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首页 2010b数学竞赛JBMO 2012-2013 problems and solutions.…

2010b数学竞赛JBMO 2012-2013 problems and solutions.pdf

2010b数学竞赛JBMO 2012-2013 problem…

上传者: 中国 2013-08-13 评分 3 0 24 3 108 暂无简介 简介 举报

简介:本文档为《2010b数学竞赛JBMO 2012-2013 problems and solutionspdf》,可适用于初中教育领域,主题内容包含SolutionsofJBMOWednesday,June,ProblemLet,andabcbepositiverealnumberssuchth符等。

SolutionsofJBMOWednesday,June,ProblemLet,andabcbepositiverealnumberssuchthatabc=ProvethatabbccaabcbacbacabcWhendoesequalityholdSolutionReplacing,,abcwith,,bccaabrespectivelyontherighthandside,thegiveninequalitybecomesbccaabbccaababcabcandequivalentlybcbccacaababaabbccwhichcanbewrittenasbccaababc,whichistrueTheequalityholdsifandonlyifbccaababc==,whichtogetherwiththegivenconditionabc=givesabc===ProblemLetthecircleskandkintersectattwodistinctpointsΑandΒ,andlettbeacommontangentofkandk,thattoucheskandkatΜandΝ,respectivelyIftAMandΜΝΑΜ=,evaluateNMBSolutionLetPbethesymmetricofAwithrespecttoM(Figure)ThenAMMP=andtAP,hencethetriangleAPNisisosceleswithAPasitsbase,soNAPNPA=WehaveBAPBAMBMN==andBANBNM=ThuswehaveNBMBNMBMNBANBAPNAPNPA====sothequadrangleMBNPiscyclic(sincethepointsBandPlieondifferentsidesofMN)HenceAPBMPBMNB==andthetrianglesAPBandMNBarecongruent(ΜΝΑΜ=ΑΜΜΡ=ΑΡ=)FromthatwegetAB=MB,iethetriangleAMBisisosceles,andsincetistangenttokandperpendiculartoAM,thecentreofkisonAM,henceAMBisarightangledtriangleFromthelasttwostatementsweinferAMB=,andsoNMBAMB==DFigureSolutionLetCbethecommonpointofMN,AB(Figure)ThenCNCBCA=andCMCBCA=SoCMCN=ButMNAM=,soCMCNAM==,thustherighttriangleACMisisosceles,henceNMBCMBBCM===FigureProblemOnaboardtherearennailseachtwoconnectedbyastringEachstringiscoloredinoneofngivendistinctcolorsForeachthreedistinctcolors,thereexistthreenailsconnectedwithstringsinthesethreecolorsCannbea)b)Solution(a)TheanswerisnoSupposeitispossibleConsidersomecolor,sayblueEachbluestringisthesideoftrianglesformedwithverticesonthegivenpointsAsthereexist==pairsofcolorsotherthanblue,andforanysuchpairofcolorstogetherwiththebluecolorthereexistsatrianglewithstringsinthesecolors,weconcludethatthereexistatleastbluestrings(otherwisethenumberoftriangleswithabluestringasasidewouldbeatmost=,acontradiction)Thesameistrueforanycolor,soaltogetherthereexistatleast=strings,whilewehavejust==ofthem(b)TheanswerisyesPutthenailsattheverticesofaregulargonandcoloreachoneofitssidesinadifferentcolorNowcoloreachdiagonalinthecoloroftheuniquesideparalleltoitItcanbecheckeddirectlythateachtripleofcolorsappearsinsometriangle(becauseofsymmetry,itisenoughtocheckonlythetriplescontainingthefirstcolor)RemarkTheargumentin(a)canbeappliedtoanyevennTheargumentin(b)canbeappliedtoanyoddnk=asfollows:firstnumberthenailsas,,,k…andsimilarlynumberthecolorsas,,,k…Thenconnectnailxwithnailybyastringofcolor(mod)xynForeachtripleofcolors()p,q,rtherearevertices,,xyzconnectedbythesethreecolorsIndeed,weneedtosolve(mod)nthesystem()()*xyp,xzq,yzrAddingallthree,weget(xyz)pqrandmultiplyingbykwegetxyz(k)(pqr)Wecannowfind,,xyzfromtheidentities()*ProblemFindallpositiveintegersx,y,zandtsuchthat=xyztSolutionReducingmodulowegetz,thereforeziseven,,=`zccNextweprovethattiseven:Obviously,tLetussupposethattisodd,say,=`tddTheequationbecomesxycd=Ifx,reducingmoduloweget,acontradictionAndif=x,wehave=ycdandreducingmoduloweobtain|()yy,ie|ywhichmeansthatyisevenThen,=`ybbWeobtain=bcd,andreducingmoduloweget())(bd,whichisfalseforall,`bdHencetiseven,,=`tdd,asclaimedNowtheequationcanbewrittenas()()==xyddxydcdcAs()gcd,=dcdcand>dc,thereexistexactlythreepossibilities:ddxddyddddyddxddxy()()()======Case()Wehave=dxyandreducingmodulo,weget(m)odx,hencexiseven,ie,=`xaa,where>a,since=awouldmean=yd,whichisimpossible(even=odd)Weobtainmod(mod),==`ddaddeeThenwehavemod(mod),==`ecdacffWeobtainmod)(=aefmod,falseInconclusion,inthiscasetherearenosolutionstotheequationCase()From=xdcweobtainxThen(mdo)dc,ie(m)odd,hencedisoddAs=dcy,wegetd,hence,=`deeAsinthepreviouscase,from=xdyreducingmoduloweobtain=xawitha(becausex)Weget=dayie=eay,hence,reducingmoduloweobtainywhichisfalse,becauseyiscongruenteitherto(ifyiseven)orto(ifyisodd)Inconclusion,inthiscasetherearenosolutionstotheequationCase()From=dcitfollowsthatthelastdigitofdis,hence,=`dkkIfc,from=kcreducingmoduloweobtain()modwhichisfalseFor=cweget=dandthesolution,,====xyztLanguage:EnglishSunday,June,ProblemFindallorderedpairs(a,b)ofpositiveintegersforwhichthenumbersabaandbabarebothpositiveintegersProblemLetABCbeanacutetrianglewithAB<ACandObethecenterofitscircumcircleωLetDbeapointonthelinesegmentBCsuchthatBAD=CAOLetEbethesecondpointofintersectionofωandthelineADIfM,NandParethemidpointsofthelinesegmentsBE,ODandAC,respectively,showthatthepointsM,NandParecollinearProblemShowthat(aba)(bab)forallpositiverealnumbersaandbsuchthatabProblemLetnbeapositiveintegerTwoplayers,AliceandBob,areplayingthefollowinggame:•Alicechoosesnrealnumbers,notnecessarilydistinct•AlicewritesallpairwisesumsonasheetofpaperandgivesittoBob(therearen(n)suchsums,notnecessarilydistinct)•BobwinsifhefindscorrectlytheinitialnnumberschosenbyAlicewithonlyoneguessCanBobbesuretowinforthefollowingcasesan=bn=cn=Justifyyouranswer(s)Forexample,whenn=,Alicemaychoosethenumbers,,,,whichhavethesamepairwisesumsasthenumbers,,,,andhenceBobcannotbesuretowinEachproblemisworthpointsTimeallowed:hoursandminutesProblemSolutionAsab=b(a)(b)anda|a,wehavea|bAsba=a(b)(a)andb|b,wehaveb|aSob|bandhenceb|•Ifb=,thena|b=givesa=Hence(a,b)=(,)istheonlysolutioninthiscase•Ifb=,thena|b=givesa=ora=Hence(a,b)=(,)and(,)aretheonlysolutionsinthiscaseTosummarize,(a,b)=(,),(,)and(,)aretheonlysolutionsProblemSolutionWewillshowthatMOPDisaparallelogramFromthisitfollowsthatM,N,ParecollinearSinceBAD=CAO=ABC,DisthefootoftheperpendicularfromAtosideBCSinceMisthemidpointofthelinesegmentBE,wehaveBM=ME=MDandhenceMDE=MED=ACBLetthelineMDintersectthelineACatDSinceADD=MDE=ACD,MDisperpendiculartoACOntheotherhand,sinceOisthecenterofthecircumcircleoftriangleABCandPisthemidpointofthesideAC,OPisperpendiculartoACThereforeMDandOPareparallelSimilarly,sincePisthemidpointofthesideAC,wehaveAP=PC=DPandhencePDC=ACBLetthelinePDintersectthelineBEatDSinceBDD=PDC=ACB=BED,weconcludethatPDisperpendiculartoBESinceMisthemidpointofthelinesegmentBE,OMisperpendiculartoBEandhenceOMandPDareparallelProblemSolutionBytheAMGMInequalitywehave:aaThereforeabaaband,similarly,bababOntheotherhand,(ab)(ba)(abab)bytheCauchySchwarzInequalityasab,andwearedoneSolutionSinceab,wehaveabaaa(a)=Thenaba=b(ab)aba=bba(b)(a)bytheAMGMInequalitySimilarly,bab(a)(b)NowusingtheseandapplyingtheAMGMInequalityanothertimeweobtain:(aba)(bab)(a)(b)(a)(b)=abSolutionWehave(aba)(bab)=((ab)ba)((ab)ab)(abab(a)(b))bytheCauchySchwarzInequalityOntheotherhand,(a)(b)abbytheAMGMInequalityandabab(a)(b)abab=(ab)(ab)abasabab,finishingtheproofProblemSolutionaYesLetabcdebethenumberschosenbyAliceAseachnumberappearsinapairwisesumtimes,byaddingallpairwisesumsanddividingtheresultby,BobobtainsabcdeSubtractingthesmallestandthelargestpairwisesumsabanddefromthisheobtainscSubtractingcfromthesecondlargestpairwisesumceheobtainseSubtractingefromthelargestpairwisesumdeheobtainsdHecansimilarlydetermineaandbbYesLetabcdefbethenumberschosenbyAliceAseachnumberappearsinapairwisesumtimes,byaddingallpairwisesumsanddividingtheresultby,BobobtainsabcdefSubtractingthesmallestandthelargestpairwisesumsabandeffromthisheobtainscdSubtractingthesmallestandthesecondlargestpairwisesumsabanddffromabcdefheobtainsceSimilarlyhecanobtainbdHeusesthesetoobtainafandbeNowad,ae,bcarethethreesmallestamongtheremainingsixpairwisesumsIfBobaddstheseup,subtractstheknownsumscdandbefromtheresultanddividesthedifferenceby,heobtainsaThenhecandeterminetheremainingnumberscNoIfAlicechoosestheeightnumbers,,,,,,,,thenBobcannotbesuretoguessthesenumberscorrectlyastheeightnumbers,,,,,,,alsogiveexactlythesamepairwisesumsasthesenumbersJMBOsolutionspdfjbmoproblemspdfjbmosolutionspdf

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