�EOB��BOC�������AOE�
�����������������SoA�B�Ceachisonadi erentquarterdisk�
Similarly�����EOD��EOA��AOD������ThereforeDwill
lieontheremainingquarterdisk�
���Thelengthsofthesidesofaquadrilateralarepositiveintegers�The
lengthofeachsidedividesthesumoftheotherthreelengths�Prove
thattwoofthesideshavethesamelength�
Solution��DuetoChaoKhekLunandLeungWaiYing�Supposethe
sidesarea�b�c�dwitha�b�c�d�Sinced�a�b�c��dand
ddividesa�b�c�wehavea�b�c��d�Noweachofa�b�cdvides
a�b�c�d�
�d�Letx�
�d�a�y�
�d�bandz�
�d�c�Then
a�b�c�dimpliesx�y�z���Soz���y���x���Then
�d�a�b�c�
�d ��
�d ��
�d ���d�
acontradiction�Thereforetwoofthesidesareequal�
��������SichuanProvinceMathCompetition�Supposethelengthsofthe
threesidesof�ABCareintegersandtheinradiusofthetriangleis��
Provethatthetriangleisarighttriangle�
Solution��DuetoChanKinHang�Leta�BC�b�CA�c�ABbe
thesidelengthsrbetheinradiusands��a�b�c����Sincethearea
��
ofthetriangleisrs�weget
p s�s�a��s�b��s�c����s�s�The
�s�a��s�b��s�c��s��s�a���s�b���s�c��
Now��a�b�c���s���s��a���s��b���s��c���b�c�a��c
a�b��a�b�c��In�mod��eachofb�c�a�c�a�b�a�b�ca
thesame�Soeithertheyarealloddoralleven�Sincetheirproduct
eventheyarealleven�Thena�b�cisevenandsisaninteger�
Thepositiveintegersx�s�a�y�s�b�z�s�csatisfyxyz
x�y�z�Supposex�y�z�Thenyz��forotherwisexyz��x
x�y�z�Thisimpliesx���y���z���s���a���b���c�
Thereforethetriangleisarighttriangle�
GeometricEquations
��������IMO�AcirclehascenteronthesideABofthecyclicquad
lateralABCD�Theotherthreesidesaretangenttothecircle�Pro
thatAD�BC�AB�
Solution�LetM
beonABsuchthatMB�BC�Then
�CMB�
������ABC
�
��CDA
�
��CDO�
ThisimpliesC�D�M�Oareconcyclic�Then
�AMD��OCD�
�DCB
�
�������DAM
�
��AMD��ADM
�
So�AMD��ADM�ThereforeAM
�ADandAB�AM
�MB
AD�BC�
��������RussianMathOlympiad�CirclesS�andS�withcentersO��O
respectivelyintersecteachotheratpointsAandB�RayO�Bintersec
S�atpointFandrayO�BintersectsS�atpointE�Thelineparal
toEFandpassingthroughBintersectsS�andS�atpointsM
an
N�respectively�Provethat�Bistheincenterof�EAFand�MN
AE�AF�
��
Solution�Since
�EAB�
� ��EO�B������O�BE������FBO���BAF�
ABbisects�EAFand�O�BE������EAB�����
� ��EAF�Now
�EBA��FBA��EBA������ ��O�BA��������O�BE������
� ��EAF�Then�EBF������EAF�whichimpliesBistheincenter
of�EAF�becausetheincenteristheuniquepointPonthebisectorof
�EAFsuchthat�EPF�����
� ��EAF��Then�AEB��BEF�
�EBM
sinceEFkMN�SoEBAM
isanisoscelestrapezoid�Hence
EA�MB�SimilarlyFA�NB�ThereforeMN
�MB�NB
�
AE�AF�
���PointCliesontheminorarcABofthecirclecenteredatO�Suppose
thetangentlineatCcutstheperpendicularstochordABthroughA
atEandthroughBatF�LetDbetheintersectionofchordABand
radiusOC�ProvethatCE�CF�AD�BDandCD��AE�BF�
Solution��DuetoWongChunWai�Notethat�EAD��ECD��FCD�
�FBDarerightangles�SoA�D�C�EareconcyclicandB�D�C�Fare
concyclic�Then�ADE��ACE��ABC��DFC�saythemeasure
oftheseanglesis��Also�BDF��BCF��BAC��DEC�say
themeasureoftheseangleis�Then
CE�CF��DEcos��DFcos����DEcos���DFcos��AD�BD�
CD
���DEsin��DFsin����DEsin���DFsin��AE�BF�
���QuadrilateralsABCPandA� B� C� P�areinscribedintwoconcentric
circles�IftrianglesABCandA� B� C�areequilateralprovethat
P� A��P� B��P� C��PA���PB���PC���
Solution�LetObethecenterofbothcirclesandEbethemidpoint
ofA� B� �From�PA� B� withmedianPE�bycosinelawwegetPA���
PB�����PE��EB����From�PC� EwithcevianPO�noteC� O�
��
�OE�bycosinelawagainwegetPC����PE����PO���OE
Puttingthesetogetherweget
PA���PB���PC�����EB���OE
����PO
���OE
�
��B� O���PO
��C� O�
���PO
��P� O���
SimilarlyP� A��P� B��PC�����PO��P� O���
Alternativelytheproblemcanbesolvedusingcomplexnumbe
WithoutlossofgeneralityletthecenterbeattheoriginA�be
rei��r�cos �isin �andPbeatRei��Let��e��i���Wehave
PA���PB���PC��
�jRei��rei�j��jRei��rei��j��jRei��rei��
�j�
��R
���Re
� Rrei
��� ������
��� ��r
�
��R
���r
��
SimilarlyP� A��P� B��P� C���R���r��
���LettheinscribedcircleoftriangleABCtouchssideBCatDsideC
atEandsideABatF�LetGbethefootofperpendicularfromD
EF�Showthat
FG
EG
�BF
CE
�
Solution��DuetoWongChunWai�LetIbetheincenterof�AB
Then�BDI�
����
�EGD�Also�DEG
�
� ��DIF
�
�DI
So�BDI��EGDaresimilar�ThenBD�ID
�DG�EG�Likewis
�CDI��FGDaresimilarandCD�ID�DG�FG�Therefore
FG
EG
�DG�EG
DG�FG
�BD�ID
CD�ID
�BD
CD
�BF
CE
�
��������IMOshortlistedproblem�LetABCDEFbeaconvexhexag
suchthat
�B��D��F�����and
AB
BC
�CD
DE
�EF
FA
���
��
Provethat
BC
CA
�AE
EF
�FD
DB
���
Solution�LetPbesuchthat�FEA��DEPand�EFA��EDP�
whereP
isontheoppositesideoflinesDE
andCD
asA�Then
�FEA��DEParesimilar�So
FA
EF
�DP
PE
and
���
EF
ED
�EA
EP
�
Since�B��D��F������weget�ABC��PDC�Also
AB
BC
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CD�EF
�DP
CD
�
Then�ABC��PDC
aresimilar�Consequentlyweget�BCA
�
�DCPand� �CB�CD�CA�CP�Since�FED��AEP�by� �
�FED��AEParesimilar�Alsosince�BCD
��ACP�by� �
�BCD��ACParesimilar�SoAE�EF�PA�FD
andBC�CA�
DB�PA�Multiplyingtheseandmovingallfactorstotheleftsidewe
getthedesiredequation�
Usingcomplexnumberswecangetanalgebraicsolution�Let
a�b�c�d�e�fdenotethecomplexnumberscorrespondingtoA�B�C�D�
E�F�respectively��Theoriginmaybetakenanywhereontheplane��
SinceABCDEFisconvex�B��Dand�Faretheargumentsofthe
complexnumbers�a�b���c�b���c�d���e�d�and�e�f���a�f��
respectively�Thenthecondition�B��D��F�����impliesthat
theproductofthesethreecomplexnumbersisapositiverealnumber�
ItisequaltotheproductoftheirabsolutevaluesAB�BC�CD�DE
andEF�FA�Since�AB�BC��CD�DE��EF�FA����wehave
a�b
c�b
�c�d
e�d
�e�f
a�f
���
So
���a�b��c�d��e�f���c�b��e�d��a�f�
��b�c��a�e��f�d���a�c��f�e��b�d��
��
Then
BC
CA
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DB
�� � � �b�c
a�c
�a�e
f�e
�f�d
b�d
� � � ����
SimilarTriangles
��������BritishMathOlympiad�P�Q�andRarearbitrarypointsont
sidesBC�CA�andABrespectivelyoftriangleABC�Provethatt
threecircumcentresoftrianglesAQR�BRP�andCPQformatriang
similartotriangleABC�
Solution�LetthecircumcentersoftrianglesAQR�BRPandCPQ
A� �B�andC� �respectively�Agooddrawingsuggeststhecirclespa
throughacommonpoint�Toprovethisletcircumcirclesoftriangl
AQRandBRPintersectatRandX�Then�QXR�������CAB
�ABC��BCAand�RXP�������ABC��CAB��BCA�
�PXQ�������QXR��RXP�������BCA�whichimplies
isonthecircumcircleoftriangleCPQ�Now
�C� A� B� ��C� A� X��XA� B�
�� ��QA� X�
� ��RA� X
�� ��QA� R��CAB�
Similarly�A� B� C���ABCand�B� C� A���BCA�Sotriangl
A� B� C�andABCaresimilar�
���HexagonABCDEFisinscribedinacirclesothatAB�CD�E
LetP�Q�RbethepointsofintersectionofACandBD�CEandD
EAandFB
respectively�ProvethattrianglesPQRandBDFa
similar�
Solution��DuetoNgKaWing�LetObethecenterofthecirc
andletL�M�NbetheprojectionsofOonBD�DF�FB�respective
ThenL�M�NaremidpointsofBD�DF�FB�respectively�LetSbet
projectionofOonAE�SinceAB�EF�wegetFB�AEandhen
��
ON
�OS�Let�AOB
��COD
��EOF����Then�RON
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� ��SON
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� ��ARB
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� ��AOB��EOF����HenceON�OR�
cos��Similarly�POL��QOM
��andOL�OP�OM�OQ�
cos�� N
extrotate�PQRaroundOatangle�sothattheimageQ�
ofQ
liesonthelineOM�theimageR�ofRliesonthelineON
andtheimageP�ofPliesonlineOL�ThenON�OR� �OL�OP��
OM�OQ��cos��So�P� Q� R� ��LMN
aresimilar�SinceL�M�N
aremidpointsofBD�DF�FB�respectivelywehave�LMN��BDF
aresimilar�Therefore�PQR��BDFaresimilar�
��������IMOshortlistedproblem�LetABCDbeacyclicquadrilateral�
LetEandFbevariablepointsonthesidesABandCD�respectively
suchthatAE�EB�CF�FD�LetPbethepointonthesegment
EFsuchthatPE�PF�AB�CD�Provethattheratiobetweenthe
areasoftrianglesAPDandBPCdoesnotdependonthechoiceofE
andF�
Solution�Let�UVW�denotetheareaof�UVW
andletd�X�YZ�
denotethedistancefromX
tolineYZ�WehaveAE�EB�CF�
FD�a�b�wherea�b���SincePE�PF�AB�CD�wehave
d�P�AD��
CD
AB�CD
d�E�AD��
AB
AB�CD
d�F�AD��
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CD
AB�CD
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AB
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b�AB
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AB
AB�CD
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AB
AB�CD
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b�CD
AB�CD
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AB�CD
�BDC��
��
SinceA�B�C�Dareconcyclicsin�BAD�sin�BCDandsin�ABC
sin�ADC�So
�APD�
�BPC�
�a�CD��ABD��b�AB��ACD�
b�CD��BAC��a�AB��BDC�
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b�CD�AB�BC�sin�ABC�a�AB�CD�BC�sin�BCD
�AD
BC
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b�sin�ABC�a�sin�BCD
�AD
BC
�
TangentLines
���TwocirclesintersectatpointsAandB�Anarbitrarylinethrough
intersectsthe�rstcircleagainatCandthesecondcircleagainatD
Thetangentstothe�rstcircleatCandtothesecondcircleat
intersectatM�TheparalleltoCM
whichpassesthroughthepoi
ofintersectionofAM
andCDintersectsACatK�ProvethatBK
tangenttothesecondcircle�
Solution�LetLbetheintersectionofAM
andCD�Since
�CMD��CAD��CMD��CAB��DAB
��CMD��BCM
��BDM
������
soA�C�M�Dareconcyclic�SinceLKkMC�
�LKC�������KCM
�������KCL��LCM
�������ACB��CAB��CBA��LBA�
SoA�B�L�Kareconcyclic�Then
�KBA��KLA��CMA��CDA��BDA�
ThereforeBKistangenttothecirclepassingthroughA�B�D�
��������IMO�Twocircles��and��arecontainedinsidethecircle
andaretangentto�atthedistinctpointsM
andN�respective
��
��passesthroughthecenterof���Thelinepassingthroughthetwo
pointsofintersectionof��and��meets�atAandB�respectively�
ThelinesMAandMBmeets��atCandD�respectively�Provethat
CDistangentto���
Solution��DuetoWongChunWai�LetX�Ybethecentersof������
respectively�Extend�� YXtomeet��atQ�JoinANtomeet��atE�
SinceABistheradicalaxisof������soAC�AM
�AE�AN�This
impliesC�M�N�Eareconcyclic�LetUbetheintersectionoflineCE
withthetangentto��atM�Then�UCM
��ENM
��ANM
�
�UMC�SoCEistangentto���SimilarlyCEistangentto���Now
YE�YQand �
CYE������ECY�����
� ��CXY
�����������CYQ���CYQ�
Theseimply�CYE��CYQarecongruent�Hence�CQY��CEY�
��� �Similarly�DQY���� �ThereforeCDistangentto���
����ProposedbyIndiafor����IMO�CirclesG�andG�toucheachother
externallyatapointW
andareinscribedinacircleG�A�B�Care
pointsonGsuchthatA�G�andG�areonthesamesideofchordBC�
whichisalsotangenttoG�andG��SupposeAW
isalsotangentto
G�andG��ProvethatW
istheincenteroftriangleABC�
Solution�LetPandQbethepointsoftangencyofG�withBCand
arcBAC�respectively�LetDbethemidpointofthecomplementary
arcBCofG�notcontainingA�andLbeapointonG�sothatDLis
tangenttoG�andintersectssegmentPC�Consideringthehomothety
withcenterQthatmapsG�ontoG�weseethatQ�P�Darecollinear
becausethetangentatP
�namelyBC�andthetangentatD
are
parallel�Since�BQD��CBD
subtendequalarcs�BQD��PBD
aresimilar�HenceDB�DP
�
DQ�DB�Bytheintersectingchord
theoremDB��DP�DQ�DL��SoDL�DB�DC�ThenDhas
thesamepowerDB��DC�withrespecttoG�andG��HenceDis
ontheradicalaxisAW
ofG�andG��SoL�W
andDW
istangent
toG�andG��
��
SinceDisthemidpointofarcBC�soAW
bisects�BAC�Also
�ABW
��BWD��BAD��WBD��CBD��CBW
andBW
bisects�ABC�ThereforeW
istheincenterof�ABC�
Comments�The�rstpartoftheproofcanalsobedonebyinversi
withrespecttothecirclecenteredatD
andofradiusDB
�
D
ItmapsarcBContothechordBC�BothG�
andG�
areinvaria
becausethepowerofDwithrespecttothemisDB��DC��Hen
W
is�xedandsoDW
istangenttobothG�andG��
Locus
���PerpendicularsfromapointPonthecircumcircleof�ABCaredraw
tolinesAB�BCwithfeetatD�E�respectively�Findthelocusoft
circumcenterof�PDEasPmovesaroundthecircle�
Solution�Since�PDB��PEB
�
�����P�D�B�E
areconcycl
Hencethecircumcircleof�PDEpassesthroughBalways�ThenP
isadiameterandthecircumcenterof�PDEisatthemidpointM
PB�LetObethecircumcenterof�ABC�thenOM
PB�Itfollow
thatthelocusofM
isthecirclewithOBasadiameter�
���SupposeAisapointinsideagivencircleandisdi erentfrom
t
center�Considerallchords�excludingthediameter�passingthrou
A�Whatisthelocusoftheintersectionofthetangentlinesatt
endpointsofthesechords�
Solution��DuetoWongHimTing�LetObethecenterandand
betheradius�LetA�bethepointonOAextendedbeyondAsu
thatOA�OA� �r��SupposeBCisonesuchchordpassingthrou
AandthetangentsatBandCintersectatD� �BysymmetryD�
onthelineODwhereDisthemidpointofBC�Since�OBD� ���
OD�OD��OB���OA�OA� ��So�OADissimilarto�OD� A
Since�ODA���� D�isonthelineLperpendiculartoOAatA� �
ConverselyforD�onLletthechordthroughAperpendicul
toOD�intersectthecircleatBandC�LetDbetheintersection
��
thechordwithOD� �Now�OAD��OD� A� aresimilarrighttriangles�
SoOD�OD� �OA�OA� �OB��OC��whichimplies�OBD��
�OCD� ���� �ThereforeD�isonthelocus�Thisshowsthelocusis
thelineL�
���Given�ABC�LetlineEFbisects�BACandAE�AF�AB�AC�
FindthelocusoftheintersectionPoflinesBEandCF�
Solution�ForsuchapointP�sinceAB�AE�AF�ACand�BAE�
�FAC�so�BAE��FACaresimilar�Then�AEP
�
�PCA�So
A�E�C�Pareconcyclic�Hence�BPC��CAE��BAC���There
forePisonthecircleCwhosepointsX
satisfy�BXC��BAC��
andwhosecenterisonthesamesideoflineBCasA�
ConverselyforPonC�LetBP�CPintersecttheanglebisector
of�BACatE�F�respectively�Since�BPC��BCA���so�EPF�
�EAC�HenceA�E�C�Pareconcyclic�So�BEA
�
�FCA�Also
�BAE��FAC�So�BAE��FACaresimilar�ThenAB�AC�
AE�AF�ThereforethelocusofPisthecircleC�
��������PutnamExam�LetC�andC�becircleswhosecentersare��
unitsapartandwhoseradiiare�and��Findthelocusofallpoints
M
forwhichthereexistspointsXonC�andYonC�suchthatM
is
themidpointofthelinesegmentXY�
Solution��DuetoPoonWaiHoi�LetO��O�bethecentersofC��C��
respectively�Ifwe�xYonC��thenasX
movesaroundC��M
will
traceacircle�Y
withradius
� �
centeredatthemidpointmY
ofO�Y�
