2001 AMC12

2001AMC122001AMC12ProblemsProblem1Thesumoftwonumbersis.Supposeisaddedtoeachnumberandtheneachoftheresultingnumbersisdoubled.Whatisthesumofthefinaltwonumbers?SolutionProblem2Letanddenotetheproductandthesum,respectively,ofthedigitsoftheinteger.Forexample,and.Supposeisatwo-digitnumbersuchthat.Whatistheunitsdigitof?SolutionProblem3ThestateincometaxwhereKristinlivesisleviedattherateofofthefirst$ofannualincomeplusofanyamountabove$.Kristinnoticedthatthestateincometaxshepaidamountedtoofherannualincome.Whatwasherannualincome?$$$$$SolutionProblem4Themeanofthreenumbersismorethantheleastofthenumbersandlessthanthegreatest.Themedianofthethreenumbersis.Whatistheirsum?SolutionProblem5Whatistheproductofallpositiveoddintegerslessthan10000?SolutionProblem6Atelephonenumberhastheform,whereeachletterrepresentsadifferentdigit.Thedigitsineachpartofthenumberareindecreasingorder;thatis,,,and.Furthermore,,,andareconsecutiveevendigits;,,,andareconsecutiveodddigits;and.Find.SolutionProblem7Acharitysellsbenefitticketsforatotalof$.Someticketssellforfullprice(awholedollaramount),andtherestsellsforhalfprice.Howmuchmoneyisraisedbythefull-pricetickets?$$$$$SolutionProblem8Whichoftheconeslistedbelowcanbeformedfromasectorofacircleofradiusbyaligningthetwostraightsides?SolutionProblem9Letbeafunctionsatisfyingforallpositiverealnumbersand.If,whatisthevalueof?SolutionProblem10Theplaneistiledbycongruentsquaresandcongruentpentagonsasindicated.ThepercentoftheplanethatisenclosedbythepentagonsisclosesttoSolutionProblem11Aboxcontainsexactlyfivechips,threeredandtwowhite.Chipsarerandomlyremovedoneatatimewithoutreplacementuntilalltheredchipsaredrawnorallthewhitechipsaredrawn.Whatistheprobabilitythatthelastchipdrawniswhite?SolutionProblem12Howmanypositiveintegersnotexceedingaremultipleoforbutnot?SolutionProblem13Theparabolawithequationandvertexisreflectedabouttheline.Thisresultsintheparabolawithequation.Whichofthefollowingequals?SolutionProblem14Giventhenine-sidedregularpolygon,howmanydistinctequilateraltrianglesintheplaneofthepolygonhaveatleasttwoverticesintheset?SolutionProblem15Aninsectlivesonthesurfaceofaregulartetrahedronwithedgesoflength1.Itwishestotravelonthesurfaceofthetetrahedronfromthemidpointofoneedgetothemidpointoftheoppositeedge.Whatisthelengthoftheshortestsuchtrip?(Note:Twoedgesofatetrahedronareoppositeiftheyhavenocommonendpoint.)SolutionProblem16Aspiderhasonesockandoneshoeforeachofitseightlegs.Inhowmanydifferentorderscanthespiderputonitssocksandshoes,assumingthat,oneachleg,thesockmustbeputonbeforetheshoe?SolutionProblem17Apointisselectedatrandomfromtheinteriorofthepentagonwithvertices,,,,and.Whatistheprobabilitythatisobtuse?SolutionProblem18Acirclecenteredatwitharadiusof1andacirclecenteredatwitharadiusof4areexternallytangent.Athirdcircleistangenttothefirsttwoandtooneoftheircommonexternaltangentsasshown.TheradiusofthethirdcircleisSolutionProblem19Thepolynomialhasthepropertythatthemeanofitszeros,theproductofitszeros,andthesumofitscoefficientsareallequal.Ifthe-interceptofthegraphofis2,whatis?SolutionProblem20Points,,,andlieinthefirstquadrantandaretheverticesofquadrilateral.Thequadrilateralformedbyjoiningthemidpointsof,,,andisasquare.Whatisthesumofthecoordinatesofpoint?SolutionProblem21Fourpositiveintegers,,,andhaveaproductofandsatisfy:Whatis?SolutionProblem22Inrectangle,pointsandlieonsothatandisthemidpointof.Also,intersectsatandat.Theareaoftherectangleis.Findtheareaoftriangle.SolutionProblem23Apolynomialofdegreefourwithleadingcoefficient1andintegercoefficientshastwozeros,bothofwhichareintegers.Whichofthefollowingcanalsobeazeroofthepolynomial?SolutionProblem24In,.Pointisonsothatand.Find.SolutionProblem25Considersequencesofpositiverealnumbersoftheforminwhicheverytermafterthefirstis1lessthantheproductofitstwoimmediateneighbors.Forhowmanydifferentvaluesofdoestheterm2001appearsomewhereinthesequence?答案：Problem1SolutionSupposethetwonumbersareand,with.Thenthedesiredsumis,whichisanswer.Problem2SolutionDenoteandasthetensandunitsdigitof,respectively.Then.Itfollowsthat,whichimpliesthat.Since,.Sotheansweris.Problem3SolutionLettheincomeamountbedenotedby.Weknowthat.Wecannowtrytosolvefor:SotheanswerisProblem4SolutionLetbethemeanofthethreenumbers.Thentheleastofthenumbersisandthegreatestis.Themiddleofthethreenumbersisthemedian,5.So,whichimpliesthat.Hence,thesumofthethreenumbersis,andtheanswerisProblem5SolutionProblem6SolutionThelastfourdigitsareeitheror,andtheotherodddigit(or)mustbe,,or.Since,thatdigitmustbe.Thusthesumofthetwoevendigitsinis.mustbe,,or,whichrespectivelyleavethepairsand,and,orand,asthetwoevendigitsin.Onlyandhassum,sois,andtherequiredfirstdigitis8,sotheansweris.Problem7SolutionLet'smultiplyticketcostsby,thenthehalfpricebecomesaninteger,andthecharitysoldticketsworthatotalofdollars.Letbethenumberofhalfpricetickets,wethenhavefullpricetickets.Thecostoffullpriceticketsisequaltothecostofhalfpricetickets.Henceweknowthathalfpriceticketscostdollars.Thenasinglehalfpriceticketcostsdollars,andthismustbeaninteger.Thusmustbeadivisorof.Keepinginmindthat,wearelookingforadivisorbetweenand,inclusive.Theprimefactorizationofis.Wecaneasilyfindoutthattheonlydivisorofwithinthegivenrangeis.Thisgivesus,hencetherewerehalfpriceticketsandfullpricetickets.Inourmodifiedsetting(withpricesmultipliedby)thepriceofahalfpriceticketis.Intheoriginalsettingthisisthepriceofafullpriceticket.Hencedollarsareraisedbythefullpricetickets.Problem8SolutionThebluelineswillbejoinedtogethertoformasinglebluelineonthesurfaceofthecone,hencewillbetheofthecone.Theredlinewillformthecircumferenceofthebase.Wecancomputeitslengthanduseittodeterminetheradius.Thelengthoftheredlineis.Thisisthecircumferenceofacirclewithradius.Thereforethecorrectansweris.Problem9Solution,sotheansweris.Problem10SolutionConsideranysingletile:Ifthesideofthesmallsquareis,thentheareaofthetileis,withcoveredbysquaresandbypentagons.Henceexactlyofanytilearecoveredbypentagons,andthereforepentagonscoveroftheplane.Whenexpressedasapercentage,thisis,andtheclosestintegertothisvalueis.Problem11SolutionImaginethatwedrawallthechipsinrandomorder,i.e.,wedonotstopwhenthelastchipofacolorisdrawn.Todrawoutallthewhitechipsfirst,thelastchipleftmustbered,andallpreviouschipscanbedrawninanyorder.Sincethereare3redchips,theprobabilitythatthelastchipofthefiveisred(andsoalsotheprobabilitythatthelastchipdrawninwhite)is.Problem12SolutionOutofthenumberstofouraredivisiblebyandthreeby,countingtwice.Henceoutofthesenumbersaremultiplesofor.Thesameisobviouslytrueforthenumberstoforanypositiveinteger.Henceoutofthenumberstotherearenumbersthataredivisiblebyor.Outofthese,thenumbers,,,,andaredivisibleby.Thereforeinthesettherearepreciselynumbersthatsatisfyallcriteriafromtheproblemstatement.Again,thesameisobviouslytrueforthesetforanypositiveinteger.Wehave,hencetherearegoodnumbersamongthenumbersto.Atthispointwealreadyknowthattheonlyanswerthatisstillpossibleis,asweonlyhavenumbersleft.Byexaminingtheremainingbyhandwecaneasilyfindoutthatexactlyofthemmatchallthecriteria,givingusgoodnumbers.Problem13SolutionWewriteas(thisispossibleforanyparabola).Thenthereflectionofis.Thenwefind.Sinceand,wehave,sotheanswerisProblem14SolutionEachofthepairsofverticesdeterminestwoequilateraltriangles,oneoneachsideofthesegment.Thiswouldgiveustriangles.However,notethattherearethreeequilateraltrianglesthathaveallthreeverticesamongtheverticesofthepolygon.Thesearethetriangles,,and.Wecountedeachofthesethreetimes(onceforeachside).Henceweovercountedby,andthecorrectnumberofequilateraltrianglesis.Problem15SolutionGivenanypathonthesurface,wecanunfoldthesurfaceintoaplanetogetapathofthesamelengthintheplane.Considerthenetofatetrahedroninthepicturebelow.Apairofoppositepointsismarkedbydots.Itisobviousthatintheplanetheshortestpathisjustasegmentthatconnectsthesetwopoints.Itslengthisthesameasthelengthofthetetrahedron'sedge,i.e..Problem16SolutionSolution1Letthespidertrytoputonallthingsinarandomorder.Eachofthepermutationsisequallyprobable.Foranyfixedleg,theprobabilitythathewillfirstputonthesockandonlythentheshoeisclearly.Thentheprobabilitythathewillcorrectlyputthingsonalllegsis.Thereforethenumberofcorrectpermutationsmustbe.Solution2Eachdressingsequencecanbeuniquelydescribedbyasequencecontainingtwos,twos,...,andtwos--thefirstoccurrenceofnumbermeansthatthespiderputsthesockontoleg,thesecondoccurrenceofmeansheputstheshoeontoleg.Ifthenumberswereallunique,theanswerwouldbe.However,since8termsappeartwice,theansweris.Problem17SolutionTheangleisobtuseifandonlyifliesinsidethecirclewithdiameter.(Thisfollowsforexamplefromthefactthattheinscribedangleishalfofthecentralangleforthesamearc.)Theareaofis,andtheareaofis.FromthePythagoreantheoremthelengthofis,thustheradiusofthecircleis,andtheareaofthehalf-circlethatisinsideis.Thereforetheprobabilitythatisobtuseis.AnswerchoiceProblem18SolutionSolution1Inthetrianglewehaveand,thusbythePythagoreantheoremwehave.Wecannowpickacoordinatesystemwherethecommontangentistheaxisandliesontheaxis.Inthiscoordinatesystemwehaveand.Letbetheradiusofthesmallcircle,andletbethe-coordinateofitscenter.Wethenknowthat,asthecircleistangenttotheaxis.Moreover,thesmallcircleistangenttobothothercircles,hencewehaveand.Wehaveand.Hencewegetthefollowingtwoequations:Simplifyingboth,wegetAsinourcasebothandarepositive,wecandividethesecondonebythefirstonetoget.Nowtherearetwopossibilities:either,or.Inthefirstcaseclearly,hencethisisnotthecorrectcase.