2007_美国数学竞赛_AMC_12B_solutions_完整答案

TheMATheMATICALASSOCIATIONOfAMeRICAamericanMathematicsCompetitions58thAnnualAmericanMathematicsContest12AMC12ContestBSolutionsPamphletWednesday,FEBRUARY21,2007ThisPamphletgivesatleastonesolutionforeachproblemonthisyear’scontestandshowsthatallproblemscanbesolvedwithouttheuseofacalculator.Whenmorethanonesolutionisprovided,thisisdonetoillustrateasignificantcontrastinmethods,e.g.,algebraicvsgeometric,computationalvsconceptual,elementaryvsadvanced.Thesesolu-tionsarebynomeanstheonlyonespossible,noraretheysuperiortoothersthereadermaydevise.Wehopethatteacherswillinformtheirstudentsaboutthesesolutions,bothasillustrationsofthekindsofingenuityneededtosolvenonroutineproblemsandasexamplesofgoodmathematicalexposition.However,thepublication,reproductionorcommunicationoftheproblemsorsolutionsoftheAMC12duringtheperiodwhenstudentsareeligibletoparticipateseriouslyjeopardizestheintegrityoftheresults.Disseminationviacopier,telephone,e-mail,WorldWideWebormediaofanytypeduringthisperiodisaviolationofthecompetitionrules.Afterthecontestperiod,permissiontomakecopiesofindividualproblemsinpaperorelectronicformincludingpostingonweb-pagesforeducationaluseisgrantedwithoutfeeprovidedthatcopiesarenotmadeordistributedforprofitorcommercialadvantageandthatcopiesbearthecopyrightnotice.Correspondenceabouttheproblems/solutionsforthisAMC12andordersforanypublicationsshouldbeaddressedto:AmericanMathematicsCompetitionsUniversityofNebraska,P.O.Box81606,Lincoln,NE68501-1606Phone:[phone];Fax:[phone];email:[e-mail]nsforthisAMC12werepreparedbytheMAA’sCommitteeontheAMC10andAMC12underthedirectionofAMC12SubcommitteeChair:Prof.DavidWells,DepartmentofMathematicsPennStateUniversity,NewKensington,PA15068Copyright©2007,TheMathematicalAssociationofAmericaSolutions200758thAMC12B21.Answer(E):Theperimeterofeachbedroomis2(12+10)=44feet,sothesurfacetobepaintedineachbedroomhasanareaof44¢8¡60=292squarefeet.Sincethereare3bedrooms,Isabellamustpaint3¢292=876squarefeet.2.Answer(B):Thestudentused120=30=4gallonsonthetriphomeand120=20=6gallonsonthetripbacktoschool.Sotheaveragegasmileagefortheroundtripwas240miles10gallons=24milespergallon:3.Answer(D):SinceOA=OB=OC,trianglesAOB,BOC,andCOAareallisosceles.Hence\ABC=\ABO+\OBC=180±¡140±2+180±¡120±2=50±:ORSince\AOC=360±¡140±¡120±=100±;theCentralAngleTheoremimpliesthat\ABC=12\AOC=50±:4.Answer(B):Because3bananascostasmuchas2apples,18bananascostasmuchas12apples.Because6applescostasmuchas4oranges,12applescostasmuchas8oranges.Therefore18bananascostasmuchas8oranges.5.Answer(D):Sarahwillreceive4.5pointsforthethreequestionssheleavesunanswered,soshemustearnatleast100¡4:5=95:5pointsonthe¯rst22problems.Because1595:5616;shemustsolveatleast16ofthe¯rst22problemscorrectly.Thiswouldgiveherascoreof100.5.6.Answer(D):Theperimeterofthetriangleis5+6+7=18,sothedistancethateachbugcrawlsis9.ThereforeAB+BD=9,andBD=4.Solutions200758thAMC12B37.Answer(E):BecauseAB=BC=EAand\A=\B=90±,quadrilateralABCEisasquare,so\AEC=90±.ABCDEAlsoCD=DE=EC,so4CDEisequilateraland\CED=60±.Therefore\E=\AEC+\CED=90±+60±=150±:8.Answer(D):Tom'sageNyearsagowasT¡N.Thesumofhisthreechildren'sagesatthattimewasT¡3N.ThereforeT¡N=2(T¡3N),so5N=TandT=N=5.Theconditionsoftheproblemcanbemet,forexample,ifTom'sageis30andtheagesofhischildrenare9,10,and11.InthatcaseT=30andN=6.9.Answer(A):Letu=3x¡1.Thenx=(u+1)=3,andf(u)=µu+13¶2+u+13+1=u2+2u+19+u+13+1=u2+5u+139:Inparticular,f(5)=52+5¢5+139=639=7:ORThevalueof3x¡1is5whenx=2.Thusf(5)=f(3¢2¡1)=22+2+1=7:10.Answer(C):Letgbethenumberofgirlsandbthenumberofboysinitiallyinthegroup.Theng=0:4(g+b).Aftertwogirlsleaveandtwoboysarrive,thesizeoftheentiregroupisunchanged,sog¡2=0:3(g+b).Thesolutionofthesystemofequationsg=0:4(g+b)andg¡2=0:3(g+b)isg=8andb=12,sotherewereinitially8girls.ORAftertwogirlsleaveandtwoboysarrive,thesizeofthegroupisunchanged.Sothetwogirlswholeftrepresent40%¡30%=10%ofthegroup.Thusthesizeofthegroupis20,andtheoriginalnumberofgirlswas40%of20,or8.Solutions200758thAMC12B411.Answer(D):Letxbethedegreemeasureof\A.ThenthedegreemeasuresofanglesB,C,andDarex=2,x=3,andx=4,respectively.Thedegreemeasuresofthefourangleshaveasumof360,so360=x+x2+x3+x4=25x12:Thusx=(12¢360)=25=172:8¼173.12.Answer(C):LetNbethenumberofstudentsintheclass.Thenthereare0:1Njuniorsand0:9Nseniors.Letsbethescoreofeachjunior.Thescorestotaled84N=83(0:9N)+s(0:1N),sos=84N¡83(0:9N)0:1N=93:Note:Inthisproblem,wecouldassumethattheclasshasonejuniorandnineseniors.Then9¢83+s=10¢84=9¢84+84ands=9(84¡83)+84=93:13.Answer(D):Thelightcompletesacycleevery63seconds.Leahseesthecolorchangeifandonlyifshebeginstolookwithinthreesecondsbeforethechangefromgreentoyellow,fromyellowtored,orfromredtogreen.Thussheseesthecolorchangewithprobability(3+3+3)=63=1=7.14.Answer(D):Letthesidelengthof4ABCbes.Thentheareasof4APB,4BPC,and4CPAare,respectively,s=2,s,and3s=2.Theareaof4ABCisthesumofthese,whichis3s.Theareaof4ABCmayalsobeexpressedas(p3=4)s2,so3s=(p3=4)s2.Theuniquepositivesolutionforsis4p3.15.Answer(E):Thetermsinvolvingoddpowersofrformthegeometricseriesar+ar3+ar5+¢¢¢.Thus7=a+ar+ar2+¢¢¢=a1¡r;