american Mathematics Competitions
58th Annual American Mathematics Contest 12
AMC 12 Contest A
Solutions Pamphlet
Tuesday, FEBRUARY 6, 2007
This Pamphlet gives at least one solution for each problem on this year’s contest and shows that all problems can be solved without the use of a calculator. When more than one solution is provided, this is done to illustrate a significant contrast in methods, e.g., algebraic vs geometric, computational vs conceptual, elementary vs advanced. These solu-tions are by no means the only ones possible, nor are they superior to others the reader may devise.
We hope that teachers will inform their students about these solutions, both as illustrations of the kinds of ingenuity needed to solve nonroutine problems and as examples of good mathematical exposition. However, the publication, reproduction or communication of the problems or solutions of the AMC 12 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination via copier, telephone, e-mail, World Wide Web or media of any type during this period is a violation of the competition rules.After the contest period, permission to make copies of individual problems in paper or electronic form including posting on web-pages for educational use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear the copyright notice.Correspondence about the problems/solutions for this AMC 12 and orders for any publications should be addressed to: American Mathematics Competitions
University of Nebraska, P.O. Box 81606, Lincoln, NE 68501-1606
Phone: 402-472-2257; Fax: 402-472-6087; email: amcinfo@unl.edu
The problems and solutions for this AMC 12 were prepared by the MAA’s Committee on the
AMC 10 and AMC 12 under the direction of AMC 12 Subcommittee Chair:
Prof. David Wells, Department of MathematicsPenn State University, New Kensington, PA 15068
Copyright © 2007, The Mathematical Association of America
Solutions2007
58thAMC12A
2
1.Answer(C):Susanpays(4)(0.75)(20)=60dollars.Pampays(5)(0.70)(20)=70dollars,soshepays70−60=10moredollarsthanSusan.2.Answer(D):Thebrickhasavolumeof40·20·10=8000cubiccentimeters.Supposethatafterthebrickisplacedinthetank,thewaterlevelrisesbyhcentimeters.Thentheadditionalvolumeoccupiedintheaquariumis100·40·h=4000hcubiccentimeters.Sincethismustbethesameasthevolumeofthebrick,wehave
8000=4000handh=2centimeters3.Answer(A):Letthesmalleroftheintegersbex.Thenthelargerisx+2.Sox+2=3x,fromwhichx=1.Thusthetwointegersare1and3,andtheirsumis4.4.Answer(A):Katerodefor30minutes=1/2hourat16mph,sosherode8miles.Shewalkedfor90minutes=3/2hoursat4mph,soshewalked6miles.Thereforeshecoveredatotalof14milesin2hours,soheraveragespeedwas7mph.5.Answer(D):Afterpayingthefederaltaxes,Mr.Publichad80%ofhisinheritancemoneyleft.Hepaid10%ofthat,or8%ofhisinheritance,instatetaxes.Hencehistotaltaxbillwas28%ofhisinheritance,andhisinheritancewas$10,500/0.28=$37,500.6.Answer(D):BecauseABCisisosceles,∠BAC=
1
(180◦−∠ABC)=70◦.2
B
40DA
Similarly,
∠DAC=
140C
1
(180◦−∠ADC)=20◦.2
Thus∠BAD=∠BAC−∠DAC=50◦.
OR
BecauseABCandADCareisoscelestriangles,applyingtheExteriorAngleTheoremtoABDgives∠BAD=70◦−20◦=50◦.
Solutions2007
58thAMC12A
3
7.Answer(C):LetDbethedifferencebetweenconsecutivetermsofthese-quence.Thena=c−2D,b=c−D,d=c+D,ande=c+2D,so
a+b+c+d+e=(c−2D)+(c−D)+c+(c+D)+(c+2D)=5c.Thus5c=30,soc=6.
Toseethatthevaluesoftheothertermscannotbefound,notethatthese-quences4,5,6,7,8and10,8,6,4,2bothsatisfythegivenconditions.
