2019年数学竞赛试题及答案

The11thRomanianMasterofMathematicsCompetitionDay1|SolutionsProblem1.AmyandBobplaythegame.Atthebeginning,Amywritesdownapositiveintegerontheboard.Thentheplayerstakemovesinturn,Bobmovesrst.Onanymoveofhis,Bobreplacesthenumbernontheblackboardwithanumberoftheformna2,whereaisapositiveinteger.Onanymoveofhers,Amyreplacesthenumbernontheblackboardwithanumberoftheformnk,wherekisapositiveinteger.Bobwinsifthenumberontheboardbecomeszero.CanAmypreventBob'swin?Russia,MaximDidinSolution.Theanswerisinthenegative.Forapositiveintegern,wedeneitssquare-freepartS(n)tobethesmallestpositiveintegerasuchthatn=aisasquareofaninteger.Inotherwords,S(n)istheproductofallprimeshavingoddexponentsintheprimeexpansionofn.WealsoagreethatS(0)=0.Nowweshowthat(i)onanymoveofhers,Amydoesnotincreasethesquare-freepartofthepositiveintegerontheboard;and(ii)onanymoveofhis,Bobalwayscanreplaceapositiveintegernwithanon-negativeintegerkwithS(k)S(n).Thus,ifthegamestartsbyapositiveintegerN,BobcanwininatmostS(N)moves.Part(i)istrivial,asthedenitionofthesquare-partyieldsS(nk)=S(n)wheneverkisodd,andS(nk)=1S(n)wheneverkiseven,foranypositiveintegern.Part(ii)isalsoeasy:if,beforeBob'smove,theboardcontainsanumbern=S(n)
b2,thenBobmayreplaceitwithn0=nb2=(S(n)1)b2,whenceS(n0)S(n)1.Remarks.(1)Tomaketheargumentmoretransparent,Bobmayrestricthimselftosubtractonlythosenumberswhicharedivisiblebythemaximalsquaredividingthecurrentnumber.Thisrestrictionhavingbeenput,onemayreplaceanynumbernappearingontheboardbyS(n),omittingthesquarefactors.Afterthischange,Amy'smovesdonotincreasethenumber,whileBob'smovesdecreaseit.Thus,Bobwins.(2)Infact,Bobmaywineveninatmost4movesofhis.Forthatpurpose,useLagrange'sfoursquarestheoreminordertoexpandS(n)asthesumofatmostfoursquaresofpositiveintegers:S(n)=a21+


+a2s.Then,oneverymoveofhis,Bobcanreplacethenumber(a21+


+a2k)b2ontheboardby(a21+


+a2k1)b2.TheonlychanceforAmytointerruptthisprocessistoreplaceacurrentnumberbyitsevenpower;butinthiscaseBobwinsimmediately.Ontheotherhand,fourisindeedtheminimumnumberofmovesinwhichBobcanguaranteehimselftowin.Toshowthat,letAmychoosethenumber7,andtakejusttherstpoweroneachofhersubsequentmoves.1Problem2.LetABCDbeanisoscelestrapezoidwithABkCD.LetEbethemidpointofAC.Denoteby!andthecircumcirclesofthetrianglesABEandCDE,respectively.LetPbethecrossingpoitnofthetangentto!atAwiththetangenttoatD.ProvethatPEistangentto.Slovenia,JakobJurijSnojSolution1.IfABCDisarectangle,thestatementistrivialduetosymmetry.Hence,inwhatfollowsweassumeAD6kBC.LetFbethemidpointofBD;bysymmetry,both!andpassthroughF.LetP0bethemeetingpointoftangentsto!atFandtoatE.WeaimtoshowthatP0=P,whichyieldstherequiredresult.Forthatpurpose,weshowthatP0AandP0Daretangentto!and,respectively.LetKbethemidpointofAF.ThenEKisamidlineinthetriangleACF,so\(AE;EK)=\(EC;CF).SinceP0Eistangentto,weget\(EC;CF)=\(P0E;EF).Thus,\(AE;EK)=\(P0E;EF),soEP0isasymmedianinthetriangleAEF.Therefore,EP0andthetangentsto!atAandFareconcurrent,andtheconcurrencypointisP0itself.HenceP0Aistangentto!.Thesecondclaimissimilar.TakingLtobethemidpointofDE,wehave\(DF;FL)=\(FB;BE)=\(P0F;FE),soP0FisasymmedianinthetriangleDEF,andhenceP0isthemeetingpointofthetangentstoatDandE.ABCDEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFP0KL!Remark.Theaboveargumentsmaycomeindierentorders.E.g.,onemaydeneP0tobethepointofintersectionofthetangentstoatDandE|henceobtainingthatP0Fisasymmedianin4DEF,thendeducethatP0Fistangentto!,andthenapplyasimilarargumenttoshowthatP0Eisasymmedianin4AEF,whenceP0Aistangentto!.Solution2.LetQbetheisogonalconjugateofPwithrespectto4AED,so\(QA;AD)=\(EA;AP)=\(EB;BA)and\(QD;DA)=\(ED;DP)=\(EC;CD).NowouraimistoprovethatQEkCD;thiswillyieldthat\(EC;CD)=\(AE;EQ)=\(PE;ED),whencePEistangentto.LetDQmeetABatX.Thenwehave\(XD;DA)=\(EC;CD)=\(EA;AB)and\(DA;AX)=\(AB;BC),hencethetrianglesDAXandABCaresimilar.Since\(AB;BE)=\(DA;AQ),thepointsQandEcorrespondtoeachotherinthesetriangles,henceQisthemidpointofDX.ThisyieldsthatthepointsQandElieonthemidlineofthetrapezoidparalleltoCD,asdesired.2ABCDEPQRXIIRemark.Thelaststepcouldbereplacedwithanotherapplicationofisogonalconjugacyinthefollowingmanner.ReectQinthecommonperpendicularbisectorofABandCDtoobtainapointRsuchthat\(CB;BR)=\(QA;AD)=\(EB;BA)and\(BC;CR)=\(QD;DA)=\(EC;CD).TheserelationsyieldthatthepointsEandRareisogonallyconjugateinatriangleBCI,whereIisthe(ideal)pointofintersectionofBAwithCD.SinceEisequidistantfromABandCD,Risalsoequidistantfromthem,whichyieldswhatweneed.(Thelaststepdeservessomeexplanation,sinceonevertexofthetriangleisideal.Suchexplanationmaybeobtainedinmanydierentways|e.g.,byashortcomputationinsines,orbynoticingthat,asintheusualcase,RisthecircumcenterofthetriangleformedbythereectionsofEinthesidelinesAB,BC,andCD.)3Solution3.(DanCarmon)LetObetheintersectionofthediagonalsACandBD.LetFbethemidpointofBD.LetSbethesecondintersectionpointofthecircumcirclesoftrianglesAOFandDOE.WewillprovethatSDandSEaretangentto;thesymmetricargumentwouldthenimplyalsothatSAandSFaretangentto.ThusS=Pand