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Aliciahadtwocontainers.ThefirstwasZfullofwaterandthesecondwasempty.Shepouredallthewaterfromthefirstcontainerintothesecond3containeratwhichpointthesecondcontainerwasZfullofwater.WhatistheratioofthevolumeofthefirstcontainertothevolumeofthesecondcontainerA|B-C-Solution3_5Letthefirstjarsvolumebe\andthesecondsbeB.Itisgiventhatil6Wefindthat16Wealreadyknowthatthisistheratioofsmallertolargervolumebecauseitislessthan
1.2019AMC10BProblem2ConsiderthestatementIfnisnotprimethenn—2isprime.Whichofthefollowingvaluesofnisacounterexampletothisstatement.A11B15C19D21E27SolutionSinceacounterexamplemustbewhennisnotprimenmustbecompositesoweeliminateAandC.Nowwesubtract2fromtheremaininganswerchoicesandweseethattheonlytime2isnotprimeiswhenn27whichis叵.2019AMC10BProblem3Inahighschoolwith5Xstudents407oftheseniorsplayamusicalinstrumentwhile307ofthenon-seniorsdonotplayamusicalinstrument.Inall
16.8/ofthestudentsdonotplayamusicalinstrument.Howmanynon-seniorsplayamusicalinstrumentA66B154C186D220E266SolutionAredballandagreenballarerandomlyandindependentlytossedintobinsnumberedwithpositiveintegerssothatforeachballtheprobabilitythatitistossedintobinAis2for卜一123Whatistheprobabilitythattheredballistossedintoahigher-numberedbinthanthegreenball233A[©gD-E-Solution1SupposethegreenballgoesinbinforsomeiLTheprobabilityofthisoccurringisGiventhisoccurstheprobabilitythattheredballgoesinagreenballgoesinbinzandtheredballgoesinabingreaterthaniisV21/r.Summingfromz=1toinfinitywegetZiSolution2infinitegeometricseriesTheprobabilitythatthetwoballswillgointoadjacentbinsis2x44x88x16…
8321286.Theprobabilitythatthetwoballsw川gointobinsthathaveadistanceof2fromeachother1111111++=—+—+—=—is2x84x168x
32166425612.Wecanseethateachtimeweaddabinbetweenthetwoballstheprobabilityhalves.Thusour1111一+--I+..一answeris61224whichconvergesinto
3.Henrydecidesonemorningtodoaworkoutandhewalkslotthewayfromhishometohisgym.Thegymis^kilometersawayfromHenryshome.Atthatpointhechangeshismindandwalkslotthewayfromwhereheisbacktowardhome.Whenhereachesthatpointhechangeshismindagainand2walks;ofthedistancefromtherebacktowardthegym.IfHenrykeepschanginghismindwhenhehaswalkedlotthedistancetowardeitherthegymorhomefromthepointwherehelastchangedhismindhewillgetveryclosetowalkingbackandforthbetweenapointIkilometersfromhomeandapoint^kilometersfromhome.Whatis以一打?2111A-B1C1-D1-E1-J4/SolutionLetthetwopointsthatHenrywalksin-betweenbe-land凡with/beingclosertohome.Inadditionletthedistancethatthepoints.landRarefromhishomebeaandbrespectively.BysymmetrythedistancefrompointBisfromthegymisthesameasthedistancefromhometopointI.Thusa-2-
6.Inadditionthedistancethathewalksasherepeatedlyheadstohomeandthentothegymish-Thereforewearelookingforthevalueofh-fi.Whenhewalksfrompoint7tohomehewalkslotthedistance3oazz一•bendingattaintA.Thereforeweknowthat
4.Similarilyweknowb-a=•2—a
4.Addingtheseequationsweget2b—a=—•2+fe—a.