8.Answer(C):Considerthetwochordswithanendpointat5.Thearcsub-tendedbytheangledeterminedbythesechordsextendsfrom10to12,sothedegreemeasureofthearcis(2/12)(360)=60.BytheCentralAngleTheorem,thedegreemeasureofthisangleis(1/2)(60)=30.Bysymmetry,thedegreemeasureoftheangleateachvertexis30.9.Answer(B):LetwbeYan’swalkingspeed,andletxandybethedistancesfromYantohishomeandtothestadium,respectively.ThetimerequiredforYantowalktothestadiumisy/w,andthetimerequiredforhimtowalkhomeisx/w.Becauseherideshisbicycleataspeedof7w,thetimerequiredforhimtoridehisbicyclefromhishometothestadiumis(x+y)/(7w).Thus
yxx+y8x+y=+=.ww7w7w
Asaconsequence,7y=8x+y,so8x=6y.Therequiredratioisx/y=6/8=
3/4.
OR
Becauseweareinterestedonlyintheratioofthedistances,wemayassumethatthedistancefromYan’shometothestadiumis1mile.Letxbehispresentdistancefromhishome.ImaginethatYanhasatwin,Nay.WhileYanwalkstothestadium,Naywalkstotheirhomeandcontinues1/7ofamilepasttheirhome.Becausewalking1/7ofamilerequiresthesameamountoftimeasriding1mile,YanandNaywillcompletetheirtripsatthesametime.Yanhaswalked
1
1−xmileswhileNayhaswalkedx+7miles,so1−x=x+17.Thusx=3/7,1−x=4/7,andtherequiredratioisx/(1−x)=3/4.
10.Answer(A):Letthesidesofthetrianglehavelengths3x,4x,and5x.The
triangleisarighttriangle,soitshypotenuseisadiameterofthecircle.Thus5x=2·3=6,sox=6/5.Theareaofthetriangleis
111824216·3x·4x=··==8..225525
OR
Solutions2007
58thAMC12A
4
Arighttrianglewithsidelengths3,4,and5hasarea(1/2)(3)(4)=6.Becausethegivenrighttriangleisinscribedinacirclewithdiameter6,thehypotenuseofthistrianglehaslength6.Thusthesidesofthegiventriangleare6/5aslongasthoseofa3–4–5triangle,anditsareais(6/5)2timesthatofa3–4–5triangle.Theareaofthegiventriangleis
26216
(6)==8..525
11.Answer(D):Agivendigitappearsasthehundredsdigit,thetensdigit,and
theunitsdigitofatermthesamenumberoftimes.Letkbethesumoftheunitsdigitsinalltheterms.ThenS=111k=3·37k,soSmustbedivisibleby37.ToseethatSneednotbedivisiblebyanylargerprime,notethatthesequence123,231,312givesS=666=2·32·37.12.Answer(E):Thenumberad−bcisevenifandonlyifadandbcareboth
oddorarebotheven.Eachofadandbcisoddifbothofitsfactorsareodd,andevenotherwise.Exactlyhalfoftheintegersfrom0to2007areodd,soeachofadandbcisoddwithprobability(1/2)·(1/2)=1/4andareevenwithprobability3/4.Hencetheprobabilitythatad−bcisevenis
11335·+·=.44448
13.Answer(B):Thepoint(a,b)isthefootoftheperpendicularfrom(12,10)
totheliney=−5x+18.Theperpendicularhasslope15,soitsequationis
1138
y=10+(x−12)=x+.
555
Thex-coordinateatthefootoftheperpendicularsatisfiestheequation
138x+=−5x+18,55
sox=2andy=−5·2+18=8.Thus(a,b)=(2,8),anda+b=10.
OR
Ifthemouseisat(x,y)=(x,18−5x),thenthesquareofthedistancefromthe
mousetothecheeseis
(x−12)2+(8−5x)2=26x2−4x+8=26(x−2)2+4.Thevalueofthisexpressionissmallestwhenx=2,sothemouseisclosesttothecheeseatthepoint(2,8),anda+b=2+8=10.