AsYmovesaroundC��mY
willtraceacircleofradius
� �
centeredat
themidpontPofO�O��Sothelocusisthesolidannuluscenteredat
Pwithinnerradius
� ��
� ���andouterradius
� ��
� ����
CollinearorConcyclicPoints
��������IMO�DiagonalsACandCEoftheregularhexagonABCDEF
aredividedbytheinnerpointsM
andN�respectivelysothat
AM
AC
�CN
CE
�r�
��
DeterminerifB�M
andNarecollinear�
Solution��DuetoLeeTakWing�LetAC�x�thenBC�x�p
CN
�xr�CM
�x���r��Let�XYZ�denotetheareaof�XY
Since�NCM
���� ��BCM
����and�BCM���CMN���BCN
so
x����r�sin���
�p�
�x�r���r�sin���
�
�
x�r
�p�
�
Cancellingx�andsolvingforr�wegetr�
� p ��
��������PutnamExam�IfA�B�C�Darefourdistinctpointssuchth
everycirclethroughAandBintersectsorcoincideswitheverycirc
throughCandD�provethatthefourpointsareeithercollinear
concyclic�
Solution�SupposeA�B�C�Dareneitherconcyclicnorcollinear�Th
theperpendicularbisectorpofABcannotcoincidewiththeperpe
dicularbisectorqofCD�Iflinespandqintersecttheircommonpoi
isthecenteroftwoconcentriccirclesonethroughAandB�theoth
throughCandD�acontradiction�Iflinespandqareparallelth
linesABandCDarealsoparallel�ConsiderpointsPandQonpan
q�respectivelymidwaybetweentheparallellinesABandCD�Clear
thecirclesthroughA�B�PandC�D�Qhavenocommonpointaga
acontradiction�
��������PutnamExam�Givenanin�nitenumberofpointsinaplan
provethatifallthedistancesbetweeneverypairareintegersthent
pointsarecollinear�
Solution�SupposetherearethreenoncollinearpointsA�B�Csu
thatAB�randAC�s�ObservethatifPisoneoftheotherpoin
thenbtthetriangleinequalityjPA�PBj�����������r�Hence
wouldbeonthelineH�
joiningA�B
orononeofthehyperbol
Hi�fX�jXA�XBj�igfori���������r��orontheperpendicul
bisectorHrofAB�SimilarlyjPA�PCj�����������s�SoPis
oneofthesetsKj�fX�jXA�XCj�jgforj���������s�Sin
��
linesAB
andACaredistincteveryintersectionHi�Kj
isonlya
�niteset�Sotherecanonlybe�nitelymanypointsthatareintegral
distancesfrom
A�B�C�acontradiction�Thereforethegivenpoints
mustbecollinear�
��������IMOshortlistedproblem�TheincircleoftriangleABCtouches
BC�CAandAB
atD�E
andF
respectively�X
isapointinside
triangleABCsuchthattheincircleoftriangleXBCtouchesBCat
DalsoandtouchesCXandXBatYandZrespectively�Provethat
EFZYisacyclicquadrilateral�
Solution�IfEFkBC�thenAB�ACandADisanaxisofsymmetry
ofEFZY�HenceEFZYisacyclicquadrilateral�IflinesEFandBC
intersectatP�thenbyMenelaus�theorem�AF�BP�CE���FB�PC�
EA����SinceBZ�BD�BF�CY�CD�CEandAF�EA���
XZ�YX�weget�XZ�BP�CY���ZB�PC�YX����Bytheconverse
oftheMenelaus�theoremZ�Y�Parecollinear�Bytheintersecting
chordtheoremPE�PF�PD��PY�PZ�HenceEFZYisacyclic
quadrilateralbytheconverseoftheintersectingchordtheorem�
��������IMO�IntheconvexquadrilateralABCD�thediagonalsACand
BD
areperpendicularandtheoppositesidesAB
andDC
arenot
parallel�SupposethepointP�wheretheperpendicularbisectorsof
AB
andDCmeetisinsideABCD�ProvethatABCD
isacyclic
quadrilateralifandonlyifthetrianglesABPandCDPhaveequal
areas�
Solution��DuetoLeungWingChung�SettheoriginatP�Suppose
AandCareontheliney�pandBandDareonthelinex�q�Let
AP�BP�r�CP�DP�s�ThenthecoordinatesofA�B�C�Dare
��p r��p��p���q�p r��q����p s��p��p���q��
p s��q���
respectively�Now�ABP��CDPhaveequalareasifandonlyif
� �� � � ��pr��p�
p
q
p r��q�
� � � ��� �
� � � �p s��p�
p
q
�ps��q�
� � � ��
��
i�e���
p r��p�
p r��q��pq������
p s��p�p s��q��pq����Sin
f�x����
p x��p�p x��q��pq���isstrictlydecreasingwhenx�
andjqj�thedeterminantsareequalifandonlyifr�
s�which
equivalenttoABCDcyclic�
��������PutnamExam�Showthatifaconvexquadrilateralwithsid
lengthsa�b�c�dandarea
p abcdhasaninscribedcirclethenitis
cyclicquadrilateral�
Solution�Sincethequadrilateralhasaninscribedcirclewehavea
c�b�d�Letkbethelengthofadiagonalandangles�andselect
sothat
k��a
��b���abcos��c��d
���cdcos�
Ifwesubtract�a�b����c�d��anddivideby�wegettheequati
� �ab���cos���cd���cos��Fromthearea�absin��cdsin���
p abcd�weget
�abcd�a
�b����cos����c�d
����cos����abcdsin�sin�
Using� �wecancancelabcdtoobtaintheequation
�����cos�����cos�����cos����cos����sin�sin
����cos�����
whichimplies��������Thereforethequadrilateraliscyclic�
ConcurrentLines
���In�ABC�supposeAB�AC�LetPandQbethefeetofthepe
pendicularsfromBandCtotheanglebisectorof�BAC�respective
LetDbeonlineBCsuchthatDA AP�ProvethatlinesBQ�P
andADareconcurrent�
Solution�LetM
betheintersectionofPCandAD�LetB�bet
mirrorimageofB
withrespecttolineAP�SinceBB� AP
an
��
AD AP�soBB�kAD�Then�BCB� ��DACaresimilar�SinceP
isthemidpointofBB� �soPCintersectsADatitsmidpointM�Now
AQ
PQ
�MC
PC
�AM
B� P�
AM
BP
�
�BPQ��MAQaresimilar�Thisimplies�BQP��MQA�Soline
BQpassesthroughM�too�
��������ChineseNationalMathCompetition�DiagonalsAC
andBD
ofacyclicquadrilateralABCDmeetsatP�Letthecircumcentersof
ABCD�ABP�BCP�CDPandDAPbeO�O��O��O�andO�respec
tively�ProvethatOP�O�O��O�Oareconcurrent�
Solution�LetlinePO�intersectthecircumcircleof�BCPandseg
mentADatpointsQandR�respectively�Now�PDR��BCA�
�PQBand�DPR��QPB�So�DRP��QBP����andPO�
AD�NextcircumcirclesofABCDandDAPsharethecommonchord
AD�soOO
AD�HencePO�
andOO
areparallel�Similarly
POandOO�areparallel�SoPO�OOisaparallelogramanddiag
onalO�OpassesthroughthemidpointGofOP�SimilarlyPO�OO�
isaparallelogram
anddiagonalO�O�
passesthroughG�Therefore
OP�O�O��O�OconcuratG�
��������IMO�LetA�B�CandDbefourdistinctpointsonalineinthat
order�ThecircleswithdiametersACandBDintersectatthepoints
XandY�ThelineXYmeetsBCatthepointZ�LetPbeapointon
thelineXYdi erentfromZ�ThelineCPintersectsthecirclewith
diameterACatthepointsCandM�andthelineBPintersectsthe
circlewithdiameterBDatthepointsBandN�Provethatthelines
AM�DNandXYareconcurrent�
Solution���DuetoYuChunLing�LetARbeparalleltoBPand
DR�beparalleltoCP�whereRandR�arepointsonlineXY�Since
BZ�ZD
�XZ�ZY�CZ�ZA�wegetBZ�AZ�CZ�DZ�Since
�CZPissimilarto�DZR�and�BZPissimilarto�AZR�so
ZP
ZR
�BZ
AZ
�CZ
DZ
�
ZP
ZR��
��
HenceRandR� mustcoincide�Therefore�BPCissimilarto�ARD
SinceXY
AD�AM
CM�CM
kDR�DN
BN
an
BN
kAR�thelinesAM�DN�XYaretheextensionsofthealtitud
of�ARD�hencetheymustbeconcurrent�
Solution���DuetoMokTzeTao�SettheoriginatZandthe
axisonlineAD�Letthecoordinatesofthecircumcentersoftriangl
AMCandBNDbe�x����and�x�����andthecircumradiiber �an
r ��respectively�ThenthecoordinatesofAandCare�x��r ����an
�x��r �����respectively�LetthecoordinatesofPbe���y���Sin
AM
CPandtheslopeofCPis�
y �
x��r �
�theequationofA
worksouttobe�x��r ��x�y �y�x� ��r� ��LetQbetheintersection
AM
withXY�thenQhascoordinates���
r� ��x� �
y �
��SimilarlyletQ�
theintersectionofDNwithXY�thenQ� hascoordinates���
r� ��x� �
y �
Sincer� ��x� ��ZX��r� ��x� ��soQ�Q� �
Solution��LetAM
intersectXYatQandDN
intersectXY
Q� �ObservethattherighttrianglesAZQ�AMC�PZCaresimilar
AZ�QZ�PZ�CZ�ThenQZ�AZ�CZ�PZ�XZ�YZ�PZ�Sim
larlyQ� Z�XZ�YZ�PZ�ThereforeQ�Q� �
���AD�BE�CFarethealtitudesof�ABC�IfP�Q�Rarethemidpoin
ofDE�EF�FD�respectivelythenshowthattheperpendicularfro
P�Q�RtoAB�BC�CA�respectivelyareconcurrent�
Solution�LetYbethefootoftheperpendicularfromQtoBCan
H
betheorthocenterof�ABC�Notethat�ACF������CAB
�ABE�Since�CEH
������CDH�C�D�H�Eareconcyclic�
�ACF��EDH�NowAH
kQY�ED
kQR�so�EDH
��RQ
Hence�RQY
��EDH
��ACF�Similarly�ABE
��FDH
�PQY�Nextsince�ACF��ABE�QYbisects�PQR�Fromthes
itfollowstheperpendicularsfromP�Q�RtoAB�BC�CAconcur
theincenterof�PQR�
��������ChineseMathOlympiadTrainingTest�ABCDEFisahexag
��
inscribedinacircle�ShowthatthediagonalsAD�BE�CFareconcur
rentifandonlyifAB�CD�EF�BC�DE�FA�
Solution��DuetoYuKaChun�SupposeAD�BE�CFconcursatX�
FromsimilartrianglesABX
andEDX�wegetAB�DE�BX�DX�
SimilarlyCD�FA�DX�FX
andEF�BC�FX�BX�Multiplying
thseweget�AB�CD�EF���DE�FA�BC����soAB�CD�EF�
BC�DE�FA�
Fortheconverseweusethesocalledmethodoffalseposition�
Suppose� �AB�CD�EF�BC�DE�FAandAD
intersectBE
atX�NowletCX
meetthecircleagainatF� �Bythe�rstpartwe
getAB�CD�EF��BC�DE�F� A�Dividingthisby� �wehave
EF� �EF�F� A�FA�IfF�isonopenarcAF�thenF� A�FAand
EF
�
EF�yieldingF� A�FA
�
��
EF� �EF�acontradiction�If
F�isontheopenarcEF�thenFA�F� AandEF��EFyielding
EF� �EF���F� A�FA�acontradiction�SoF� �F�
AlternativelywecanuseCeva�stheoremanditsconverse�Let
ACandBEmeetatG�CEandADmeetatH�EAandCFmeetat
I�Leth�kbethedistancesfromA�CtoBE�respectively�Then
AG
CG
�h k�
ABsin�ABG
BCsin�CBG
�
Similarly C
HEH
�CDsin�CDH
DEsin�EDH
and
EI
AI
�EFsin�EFI
FAsin�AFI
�
Now�ABG��EDH��CBG��EFI��CDH��AFI�ByCeva�s
theoremanditsconverseAD�BE�CFareconcurrentifandonlyif
��
AG�CH�EI
CG�EH�AI
�AB�CD�EF
BC�DE�FA
�
���AcircleintersectsatriangleABCatsixpointsA��A��B��B��C��C��
wheretheorderofappearancealongthetriangleisA�C��C��B�A��A��
��
C�B��B��A�SupposeB�C��B�C�meetsatXC�A��C�A�meets
YandA�B��A�B�meetsatZ�ShowthatAX�BY�CZareconcurre n
Solution�LetDbetheintersectionofAXandB�C��SinceAX�B�C
B�C�areconcurrentby�thetrigonometricformof�Ceva�stheorem
��
DC��B�B��AC�
DB��AB��C�C�
�sinC�AD�sinB�B�C��sinB�C�B�
sinDAB��sinC�B�C��sinB�C�C�
�
Then
sinBAX
sinXAC
�sinC�AD
sinDAB�
�sinC�B�C��sinB�C�C�
sinB�B�C��sinB�C�B�
�Similarly
sinCBY
sinYBA
�sinA�C�A��sinC�A�A�
sinC�C�A��sinC�A�C�
�
sinACZ
sinZCB
�sinB�A�B��sinA�B�A�
sinA�A�B��sinA�B�A�
�
Using�C�B�C�
��C�A�C�
andsimilarangleequalityweseeth
theproductofthethreeequationsinvolvingX�Y�Zaboveisequal
��BytheconverseofthetrigonometricformofCeva�stheoremwes
thatAX�BY�CZareconcurrent�
��������IMOshortlistedproblem�Acirclepassingthroughvertices
andCoftriangleABCintersectssidesAB
andACatC�andB
respectively�ProvethatBB� �CC�andHH�areconcurrentwhere
andH� aretheorthocentersoftrianglesABCandAB� C� �respective
Solution��DuetoLamPoLeung�Letd�X�L�denotethedistan
fromapointXtoalineL�Fortheproblemwewillusethefollowi
lemma�
Lemma�LetlinesL��L�intersectatP�formingfourangleswithve
texP��SupposeH�H�lieonanoppositepairoftheseangles�
d�H�L���d�H� �L���d�H�L���d�H� �L���thenH�P�H� arecollinea
Proof�LetHH�intersectL��L�atX�Y�respectively�Then
HH�
H� X�
HX
H� X���
d�H�L��
d�H� �L��
��
�
d�H�L��
d�H� �L��
���
HY
H� Y���
HH�
H� Y�
��
SoX�YisonbothL�andL��henceitisP�ThereforeH�P�H� are
collinear�
FortheproblemletBB� �CC�intersectatP�Since�ABH
�
�����A
�
�AC� H� �soBH
kC� H� �SimilarlyCH
kB� H� �Let
BH�CC� intersectatLandCH�BB�intersectatK�Now
�PBH��ABH��C� BP�������A���B� CP
��ACH��B� CP��PCH�
SoK�B�C�Lareconcyclic�Then�LHK��BHCaresimilar�Also
�BHC��B� H� C�aresimilarbecause
�CBH������ACB������AC� B� ��C� B� H�
andsimilarly�BCH
�
�B� C� H� �Therefore�LHK��B� H� C�are
similar�SoKH�B� H� �LH�C� H� �SinceBHkC� H� andCHkB� H� �
sod�H�BB� ��d�H� �BB� ��
d�H�CC� ��d�H� �CC� ��Bythelemma
HH� alsopassesthroughP�
PerpendicularLines
��������APMO�LetABCbeatriangleandDthefootofthealtitude
fromA�LetEandFbeonalinepassingthroughDsuchthatAE
isperpendiculartoBE�AFisperpendiculartoCF�andEandFare
di erentfromD�LetM
andNbethemidpointsofthelinesegments
BCandEF�respectively�ProvethatANisperpendiculartoNM�
Solution��DuetoCheungPokMan�Therearemanydi erentpic
turessoitisbettertousecoordinategeometrytocoverallcases�
SetAattheoriginandlety�b���betheequationoftheline
throughD�E�F��Notethecaseb��impliesD
�E
�F�which
isnotallowed��LetthecoordinatesofD�E�Fbe�d�b���e�b���f�b��
respectively�SinceBE AEandslopeofAEisb�e�sotheequa
tionoflineAEisex�by��b��e�����Similarlytheequationof
lineCFisfx�by��b��f����andtheequationoflineBCis
dx�by��b��d�����
��
FromthesewefoundthecoordinatesofB�Care�d�e�b�de b��� d
f�b�df b��respectively�ThenthecoordinatesofM�Nare�d�e�f
�
� b
de�df
�b
���e
�f�
�b��respectively�SotheslopeofANis�b��e�f�andt
slopeofMN
is��de�df
�b
��d��e�f�b�Theproductoftheseslopes
���ThereforeAN MN�
��������APMO�LetABCbeatriangle�LetM
andN
bethepoin
inwhichthemedianandtheanglebisectorrespectivelyatAme
thesideBC�LetQandPbethepointsinwhichtheperpendicular
N
toNAmeetsMAandBA�respectivelyandOthepointinwhi
theperpendicularatPtoBAmeetsANproduced�ProvethatQO
perpendiculartoBC�
Solution���DuetoWongChunWai�SettheoriginatN
andt
xaxisonlineNO�LettheequationoflineABbey�ax�b�th
theequationoflinesACandPOarey��ax�bandy��
� ax�
respectively�LettheequationofBC
bey�
cx�ThenB
hasc
ordinates�
bc�a
�bc
c�a
��Chascoordinates��
bc�a
��bc
c�a
��M
h
coordinates�
ab
c��a��
abc
c��a���Ahascoordinates��
b a����Ohasc
ordinates�ab���andQhascoordinates���
ab c��ThenBChasslope
andQOhasslope�
� c�ThereforeQO BC�
Solution���DuetoPoonWaiHoi�ThecaseAB
�ACisclea
WithoutlossofgeneralitywemayassumeAB�AC�LetANinterse
thecircumcircleof�ABCatD�Then
�DBC��DAC�
� ��BAC��DAB��DCB�
SoDB�DCandMD BC�
WithAasthecenterofhomothetyshrinkDtoO�BtoB� and
toC� �Then�OB� C� �� ��BAC��OC� B�andBCkB� C� �LetB�
cutPN
atK�Then�OB� K��DAB��OPK�SoP�B� �O�K
a
concyclic�Hence�B� KO��B� PO����andsoB� K�C� K�Sin
��
BCkB� C� �thisimpliesA�K�M
arecollinear�ThereforeK
�Q�
Since�B� KO����andBCkB� C� �wegetQO BC�
���LetBB�andCC�bealtitudesoftriangleABC�AssumethatAB��
AC�LetM
bethemidpointofBC�HtheorthocenterofABCandD
theintersectionofB� C�andBC�ProvethatDH AM�
Solution�LetA�bethefootofthealtitudefrom
AtoBC�Since
A� �B� �C� Mlieontheninepointcircleof�ABC�sobytheintersecting
chordtheoremDB� �DC� �DA� �DM�Since�AC� H������AB� H�
pointsA�C� �H�B�lieonacircle��
withthemidpointX
ofAH
as
center�Since�HA� M
���� �sothecircle��
throughH�A� �M
has
themidpointYofHM
ascenter�SinceDB� �DC��DA� �DM�the
powersofDwithrespectto��and��arethesame�SoD�andH�
areontheradicalaxisof������ThenDH
XY�Bythemidpoint
theoremXYkAM�ThereforeDH AM�
��������ChineseTeamSelectionTest�ThesemicirclewithsideBCof
�ABCasdiameterintersectssidesAB�ACatpointsD�E�respec
tively�LetF�GbethefeetoftheperpendicularsfromD�Etoside
BCrespectively�LetM
betheintersectionofDGandEF�Provethat
AM
BC�
Solution�LetHbethefootoftheperpendicularfromAtoBC�Now
�BDC������BEC�SoDF�BDsinB
�BCcosBsinB
and
similarlyEG�BCcosCsinC�Now
GM
MD
�EG
FD
�cosCsinC
cosBsinB
�cosC
cosB
AB
AC
�
SinceBH�ABcosB�HG�AEcosC�weget
BH
HG
�ABcosB
AEcosC
�ACcosB
ADcosC
and
BH
HG
GM
MD
DA
AB
���
BytheconverseofMenelaus�theoremon�BDG�pointsA�M�Hare
collinear�ThereforeAM
BC�
��������IMO�AcirclewithcenterOpassesthroughtheverticesAand
CoftriangleABCandintersectsthesegmentsABandACagainat
��
distinctpointsK
andN�respectively�Thecircumcirclesoftriang l
ABCandKBN
intersectatexactlytwodistinctpointsB
andM
ProvethatOM
MB�
Solution�LetCM
intersectthecirclewithcenterOatapoint
Since�BMC�������BAC�������KAC��KLC�soBM
paralleltoKL�Now
�LKM
��LKN��NKM
��LCN��NBM
�������BMC��BAC��KLM�
ThenKM
�LM�SinceKO�LO�soOM
KL�HenceOM
BM
��������ChineseSenoirHighMathCompetition�Acirclewithcenter
isinternallytangenttotwocirclesinsideitatpointsSandT�Suppo
thetwocirclesinsideintersectatM
andNwithNclosertoST�Sho
thatOM
MNifandonlyifS�N�Tarecollinear�
Solution��DuetoLeungWaiYing�Considerthetangentlinesat
andatT��SupposetheyareparallelthenS�O�Twillbecollinear
thatM
andN
willbeequidistantfromST�contradictingN
isclos
toST��LetthetangentlinesmeetatK�then�OSK������OT
impliesO�S�K�TlieonacirclewithdiameterOK�AlsoKS��KT
impliesK
isontheradicalaxisMN
ofthetwoinsidecircles�
M�N�Karecollinear�
IfS�N�Tarecollinearthen
�SMT��SMN��TMN��NSK��KTN�������SKT
SoM�S�K�T�Oareconcyclic�Then�OMN
��OMK��OSK
��� �
ConverselyifOM
MN�then�OMK
������OSK
impl
M�S�K�T�Oareconcyclic�Then
�SKT�������SMT
�������SMN��TMN
�������NSK��KTN�
��
Thus�TNS�������NSK��SKT��KTN������Therefore
S�N�Tarecollinear�
���AD�BE�CFarethealtitudesof�ABC�LinesEF�FD�DEmeetlines
BC�CA�ABinpointsL�M�N�respectively�ShowthatL�M�N
are
collinearandthelinethroughthemisperpendiculartothelinejoining
theorthocenterHandcircumcenterOof�ABC�
Solution�Since�ADB������AEB�A�B�D�Eareconcyclic�By
theintersectingchordtheoremNA�NB�ND�NE�Sothepowerof
N
withrespecttothecircumcirclesof�ABC��DEFarethesame�
HenceNisontheradicalaxisofthesecircles�SimilarlyL�M
arealso
onthisradicalaxis�SoL�M�Narecollinear�
Sincethecircumcircleof�DEFistheninepointcircleof�ABC�
thecenterNoftheninepointcircleisthemidpointofHandO�Since
theradicalaxisisperpendiculartothelineofcentersOandN�sothe
linethroughL�M�NisperpendiculartothelineHO�
GeometricInequalities�Maximum�Minimum
���������IMO�LetP��P������P�n��bedistinctpointsonsomehalfof
theunitcirclecenteredattheoriginO�Showthat
j��OP����OP�����������
OP�n��j���
Solution�Whenn���thenj��OP�j���Supposethecasen�kistrue�
Forthecasen�k���wemayassumeP��P������P�k��arearranged
clockwise�Let�� OR���OP�����������
OP�k��
and� OS���OP�������
OP�k���
Bythecasen�k�j�� ORj���Also�� ORliesinside�P�OP�k���Since
j��OP�j���j�����
OP�k��j�OSbisects�P�OP�k���Hence�ROS���� �
Thenj��OP�����������
OP�k��j�j�� OR�� OSj�j�� ORj���
����Lettheanglebisectorsof�A��B��CoftriangleABCintersectits
circumcircleatP�Q�R�respectively�Provethat
AP�BQ�CR�BC�CA�AB�
��
Solution��DuetoLauLapMing�Since�ABQ��CBQ�weha
AQ�CQ�Bycosinelaw
AQ
��AB
��BQ
���AB�BQcos�ABQ
CQ
��CB
��BQ
���CB�BQcos�CBQ�
IfAB
��CB�thensubtractingtheseandsimplifyingwegetAB
CB��BQcos�ABQ��BQ�IfAB�CB�thenBQisadiamet
andweagaingetAB�CB��AB��BQ�SimilarlyBC�AC��C
andCA�BA��AP�Addingtheseinequalitiesanddividingby�w
getthedesiredinequality�
���������APMO�LetABCbeatriangleinscribedinacircleandletl a
ma�Ma�lb�mb�Mb�lc�mc�Mc�wherema�mb�mcarethelengt
oftheanglebisectors�internaltothetriangle�andMa�Mb�Mca
thelengthsoftheanglebisectorsextendeduntiltheymeetthecirc
Provethat
l asin
�A
�
l b
sin
�B
�
l c
sin
�C
���
andthatequalityholdsi ABCisequilateral�
Solution��DuetoFungHoYin�LetA� bethepointtheanglebisect
of�Aextendedtomeetthecircle�Applyingsinelawto�ABA� �w
getAB�sinC�Ma�sin�B�A ���Applyingsinelawto�ABD�weg
AB�sin�C�A ���ma�sinB�So
l a�
ma
Ma
�
sinBsinC
sin�B�A ��sin�C�A ��
�sinBsinC�
BytheAMGMinequality
l asin
�A
�
l b
sin
�B
�
l c
sin
�C
�sinBsinC
sin
�A
�sinCsinA
sin
�B
�sinAsinB
sin
�C
�
withequalityifandonlyifsinA�sinB�sinCandC�A �
�B�A �
������� �whichisequivalentto�A��B��C�
��
�����MathematicsMagazineProblem
�����LetIandO
betheincen
terandcircumcenterof�ABC�respectively�Assume�ABCisnot
equilateral�soI��O��Provethat
�AIO����ifandonlyif
�BC�AB�CA�
Solution��DuetoWongChunWai�LetDbetheintersectionofrayAI
andthecircumcircleof�ABC�ItiswellknownthatDC�DB�DI�
�DC�DBbecause�CAD��BADandDB�DIbecause�BID�
�BAD��ABI��CAD��CBI��DBC��CBI��DBI��Since
ABDCisacyclicquadrilateralbyPtolemy�stheoremAD�BC�
AB�DC�AC�DB��AB�AC��DI�ThenDI�AD�BC��AB�AC��
Since�AODisisosceles�AIO����ifandonlyifDI�AD���which
isequivalentto�BC�AB�AC�
Comments�Inthesolutionabovewesee�AIO����ifandonlyif
�BC�AB�AC�Alsotheconverseofthewellknownfactistrue
i�e�thepointIonADsuchthatDC�DB�DIistheincenterof
�ABC�Thisisbecause�BID��DBIifandonlyif�CBI��ABI�
since�DBC��BADalways��
����SquaresABDEandACFGaredrawnoutside�ABC�LetP�Qbe
pointsonEGsuchthatBPandCQareperpendiculartoBC�Prove
thatBP�CQ�BC�EG�Whendoesequalityhold�
Solution�LetM�N�ObemidpointsofBC�PQ�EG�respectively�Let
HbethepointsothatHEAGisaparallelogram�Translatingby� GA�
thenrotatingby���aboutA��GHAwillcoincidewith�ABCandO
willmovetoM�SoHA�BC�HA BC�OE�OG�MA�EG
MA�LetLbeonMN
suchthatALkEG�SinceNLkPB�PB
BC�BC HA�soLNOAisaparallelogram�ThenAO�LN�Since
MA EG�soMA AL�whichimpliesML�MA�Therefore
BP�CQ��MN���LN�ML�
���AO�MA����BM
�OE��BC�EG�
EqualityholdsifandonlyifLcoincideswithA�i�e�AB�AC�
��
����PointPisinside�ABC�DeterminepointsDonsideABandE
sideACsuchthatBD�CEandPD�PEisminimum�
Solution�TheminimumisattainedwhenADPEisacyclicqua d
lateral�ToseethisconsiderthepointG
suchthatG
liesont
oppositesideoflineAC
asB��ABP
�
�ACG
andCG
�
B
LetE
betheintersectionoflinesAC
andPG�LetD
bethei
tersectionofAB
withthecircumcircleofAPE�SinceADPE
is
cyclicquadrilateral�BDP
�
�AEP
�
�CEG�Usingthede�n
tionofG�wehave�BDP��CEGarecongruent�SoBD�CEan
PD�PE�GE�PE�GP�
ForotherD� �E�onsidesAB�AC�respectivelysuchthatBD�
CE� �bythede�nitionofG�wehave�BPD� ��CGE�arecongruen
ThenPD� �GE�andPD� �PE� �GE� �PE� �GP�
SolidorSpaceGeometry
�����ProposedbyItalyfor����IMO�Whichregularpolygonscanbeo
tained�andhow�bycuttingacubewithaplane�
Solution��DuetoFanWaiTongKeeWingTaoandTamSiuLun
Observethatiftwosidesofapolygonisonafaceofthecubeth
thewholepolygonliesontheface�Sinceacubehas�faceson
regularpolygonwith���or�sidesarepossible�Letthevertic
ofthebottomfaceofthecubebeA�B�C�Dandtheverticesont
topfacebeA� �B� �C� �D�withA�ontopofA�B�ontopofBand
on�ThentheplanethroughA�B� �D� cutsanequilateraltriangle�T
perpendicularbisectingplanetoedgeAA�cutsasquare�Thepla
throughthemidpointsofedgesAB�BC�CC� �C� D� �D� A� �A� Acuts
regularhexagon�Finallyaregularpentagonisimpossibleotherwi
the�vesideswillbeon�vefacesofthecubeimplyingtwooft
sidesareonparallelplanesbutnotwosidesofaregularpentagona
parallel�
���������IsraeliMathOlympiad�Fourpointsaregiveninspaceingener
position�i�e�theyarenotcoplanarandanythreearenotcollinea
��
Aplane�iscalledanequalizingplaneifallfourpointshavethesame
distancefrom��Findthenumberofequalizingplanes�
Solution�Thefourpointscannotalllieononesideofanequalizing
planeotherwisetheywouldlieinaplaneparalleltotheequalizing
plane�Henceeitherthreelieononesideandoneontheotherortwo
lieoneachside�Intheformercasethereisexactlyoneequalizing
planewhichisparalleltotheplanePcontainingthethreepointsand
passingthroughthemidpointofthesegmentjoiningthefourthpointx
andthefootoftheperpendicularfromxtoP�Inthelattercaseagain
thereisexactlyoneequalizingplane�Thetwopairofpointsdetermine
twoskewlinesinspace�Considerthetwoplaneseachcontainingone
ofthelineandisparalleltotheotherline�Theequalizingplaneis
theplanemidwaybetweenthesetwoplane�Sincethereare�����
waysofdividingthefourpointsintothesetwocasesthereareexactly
�equalizingplanes�
��
SolutionstoNumberTheoryProblems
Digits
���������PutnamExam�Provethateverypositiveintegerhasamultip
whosedecimalrepresentationinvolvesalltendigits�
Solution�Letnbeapositiveintegerandp��������������k�whe
kissolargethat��k�n�Thenthenconsecutiveintegersp���p
������p�nhavedecimalrepresentationsbeginningwith�����������
andoneofthemisamultipleofn�
����Doesthereexistapositiveintegerasuchthatthesumofthedig
�inbase���ofais����andthesumofthedigits�inbase���ofa�
������
Solution�Yes�Infactthereissuchanumberwhosedigitsconsist
��sand��s�Letk������Considera�����
�����
��������k�Th
thesumofthedigitsofaisk�Now
a����
��
���
��
�������
�k��
��
X��i�j�k
���i��j�
Observethattheexponentarealldi erentbytheuniquenessofba
�representation�Thereforethesumofthedigitsofa�inbase��
k��Ck ��k��
�����ProposedbyUSSRfor����IMO�Letan
bethelastnonzerodig
inthedecimalrepresentationofthenumbern��Doesthesequen
a��a������an����becomeperiodicaftera�nitenumberofterms�
Solution�SupposeafterNtermsthesequencebecomesperiodicwi
periodT�Thenai�jT
�aifori�N�j����������Bythepigeonho
principletherearetwonumbersamong��N�����N�����N������th
havethesameremainderwhendividedbyT�say��m
���k�modT
withN�m�k�Then��k���m
�jTforsomeintegerj�
��
Observethat��k��
��k���k����impliesa��k
�
a��k���Let
n���k���jT�Since��k���N�an���a��k�jT
�a��k���jT
�an�
Since�n����������k���m��n����������������n���so�an
�
�an��
�an
�mod����Thisimpliesan
���Howeverintheprime
factorizationofn��theexponentof�isgreaterthantheexponentof
�whichimpliesan
isevenacontradiction�
ModuloArithmetic
���������PutnamExam�Provethatthenumberofoddbinomialcoe�
cientsinanyrowofthePascaltriangleisapowerof��
Solution�Byinduction���x��
m
���x
�m
�mod���Ifwewriten
inbase�sayn��a�
��a�
������ak�wheretheai�saredistinct
nonnegativeintegersthen
���x�n����x��
a�
������x��
ak
����x
�a��������x
�ak��mod���
Inexpandingtheexpressioninfrontof�mod��wegetthesumofxnS�
whereforeachsubsetSoff��������kg�nS�
X i�S
�ai�Sincethereare�k
subsetsoff��������kg�thereareexactly�ktermseachwithcoe�cient
��Thisimpliesthereareexactly�koddbinomialcoe�cientsinthe
nthrowofthePascaltriangle�
����Leta��a��a������a��andb ��b��b������b��betwopermutationsofthe
naturalnumbers�������������Showthatifeachofthenumbersa�b ��
a�b ��a�b ������a��b ��isdividedby��thenatleasttwoofthemwill
havethesameremainder�
Solution�Supposea�b ��a�b ������a��b ��havedistinctremainderswhen
dividedby���Bysymmetrywemayassumea�b ����mod����Let
x��a�b ������a��b ����Ononehandx��������mod����Onthe
otherhandsinceaibi����mod����fori����������wegeta�
�
���b ��Sox��a����a����b ����b�����������������mod����a
contradiction�
��
���������CzechSlovakMatc h�Leta��a�����beasequencesatisfyinga�
��a���and
an������n
��an������n
��an
foralln���Dothereexistindicesp�qandrsuchthatapaq�ar�
Solution��DuetoLauLapMing�The�rstfewtermsare����������
���������Sincethedi erencesofconsecutivetermsaremultiplesof
wesuspectan
���mod��foralln�Clearlya��a����mod���
an�an�����mod���then
an������n
�������n
��������mod���
Sobyinductionallan
���mod���Thenapaq��arforallp�q�r
�����mod���
PrimeFactorization
�����AmericanMathematicalMonthlyProblemE�����LetAn
bethes
ofpositiveintegerswhicharelessthannandarerelativelyprimeto
Forwhichn���dotheintegersinAnformanarithmeticprogressio
Solution�SupposeAn
isanarithmeticprogression�Ifnisoddan
n���then����AnimpliesAn�f��������n��g�whichimpliesn
prime�Ifnisevenandnotdivisibleby�then����An����Animp
An
�f����������n��g�whichimpliesnisapowerof��Finally
nisevenanddivisibleby�thenletpbethesmallestprimen
dividingn�Eitherp���mod��orp���mod���Intheform
casesince��parethe�rsttwoelementsofAn
andn���An�
��k�p����
n��forsomek�Thisimpliesn���mod���
contradiction�Sop���mod���Then�p��isdivisibleby�an
so�p����An�ConsequentlyAn
�f��pg�whichimpliesn��
consideringtheprimefactorizationofn�ThereforeAnisanarithmet
progressionifandonlyifnisaprimeapowerof�orn���
���������IMO�Provethatthesetofintegersoftheform
�k���k
��������containsanin�nitesubsetinwhicheverytwomembersa
relativelyprime�
���
Solution�Weshallgivearecipeforactuallyconstructinganin�nite
setofintegersoftheformai��ki
���i���������eachrelatively
primetoalltheothers�Leta���������Supposewehavenpairwise
relativelyprimenumbersa���k����
a���k��������an��kn
���
Weformtheproducts�a�a����an�whichisodd�Nowconsiderthe
s��numbers��������������s�Atleasttwoofthesewillbecongruent
�mods�say������mods��orequivalently�����������ms
forsomeintegerm�Theoddnumbersdoesnotdivide���soitmust
divide�������hence�������lsforsomeintegerl�Since������
isdivisiblebysandsisodd������isrelativelyprimetos�This
implies���������ki��fori���������n�Sowemayde�nean���
�������Thisinductiveconstructioncanberepeatedtoformanin�nite
sequence�
Comments�ByEuler�stheoremwemaytaketheexponent��tobe
��s��theEuler�functionofs�whichequalsthenumberofpositivein
tegerslessthansthatarerelativelyprimetos�then��s ���mods��
���������ChineseMathOlympiadTrainingTest�Determinethesmallest
valueofthenaturalnumbern��withthepropertythatwhenever
thesetSn
�f��������ngispartitionedintotheunionoftwosub
setsatleastoneofthesubsetscontainsthreenumbersa�bandc�not
necessarilydistinct�suchthatab�c�
Solution��DuetoLamPeiFung�We�rstshowthat������hasthe
propertythenwewillshowitistheleastsolution�
SupposeS��ispartitionedintotwosubsetsX��X��Withoutloss
ofgeneralitylet�beinX��If����isinX��thenwearedone�
Otherwise�isinX��If�����isinX��thenwearedone�Otherwise
��isinX��If�������isinX��thenwearedone�Otherwise��is
inX��Finallyeither��������isinX�or��������isinX��
Ineithercasewearedone�
Toshow���isthesmallestwewillshowthatS��canbepar
titionedintotwosubsetseachofwhichdoesnotcontainproductsof
itselements�De�neCtobe!prime"inS��ifCisnottheproduct
ofelementsofS���The!primes"inS��consistof����p��pwhere
p����isausualprimenumber�SincethesmallestprimeinS��is
���
�nonumberinS��istheproductofmorethanfour!primes"�P
allthe!primes"andnumbersthatcanbewrittenasproductsoffo
!primes"inonesubsetX��andletX��S��nX��
NoproductsinX�areinX�becausenumbersinX�haveatlea
two!prime"factorssotheirproductscanbewrittenwithatleastfo
!prime"factors�Nextlookingattheproductof����p��p�poddprim
�����weseethataproductoftwo!primes"cannotbefactoredin
aproductoffour!primes"�SonoproductsinX�areinX��
BasenRepresentations
���������IMO�Canyouchoose����pairwisedistinctnonnegativeintege
lessthan���suchthatnothreeareinarithmeticprogression�
Solution�Weconsiderthegreedyalgorithmforconstructingsuch
sequence�startwith��andateachstepaddthesmallestinteg
whichisnotinarithmeticprogressionwithanytwoprecedingterm
Weget�����������������������������Inbase�thissequenceis
����������������������������������������
�Notethissequenceisthenonnegativeintegersinbase���Since����
base�is�����������soswitchingthisfrombase�tobase��weg
the����thtermofthesequenceis����������Toseethissequen
workssupposex�y�zwithx�y�zarethreetermsofthesequen
inarithmeticprogression�Considertherightmostdigitinbase�whe
xdi ersfromy�thenthatdigitforzisa�acontradiction�
�����AmericanMathematicalMonthlyProblem
�����Letpbeanod
primenumberandrbeapositiveintegernotdivisiblebyp�Fora
positiveintegerkshowthatthereexistsapositiveintegermsuchth
therightmostkdigitsofmr�whenexpressedinthebasep�areall�
Solution�Weprovebyinductiononk�Fork���takem���Nex
supposemr�inbasep�endsink��si�e�
mr���p�����pk����apk�higherterms��
���
Clearlygcd�m�p����Then
�m�cpk�r�mr�rmr��cpk�����crpkri
���p�����pk����a�rmr��c�pk�higherterms�
Sincegcd�mr�p����thecongruencea�rmr��c���modp�issolv
ableforc�Ifc �isasolutionthen�m�c �pk�rwillendin�k�����s
asrequired�
�����ProposedbyRomaniafor����IMO�Showthatthesequencefang
de�nedbyan
��n
p ��forn�����������wherethebracketsdenote
thegreatestintegerfunction�containsanin�nitenumberofintegral
powersof��
Solution�Write
p �inbase�asb ��b�b �b �����whereeachb i��or��
Since
p �isirrationaltherearein�nitelymanyb k���Ifb k���then
inbase��k��p ��b ����bk���bk���Letm���k��p ���then
�k��p ������k��p ���m��k��p ��
� ��
Multiplyingby
p �andadding
p ��weget�k��m���
p ���k�
p� ��
Then��m���
p ����k�
Representations
����Findall�even�naturalnumbersnwhichcanbewrittenasasumof
twooddcompositenumbers�
Solution�Letn���anddbeitsunitsdigit�Ifd���thenn�
����n����willdo�Ifd���thenn�����n����willdo�Ifd���
thenn����n���willdo�Ifd���thenn�����n����willdo�If
d���thenn�����n����willdo�Forn����directcheckingshows
only��������������������������������������
canbesoexpressed�
����Findallpositiveintegerswhichcannotbewrittenasthesumoftwo
ormoreconsecutivepositiveintegers�
���
Solution��DuetoCheungPokMan�Foroddintegern��k���
n�k��k����Forevenintegern���supposen�m��m���
�����m�r����m�r��r�����withm�r���Then�m�r�r���
andoneof�m�r�r��isodd�Sonmusthaveanodddivisorgreat
than��Inparticularn��j�j�����������cannotbewritten
thesumofconsecutivepositiveintegers�Fortheotherevenintege
n��j��k���withj�k���If�j�k�thenn���j�k����j�k���
������j�k��If�j�k�thenn��k��j�����k��j���������k��j
�����ProposedbyAustraliafor����IMO�Observethat����������
IsthereanintegerNwhichcanbewrittenasasumof����consecuti
positiveintegersandwhichcanbewrittenasasumof�morethanon
consecutiveintegersinexactly����ways�
Solution�ForsuchNwehaveN�
���� X i�
��m�i�������m�������
Nisoddandisdivisibleby����������Alsothereareexactly��
positiveintegerpairs�n�k�suchthatN�
k X i��
�n�i��
�k����n�� k
�
Hence�Ncanbefactoredas�k�����n�k�inexactly����ways��No
if�N�abwith��a�b�thenn����b�a����k�a����Th
means�Nhasexactly���������������positivedivisors�No
write�Ninprimefactorizationas���e�
����e�
�����Thenweg
�����������e �����e��������Sofe��e�g�f������g�Therefor
N�����������or�����������Asallthestepscanbereversedthe
aretheonlyanswers�
����Showthatifp��isprimethenpncannotbethesumoftwopositi
cubesforanyn���Whataboutp��or��
Solution�Supposenisthesmallestpositiveintegersuchthatpnist
sumoftwopositivecubessaypn�a��b���a�b��a��ab�b���Th
a�b�pkanda��ab�b��pn�k�Sincea�b���sok���Sin
a��ab�b��ab���son�k�Now�ab��a�b����a��ab�b��
p�k�pn�k
and��k�n�sopj�ab�Sincep���sopjaorpj
Sincea�b�pk�sopjaandpjb�saya�pAandb�pB�Th
A��B��pn���contradictingthesmallestpropertyofn�
���
Forp���supposea��b���n�Ifa�b���then�ja��jband
�a�����b �����n���Soa�b��kandn��k���
Forp���supposea��b���n�Ifa�b��kanda��ab�b��
�n�k����then�j�abimplies�ja��jband�a�����b �����n���
Otherwisewehave��a��ab�b��abandthena�b���Sointhis
casea�bare���k��kandn��k���
�����DuetoPaulErd�osandM�Sur�anyi�Provethateveryintegerkcanbe
representedinin�nitelymanywaysintheformk������������m�
forsomepositiveintegermandsomechoiceofsigns�or��
Solution�We�rstshoweveryintegercanbesorepresentedinatleast
oneway�Ifkcanberepresentedthenchangingallthesignswesee
�kalsocanberepresented�Soitsu�cestodothenonnegativecases�
Thekeyobservationistheidentity
�m���
���m���
���m���
���m���
����
Now�������������������������������������
����������������������Bytheidentityifkcanberepresented
thenk��canberepresented�Sobyinductioneverynonnegative
integer�andhenceeveryinteger�canberepresented�Toseethere
arein�nitelymanysuchrepresentationsweusetheidentityagain�
Observe�������m������m������m������m������m�
�����m������m������m�����Soforeveryrepresentationwe
canadd�moretermstogetanotherrepresentation�
���������IMOshortlistedproblem�A�nitesequenceofintegersa��a������
an
iscalledquadraticifforeachi�f��������ng�ja i�ai��j�i��
�a�Provethatforanytwointegersbandc�thereexistsanatural
numbernandaquadraticsequencewitha��bandan�c�
�b�Findtheleastnaturalnumbernforwhichthereexistsaquadratic
sequencewitha���andan������
Solution�Part�a�followsfromthelastproblembylettingk�c�
b�Forpart�b�considerak
insuchaquadraticsequence�Wehave
���
ak
�����������k�
�k�k�����k������Soa��
������Al
ak
�����������k�
�mod���Since�������������
isod
n����Toconstructsuchaquadraticsequencewithn�����rstno
�������������������Nowwewrite�����������������
����������Then
�����������������������������
����
����
��������
����Provethateveryintegergreaterthan��canberepresentedasasum
threeintegers��whicharepairwiserelativelyprimeandshowth
��doesnothavethisproperty�
Solution��DuetoChanKinHangandNgKaWing�Letk���Fro
����������wesee������k���worksfor�k�From��������
wesee������k���worksfor�k���From����������wes
������k���worksfor�k���
For�k���wesplitintocases��k� ��and��k� ���Weha
��k� �������k� ������k� ���and��k� �������k� ������k� ��
For�k���wesplitintocases��k� ��and��k� ���Weha
��k� �������k� ������k� ���and��k� �������k� ������k� ��
For�k���wesplitintocases��k� ��and��k� ����Weha
��k� �������k� ������k� ���and��k� ��������k� ������k� ��
Finally��doesnothavetheproperty�Otherwise���a�b�
wherea�b�carerelativelyprimeanda�b�c�Thena�b�careodd�
a���then������a�b�c�������showsthisisimpossib
Ifa���thenb���c��anda�b�c�������againimpossible
ChineseRemainderTheorem
���������ChineseTeam
SelectionTest�De�nexn
�
�xn����for
positiveintegersn�Provethatanintegervaluecanbechosenforx�
thatx���isdivisibleby�����
���
Solution�Lety n�
xn
�n�theny n�y n���
� �n�whichimplies
y n�y ��
� ��
� �������
� �n�
Thisgivesxn��x�����n���Wewantx�����x����������tobe
divisibleby������������whichmeans
x����mod���x����mod���x�����mod����
Since������arepairwiserelativelyprimebytheChineseremainder
theoremsuchx�exists�
�����ProposedbyNorthKoreafor����IMO�DoesthereexistasetM
withthefollowingproperties�
�a�ThesetM
consistsof����naturalnumbers�
�b�EveryelementinM
andthesumofanynumberofelementsinM
havetheformmk�wherem�karepositiveintegersandk���
Solution��DuetoCheungPokMan�Letn��������������Choose
ndistinctprimenumbersp ��p������pn�Letd��e��e��e�
���nen�where
e iisasolutionofthenequationsx����modp i�andx���modp j�
forevery��j�n�j��i��Sincethep i�sarepairwiserelatively
primesuchasolutionexistsbytheChineseremaindertheorem��Since
e ��e������en���modp ���disap ithpower�Sincee ����e ������en�
��modp ����disap �thpowerandsoon�Itfollowsd��d���������d
areallperfectpowersandanysumofthemisamultipleofd�lessthan
orequaltond�henceisalsoaperfectpower�
Divisibility
����Findallpositiveintegersa�bsuchthatb��and�a��isdivisibleby
�b���
Solution�Sinceb�
��so�b���
�a���henceb�
a�Leta�
qb�r���r�b�thenbydivisionweget
�a��
�b��
��a�b��a��b������a�qb�
�r��
�b��
�
���
Since��
�r��
�b��
���therearenosolutions�
����Showthattherearein�nitelymanycompositensuchthat�n����n
isdivisiblebyn�
Solution�Weusethefactx�yjxk�ykforpositiveintegerk�Consid
n���
t
���
t
fort���������Byinductionwecanshow�tj��t
��n�����
t��AlternativelybyEuler�stheorem��
t
�����t ��
��mod�t���Thenn����tk�Son���
t
���
t
j���
t�k����t�k
�n����n���
����Provethattherearein�nitelymanypositiveintegersnsuchthat�n�
isdivisiblebyn�Findallsuchn�sthatareprimenumbers�
Solution�Lookingatthecasesn��to��suggestforn��k�k
����������weshouldhavenj�n���Thecasek��isclear�Suppo
casekistrue�Now��
k��
������k
������k����
k
����Bycase
��k
���mod���so��
k����k������������������mod���
��k��
��isdivisibleby�k���completingtheinduction�
Ifaprimendivides�n���thenbyFermat�slittletheoremnj�n
��too�Thennj��n������n������son���
���������RomanianMathOlympiad�Findallpositiveintegers�x�n�su
thatxn��n��isadivisorofxn����n�����
Solution��DuetoChengKeiTsiandLeungWaiYing�Forx�
���n��n�����n����n������n��n���Forx������n��n���
�n����n������n��n���Forx������n��n�����n��
�n��������n��n����Sotherearenosolutionswithx������
Forx���ifn���thenwegetx�xn��n����xn����n���
Now
xn����n
����
��x����xn��n���
�xn���n���x����n��
��x����xn��n���
���
becauseforn���xn���n���x��n���x���x����andfor
n���xn���n���x�x��n����n������Henceonlyn��
andx��arepossible�Nowxn
��n���x��isadivisorof
xn����n�����x�����x����x������ifandonlyifx��
isadivisorof���Sincex�����x��or���Sothesolutionsare
�x�y�������and�������
���������BulgarianMathCompetition�Findallpairsofpositiveintegers
�x�y�forwhich
x��y�
x�y
isanintegeranddivides�����
Solution�Suppose�x�y�issuchapair�Wemayassumex�y�oth
erwiseconsider�y�x��Thenx��y�
�k�x�y��wherekj�����
���������Ifp��or�or��dividesk�thenbythefactthat
primep���mod��dividingx��y�
impliespdividesxory�we
maycancelp�togetanequationx� ��y� �
�
k��x��y ��withk�
notdivisibleby�����
Sincex� ��y� �
�
x� �
�
x�
�
x��y ��
wemusthavex� ��y� �
�
��x��y ���Completingsquaresweget
��x��������y �����
����whichgives�x��y��������or������
Itfollows�x�y����c�c����c�c���c��c���c��c��wherecisapositive
divisorof�������
���������RussianMathOlympiad�Isthereasequenceofnaturalnumbers
inwhicheverynaturalnumberoccursjustonceandmoreoverforany
k����������thesumofthe�rstktermsisdivisiblebyk�
Solution�Leta�
�
��Supposea������ak
hasbeenchosentohave
theproperty�Letnbethesmallestnaturalnumbernotyetappeared�
BytheChineseremaindertheoremthereisanintegerm
suchthat
m��a������ak�modk���andm��a������ak�n�modk����
Wecanincreasem
byalargemultipleof�k����k���toensureit
ispositiveandnotequaltoanyoneofa������ak�Letak���m
and
ak���n�Thesequenceconstructedthiswayhavetheproperty�
���������PutnamExam�LetA���andA����Forn���thenumber
An
isde�nedbyconcatenatingthedecimalexpansionsofAn��
and
An��fromlefttoright�ForexampleA��A�A�����A�A�A��
���
����A��AA��������andsoforth�DetermineallnsuchthatA
isdivisibleby���
Solution�TheFibonaccinumbersFn
isde�nedbyF����F���an
Fn
�Fn���Fn��forn���NoteAn
hasFn
digits�Sowehavet
recursionAn���Fn�
�An���An�������Fn�
�An���An���mod��
ByinductionthesequenceFn
�mod��is����������������The�r
eighttermsofAn�mod���are������������������Notethenumbe
starttorepeatafterthesixthterm��Infacttherecursionimpl
An���An�mod���byinduction�SoAnisdivisibleby��ifandon
ifn��k��forsomepositiveintegerk�
���������BulgarianMathCompetition�Ifk���showthatkdoesn
divide�k�����Usethisto�ndallprimenumberspandqsuchth
�p��qisdivisiblebypq�
Solution�Supposekj�k����forsomek���Thenkisodd�Wri
k�pe��
���perr�wherep i�saredistinctprimes�Letp i����mi
witheq iodd�Letmj�minfm������mrg�Sincemi���modmj��w
getpeii
���modmj�andsok��mjq��forsomepositiveinteg
q�Sincepj
jkandkj�k�����so��
m
jq
��k��
����modp j
Then�pj�� q����m
jq�qj
����modpj�becauseq jisodd�Howeve
�pj�����modpj�byFermat�slittletheoremsincegcd���pj����
�pj�� q���modpj��acontradiction�
Supposep�qareprimeand�p��qisdivisiblebypq�Then�p
��q�modp��Ifp�qareoddthen�p���modp�byFermat�slitt
theoremand�q���p����modp��So�pq�����p����modp
Similarly�pq����modq��Then�pq������modpq��contradic
ingthe�rstpartoftheproblem�Ifp���thenq��orq���
q���then�����q����modq�byFermat�slittletheoremwhi
impliesq���Thereforethesolutionsare�p�q�������������������
����Showthatforanypositiveintegern�thereisanumberwhosedecim