4.Multiplyingby1weget6J86一°=6+3,一〃soLetSbethesetofallpositiveintegerdivisorsofmanynumbersaretheproductoftwodistinctelementsofSA98B100C117D119E121Solution1TofindthenumberofnumbersthataretheproductoftwodistinctelementsofSwefirstsquareancjfactorit.Factoringwefind1000002=210・5皿Therefore1000003has10+110+1=121distinctfactors.EachofthesecanbeachievedbymultiplyingtwofactorsofS.Howeverthefactorsmustbedistinctsoweeliminate1andaswellas2“and511sotheansweris121—I
117.Solution2Theprimefactorizationof100000is2•;.Thuswechoosetwonumbers2a56and2c5dwhere045and°,b*Lwhoseproductis2a声qwhere0a4-r11and06+d
10.ConsiderKXMMX2=205:r.Thenumberofdivisorsis10+110+1=
121.Howeversomeofthedivisorsof9n51cannotbewrittenasaproductoftwodistinctdivisorsofi-namely:1=205°910511211and
510.Thisgives121—1-117candidatenumbers.ItisnottoohardtoshowthateverynumberoftheformwhereV〃VWandP・1arenotboth0or10canbewritten[117asaproductoftwodistinctelementsinS.HencetheanswerisI_____—AsshowninthefigurelinesegmentAI\strisectedbypointsandsothatAB=BC=CD=
2.Threesemicirclesofradius1AEBBFCandCGD.havetheirdiameterson/^andaretangenttolineFf7atE/and^.respectively.AcircleofradiushasitscenteronFTheareaoftheregioninsidethecirclebutoutsidethethreesemicirclesshadedinthefigurecanbe—7T-+d.jexpressedintheformbwherefLanddarepositiveintegersandnandbarerelativelyprime.Whatis1+4-c4-dA13B14C15D16E17SolutionDividethecircleintofourparts:Thetopsemicircle:Athebottomsectorwitharclength120degrees:BthetriangleformedbytheradiiofAandthechord:CandthefourpartswhicharethecornersofacircleinscribedinasquareD.TheareaisjustA+B-C+D.AreaofA:21r47rAreaofB:3AreaofC:Radiusof2distanceof1toBCcreates230-60-90trianglessoareaofitis人以♦1/2=v/3AreaofD:4*1—1/4Totalsum:9一《+47+3+3+4=Debraflipsafaircoinrepeatedlykeepingtrackofhowmanyheadsandhowmanytailsshehasseenintotaluntilshegetseithertwoheadsinarowortwotailsinarowatwhichpointshestopsflipping.WhatistheprobabilitythatshegetstwoheadsinarowbutsheseesasecondtailbeforesheseesasecondheadAJB C]D:Eg3b24io12bSolutionWefirstwanttofindoutthesequencesofcoinflipsthatsatisfythisequation.SinceDebraseestwotailsbeforetwoheadsherfirstflipcantbeheadsasthatwouldmeanshewouldeitherendattailsorseetwoheadsbeforesheseestwotails.Thereforeherfirstflipmustbetails.Ifyoucalculatetheshortestwayshecangettwoheadsinarowandseetwotailsbeforesheseestwo11headsitwouldbeTHTHHwhichwouldbe9ror
39.Followingthisshe1canprolonghercoinflippingbyaddinganextraTHwhichisanextra1chance.Sinceshecandothisindefinitelythisisaninfinitegeometric自|Bsequencewhichmeanstheansweris1_7orII.RaashanSylviaandTedplaythefollowinggame.Eachstartswith$
1.Abellringsevery15secondsatwhichtimeeachoftheplayerswhocurrentlyhavemoneysimultaneouslychoosesoneoftheothertwoplayersindependentlyandatrandomandgives$Itothatplayer.Whatistheprobabilitythatafterthebellhasrung2fl9timeseachplayerwillhave$IForexampleRaashanandTedmayeachdecidetogive*ItoSylviaandSylviamaydecidetogiveherherdollartoTedatwhichpointRaashanwillhaveSylviawillhave$2andTedwillhave$1andthatistheendofthefirstroundofplay.InthesecondroundRashaanhasnomoneytogivebutSylviaandTedmightchooseeachothertogivetheir+Itoandtheholdingswillbethesameattheendofthesecondround.A;B1C1D;EI4•5ju5SolutionAftereachbellsringtherearetwosituations:eithertheyeachhaveSleachoroneofthemhas$2anotherhasSIandthethirdhasSo.ineachofthesecasesweneedtocalculatetheprobabilityofreturningtothe1-1-lstate.Eachplayerhas$
1.WLOGletRaashangivehisdollartoSylvia.ThenSylviamustgiveherdollartoTedandTedmustgivehisdollarto111Raashanwhichhappenswith22-Iprobability.Oneplayerhas$2anotherhasSiandthethirdhasSO.WLOGletRaashanhave$2Sylviahave$1andTedhave$〕.ThenRaashanmustgivehisdollartoSylviaandSylviamustgiveherdollartoTedwhichhappenswith11122Iprobability.Sincetheprobabilityofreturningtothe1-1-lstateis1nomatterwhatthesituationistheprobabilitythateachplayerw川have$latterthebellrings2011KimesisPoints.
6.