Solutions2007
58thAMC12A
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14.Answer(C):If45isexpressedasaproductoffivedistinctintegerfactors,
theabsolutevalueoftheproductofanyfourisatleast|(−3)(−1)(1)(3)|=9,sonofactorcanhaveanabsolutevaluegreaterthan5.Thusthefactorsofthegivenexpressionarefiveoftheintegers±1,±3,and±5.Theproductofallsixoftheseis−225=(−5)(45),sothefactorsare−3,−1,1,3,and5.Thecorrespondingvaluesofa,b,c,d,andeare9,7,5,3,and1,andtheirsumis25.15.Answer(E):Themeanoftheaugmentedsetis(28+n)/5.Ifn<6,the
medianofthatsetis6,so28+n=5·6,andn=2.If6 formanincreasingarithmeticsequence.Thereare8possiblesequenceswithacommondifferenceof1,sincethefirsttermcanbeanyofthedigits0through7.Thereare6possiblesequenceswithacommondifferenceof2,4withacommondifferenceof3,and2withacommondifferenceof4.Hencethereare20possiblearithmeticsequences.Eachofthe4setsthatcontain0canbearrangedtoform2·2!=4differentnumbers,andthe16setsthatdonotcontain0canbearrangedtoform3!=6differentnumbers.Thusthereareatotalof4·4+16·6=112numberswiththerequiredproperties.17.Answer(B):Squarebothsidesofbothgivenequationstoobtain sin2a+2sinasinb+sin2b=5/3and cos2a+2cosacosb+cos2b=1. Thenaddcorrespondingsidesoftheresultingequationstoobtain 8 sin2a+cos2a+sin2b+cos2b+2(sinasinb+cosacosb)=. 3 Becausesin2a+cos2a=sin2b+cos2b=1,itfollowsthat cos(a−b)=sinasinb+cosacosb= 1.3 Oneorderedpair(a,b)thatsatisfiesthegivenconditionisapproximately(0.296,1.527). 18.Answer(D):Becausef(x)hasrealcoefficientsand2iand2+iarezeros,so aretheirconjugates−2iand2−i.Therefore f(x)=(x+2i)(x−2i)(x−(2+i))(x−(2−i))=(x2+4)(x2−4x+5) =x4−4x3+9x2−16x+20. Hencea+b+c+d=−4+9−16+20=9. Solutions2007 OR 58thAMC12A 6 Asinthefirstsolution, f(x)=(x+2i)(x−2i)(x−(2+i))(x−(2−i)), so a+b+c+d=f(1)−1=(1+2i)(1−2i)(−1−i)(−1+i)−1=(1+4)(1+1)−1=9.19.Answer(E):LethbethelengthofthealtitudefromAinABC.Then 11 ·BC·h=·223·h,22 soh=18.ThusAisononeofthelinesy=18ory=−18.LineDEhasequationx−y−300=0.LetAhavecoordinates(a,b).Bytheformulafor√thedistancefromapointtoaline,thedistancefromAtolineDEis|a−b−300|/2.TheareaofADEis 2007= 7002= 1|a−b−300|1|a±18−300|√ √√··DE=··92.2222Thusa=±18±1556+300,andthesumofthefourpossiblevaluesofais 4·300=1200. OR Asabove,concludethatAisononeofthelinesy=±18.Bysimilarreasoning,Aisononeoftwoparticularlinesl1andl2paralleltoDE.ThereforetherearefourpossiblepositionsforA,determinedbytheintersectionsofthelinesy=18andy=−18witheachofl1andl2.Lettheliney=18intersectl1andl2inpoints(x1,y1)and(x2,y2),andlettheliney=−18intersectl1andl2inpoints(x3,y3)and(x4,y4).Thefourpointsofintersectionaretheverticesofaparallelogram,andthecenteroftheparallelogramhasx-coordinate(1/4)(x1+x2+x3+x4).Thecenteristheintersectionoftheliney=0andlineDE.BecauselineDEhasequationy=x−300,thecenteroftheparallelogramis(300,0).Thusthesumofallpossiblex-coordinatesofAis4·300=1200. 20.Answer(B):Removingthecornersremovestwosegmentsofequallength fromeachx.Then√edgeofthecube.Callthatlength√eachoctagonhassidelength2x,andthecubehasedgelength1=2+2x,so √2−21 √=x=. 