(Note:Thiscasecorrespondstotheothercirclethatistangenttobothgivencirclesandtheaxis-alargecirclewhosecenterissomewheretotheleftof.)Thesecondcasesolvesto.Wethenhave,hence.Solution2Thehorizontallineistheequivalentofacircleofcurvature,thuswecanapplyDescartes'CircleFormula.Thefourcircleshavecurvatures,and.WehaveSimplifying,wegetObviouslycannotequal,therefore.Problem19SolutionWearegiven.SotheproductoftherootsisbyVieta'sformulas.Thesealsotellusthatistheaverageofthezeros,so.Wearealsogiventhatthesumofthecoefficientsis,so.Sotheansweris.Problem20SolutionWealreadyknowtwoverticesofthesquare:and.Thereareonlytwopossibilitiesfortheotherverticesofthesquare:eithertheyareand,ortheyareand.Thesecondcasewouldgiveusoutsidethefirstquadrant,hencethefirstcaseisthecorrectone.Asisthemidpointof,wecancompute,and.Problem21SolutionWecanrewritethethreeequationsasfollows:Let.Weget:Clearlydivides.Ontheotherhand,cannotdivide,asitthenwoulddivide.Similarly,cannotdivide.Hencedividesbothand.Thisleavesuswithonlytwocases:and.Thefirstcasesolvesto,whichgivesus,butthen.(Wedonotneedtomultiply,itisenoughtonotee.g.thatthelefthandsideisnotdivisibleby.)Thesecondcasesolvesto,whichgivesusavalidquadruple,andwehave.Problem22SolutionSolution1Notethatthetrianglesandaresimilar,astheyhavethesameangles.Hence.Also,trianglesandaresimilar,hence.Wecannowcomputeas.Wehave:▪.▪isof,asthesetwotriangleshavethesamebase,andisof,thereforealsotheheightfromontoisoftheheightfrom.Hence.▪isof,asthebaseisofthebase,andtheheightfromisoftheheightfrom.Hence.▪isofforsimilarreasons,hence.Therefore.Solution2Asintheprevioussolution,wenotethesimilartrianglesandprovethatisinandinof.Wecanthencomputethat.Asisthemidpointof,theheightfromontoisoftheheightfromonto.Thereforewehave.Problem23SolutionLetthepolynomialbeandletthetwointegerzerosbeand.Wecanthenwriteforsomeintegersand.Ifacomplexnumberwithisarootof,itmustbetherootof,andtheotherrootofmustbe.Wecanthenwrite.Wecannowexamineeachofthefivegivencomplexnumbers,andfindtheoneforwhichthevaluesandareintegers.Thisis,forwhichwehaveand.(Asanexample,thepolynomialhaszeroes,,and.)Problem24SolutionWestartwiththeobservationthat,and.Wecandrawtheheightfromonto.Inthetriangle,wehave.Hence.Bythedefinitionof,wealsohave,therefore.Thismeansthatthetriangleisisosceles,andas,wemusthave.Thenwecompute,thusandthetriangleisisoscelesaswell.Hence.Nowwecannotethat,hencealsothetriangleisisoscelesandwehave.Combiningtheprevioustwoobservationswegetthat,andas,thismeansthat.Finally,weget.Problem25SolutionItneverhurtstocomputeafewtermsofthesequenceinordertogetafeelhowitlookslike.Inourcase,thedefinitionisthat.Thiscanberewrittenas.Wehaveand,andwecompute:Atthispointweseethatthesequencewillbecomeperiodic:wehave,,andeachsubsequenttermisuniquelydeterminedbytheprevioustwo.Henceifappears,ithastobeoneofto.As,weonlyhavefourpossibilitiesleft.Clearlyfor,andfor.Theequationsolvesto,andtheequationto.Notwovaluesofwejustcomputedareequal,andthereforetherearedifferentvaluesofforwhichthesequencecontainsthevalue.