13.land81211%oncircleuintheplane.Supposethatthetangentlinestomat/land^intersectatapointonthe.r-axis.WhatistheareaofaSolutionFirstobservethatthetwotangentlinesareofidenticallength.ThereforesupposetheintersectionwasUsingPythagoreanTheoremgivesj*=
5.Noticeduetotherightanglesformedbyaradiusanditstangentlinethatthequadrilateralkitedefinedbycirclecenter「1Randacyclicquadrilateral.ThereforewecanusePtolemystheorem:2170T=d•\ZllOwherer/representsthedistancebetweencirclecenterand5I.Therefored—\f\7x.UsingPythagoreanTheoremon
5.0eitheroneoflorRandthecirclecenterwerealizethat170I
17.r71Catwhichpoint8sotheansweris》.n++4r上=7Defineasequencerecursivelyby上=andn”+oforallnonnegativeintegersnLetmbetheleastpositiveintegersuchthatxmV4+—.920InwhichofthefollowingintervalsdoesmlieA
[926]B
[2780]C
[81242]D|243728|E[
729.oo]SolutionWefirstprovethatJ\tIforallnItoyinductionfrom+5]n+4-4xn+6xn-4xn+5provearedecreasingby史+5以+4—小Hn+641n.1estimatethevalueof小+i-by--〈aredecreasingI^arealsodecreasingsowehaveHi910and而一一可Vj一4$加一4廉也leadstoggin10正广=指口工一分—《-rz0-4=-r_...m1101111Theproblem111q111inIn—%1/9In—%1/10Asestimations9and1In2%
0.7wecanestimatethat126nI1iChooseHowmanysequencesofOsandIsoflengthIaretherethatbeginwitha1endwitha0containnotwoconsecutiveIisandcontainnothreeconsecutiveIsA55B60C65D70E75Solution1RecursionWecandeducethatanyvalidsequenceoflength//willstartwitha0followedbyeither10or110”.Becauseofthiswecandefinearecursivefunction:/n=/n-3+/n-2Thisisbecauseforanyvalidsequenceoflengthnyoucanappendeither10orH110andtheresultingsequencewouldstillsatisfythegivenconditions./5=land/6=2soyoufollowtherecursionupuntil/19=65叵-SolutionbyMagentaCobraSolution2CaseworkAfteranygivenzerothenextzeromustappearexactlytwoorthreespotsdowntheline.Andwestartedatposition1andendedatposition19sowemovedover
18.Thereforewemustaddaseriesof2*sand3stoget
18.HowcanwedothisOption1:nine2sthereisonly1waytoarrangethis.Option4:six3sthereisonly1waytoarrangethis.Sumthefournumbersgivenabove:1+28+35+1=6560%ofseniorsdonotplayamusicalinstrument.Ifwedenotexasthenumber3468―+—•500-x500ofseniorsthen510KMM-x+150-Thusthereare500-r—2Qflnon-seniors.Since70%ofthenon-seniors7220・—=[8154playamusicalinstrument1_;2019AMC10BProblem4AlllineswithequationMsuchthatformanarithmeticprogressionpassthroughacommonpoint.WhatarethecoordinatesofthatpointA-12B01C1-2Dl0E12SolutionIfalllinessatisfytheequationthenwecanjustpluginvaluesforabandcthatformanarithmeticprogression.Letsdoa=1b=2c=3anda=1b=3andc=
5.Thenthetwolineswegetare:r*2i/—3/♦3-5jseelimination:y--Plugthisintooneofthepreviouslines..r4-4-3-t-IThustheProblem5TriangleARCX\esinthefirstquadrant.Points儿BandCarereflectedacrossthelineVJtopointsAHandCrespectively.AssumethatnoneoftheverticesofthetrianglelieonthelineU-J.WhichofthefollowingstatementsisnotalwaystrueA.TriangleAfBfC]\esinthefirstquadrant.BjTHangles\Rand.VfiChavethesamearea.CTheslopeoflineAVis—
1.DTheslopesoflinesandf/Carethesame.ELines.4/and\fRareperpendiculartoeachother.SolutionLetsanalyzealloftheoptionsseparately.ClearlyAistruebecauseacoordinateinthefirstquadrantwillhave++anditsinversewouldalsohave++Thetriangleshavethesameareaitsthesametriangle.IfcoordinateAhasxythenitsinversewillhaveyx.x-y/y-x=-1sothisistrue.LikewiseifcoordinateAhasx1y1andAA*hasaslopeof-1thencoordinateBwithx2y2willalsohaveaslopeof-
