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# 2003年加拿大数学奥林匹克

Solutionstothe2003CMO

writtenMarch26,2003

Solution

Theminutehandmakesafullrevolutionof360?every60minutes,soaftermminutes

m=6mdegrees.Thehourhandmakesafullrevolutioneveryithassweptthrough360

60m=m/2degrees.12hours(720minutes),soaftermminutesithassweptthrough360

720Sincebothhandsstartedinthesamepositionat12:00,theanglebetweenthetwo

handswillbe1?if6m?m/2=±1+360kforsomeintegerk.Solvingthisequationweget5k±2720k±2=65k+.m=1111

Since1≤m≤720,wehave1≤k≤11.Sincemisaninteger,5k±2mustbedivisibleby11,say5k±2=11q.Then

5k=11q±2?k=2q+q±2.5

Ifisnowclearthatonlyq=2andq=3satisfyalltheconditions.Thusk=4ork=7andsubstitutingthesevaluesintotheexpressionformwe?ndthattheonlypossiblevaluesofmare262and458.

2.Findthelastthreedigitsofthenumber20032002

Solution

20012001.Wemust?ndtheremainderwhen20032002isdividedby1000,whichwillbethe2001isdividedby1000,since2003≡3(mod1000).sameastheremainderwhen32002

Todothiswewill?rst?ndapositiveintegernsuchthat3n≡1(mod1000)andthentrytoexpress20022001intheformnk+r,sothat

200320022001≡3nk+r≡(3n)k·3r≡1k·3r≡3r(mod1000).

Since32=10?1,wecanevaluate32musingthebinomialtheorem:

32m=(10?1)m=(?1)m+10m(?1)m?1+100m(m?1)(?1)m?2+···+10m.2

Afterthe?rst3termsofthisexpansion,allremainingtermsaredivisibleby1000,solettingm=2q,wehavethat

34q≡1?20q+100q(2q?1)(mod1000).(1)

Usingthis,wecancheckthat3100≡1(mod1000)andnowwewishto?ndtheremainderwhen20022001isdividedby100.

Now20022001≡22001(mod100)≡4·21999(mod4·25),sowe’llinvestigatepowersof2modulo25.Notingthat210=1024≡?1(mod25),wehave

21999=(210)199·29≡(?1)199·512≡?12≡13(mod25).

Thus22001≡4·13=52(mod100).Therefore20022001canbewrittenintheform100k+52forsomeintegerk,so

200320022001≡352(mod1000)≡1?20·13+1300·25≡241(mod1000)

2001usingequation(1).Sothelast3digitsof20032002are241.

3.Findallrealpositivesolutions(ifany)to

x3+y3+z3=x+y+z,and

x2+y2+z2=xyz.

Solution1

Letf(x,y,z)=(x3?x)+(y3?y)+(z3?z).The?rstequationaboveisequivalenttof(x,y,z)=0.Ifx,y,z≥1,thenf(x,y,z)≥0withequalityonlyifx=y=z=1.Butifx=y=z=1,thenthesecondequationisnotsatis?ed.Soinanysolutiontothesystemofequations,atleastoneofthevariablesislessthan1.Withoutlossofgenerality,supposethatx<1.Then

x2+y2+z2>y2+z2≥2yz>yz>xyz.

Thereforethesystemhasnorealpositivesolutions.

Solution2

Wewillshowthatthesystemhasnorealpositivesolution.Assumeotherwise.

y2z2?4y2?4z2≥0.

Dividingthroughby4y2z2yields

1111≥2+2≥2.4yzy

Hencey2≥4andsoy≥2,ybeingpositive.Asimilarargumentyieldsx,y,z≥2.Butthe?rstequationcanbewrittenas

x(x2?1)+y(y2?1)+z(z2?1)=0,

Solution3

Applyingthearithmetic-geometricmeaninequalityandthePowerMeanInequalitiestox,y,zwehave

????222333√x+y+zx+y+z3x+y+z3≤≤.xyz≤333

LettingS=x+y+z=x3+y3+z3andP=xyz=x2+y2+z2,thisinequalitycanbewritten????√PS33S≤.P≤≤333????√3NowP≤PimpliesP2≤P3/27,soP≥27.AlsoS≤3SimpliesS3/27≤S/3,333????√√33S.soS≤3.ButthenP≥3and3

3≤1whichisinconsistentwithP≤3S

3

Thereforethesystemcannothavearealpositivesolution.

Solution1

Solution2

AX=2AP,

Now

XY=AY?AX=2(AQ?AP)=2PQand,similarly,YZ=2QR.

ThereforeXY:YZ=PQ:QR.ButO1P||O2Q||O3R,soPQ:QR=O1O2:O2O3.Sincethecentresofthecirclesare?xed,theratioXY:YZ=O1O2:O2O3doesnotdependonthechoiceofl.

IfX,Y,ZdonotalllieonthesamesideofAB,wecanobtainthesameresultwithasimilarproof.Forinstance,ifXandYareoppositesidesofAB,thenwewillhaveXY=AY+AX,butsinceinthiscasePQ=AQ+AP,itisstillthecasethatXY=2PQandresultstillfollows,etc.AY=2AQandAZ=2AR.

5.LetSbeasetofnpointsintheplanesuchthatanytwopointsofSareatleast1unitapart.Provethere√isasubsetTofSwithatleastn/7pointssuchthatanytwopointsofTareatleast3unitsapart.

Solution

WewillconstructthesetTinthefollowingway:AssumethepointsofSareinthexy-planeandletPbeapointinSwithmaximumy-coordinate.ThispointPwillbeamemberof√thesetTandnow,fromS,wewillremovePandallpointsinSwhicharelessthan3unitsfromP.Fromtheremainingpointswechooseonewithmaximumy-coordinatetobeamemberofTandremovefromSallpointsatdistancelessthan√3unitsfromthisnewpoint.Wecontinueinthisway,√untilallthepointsofSareexhausted.ClearlyanytwopointsinTareatleast3unitsapart.ToshowthatThasatleastn/7points,wemustprovethatateachstagenomorethan6otherpointsareremovedalongwithP.