2006_美国数学竞赛_AMC_12a_solutions_完整答案

TheMATHEMATICALASSOCIATIONOFAMERICAamericanMathematicsCompetitions57thAnnualAmericanMathematicsContest12aMC12–ContestaSolutionsPamphletTuesday,January31,2006ThisPamphletgivesatleastonesolutionforeachproblemonthisyear’scontestandshowsthatallproblemscanbesolvedwithouttheuseofacalculator.Whenmorethanonesolutionisprovided,thisisdonetoillustrateasignificantcontrastinmethods,e.g.,algebraicvsgeometric,computationalvsconceptual,elementaryvsadvanced.Thesesolu-tionsarebynomeanstheonlyonespossible,noraretheysuperiortoothersthereadermaydevise.Wehopethatteacherswillinformtheirstudentsaboutthesesolutions,bothasillustrationsofthekindsofingenuityneededtosolvenonroutineproblemsandasexamplesofgoodmathematicalexposition.However,thepublication,reproductionorcommunicationoftheproblemsorsolutionsoftheAMC12duringtheperiodwhenstudentsareeligibletoparticipateseriouslyjeopardizestheintegrityoftheresults.Disseminationatanytimeviacopier,telephone,email,theWorldWideWebormediaofanytypeisaviolationofthecompetitionrulesCorrespondenceabouttheproblemsandsolutionsforthisAMC12andordersforanyofthepublicationslistedbelowshouldbeaddressedto:AmericanMathematicsCompetitionsUniversityofNebraska,P.O.Box81606Lincoln,NE68501-1606Phone:[phone];Fax:[phone];email:[e-mail]nsforthisAMC12werepreparedbytheMAA’sCommitteeontheAMC10andAMC12underthedirectionofAMC12SubcommitteeChair:Prof.DavidWells,DepartmentofMathematicsPennStateUniversity,NewKensington,PA15068Copyright©2006,TheMathematicalAssociationofAmericaSolutions200657thAMC12A21.(A)Fivesandwichescost5¢3=15dollarsandeightsodascost8¢2=16dollars.Togethertheycost15+16=31dollars.2.(C)Bythede¯nitionwehaveh­(h­h)=h­(h3¡h)=h3¡(h3¡h)=h:3.(B)Maryis(3=5)(30)=18yearsold.4.(E)Thelargestpossiblesumofthetwodigitsrepresentingtheminutesis5+9=14,occurringat59minutespasteachhour.Thelargestpossiblesingledigitthatcanrepresentthehouris9.Thisexceedsthelargestpossiblesumoftwodigitsthatcanrepresentthehour,whichis1+2=3.Therefore,thelargestpossiblesumofallthedigitsis14+9=23,occurringat9:59.5.(D)Eachsliceofplainpizzacost$1.Davepaid$2fortheanchoviesinadditionto$5forthe¯veslicesofpizzathatheate,foratotalof$7.Dougpaidonly$3forthethreeslicesofpizzathatheate.HenceDavepaid7¡3=4dollarsmorethanDoug.6.(A)LetErepresenttheendofthecutonDC,andletFrepresenttheendofthecutonAB.Forasquaretobeformed,wemusthaveDE=y=FBandDE+y+FB=18;soy=6:Therectanglethatisformedbythiscutisindeedasquare,sincetheoriginalrectanglehasarea8¢18=144,andtherectanglethatisformedbythiscuthasasideoflength2¢6=12=p144.AyyyDEFCBAyyDEFFECB7.(B)LetDaniellebexyearsold.Sallyis40%younger,sosheis0:6xyearsold.Maryis20%olderthanSally,soMaryis1:2(0:6x)=0:72xyearsold.Thesumoftheiragesis23:2=x+0:6x+0:72x=2:32xyears,sox=10.ThereforeMary'sageis0:72x=7:2years,andshewillbe8onhernextbirthday.Solutions200657thAMC12A38.(C)Firstnotethat,ingeneral,thesumofnconsecutiveintegersisntimestheirmedian.Ifthesumis15,wehavethefollowingcases:ifn=2,thenthemedianis7.5andthetwointegersare7and8;ifn=3,thenthemedianis5andthethreeintegersare4,5,and6;ifn=5,thenthemedianis3andthe¯veintegersare1,2,3,4,and5.Becausethesumoffourconsecutiveintegersiseven,15cannotbewritteninsuchamanner.Also,thesumofmorethan¯veconsecutiveintegersmustbemorethan1+2+3+4+5=15.Hencethereare3setssatisfyingthecondition.Note:Itcanbeshownthatthenumberofsetsoftwoormoreconsecutivepositiveintegershavingasumofkisequaltothenumberofoddpositivedivisorsofk,excluding1.9.(A)Letpbethecost(incents)ofapencil,andletsbethecost(incents)ofasetofonepencilandoneeraser.BecauseOscarbuys3setsand10extrapencilsfor$1.00,wehave3s+10p=100:Thus3sisamultipleof10thatislessthan100,sosis10,20,or30.Thecorrespondingvaluesofpare7,4,and1.Sincethecostofapencilismorethanhalfthecostoftheset,theonlypossibilityiss=10.10.(E)Supposethatk=p120¡pxisaninteger.Then0·k·p120,andbecausekisaninteger,wehave0·k·10.Thusthereare11possibleintegervaluesofk.Foreachsuchk,thecorrespondingvalueofxis¡120¡k2¢2.Because¡120¡k2¢2ispositiveanddecreasingfor0·k·10,the11valuesofxaredistinct.11.(C)Theequation(x+y)2=x2+y2isequivalenttox2+2xy+y2=x2+y2,whichreducestoxy=0.Thusthegraphoftheequationconsistsofthetwolinesthatarethecoordinateaxes.12.(B)Thetopofthelargestringis20cmaboveitsbottom.Thatpointis2cmbelowthetopofthenextring,soitis17cmabovethebottomofthenextring.Theadditionaldistancestothebottomsoftheremainingringsare16cm;15cm;:::;1cm.Thusthetotaldistanceis20+(17+16+¢¢¢+2+1)=20+17¢182=20+17¢9=173cm:ORTherequireddistanceisthesumoftheoutsidediametersofthe18ringsminusa2-cmoverlapforeachofthe17pairsofconsecutiverings.Thisequals(3+4+5+¢¢¢+20)¡2¢17=(1+2+3+4+5+¢¢¢+20)¡3¡34=20¢212¡37=173cm.Solutions200657thAMC12A413.(E)Letr,s,andtbetheradiiofthecirclescenteredatA,B,andC,respec-tively.Thenr+s=3,r+t=4,ands+t=5,fromwhichr=1,s=2,andt=3.Thusthesumoftheareasofthecirclesis¼(12+22+32)=14¼:14.(C)IfadebtofDdollarscanberesolvedinthisway,thenintegerspandgmustexistwithD=300p+210g=30(10p+7g):Asaconsequence,Dmustbeamultipleof30,sonopositivedebtoflessthan$30canberesolved.Adebtof$30canberesolvedsince30=300(¡2)+21