representationcontainsndigitseachofwhichis�or�andwhich
divisibleby�n�
Solution�Wewillprovethatthe�nnumberswithndigitsof��s
��shavedi erentremainderswhendividedby�n�Henceoneofthem
���
divisibleby�n�Forn���thisisclear�Supposethisistrueforn�k�
Nowifa�bare�k���digitnumberswhereeachdigitequals�or�
anda�b�mod�k����thentheunitsdigitsofa�barethesame�If
a���a� �i�b���b��i�whereiistheunitsdigitthen�k��divides
a�b����a� �b��isequivalentto�kdividesa� �b��Sincea� �b�are
kdigitnumbers�withdigitsequal�or��wehavea��b��Soa�b�
completingtheinduction�
����Forapositiveintegern�letf�n�bethelargestintegerksuchthat�k
dividesnandg�n�bethesumofthedigitsinthebinaryrepresentation
ofn�Provethatforanypositiveintegern�
�a�f�n���n�g�n��
�b��divides
� �n n� ���n��
n�n�
ifandonlyifnisnotapowerof��
Solution��DuetoNgKaManandPoonWingChi��a�Writenin
base�as�arar�����a�� ��Then
ai��ar���ai��ai����ar���ai������
n �i� �� n �i
��
� �
So
g�n��
r X i��
ai�
r X i��
� n �i� �� n �i
��
�� �n�
r X i��
n �i� �n�f�n���
�b�LetMn���n����n����Sinceg��n��g�n��using�a�weget
f�Mn��f���n�����f�n����g�n��g��n��g�n��
Sothelargestksuchthat�kdividesMn
isk�g�n��Now�divides
Mnifandonlyifg�n����whichisequivalenttonnotbeingapower
of��
�����ProposedbyAustraliafor����IMO�Provethatforanypositivein
tegerm�thereexistanin�nitenumberofpairsofintegers�x�y�such
that
���
�a�xandyarerelativelyprime�
�b�ydividesx��m�
�c�xdividesy��m�
Solution�Note�x�y�������issuchapair�Nowif�x�y�iss u
apairwithx�y�thenconsider�y�z��wherey��m
�xz�Th
everycommondivisorofzandyisadivisorofm�andhenceofx�
gcd�z�y����Now
x��z
��m���y
��m��x
�m�y
��my
��m�x
��m�
isdivisiblebyy�Sincegcd�x�y����yjz��m�so�y�z�isanoth
suchpairwithy�y��x�z�Thiscanberepeatedin�nitelyma
times�
����Findallintegersn��suchthat�n��n������n���n
isdivisib
byn�
Solution�Foroddn��j�����since�n�k�n�kn
�modn�f
��k�j�so�n��n������n���n
isdivisiblebyn�Forevenn�wri
n��st�wheretisodd�Then�sj�n��n������n���n�Nowif
isevenandlessthann�then�sjkn�Ifkisoddandlessthann�th
byEuler�stheoremk�s��
���mod�s��sokn
���mod�s��Th
���n��n������n���n
�n �
�mod�s��whichimplies�s��jn
contradiction�Soonlyoddn��hastheproperty�
���������PutnamExam�Showthatifnisanintegergreaterthan�th
ndoesnotdivide�n���
Solution�Supposenj�n��forsomen���Since�n��isodd
nisodd�Letpbethesmallestprimedivisorofn�Thenpj�n���
�n���modp��ByFermat�slittletheorem�p�����modp��Let
bethesmallestpositiveintegersuchthat�k���modp��Thenkj
�becauseotherwisen�kq�rwith��r�kand���n���k�q�r
�r�modp��contradictingkbeingsmallest��Similarlykjp���
kjgcd�n�p����Nowd�gcd�n�p���mustbe�sincedjn�d�p�
andpisthesmallestprimedivisorofn�Sok�
�and��
�k
��modp��acontradiction�
���
�����ProposedbyRomaniafor����IMO�Fork���letn��n������nkbe
positiveintegerssuchthat
n�
� � ��n�
����n�
� � ��n�
��������nk
� � ��nk�
�
����n�
� � ��nk
����
Provethatn��n������nk���
Solution�Observethatifni��forsomei�thenni��willequal�
andthechaine ectcausesallofthemtobe��Soassumenoniis��
Letp kbethesmallestprimenumberdividingnk�Thenp kj�nk�
�
���
So�nk�
�
���modp k��Letmk
bethesmallestpositiveintegerm
suchthat�m���modp k��Thenmk
jnk��
andmk
jpk��by
Fermat�slittletheorem�Inparticular��mk
�p k���p kandso
thesmallestprimedivisorp k��ofnk��islessthanp k�Thenweget
thecontradictionthatp k�p k�������p ��p k�
���������APMO�Determinethelargestofallintegernwiththeproperty
thatnisdivisiblebyallpositiveintegersthatarelessthan�p n�
Solution��DuetoLauLapMing�Thelargestnis����Since����
�������and��
�p ����
�����hastheproperty�Nextifn
hasthepropertyandn�
����then�������dividen�Hencen�
�����������Then�������dividen�son��������������
���������Then�������������dividen�son������������
������������Letkbetheintegersuchthat��k�
�p n���k���
Then�k��k��k���k���k���k���k���kdividen�andwegetthefollowing
contradiction
n��k�k�k��k��k��k��k��k
���k�����k�����k������k������k������k���
�k���n�
���������UkrainianMathOlympiad�Findthesmallestintegernsuchthat
amonganynintegers�withpossiblerepetitions�thereexist��integers
whosesumisdivisibleby���
Solution�Takingseventeen��sandseventeen��sweseethatthe
smallestsuchintegerncannotbe��orless�Wewillshow��isthe
���
answer�Considerthestatement!amongany�k��integersthereex
kofthemwhosesumisdivisiblebyk�"Wewill�rstshowthatift
statementistruefork�k�andk��thenitistruefork�k�k��
Supposeitistruefork�k�andk��Sincethecasek�k�istru
for�k�k���integerswecantakeout�k���ofthemandpickk�
themwithsumdivisiblebyk�toformagroup�Thenreturntheoth
k���integerstotheremainingintegersandrepeatthetakingan
picking�Totallywewewillget�k���groups�Sincethecasek�
istruefromthe�k���sumss ������s�k���ofthegroupsconsideri
thenumbersdi�s i�gcd�k��k���wecangetk�ofthemwhosesum
divisiblebyk��Theunionofthek�groupswithsums i�sconsists
k�k�numberswhosesumisthendivisiblebyk�k��
To�nishtheproblemsince���
�����wehavetoshowt
statementistruefork��and��Among�������numbe
therearetwooddortwoevennumberstheirsumiseven�Amo
�������integersconsider�mod��oftheintegers�If���ea
appearsthenthesumofthosethreewillbe��mod��otherwisethe
aretwochoicesfor�integersandthreeofthemwillbecongruent�m
��theirsumis��mod���
Comments�Thestatementistrueforeverypositiveintegerk�Allw
havetoconsideristhecasek�pisprime�Suppose�p��intege
aregiven�Thereare
m�
� �p��
p
� ���p�����p�������p���
�p����
waysinpickingpofthem�Ifnopofthemhaveasumdivisibleby
thenconsider
S�
X �a������ap�p���
wherethesum
isoverallm
pickingsa������ap�ByFermat�slitt
theorem
S���������m����modp��
Ontheotherhandinexpansionthetermsae��
���a
epp
haveexpone
sume ������e p�p���Hencethenumbersofnonzeroexponents
���
inthetermswillbepositiveintegersj�p���Sincep�jofthee iis
�thecoe�cientoftheterminthefullexpansionofSis
� �p���j
p�j
��p��
e ������ep
� ���p���j����p����p�j���
�p�j��
�p��
e ������ep
� �
whichisdivisiblebyp�Soallcoe�cientsaredivisiblebyp�hence
S���modp��acontradiction�
PerfectSquares�PerfectCubes
����Leta�b�cbepositiveintegerssuchthat
� a�
� b�
� c�Ifthegreatest
commondivisorofa�b�cis�thenprovethata�bmustbeaperfect
square�
Solution�Byalgebra
� a�
� b�
� cisequivalentto
a�c
c
�
cb�c
�Sup
pose
a�c
c
�
cb�c
�p q�wherep�qarepositiveintegersandgcd�p�q��
��Then
ap�q
�c qand
bp�q
�c pbysimplealgebra�So
a
p�p�q�
�
b
q�p�q�
�
c pq�
Nowgcd�p�q���impliesgcd�p�p�q��q�p�q��pq����Sincegcd�a�b�c�
�
��wehavea�
p�p�q��b�
q�p�q�andc�
pq�Therefore
a�b��p�q���
���������E�otv�osK�ursch�akMathCompetition�Letnbeapositiveinteger�
Showthatif���p��n���isanintegerthenitisasquare�
Solution�If���p��n����m�anintegerthen����n������m�
����Thisimpliesmisevensaym��k�So��n��k���k�Thisimplies
kisevensayk��j�Then�n��j�j����Sincegcd�j�j������
eitherj�
�x��j���
y�
orj�
x��j���
�y��Intheformer
caseweget���y��mod���whichisimpossible�Inthelattercase
m��k��j��x�isasquare�
���
���������PutnamExam�Prov ethatforanyintegersa�b�c�thereexists
positiveintegernsuchthat
p n��an��bn�cisnotaninteger�
Solution�LetP�x��x��ax��bx�candn�jbj���Obser
thatP�n��P�n����mod���SupposebothP�n�andP�n���a
perfectsquares�Sinceperfectsquaresarecongruentto�or��mod�
soP�n��P�n����mod���HoweverP�n����P�n���n��
isnotdivisibleby�acontradiction�SoeitherP�n�orP�n���
notaperfectsquare�Thereforeeither
p P�n�or
p P�n���isnot
integer�
���������IMOshortlistedproblem�Letkbeapositiveinteger�Proveth
therearein�nitelymanyperfectsquaresoftheformn�k���where
isapositiveinteger�
Solution�Itsu�cestoshowthereisasequenceofpositiveintege
aksuchthata� k����mod�k�andtheak�shavenomaximum�L
a��a��a�����Fork���supposea� k
����mod�k��Th
eithera� k����mod�k���ora� k��k���mod�k����Intheform
caseletak���ak�Inthelattercaseletak���ak��k���Thensin
k��andakisodd
a� k���a
� k��kak���k���a
� k��kak�a
� k��k����mod�k
���
Sincea� k��k��forallk�thesequencehasnomaximum�
����Leta�b�cbeintegerssuchthat
a b�
b c�
c a���Provethatabcist
cubeofaninteger�
Solution�Withoutlossofgeneralitywemayassumegcd�a�b�c��
�Otherwiseifd�gcd�a�b�c��thenfora� �a�d�b� �b�d�c� �c�d�t
equationstillholdsfora� �b��c� anda� b� c� isacubeifandonlyifabcis
cube��Multiplyingbyabc�wegetanewequationa�c�b�a�c�b��ab
Ifabc����thenwearedone�Otherwiseletpbeaprimed
visorofabc�Sincegcd�a�b�c����thenewequationimpliesthat
dividesexactlytwoofa�b�c�Bysymmetrywemayassumepdivid
���
a�b�butnotc�Supposethelargestpowersofpdividinga�barem�n�
respectively�
Ifn��m�thenn����m
andpn��
ja�c�b�c��abc�Hence
pn��jc�b�forcingpjc�acontradiction�Ifn��m�thenn��m��and
p�m��jc�b�b�a��abc�Hencep�m��ja�c�forcingpjc�acontradiction�
Thereforen��mandabc�
Y pjabc
p�m
isacube�
DiophantineEquations
����Findallsetsofpositiveintegersx�yandzsuchthatx�y�zand
xy�yz�zx�
Solution��DuetoCheungPokMan�Since������������������
wehaveyz�zyify���Hencetheequationhasnosolutionify���
Since��x�y�theonlypossiblevaluesfor�x�y�are�����������and
������Theseleadtotheequations����z����z�zand���z�z��
Thethirdequationhasnosolutionsince�z�z�forz��and�������
isnotasolutiontoxy�yz�zx�Thesecondequationhasnosolution
eithersince�z�
z�The�rstequationleadstotheuniquesolution
��������
�����DuetoW�Sierpinskiin�����Findallpositiveintegralsolutionsof
�x��y��z�
Solution�Wewillshowthereisexactlyonesetofsolutionnamely
x�y�z���Tosimplifytheequationweconsidermodulo��We
have�����y��x��y��z�����z�mod���Itfollowsthatz
mustbeevensayz��w�Then�x��z��y���w
��y���w��y��
Now�w��yand�w��yarenotbothdivisibleby�sincetheirsum
isnotdivisibleby��So�w��y�
�xand�w��y�
��Then
����w
�����y
���mod��and����w
�����y
���mod���From
thesewegetwisoddandyiseven�Ify��then���w��y��x��
or��mod��acontradiction�Soy���Then�w��y��implies
w��andz���Finallywegetx���
���
�����DuetoEuleralso����MoscowMathOlympiad�Ifn���thenpr o
that�ncanberepresentedintheform�n��x��y�withx�yod
positiveintegers�
Solution�Afterworkingoutsolutionsforthe�rstfewcasesapatte
beginstoemerge�If�x�y�isasolutiontocasen�thenthepatte
suggeststhefollowing�If�x�y���isoddthen��x�y����j�x�yj�
shouldbeasolutionforthecasen���If�x�y���iseventh
�jx�yj�����x�y����shouldbeasolutionforthecasen���Beforew
con�rmthisweobservethatsince�x�y����jx�yj���max�x�y�
oddexactlyoneof�x�y����jx�yj��isodd�Similarlyexactlyo
of��x�y����j�x�yj��isodd�Alsoif�x�y�isasolutionandone
x�yisoddthentheotherisalsoodd�
Nowwecon�rmthepatternbyinduction�Forthecasen�
�x�y�������with���������leadstoasolution�����forca
n���Supposeincasen�wehaveasolution�x�y��If�x�y���
oddthen�� x
�y
�
� � �� j�x�yj
�
� � ���x
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� � �� �x�y
�
� � ���x
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���n
���Therefo r
thepatternistrueforallcasesbyinduction�
���������IMOshortlistedproblem�Findallpositiveintegersxandysu
thatx�y��z��xyz�wherezisthegreatestcommondivisorof
andy�
Solution�Suppose�x�y�isapairofsolution�Letx�az�y�b
wherea�barepositiveintegers�andgcd�a�b��
���Theequati
impliesa�b�z�z�
�abz��Hencea�czforsomeintegercan
wehavec�b��z�cbz��whichgivesc�
b��z
bz���
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thatb��or�so�x�y�������or������
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Ingeneralcz
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z��z��
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sob�z�Itfollowsthatc�
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���soc���Nowbisan
integersolutionofw��z�w�z�����Sothediscriminantz��z��
isasquare�Howeveritisbetween�z�����and�z����acontradiction�
Thereforetheonlysolutionsare�x�y�������������������and������
����Findallpositiveintegralsolutionstotheequationxy�yz�zx�
xyz���
Solution�Bysymmetrywemayassumex�y�z�Dividingboth
sidesbyxyz�weget
� z�
� y�
� x���
�xyz
�So
����
�xyz
�� z�
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� x�
� x�
Thenx��or��Ifx���thentheequationimpliesy�z���Ifx���
then
� z�
� y�
� ��
� yz
�So
� ��
� ��
� yz
�� z�
� y�
� y�Theny���Sim
plecheckingsyieldy���z���Thereforetherequiredsolutionsare
�x�y�z����������������������������������������������������������
����Showthatiftheequationx��y����
xyzhaspositiveintegral
solutionsx�y�z�thenz���
Solution��DuetoChanKinHang�Supposetheequationhaspositive
integralsolutionsx�y�zwithz����Thenx��y�forotherwise�x����
x�zwouldgivex��z�����andsox���z����Astheequationis
symmetricinx�y�wemayassumex�y�Amongthepositiveintegral
solutions�x�y�z�withx�yandz����let�x��y��z��beasolution
withx�
leastpossible�Nowx��y �z �x��y� �����
�hasx�
as
aroot�Theotherrootisx�
�y �z ��x�
��y� �����x��Wehave
��x���y� �����x���y� ������y �����y ��Now�y��x��z��isalso
apositiveintegralsolutionwithy ��x�andz �����Howevery ��x�
contradictsx�beingleastpossible�
���
���������CzechSlovakMatch�Findallpairsofnonnegativeintegersxa n
ywhichsolvetheequationpx�yp���wherepisagivenoddprim
Solution�If�x�y�isasolutionthen
px�yp����y����yp�������y
��y���
andsoy���pn
forsomen�Ifn���then�x�y�������andpm
bearbitrary�Otherwise
px��pn���p��
�pnp�p�pnp�� �
� p �� pnp�� �����
� p p��� p
�n�p�pn�
Sincepisprimeallofthebinomialcoe�cientsaredivisiblebyp�Hen
alltermsaredivisiblebypn���andallbutthelastbypn���Therefo
thehighestpowerofpdividingtherightsideispn��andsox�n�
Wealsohave
��pnp�p�pnp�� �
� p �� pnp�� �����
� p p��� p
�n�
Forp�
��thisgives��
��n
�����n�whichimpliesn�
�an
�x�y��������Forp���
� p p��
� isnotdivisiblebyp��soeveryter
butthelatontherightisdivisiblebyp�n���whilethelasttermisno
Since�isdivisiblebyp�n���thisisacontradiction�
Thereforetheonlysolutionsare�x�y�������foralloddprim
pand�x�y�������forp���
����Findallintegersolutionsofthesystemofequations
x�y�z��
and
x��y
��z
����
Solution�Suppose�x�y�z�isasolution�Fromtheidentity
�x�y�z����x
��y
��z
�����x�y��y�z��z�x��
weget��
���z����x����y��Since��
���z�����x�
���y��Checkingthefactorizationof�weseethatthesolutionsa
�����������������������������������
���
SolutionstoCombinatoricsProblems
CountingMethods
���������ItalianMathematicalOlympiad�Givenanalphabetwiththree
lettersa�b�c��ndthenumberofwordsofnletterswhichcontainan
evennumberofa�s�
Solution���DuetoChaoKhekLunandNgKaWing�Foranon
negativeeveninteger�k�n�thenumberofnletterwordswith�ka�s
isCn �k�n��k�Theansweristhesumofthesenumberswhichcanbe
simpli�edto������n������n���usingbinomialexpansion�
Solution���DuetoTamSiuLung�LetSnbethenumberofnletter
wordswithevennumberofa�sandTnbethenumberofnletterwords
withoddnumberofa�s�ThenSn�Tn
��n�AmongtheSn
words
thereareTn��wordsendedinaand�Sn��wordsendedinborc�So
wegetSn�Tn����Sn���SimilarlyTn�Sn����Tn���Subtracting
thesewegetSn�Tn�Sn���Tn���SoSn�Tn�S��T��������
ThereforeSn���n������
����FindthenumberofnwordsfromthealphabetA�f�����g�ifany
twoneighborscandi erbyatmost��
Solution�Letxn
bethenumberofnwordssatisfyingthecondition�
Sox�
���x�
���Lety nbethenumberofnwordssatisfyingthe
conditionandbeginningwith���Byinterchanging�and�y nisalso
thenumberofnwordssatisfyingtheconditionandbeginningwith
���Consideringa��or�infrontofannwordwegetxn��
�
�xn��ynandy n���xn�y n�Solvingfory ninthe�rstequationthen
substitutingintothesecondequationwegetxn����xn���xn���
Forconveniencesetx��x���x����Sincer���r����hasroots
��
p �andx����x����wegetxn
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p ��n����
p ��n�
where�����
p���������
p �����Thereforexn�����
p��n���
���
p ��n������
���������RomanianMathOlympiad�LetA��A������An
bepointsona
circle�Findthenumberofpossiblecoloringsofthesepointswithp
colorsp���suchthatanytwoneighboringpointshavedistinctcolors�
���
Solution�LetCn
betheanswerfornpoints�WehaveC��p�C�
p�p���andC�
�p�p����p����Forn��pointsifA�
andA
havedi erentcolorsthenA������AncanbecoloredinCnwayswh
An��
canbecoloredinp��ways�IfA�
andAn
havethesam
colorthenA������An
canbecoloredinCn��
waysandAn��
c
becoloredinp��ways�SoCn��
��p���Cn��p���Cn��
f
n���whichcanbewrittenasCn���Cn
��p����Cn�Cn�
ThisimpliesCn���Cn
��p���n���C��C���p�p���n�Th
Cn��p���n�����n�p���forn��byinduction�
PigeonholePrinciple
���������AustrianPolishMathCompetition�Doesthesetf�����������
containasubsetAconsistingof����numberssuchthatx�Aimpl
�x��A�
Solution�LetA�bethesubsetofS�f������������gcontaining
numbersoftheform�nk�wherenisanonnegativeintegerandk
anoddpositiveinteger�ThennotwoelementsofA�haveratio��
simplecountshowsA�has����elements�Nowforeachx�A��for
asetSx�fx��xg�S�NotetheunionofallSx�scontainsS�Sobyt
pigeonholeprincipleanysubsetofShavingmorethan����elemen
mustcontainapairinsomeSx�henceofratio��Sonosubsetof��
numbersinShastheproperty�
���������PolishMathOlympiad�Supposeatrianglecanbeplacedinsi
asquareofunitareainsuchawaythatthecenterofthesquareisn
insidethetriangle�Showthatonesideofthetrianglehaslengthle
than��
Solution��DuetoToKarKeung�Throughthecentercofthesquar
drawalineL�paralleltotheclosestsideofthetriangleandaseco
lineL�perpendiculartoL�atc�ThelinesL�andL�dividethesqua
intofourcongruentquadrilaterals�Sincecisnotinsidethetriang
thetrianglecanlieinatmosttwo�adjacent�quadrilaterals�Byt
pigeonholeprincipletwooftheverticesofthetrianglemustbelong
thesamequadrilateral�Nowthefurthestdistancebetweentwopoin
���
inthequadrilateralisthedistancebetweentwooppositeverticeswhich
isatmost��Sothesideofthetrianglewithtwoverticeslyinginthe
samequadrilateralmusthavelengthlessthan��
����Thecellsofa���squarearecoloredwithtwocolors�Provethat
thereexistatleast��rectangleswithverticesofthesamecolorand
withsidesparalleltothesidesofthesquare�
Solution��DuetoWongChunWai�Letthecolorsbeblackandwhite�
Forarowsupposetherearekblackcellsand��kwhitecells�Then
thereareCk ��C
��k
�
�k���k�����pairsofcellswiththesame
color�Sothereareatleast������pairsofcellsonthesamerowwith
thesamecolor�NextthereareC� ����pairsofcolumns�Sothereare
�������combinationsofcolorandpairofcolumns�Forcombination
i��to��iftherearej ipairsinthesamecombinationthenthere
areatleastj i��rectanglesforthatcombination�Sincethesumof
thej i�sisatleast��sothereareatleast
� X i�
��ji������������
suchrectangles�
����Forn���let�nchesspiecesbeplacedatthecentersof�nsquaresof
ann�nchessboard�Showthattherearefourpiecesamongthemthat
formedtheverticesofaparallelogram�If�nisreplacedby�n���is
thestatementstilltrueingeneral�
Solution��DuetoHoWingYip�Letm
bethenumberofrowsthat
haveatleast�pieces��Theneachoftheremainingn�mrowscontains
atmost�piece��Foreachofthesemrowslocatetheleftmostsquare
thatcontainsapiece�Recordthedistances�i�e�numberofsquares�
betweenthispieceandtheotherpiecesonthesamerow�Thedistances
canonlybe��������n��becausetherearencolumns�
Sincethenumberofpiecesinthesemrowsaltogetherisatleast
�n��n�m��n�m�thereareatleast�n�m��m�ndistances
recordedaltogetherforthesemrows�Bythepigeonholeprincipleat
leasttwoofthesedistancesarethesame�Thisimpliesthereareat
leasttworowseachcontaining�piecesthatareofthesamedistance
apart�These�piecesyieldaparallelogram�
���
Forthesecondquestionplacing�n��piecesonthesquares
the�rstrowand�rstcolumnshowstherearenoparallelograms�
����Thesetf����������gispartitionedintothreesubsets�Showthat
leastoneofthesubsetscontainsthreedi erentnumbersa�b�cs u
thata�b�c�
Solution�BythepigeonholeprincipleoneofthesubsetssayX�mu
containatleast����elementssayx�
�x�
�����x���Formt
di erencesx��x��x��x������x���x�andremovex��becausea�b
aretobedi erent�ifitappearsonthelist�Ifoneoftheremaini
di erencesbelongstoX�thenwearedone�
Otherwisebythepigeonholeprincipleagainoneofthesubse
sayY���X��mustcontainatleast����elementsfromthesedi erenc
y j�xi j�x��sayy ��y ������y ��Considerthedi erencesy �
y ��y��y ������y��y �andremovey �andxi �
iftheyappearont
list�Ifoneofthesedi erencesbelongtoY�thenwearedone�Ifo
ofthemsayy j�y ��xi j�xi ����xi ��xi j��belongtoX�thenl
xi ��xi j�xi j�xi �aredi erentelementsofXand�xi j�xi ���xi ��x
andwearedone�
Thuswemayassume�ofthesedi erencesz k�y jk
�y ��belo
totheremaingingsubsetZandsayz ��z ������z ��Form
t
di erencez ��z ��z��z ��z�z ��z��z �andremovez ��yj ��xi j�
ifth
appearonthelist�Theremainingdi erencez k�z ��y jk
�y j�
xi jk
�y ij�
mustbelongtooneofX�YorZ�Asabovewegetthr
distinctelementsa�b�cinoneofX�YorZsuchthata�b�c�
InclusionExclusionPrinciple
����Letm
�n���FindthenumberofsurjectivefunctionsfromBm
f��������mgtoBn�f��������ng�
Solution�Fori���������n�letAibethesetoffunctionsf�Bm
���
Bn
suchthati��f��������f�m��Bytheinclusionexclusionprinciple
jA������Anj
�
X��i�n
jAij�
X��i�j�n
jAi�Ajj�
X��i�j�k�n
jAi�A��A�j����
�� n �� �n
���m
�� n �� �n
���m
�� n �� �n
���m
�����
ThenumberofsurjectionsfromBm
toBn
is
nm
�jA������Anj�
n X i��
����i
� n i� �n�i�m�
����LetAbeasetwith�elements�Findthemaximalnumberof�element
subsetsofA�suchthattheintersectionofanytwoofthemisnota�
elementset�
Solution�LetjSjdenotethenumberofelementsinasetS�Let
B������Bn�AbesuchthatjBij���jBi�Bjj���fori�j�������n�
Ifa�AbelongstoB������Bk�thenjBi�Bjj��fori�j�������k�
Since��jAj�jB������Bkj����k�wegetk���Fromthiswe
seethateveryelementofAisinatmost�Bi�s�Then�n�����so
n���Toshow�ispossiblejustconsider
B��f�����g�B��f�����g�B��f�����g�B�f�����g�
B��f�����g�B��f�����g�B��f�����g�B��f�����g�
�����a������HongKongChinaMathOlympiad�Studentshavetakena
testpaperineachofn�n���subjects�Itisknownthatforany
subjectexactlythreestudentsgetthebestscoreinthesubjectand
foranytwosubjectsexcatlyonestudentgetsthebestscoreinevery
oneofthesetwosubjects�Determinethesmallestnsothattheabove
conditionsimplythatexactlyonestudentgetsthebestscoreinevery
oneofthensubjects�
���
�b������AustrianPolishMathCompetition�Thereare����club
Eachhas��members�Ifeverytwoclubshaveexactlyonecomm
memberthenprovethatall����clubshaveacommonmember�
Solution��a��DuetoFanWaiTong�Fori���������n�letSi
thesetofstudentswhogetthebestscoreintheithsubject�Suppo
nobodygetsthebestscoreineveryoneofthensubjects�Letx
onestudentwhoisbestinmostnumberofsubjectssaym�m�
subjects�WithoutlossofgeneralitysupposexisinS��S������S
Fori���������m�letS� i�Sinfxg�ThenthemsetsS� iarepairwi
disjointandsoeachsharesa�distinct�commonmemberwithSm�
SinceSm��hasthreememberssom���Thismeanseachstudent
bestinatmostthreesubjects�Bytheinclusionexclusionprinciple
jS��S������Snj
�
X��i�n
jSij�
X��i�j�n
jSi�Sjj�
X��i�j�k�n
jSi�Sj�Skj
��n�
� n �� �
jS��S������Snj�
whichimpliesn���Thereforeifn���thenthereisatleasto
studentwhogetthebestscoreineveryoneofthensubjects�There
exactlyonesuchstudentsbecauseonlyonestudentgetsthebestsco
inapairofsubjects�
Finallywegiveanexampleofthecasen��withnobodybest
allsubjects�
S��fx��x��x�g�S��fx��x�x�g�S��fx��x�x�g�
S�fx��x��x�g�S��fx��x��x�g�S��fx��x��x�g�
S��fx��x�x�g�
�b�Letn�����andk����LetC��C������Cn
bethenclubs�F
eachmemberofC��formalistoftheindicesoftheotherclubsth
thismemberalsobelongsto�SinceC�
andanyotherclubCiha
exactlyonecommonmembertheklistsofthekmembersofC�a
���
disjointandtogethercontainallintegersfrom�ton�Bythepigeonhole
principleoneofthelistssayx�slistwillcontainatleastm�dn��
k
e
numbers��Thenotationmeansmistheleastintegergreaterthanor
equalto
n��
k
��
Nextwewillshowthisxisamemberofallnclubs�Supposexis
notamemberofsomeclubCi�Theneachofthem��clubsthatx
belongtowillshareadi erentmemberwithCi�otherwisetwoofthe
m��clubswillshareamemberyinCiandalsox�acontradiction��
SinceCihaskmemberssok�m���n��k
���whichimplies
k��k���n�Sincek��k���
�����
n�
�����thisisa
contradiction�Soxmustbeamemberofallnclubs�
Comments�Itisclearthatthetwoproblemsareessentiallythesame�
Asthenumberofmembersinthesetsgetslargetheinclusionexclusion
principlein�a�willbelesse ective�Theargumentinpart�b�ismore
convenientandshowsthatfornsetseachhavingkmembersandeach
pairhavingexactlyonecommonmemberifn�k��k���thenalln
setshaveacommonmember�
CombinatorialDesigns
���������ByelorussianMathOlympiad�Inthebegining��beetlesare
placedatdi erentsquaresofa���squareboard�Ineachmoveevery
beetlecreepstoahorizontalorverticaladjacentsquare�Ifnobeetle
makeseithertwohorizontalmovesortwoverticalmovesinsuccession
showthataftersomemovestherewillbeatleasttwobeetlesinthe
samesquare�
Solution��DuetoCheungPokManandYungFai�Assignanordered
pair�a�b�toeachsquarewitha�b�����������Dividethe��squares
into�types�TypeAconsistsofsquareswithbothaandboddtype
BconsistsofsquareswithbothaandbevenandtypeCconsistsof
theremainingsquares�ThenumbersofsquaresofthetypesA�Band
Care����and��respectively�
Assumenocollisionoccurs�Aftertwosuccessivemovesbeetlesin
typeAsquareswillbeintypeBsquares�Sothenumberofbeetlesin
���
typeAsquaresareatmost��atanytime�Thenthereareatmost
beetlesintypeAortypeBsquaresatanytime�Alsoafteronemov
beetlesintypeCsquareswillgototypeAortypeBsquares�Sothe
areatmost��beetlesintypeCsquaresatanytime�Hencetherea
atmost��beetlesontheboardacontradiction�
���������GreekMathOlympiad�Linesl ��l������lk
areonaplanesu
thatnotwoareparallelandnothreeareconcurrent�Showthatw
canlabeltheCk �
intersectionpointsoftheselinesbythenumbe
��������k��sothatineachofthelinesl ��l������lk
thenumbe
��������k��appearexactlyonceifandonlyifkiseven�
Solution��DuetoNgKaWing�Ifsuchlabelingexistsforaninteg
k�thenthelabel�mustoccuronceoneachlineandeachpointlabel
�liesonexactly�lines�Hencetherearek����si�e�kiseven�
Converselyifkiseventhenthefollowinglabelingworks�f
��i�j�k���givetheintersectionoflinesl iandl jthelab
i�j��wheni�j�k�thelabeli�j�kwheni�j�k�Fort
intersectionoflinesl kandl i�i���������k���givethelabel�i�
when�i�k�thelabel�i�kwhen�i�k�
Alternativelywecanmakeuseofthesymmetryofanoddnum
bersidedregularpolygontoconstructthelabelingasfollows�for
evenconsiderthek��sidedregularpolygonwiththeverticeslabel
��������k���For��i�j�k���theperpendicularbisector
thesegmentjoiningverticesiandjpassesthroughauniqueverte
givetheintersectionoflinesl iandl jthelabelofthatvertex�Fort
intersectionoflinesl kandl i�i���������k���givethelabeli�
���������TournamentsoftheTowns�Inalotterygameapersonmu
selectsixdistinctnumbersfrom������������toputonaticket�T
lotterycommiteewillthendrawsixdistinctnumbersrandomlyfro
�������������Anyticketwithnumbersnotcontaininganyoftheses
numbersisawinningticket�Showthatthereisaschemeofbuyi
�ticketsguaranteeingatleastawinningticketbut�ticketsisn
enoughtoguaranteeawinningticketingeneral�
���
Solution�Considerthenineticketswithnumbers
��������������
��������������
��������������
��������������������
��������������������
��������������������
��������������������
��������������������
��������������������
Forthe�rstthreeticketsiftheyarenotwinningthentwoofthe
sixnumbersdrawnmustbeamong����������Forthenextthreetick
etsiftheyarenotwinningthentwoofthesixsumbersmustbe
�������������Forthelastthreeticketsiftheyarenotwinningthen
threeofthesixnumbersmustbeamong�������������Sinceonlysix
numbersaredrawnatleastoneofthenineticketsisawinningticket�
Foranyeightticketsifonenumberappearsinthreeticketsthen
thisnumberandonenumberfromeachofthe�veremainingtickets