22+2Eachremovedcornerisatetrahedronwhosealtitudeisxandwhosebaseisan isoscelesrighttrianglewithleglengthx.Thusthetotalvolumeoftheeighttetrahedrais√ √3112110−728··x·x=2−2=.3263 Solutions2007 58thAMC12A 7 21.Answer(A):Theproductofthezerosoffisc/a,andthesumofthezeros is−b/a.Becausethesetwonumbersareequal,c=−b,andthesumofthecoefficientsisa+b+c=a,whichisthecoefficientofx2.Toseethatnoneofthe√otherchoicesiscorrect,letf(x)=−2x2−4x+4.Thezerosoffare−1±3,sothesumofthezeros,theproductofthezeros,andthesumofthecoefficientsareall−2.However,thecoefficientofxis−4,they-interceptis4,√ thex-interceptsare−1±3,andthemeanofthex-interceptsis−1.22.Answer(D):Ifn≤2007,thenS(n)≤S(1999)=28.Ifn≤28,then S(n)≤S(28)=10.Thereforeifnsatisfiestherequiredconditionitmustalsosatisfy n≥2007−28−10=1969. Inaddition,n,S(n),andS(S(n))allleavethesameremainderwhendividedby9.Because2007isamultipleof9,itfollowsthatn,S(n),andS(S(n))mustallbemultiplesof3.Therequiredconditionissatisfiedby4multiplesof3between1969and2007,namely1977,1980,1983,and2001. Note:Thereappeartobemanycasestocheck,thatis,allthemultiplesof3between1969and2007.However,for1987≤n≤1999,wehaven+S(n)≥1990+19=2009,sothesenumbersareeliminated.Thusweneedonlycheck1971,1974,1977,1980,1983,1986,2001,and2004. 23.Answer(A):LetA=(p,logap)andB=(q,2logaq).ThenAB=6= |p−q|.BecauseABishorizontal,logap=2logaq=logaq2,sop=q2.Thus|q2−q|=6,andtheonlypositivesolutionisq=3.NotethatC=(q,3logaq),√66 soBC=6=logaq,fromwhicha=q=3anda=3.24.Answer(D):NotethatF(n)isthenumberofpointsatwhichthegraphs ofy=sinxandy=sinnxintersecton[0,π].Foreachn,sinnx≥0oneachinterval[(2k−2)π/n,(2k−1)π/n]wherekisapositiveintegerand2k−1≤n.Thenumberofsuchintervalsisn/2ifnisevenand(n+1)/2ifnisodd.Thegraphsintersecttwiceoneachintervalunlesssinx=1=sinnxatsomepointintheinterval,inwhichcasethegraphsintersectonce.Thislastequationissatisfiedifandonlyifn≡1(mod4)andtheintervalcontainsπ/2.Ifniseven,thiscountdoesnotincludethepointofintersectionat(π,0).ThereforeF(n)=2(n/2)+1=n+1ifniseven,F(n)=2(n+1)/2=n+1ifn≡3(mod4),andF(n)=nifn≡1(mod4).Hence 200720072007−1(2006)(3+2008) F(n)=(n+1)−=−501=2,016,532. 42n=2n=2 Solutions2007 58thAMC12A 8 25.Answer(E):Foreachpositiveintegern,letSn={k:1≤k≤n},andlet cnbethenumberofspacysubsetsofSn.Thenc1=2,c2=3,andc3=4.Forn≥4,thespacysubsetsofSncanbepartitionedintotwotypes:thosethatcontainnandthosethatdonot.ThosethatdonotcontainnarepreciselythespacysubsetsofSn−1.Thosethatcontainndonotcontaineithern−1orn−2andarethereforeinone-to-onecorrespondencewiththespacysubsetsofSn−3.Itfollowsthatcn=cn−3+cn−1.Thusthefirsttwelvetermsinthesequence(cn)are2,3,4,6,9,13,19,28,41,60,88,129,andtherearec12=129spacysubsetsofS12. OR NotethateachspacysubsetofS12containsatmost4elements.Foreachsuchsubseta1,a2,...,ak,letb1=a1−1,bj=aj−aj−1−3for2≤j≤k,andbk+1=12−ak.Thenbj≥0for1≤j≤k+1,and b1+b2+···+bk+1=12−1−3(k−1)=14−3k. 2k Thenumberofsolutionsfor(b1,b2,...,bk+1)is14−for0≤k≤4,sotheknumberofspacysubsetsofS12is 14121086 ++++=1+12+45+56+15=129.01234 The American Mathematics Competitions are Sponsored by The Mathematical Association of AmericaThe Akamai FoundationContributorsAmerican Mathematical Association of Two Year CollegesAmerican Mathematical SocietyAmerican Society of Pension Actuaries American Statistical AssociationArt of Problem SolvingAwesome MathCanada/USA Mathcamp Canada/USA MathpathCasualty Actuarial SocietyClay Mathematics InstituteInstitute for Operations Research and the Management SciencesL. G. Balfour CompanyMu Alpha ThetaNational Assessment & TestingNational Council of Teachers of MathematicsPedagoguery Software Inc.Pi Mu Epsilon Society of ActuariesU.S.A. Math Talent SearchW. H. Freeman and CompanyWolfram Research Inc.
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