maybethesixnumbersdrawnresultinginnowinningtickets�
Soofthe��numbersontheeightticketswemayassume�atleast�
��appearedexactly�timessaytheyare�����������Considerthetwo
ticketswith�onthem�Theremaining��numbersonthemwillmiss
�atleast�oneofthenumbers�����������say���Now��appearsin
twoothertickets�Then����andonenumberfromeachofthefour
remainingticketsmaybethesixnumbersdrawnbythecommittee
resultinginnowinningtickets�
���������ByelorussianMathOlympiad�Bydividingeachsideofanequi
lateraltriangleinto�equalpartsthetrianglecanbedividedinto��
smallerequilateraltriangles�Abeetleisplacedoneachvertexofthese
trianglesatthesametime�Thenthebeetlesmovealongdi erentedges
withthesamespeed�Whentheygettoavertextheymustmakea
���or����turn�Provethatatsomemomenttwobeetlesmustmeet
atsomevertex�Isthestatementtrueif�isreplacedby��
Solution�Weputcoordinatesattheverticessothat�a�b��for��b�
a���correspondstothepositionof
� a b� inthePascaltriangle�First
markthevertices
������������������������������������������������������������
���
Afteronemoveifnobeetlesmeetthenthe��beetlesatthemark
verticeswillmoveto��unmarkedverticesand��otherbeetlesw
movetothemarkedvertices�Afteranothermovethese��beetlesw
beatunmarkedvertices�Sincethereareonly��unmarkedvertice
twoofthemwillmeet�
If�isreplacedby�thendividetheverticesintogroupsasfollow
f�����������������g�f�����������������g�
f�����������������������g�f�����������������g�
f�����������������������g�f�����������������������g�
Letthebeetlesineachgroupmoveinthecounterclockwisedirecti
alongtheverticesinthegroup�Thenthebeetleswillnotmeetata
moment�
Covering�ConvexHull
���������AustralianMathOlympiad�Therearenpointsgivenonapla
suchthattheareaofthetriangleformedbyevery�ofthemisatmo
��Showthatthenpointslieonorinsidesometriangleofareaatmo
�� Solution��DuetoLeeTakWing�LetthenpointsbeP��P������P
Suppose�PiPjPk
havethemaximumareaamongalltriangleswi
verticesfrom
thesenpoints�NoPlcanlieontheoppositeside
thelinethroughPiparalleltoPjPkasPjPk�otherwise�PjPkPlh
largerareathan�PiPjPk�SimilarlynoPlcanlieontheoppositesi
ofthelinethroughPjparalleltoPiPkasPiPkorontheoppositesi
ofthelinethroughPkparalleltoPiPjasPiPj�Thereforeeachoft
npointslieintheinteriororontheboundaryofthetrianglehavi
Pi�Pj�Pk
asmidpointsofitssides�Sincetheareaof�PiPjPk
is
most�sotheareaofthistriangleisatmost��
���������PutnamExam�Showthatanycontinuouscurveofunitleng
canbecoveredbyaclosedrectanglesofarea����
���
Solution�Placethecurvesothatitsendpointsliesonthexaxis�
Thentakethesmallestrectanglewithsidesparalleltotheaxeswhich
coversthecurve�Letitshorizontalandverticaldimensionsbeaand
b�respectively�LetP�andP�beitsendpoints�LetP��P��P��Pbe
thepointsonthecurveintheordernamedwhichlieoneoneachof
thefoursidesoftherectangle�ThepolygonallineP�P�P�P�PP�has
lengthatmostone�
Thehorizontalprojectionsofthesegmentsofthispolygonalline
adduptoatleasta�sincethelinehaspointsontheleftandright
sidesoftherectangle�Theverticalprojectionsofthesegmentsofthis
polygonallineadduptoatleast�b�sincetheendpointsareonthe
xaxisandthelinealsohaspointsonthetopandbottomsideofthe
rectangle�
Sothepolygonallinehaslengthatleast
p a���b����Bythe
AMGMinequality�ab�a���b���andsotheareaisatmost����
���������PutnamExam�LetFbea�nitecollectionofopendiscsinthe
planewhoseunioncoversasetE�Showthatthereisapairwisedisjoint
subcollectionD������Dn
inFsuchthattheunionof�D�������Dn
coversE�where�DisthediscwiththesamecenterasDbuthaving
threetimestheradius�
Solution�WeconstructsuchDi�sbythegreedyalgorithm�LetD�
beadiscoflargestradiusinF�SupposeD������Djhasbeenpicked�
ThenwepickadiscDj��
disjointfrom
eachofD������Djandhas
thelargestpossibleradius�SinceFisa�nitecollectionthealgorithm
willstopata�naldiscDn�ForxinE�supposexisnotintheunion
ofD������Dn�ThenxisinsomediscDofradiusrinF�NowD
is
notoneoftheDj�simpliesitintersectssomediscDjofradiusr j�r�
Bythetriangleinequalitythecentersisatmostr�r junitsapart�
ThenDiscontainedin�Dj�Inparticularxisin�Dj�ThereforeEis
containedintheunionof�D�������Dn�
���������IMO�Determineallintegersn��forwhichthereexistnpoints
A��A������An
intheplaneandrealnumbersr ��r������rn
satisfying
thefollowingtwoconditions�
���
�a�nothreeofthepointsA��A������An
lieonaline�
�b�foreachtriplei�j�k���i�j�k�n�thetriangleAiAjAkh
areaequaltor i�r j�r k�
Solution��DuetoHoWingYip�Forn���noteA��������A�
������A�
�������A
�������r ��r ��r ��r ����satisfyt
conditions�Nextwewillshowtherearenosolutionsforn���Suppo
thecontraryconsidertheconvexhullofA��A��A��A�A���Thisist
smallestconvexsetcontainingthe�vepoints��Therearethreecase
TriangularCase�WemayassumethepointsarenamedsoA��A��A
aretheverticesoftheconvexhullwithA�A�
insidesuchthatA
isoutside�A�A�AandAisoutside�A�A�A��Denotethearea
�XYZby�XYZ��Wegetacontradictionasfollows�
�A�AA����A�A�A����r��r �r ����r��r ��r ��
��r��r ��r ���r��r ��r ��
��A�A�A���A�A�A����A�A�A���
PentagonalCase�Wemayassumer ��
minfr��r��r��r�r�g�Dra
lineLthroughA�paralleltoA�A�Since�A�A�A��r ��r ��r
r ��r ��r ��A�A�A��A�
isonlineLoronthehalfplaneof
oppositeA��AandsimilarlyforA��SinceA��A��A�cannotallbe
L�weget�A�A�A������contradictingconvexity�
QuadrilateralCase�WemayassumeA�isinsidetheconvexhull�Fir
obsevethatr ��r ��R��r �Thisisbecause
�r��r ��r ����r��r �r ����r��r ��r ���r��r �r ��
istheareaSoftheconvexhull�So�S���r ��r ��r ��r ��Also
S��A�A�A����A�A�A����A�AA����AA�A��
���r ��r �r ��r ��r ��
Fromthelastequationwegetr ����r ��r ��r ��r �����S����
NextobservethatA��A��A�
notcollinearimpliesoneside
�A�A�A�islessthan�����ThenoneofthequadrilateralsA�A�A�A
���
orA�A�A�A�isconvex�Bythe�rstobservationofthiscaser ��r ��
r ��r i�wherer i�r orr ��Sincer ��r ��r ��r �wegetr ��r �or
r �SimilarlyconsideringA��A��Anotcollinearwealsogetr ��r �
orr ��Thereforethreeofthenumbersr ��r��r��r�r�arenegativebut
theareaofthecorrespondingtriangleispositiveacontradiction�
���������IMO�Determineall�nitesetsSofatleastthreepointsinthe
planewhichsatisfythefollowingcondition�foranytwodistinctpoints
AandBinS�theperpendicularbisectorofthelinesegmentABisan
axisofsymmetryofS�
Solution�Clearlynothreepointsofsuchasetiscollinear�otherwise
consideringtheperpendicularbisectorofthetwofurthestpointsofS
onthatlinewewillgetacontradiction��LetH
betheconvexhull
ofsuchasetwhichisthesmallestconvexsetcontainingS�SinceSis
�nitetheboundaryofHisapolygonwiththeverticesP��P������Pn
belongingtoS�LetPi�
Pj
ifi�j�modn��Fori���������n�
theconditiononthesetimpliesPiisontheperpendicularbisectorof
Pi��Pi���SoPi��Pi�PiPi���Consideringtheperpendicularbisector
ofPi��Pi���weseethat�Pi��PiPi����PiPi��Pi���Sotheboundary
ofHisaregularpolygon�
NexttherecannotbeanypointPofSinsidetheregularpolygon�
�ToseethisassumesuchaPexists�Placeitattheoriginandthe
furthestpointQofSfromPonthepositiverealaxis�Sincetheorigin
Pisintheinterioroftheconvexpolygonnotalltheverticescanlie
onortotherightoftheyaxis�SothereexistsavertexPj
tothe
leftoftheyaxis�SincetheperpendicularbisectorofPQisanaxisof
symmetrythemirrorimageofPjwillbeapointinSfurtherthanQ
fromP�acontradiction��SoSisthesetofverticesofsomeregular
polygon�Converselysuchasetclearlyhastherequiredproperty�
Comments�Theo�cialsolutionisshorterandgoesasfollows�Suppose
S�fX������Xngissuchaset�ConsiderthebarycenterofS�whichis
thepointGsuchthat
�� OG�
��OX���������OXn
n
�
���
�Thecasen��yieldsthemidpointofsegmentX�X�andthec a
n��yieldsthecentroidoftriangleX�X�X���Notethebarycent
doesnotdependontheorigin�Toseethissupposewegetapoint
usinganotheroriginO� �i�e�
���O� G�istheaverageof
���O� Xifori�������
Subtractingthetwoaveragesweget�� OG�
���O� G��
�� OO� �Adding
��� O�G
tobothsidesweget�� OG�
��OG� �soG�G� �
BytheconditiononS�afterre#ectionwithrespecttotheperpe
dicularbisectorofeverysegmentXiXj�thepointsofSarepermut
only�SoGisunchangedwhichimpliesGisoneverysuchperpe
dicularbisector�HenceGisequidistantfromallXi�s�Thereforet
Xi�sareconcyclic�ForthreeconsecutivepointsofS�sayXi�Xj�X
onthecircleconsideringtheperpendicularbisectorofsegmentXiX
wehaveXiXj�XjXk�ItfollowsthatthepointsofSarethevertic
ofaregularpolygonandtheconverseisclear�
���
SolutionstoMiscellaneousProblems
���������RussianMathOlympiad�Therearenseatsatamerrygoaround�
Aboytakesnrides�Betweeneachridehemovesclockwiseacertain
number�lessthann�ofplacestoanewhorse�Eachtimehemovesa
di erentnumberofplaces�Findallnforwhichtheboyendsupriding
eachhorse�
Solution�Thecasen��works�Ifn��isoddtheboy�stravel
���������n����n�n�����placesbetweenthe�rstandthelast
rides�Sincen�n�����isdivisiblebyn�hislastridewillrepeatthe�rst
horse�Ifniseventhisispossiblebymovingforward��n�����n�
������n��placescorrespondingtohorses����n���n�������n ����
���������IsraeliMathOlympiad�Twoplayersplayagameonanin�nite
boardthatconsistsof���squares�PlayerIchoosesasquareand
marksitwithanO�ThenplayerIIchoosesanothersquareandmarks
itwithX�Theyplayuntiloneoftheplayersmarksaroworacolumn
of�consecutivesquaresandthisplayerwinsthegame�Ifnoplayer
canachievethisthegameisatie�ShowthatplayerIIcanprevent
playerIfromwinning�
Solution��DuetoChaoKhekLun�Dividetheboardinto���blocks�
Thenbisecteach���blockintotwo���tilessothatforeverypair
ofblockssharingacommonedgethebisectingsegmentinonewillbe
horizontalandtheothervertical�Sinceevery�veconsecutivesquares
ontheboardcontainsatileafterplayerIchoseasquareplayerII
couldpreventplayerIfromwinningbychoosingtheothersquarein
thetile�
���������USAMO�Acalculatorisbrokensothattheonlykeysthatstill
workarethesincostansin���cos���andtan��buttons�Thedis
playinitiallyshows��Givenanypositiverationalnumberq�showthat
pressingsome�nitesequenceofbuttonswillyieldq�Assumethatthe
calculatordoesrealnumbercalculationswithin�niteprecision�All
functionsareintermsofradians� ��
�
Solution�Wewillshowthatallnumbersoftheform
p m�n�whe
m�narepositiveintegerscanbedisplayedbyinductiononk�m�
�Sincer�s�
p r��s��theseincludeallpositiverationals��
Fork���pressingcoswilldisplay��Supposethestateme
istrueforintegerlessthank�Observethatifxisdisplayedth
usingthefacts �
tan��ximpliescos���sin ��
������ an
tan������� ����x�Sowecandisplay��x�Thereforetodispl
p m�nwithk�
m�n�wemayassumem
�
n�Bytheindu
tionstepn�kimplies
p �n�m��m
canbedisplayed�Thenusi
��tan��p �n�m��m
andcos��
p m�n�wecandisplay
p m�
Thiscompletestheinduction�
���������E�otv�osK�ursch�akMathCompetition�Eachofthreeschools
attendedbyexactlynstudents�Eachstudenthasexactlyn��a
quaintancesintheothertwoschools�Provethatonecanpickthr
studentsonefromeachschoolwhoknowoneanother�Itisassum
thatacquaintanceismutual�
Solution��DuetoChanKinHang�Considerastudentwhohast
highestnumbersayk�ofacquaintancesinanotherschool�Callth
studentx�hisschoolX
andthekacquaintancesinschoolY�Sin
n���n�k�xmusthaveatleastoneacquaintancesayz�inthethi
schoolZ�NowzhasatmostkacquaintancesinschoolXandhence
hasatleast�n����kacquaintancesinschoolY�Addingthenumber
acquaintancesofxandzinschoolY�wegetk��n����k�n���
andsoxandzmusthaveacommonacquaintanceyinschoolY�
����Isthereawaytopack��������bricksintoa��������box�
Solution�Assigncoordinate�x�y�z�toeachofthecellswherex�y�z
����������Letthecell�x�y�z�begivencolorx�y�z�mod���No
each�����brickcontainsall�colorsexactlyonce�Ifthepacki
ispossiblethenthereareexactly���cellsofeachcolor�However
directcountingshowsthereare���cellsofcolor�acontradiction�
suchapackingisimpossible�
���
����Isitpossibletowriteapositiveintegerintoeachsquareofthe�rst
quadrantsuchthateachcolumnandeachrowcontainseverypositive
integerexactlyonce�
Solution�Yesitispossible�De�neA�����andAn���
� Bn
An
An
Bn
� �
wheretheentriesofBn
arethoseofAn
plus�n���So
A������
A��
� ��
�
�� �
A��
� B ��
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�� C A�
����
NotethatifeverycolumnandeveryrowofAn
contain���������n��
exactlyoncetheneverycolumnandeveryrowofBn
willcontain
�n����������nexactlyonce�SoeverycolumnandeveryrowofAn��
willcontain���������nexactlyonce�Now�llthe�rstquadrantusing
theAn�s�
����Therearenidenticalcarsonacirculartrack�Amongallofthemthey
havejustenoughgasforonecartocompletealap�Showthatthereis
acarwhichcancompletealapbycollectinggasfromtheothercars
onitswayaroundthetrackintheclockwisedirection�
Solution��DuetoChanKinHang�Thecasen��isclear�Suppose
thecasen�kistrue�Forthecasen�k����rstobservethatthere
isacarAwhichcanreachthenextcarB��Ifnocarcanreachthe
nextcarthenthegasforallcarswouldnotbeenoughforcompleting
alap��LetusemptythegasofBintoAandremoveB�Thenthek
carsleftsatisfythecondition�Sothereisacarthatcancompletea
lap�Thissamecarwillalsobeabletocompletethelapcollectinggas
fromothercarswhenBisincludedbecausewhenthiscargetstocar
A�thegascollectedfromcarAwillbeenoughtogetittocarB�
���������RussianMathOlympiad�Attheverticesofacubearewritten
eightpairwisedistinctnaturalnumbersandoneachofitsedgesis
writtenthegreatestcommondivisorofthenumbersattheendpoints
���
oftheedge�Canthesumofthenumberswrittenattheverticesbet
sameasthesumofthenumberswrittenattheedges�
Solution�Observethatifa�b�thengcd�a�b��bandgcd�a�b��a�
So�gcd�a�b��a�b�Ifthesumofthevertexnumbersequalsthesu
oftheedgenumbersthenwewillhavegcd�a�b���a�b���foreve
pairofadjacentvertexnumberswhichimpliesa��borb��aatt
twoendsofeveryedge�Ateveryvertexthereare�adjacentvertice
Thea��borb��aconditionimpliestwooftheseadjacentvert
numbersmustbethesameacontradiction�
����Canthepositiveintegersbepartitionedintoin�nitelymanysubse
suchthateachsubsetisobtainedfromanyothersubsetbyaddingt
sameintegertoeachelementoftheothersubset�
Solution�Yes�LetAbethesetofpositiveintegerswhoseodddig
positions�fromtheright�arezeros�LetBbethesetofpositiveintege
whoseevendigitpositions�fromtheright�arezeros�ThenAand
arein�nitesetandthesetofpositiveintegersistheunionofa�B
fa�b�b�BgasarangesovertheelementsofA��Forexamp
����������������������B��
���������RussianMathOlympiad�Isitpossibleto�llinthecellsof
���tablewithpositiveintegersrangingfrom�to��insuchaw
thatthesumoftheelementsofevery���squareisthesame�
Solution�Place�����������������onthe�rstfourthandseven
rows�Place�����������������onthesecond�fthandeigthrow
Place�����������������onthethirdsixthandninthrows�Th
every���squarehassum���Considerthistableandits���rotatio
Foreachcell�llitwiththenumber�a�b���whereaisthenumb
inthecelloriginallyandbisthenumberinthecellafterthetable
rotatedby��� �Byinspection�to��appearsexactlyonceeachan
every���squarehassum��������������
���������GermanMathematicalOlympiad�Showthatforeverypositi
integern���thereexistsapermutationp��p������pn
of�������
suchthatp k��dividesp��p������pkfork���������n���
���
Solution��Thecasesn��������suggestthefollowingpermutations��
Forevenn��m�considerthepermutation
m�����m���������m�m�m�
Foroddn��m���considerthepermutation
m�����m���������m�m�m��m���
Ifk��j������j�m�then�m�����������m�j��j�m�j��If
k��j����j�m�then�m�����������m�j��j�j�m�j����
����Eachlatticepointoftheplaneislabeledbyapositiveinteger�Each
ofthesenumbersisthearithmeticmeanofitsfourneighbors�above
belowleftright��Showthatallthenumbersareequal�
Solution�Considerthesmallestnumbermlabelledatalatticepoint�
Ifthefourneighboringnumbersarea�b�c�d�then�a�b�c�d����m
anda�b�c�d�m
implya�b�c�d�m�Sinceanytwolattice
pointscanbeconnectedbyhorizontalandverticalsegmentsifone
endislabelledm�thenalongthispathallnumberswillequaltom�
Thereforeeverynumberequalsm�
���������TournamentoftheTowns�Inapartynboysandngirlsare
paired�Itisobservedthatineachpairthedi erenceinheightisless
than��cm�Showthatthedi erenceinheightofthekthtallestboy
andthekthtallestgirlisalsolessthan��cmfork���������n�
Solution��DuetoAndyLiuUniversityofAlbertaCanada�Letb ��
b ������b nbetheheightsoftheboysandg ��g ������g n
bethoseofthegirls�Supposeforsomek�jb k�g kj����Inthecase
g k�b k����wehaveg i�b j�g k�b k���for��i�kand
k�j�n�Considerthegirlsofheightg i�where��i�kandthe
boysofheightb j�wherek�j�n�Bythepigeonholeprincipletwo
ofthesen��peoplemustbepairedoriginally�Howeverforthatpair
g i�b j���contradictsthehypothesis��Thecaseb k�g k���is
handledsimilarly��Sojb k�g kj���forallk�
���
���������LeningradMathOlympiad�Onemayperformthefollowingt w
operationsonapositiveinteger�
�a�multiplyitbyanypositiveintegerand
�b�deletezerosinitsdecimalrepresentation�
ProvethatforeverypositiveintegerX�onecanperformasequence
theseoperationsthatwilltransformXtoaonedigitnumber�
Solution�BythepigeonholeprincipleatleasttwooftheX��num
bers
��������������������
z�
X��digits
havethesameremainderwhendividedbyX�Sotakingthedi eren
oftwoofthesenumberswegetanumberoftheform�����������
whichisamultipleofX�Performoperation�a�onX
togetsuch
multiple�Thenperformoperation�b�todeletethezeros�ifany��
thenewnumberhasmorethanonedigitswedothefollowingstep
���multiplyby��togetanumber�������������deletethezeroan
multiplyby�togetanumber�������������deletethezerostog
������now�������������������������������anddele
zerowegetthesingledigit��
���������IMOshortlistedproblem�Fourintegersaremarkedonacirc
Oneachstepwesimultaneouslyreplaceeachnumberbythedi eren
betweenthisnumberandnextnumberonthecircleinagivendirecti
�thatisthenumbersa�b�c�darereplacedbya�b�b�c�c�d�d�a
Isitpossibleafter����suchstepstohavenumbersa�b�c�dsuchth
thenumbersjbc�adj�jac�bdj�jab�cdjareprimes�
Solution��DuetoNgKaManandNgKaWing�Iftheinitialnumbe
area�w�b�x�c�y�d�z�thenafter�stepsthenumberswillb
a���w��x��y��z��
b���x��y��z��w��
c���y��z��w��x��
d���z��w��y��z��
Fromthatpointona�b�c�dwillalwaysbeevensojbc�adj�jac
bdj�jab�cdjwillalwaysbedivisibleby��
���
���������NanchangCityMathCompetition�Thereare����coinsona
table�Someareplacedwiththeheadsidesupandsomethetailsides
up�Agroupof����personswillperform
thefollowingoperations�
the�rstpersonisallowedturnoveranyonecointhesecondpersonis
allowedturnoveranytwocoins����thekthpersonisallowedturn
overanykcoins����the����thpersonisallowedtoturnoverevery
coin�Provethat
���nomatterwhichsidesofthecoinsareupinitiallythe����persons
cancomeupwithaprocedureturningallcoinsthesamesidesup
attheendoftheoperations
���intheaboveprocedurewhethertheheadorthetailsidesturned
upattheendwilldependontheinitialplacementofthecoins�
Solution��DuetoChanKinHang����Thenumber����maynot
bespecial�Soletusreplaceitbyavariablen�Thecasesn��and
�aretruebutthecasen��isfalse�whenbothcoinsareheads
upinitially��Sowesuspectthestatementistrueforoddnanddo
inductiononk�wheren��k���Thecasesk����aretrue�Suppose
thecasekistrue�Forthecasek���wehaven��k��coins�
Firstsupposeallcoinsarethesamesideupinitially�Fori�
��������k�lettheithperson#ipanyicoinsandletthe��k���i�th
person#ipstheremaining�k���icoins�Theneachcoinis#ipped
k��timesandattheendallcoinswillbethesamesideup�
Nextsupposenotallcoinsarethesamesidesupinitially�Then
thereisonecoinheadupandanothertailup�Markthesetwocoins�
Letthe�rst�k��persons#iptheother�k��coinsthesameside
upbythecasek�Thenthereareexactly�kcoinsthesamesideup
andonecoinoppositesideup�The�kthperson#ipsthe�kcoinsthe
samesideupandthe�k��stperson#ipsallcoinsandthissubcase
issolved�
Sothek��caseistrueineitherwayandtheinductionstepis
completeinparticularcasen�����istrue�
���Iftheproceduredoesnotdependontheinitialplacementthen
insomeinitialplacementsofthecoinsthecoinsmaybe#ippedwith
���
allheadsupandmayalsobe#ippedwithalltailsup�Reversingt
#ippingsontheheadsupcasewecanthengofromallcoinsheadsu
toalltailsupin���������������#ippings�Howeverforeachco
togofromheaduptotailupeachmustbe#ippedanoddnumber
timesandthe����coinsmusttotaltoanoddnumberof#ippings
contradiction�
�����ProposedbyIndiafor����IMO�Showthatthereexistsaconv
polygonof����sidessatisfyingthefollowingconditions�
�a�itssidesare��������������insomeorder�
�b�thepolygoniscircumscribableaboutacircle�
Solution�Forn��������������de�ne
xn�
� � �n��
ifn�����mod��
���
ifn���mod��
���
ifn���mod��
andan�xn�xn��withx�����x��Thesequencean
is
����������������������������������������
ConsideracirclecenteredatOwithlargeradiusrandwindapolygon
lineA�A����A����A����withlengthAiAi���aiaroundthecircle
thatthesegmentsAiAi��aretangenttothecircleatsomepoint
withAiPi�xiandPiAi���xi���ThenOA��
p x� ��r��OA���
De�ne
f�r���tan��x� r��tan��x� r������tan��x����
r
���A�OP���P����OA�������P�OP���P�OP�
������P����OP�����
Nowfiscontinuouslimr���f�r�������and
limr���f�r����Byt
intermediatevaluetheoremthereexistsrsuchthatf�r�����F
suchr�A����willcoincidewithA��resultinginthedesiredpolygon
���
Comments�Thekeyfactthatmakesthepolygonexistsisthatthere
isapermutationa��a������a����of������������suchthatthesystem
ofequations
x��x��a��x��x��a������x�����x��a����
havepositiverealsolutions�
����Thereare��white��black��redchipsonatable�Inonestepyou
maychoose�chipsofdi erentcolorsandreplaceeachonebyachipof
thethirdcolor�Canallchipsbecomethesamecoloraftersomesteps�
Solution�Write�a�b�c�forawhitebblackcredchips�Sofrom
�a�b�c��inonestepwecanget�a���b���c���or�a���b���c���
or�a���b���c����Observethatinall�casesthedi erence
�a�����b�����a�����b�����a�����b����a�b�mod���
Soa�b�mod��isaninvariant�Ifallchipsbecomethesamecolor
thenattheendwehave��������or��������or���������Soa�b�
��mod��attheend�Howevera�b����������mod��inthe
beginning�Sotheanswerisno�
����Thefollowingoperationsarepermittedwiththequadraticpolynomial
ax��bx�c�
�a�switchaandc
�b�replacexbyx�t�wheretisarealnumber�
Byrepeatingtheseoperationscanyoutransformx��x��intox��
x���
Solution�Considerthediscriminant$�b���ac�Afteroperation�a�
$�b���ca�b���ac�Afteroperation�b�a�x�t���b�x�t��c�
ax����at�b�x��at��bt�c�and$���at�b����a�at��bt�c��
b���ac�So$isaninvariant�Forx��x���$���Forx��x���
$���Sotheanswerisno�
����Fivenumbers���������arewrittenonablackboard�Astudentmay
eraseanytwoofthenumbersaandbontheboardandwritethe
���
numbersa�bandabreplacingthem�Ifthisoperationisperformedr
peatedlycanthenumbers����������������everappearontheboar
Solution�Observethatthenumberofmultiplesof�amongthe�
numbersontheblackboardcannotdecreaseaftereachoperation�
a�baremultiplesof�thena�b�abwillalsobemultiplesof��Ifo
ofthemisamultipleof�thenabwillalsobeamultipleof���T
numberofmultiplesof�canincreaseinonlyonewaynamelywheno
ofaorbis��mod��andtheotheris��mod���thena�b���mod
andab���mod���Nownotethereisonemultipleof�inf��������
andfourmultiplesof�inf����������������g�Sowhenthenumber
multiplesof�increasestofourthe�fthnumbermustbe��mod�
Since�����mod���so����������������canneverappearont
board�
����Nine���cellsofa�����squareareinfected�Inoneunittimet
cellswithatleast�infectedneighbors�havingacommonside�becom
infected�Cantheinfectionspreadtothewholesquare�Whatifni
isreplacedbyten�
Solution��DuetoCheungPokMan�Colortheinfectedcellsblackan
recordtheperimeteroftheblackregionateveryunittime�Ifac
hasfourthreetwoinfectedneighborsthenitwillbecomeinfect
andtheperimeterwilldecreaseby���respectivelywhenthatc
iscoloredblack�Ifacellhasoneornoinfectedneighborsthenitw
notbeinfected�Observethattheperimeteroftheblackregioncann
increase�Sinceinthebeginningtheperimeteroftheblackregion
atmost�������anda�����blackregionhasperimeter��t
infectioncannotspreadtothewholesquare�
Ifnineisreplacedbytenthenitispossibleasthetendiagon
cellswheninfectedcanspreadtothewholesquare�
���������ColombianMathOlympiad�Weplaythefollowinggamewi
anequilateraltriangleofn�n�����dollarcoins�withncoinsonea
side��Initiallyallofthecoinsareturnedheadsup�Oneachturnw
mayturnoverthreecoinswhicharemutuallyadjacent�thegoalis
���
makeallofthecoinsturnedtailsup�Forwhichvaluesofncanthisbe
done�
Solution�Thiscanbedoneonlyforalln�����mod���Belowby
atrianglewewillmeanthreecoinswhicharemutuallyadjacent�For
n���clearlyitcanbedoneandforn���#ipeachofthefour
triangles�Forn�����mod��andn���#ipeverytriangle�Then
thecoinsatthecornersare#ippedonce�Thecoinsonthesides�not
corners�are#ippedthreetimeseach�Soallthesecoinswillhavetails
up�Theinteriorcoinsare#ippedsixtimeseachandhaveheadsup�
Sincetheinteriorcoinshavesidelengthn���bytheinductionstep
allofthemcanbe#ippedsotohavetailsup�
Nextsupposen���mod���Colortheheadsofeachcoinred
whiteandbluesothatadjacentcoinshavedi erentcolorsandany
threecoinsinarowhavedi erentcolors�Thenthecoinsinthecorner
havethesamecolorsayred�Asimplecountshowsthatthereare
onemoreredcoinsthanwhiteorbluecoins�Sothe�oddoreven�
paritiesoftheredandwhitecoinsaredi erentinthebeginning�As
we#ipthetrianglesateachturneither�a�bothredandwhitecoins
increaseby�or�b�bothdecreaseby�or�c�oneincreasesby�and
theotherdecreasesby��Sotheparitiesoftheredandwhitecoins
staydi erent�Inthecaseallcoinsaretailsupthenumberofredand
whitecoinswouldbezeroandtheparitieswouldbethesame�Sothis
cannothappen�
���������ChineseTeamSelectionTest�Everyintegeriscoloredwithone
of���colorsandall���colorsareused�Forintervals�a�b���c�d�having
integersendpointsandsamelengthsifa�chavethesamecolorand
b�dhavethesamecolorthentheintervalsarecoloredthesameway
whichmeansa�xandc�xhavethesamecolorforx���������b�a�
Provethat�����and����havedi erentcolors�
Solution�Wewillshowthatx�yhavethesamecolorifandonlyif
x�y�mod�����whichimplies�����and����havedi erentcolors�
Letthecolorsbe�����������andletf�x�bethecolor�number�
ofx�Sinceall���colorswereusedthereisanintegermisuchthat
f�mi��ifori�������������LetM
�min�m��m������m����������
���
Considera�xedintegera�M
andanarbitrarypositiveinteg
n�Sincethereare����waysofcoloringapairofintegersatleasttw
ofthepairsa�i�a�i�n�i����������������arecoloredthesam
waywhichmeansf�a�i ���f�a�i ��andf�a�i ��n��f�a�i ��
forsomeintegersi ��i�suchthat��i ��i �������Letd�i ��
Sincethereare�nitelymanycombinationsoforderedpairs�i��d�an
nisarbitrarytherearein�nitelymanyn�ssayn��n������havingt
samei ��sandd�s�
Sincethesenk�smaybearbitrarilylargetheunionoftheinterva
�a�i ��a�i ��nk�willcontaineveryintegerx�a������Soforeve
suchx�thereisaninterval�a�i ��a�i ��nk�containingx�Sin
f�a�i ���f�a�i ��d�andf�a�i ��nk��f�a�i ��d�nk��
intervals�a�i ��a�i ��nk���a�i ��d�a�i ��d�nk�arecolor
thesameway�Inparticularf�x��f�x�d��Sof�x�hasperiod
whenx�a������Sincea�M����colorsareusedfortheintege
x�a�����andsod�����Considertheleastpossiblesuchperi
d�
Nextbythepigeonholeprincipletwooff�a�������f�a�����
�������f�a����������arethesamesayf�b��f�c�witha�����
b�c�a����������Foreveryx�a����������choosealar
integermsothatxisin�b�b�md��Sincef�b�md��f�b��f�c�
f�c�md��intervals�b�b�md���c�c�md�arecoloredthesameway�
particularf�x��f�x�c�b��Sof�x�hasperiodc�b����wh
x�a����������Sotheleastperiodoff�x�forx�a�������
mustbe����Finallysinceacanbeascloseto�aswelikefmu
haveperiod���onthesetofintegers�Sinceall���colorsareused
twoof���consecutiveintgerscanhavethesamecolor�
���