1.Thisistrue.ByprocessofeliminationthisistheanswerbutifcoordinateAhasx1y1andcoordinateBhasx2y2thentheirinverseswillbey1x1y2x2anditisnotnecessarilytruethaty2-y1/x2-x1=-y2-y1/x2-x
1.Negativeinversesofeachother.ClearlytheanswerisE.2019AMC10BProblem6Thereisarealnsuchthatn+l!+n+2!=〃!•IIC.WhatisthesumofthedigitsofnA3B8C10D11E12Solutionn+ln!+n+2n+1n!=440-n!n![n+1+n+2n+1]=440-n!n+l+n2+3n+2=44h2+4n-437=0-4±v/16+137-1-4±1238za--——*丁n192221+9=|C102019AMC10BProblem7Eachpieceofcandyinastorecostsawholenumberofcents.Casperhasexactlyenoughmoneytobuyeither12piecesofredcandy14piecesofgreencandy15piecesofbluecandyornpiecesofpurplecandy.Apieceofpurplecandycosts20cents.WhatisthesmallestpossiblevalueofnA18B21C24D25E28SolutionIfhehasenoughmoneytobuy12piecesofredcandy14piecesofgreencandyand15piecesofbluecandythentheleastmoneyhecanhaveis11151=
420.Sinceapieceofpurplecandycosts20centsthe420leastvalueofncanbe202019AMC10BProblem8Thefigurebelowshowsasquareandfourequilateraltriangleswitheachtrianglehavingasidelyingonasideofthesquaresuchthateachtrianglehassidelength2andthethirdverticesofthetrianglesmeetatthecenterofthesquare.Theregioninsidethesquarebutoutsidethetrianglesisshaded.Whatistheareaoftheshadedregion[Asymptotediagramneeded]A4B12-l/3C3v/3DE16-\/3SolutionWenoticethatthesquarecanbesplitintoIcongruentsmallersquareswiththealtitudeoftheequilateraltrianglebeingthesideofthesquare.ThereforetheareaofeachshadedpartthatresideswithinasquareisthetotalareaofthesquaresubtractedfromeachtriangleNotethatithasalreadybeensplitinhalf.Whenwesplitanequilateraltriangleinhalfweget^triangleswitha30-6090relationship.Thereforewegetthatthealtitudeandasidelengthofasquareisv.Wecanthencomputetheareaofthetwotrianglesusingthebase-height-arearelationshipandget
2.TheareaofthesmallsquaresisthealtitudesquaredwhichisThereforetheareaoftheshadedregionineachofthefoursquaresis3-\/3Sincetherearefourofthesesquareswemultiplythisby4togetasouranswer.ThisischoiceThefunction/isdefinedby丁=\.xIforallrealnumberswheredenotesthegreatestintegerlessthanorequaltotherealnumberr.Whatistherangeof7A-
1.0BThrmHi4iiociimmiiv**C-
1.01D0ETWofnockiH*taiv««intHtycSolutionThereare4casesweneedtotesthere:xisapositiveinteger.WLOGassumex=
1.Thenfl=1-1=
0.xisapositivefraction.WLOGassumex=
0.
5.Thenf
0.5=0-0=fi.xisanegativeinteger.WLOGassumex=-
1.Thenf-1=1-1=
0.xisanegativefraction.WLOGassumex=-
0.
5.Thenf-
0.5=0-1=-
1.Thustherangeoffunctionfis2019AMC10BProblem10InagivenplanepointslandHarelOunitsapart.HowmanypointsCarethereintheplanesuchthattheperimeterofisA
13.is5iunitsandtheareaofA4/rislOUsquareunitsA0B2C4D8EinfinitelymanySolutionNoticethatwhateverpointwepickforCABwillbethebaseofthetriangle.WLOGpointsAandBare00and010[noticethatforanyothercombinationofpointswecanjustrotatetheplanetobethesamething].WhenwepickpointCwehavetomakesuretheyvalueofCis20becausethatstheonlywaytheareaofthetrianglecanbe
100.WefigurethattheonethingweneedtotesttoseeifthereissuchatriangleiswhentheperimeterisminimizedandthevalueofCisx
20.ThusweputCinthemiddlesopointCis
520.WecaneasilyseethatACandBCwillbothbe\Z2P+52*\/
425.Theperimeterofthisminimizedtriangleis2v125I11whichislargerthan
50.Sincetheminimizedperimeterisgreaterthan50thereisnotrianglethatsatisfiestheconditiongivingusTwojarseachcontainthesamenumberofmarblesandeverymarbleiseitherblueorgreen.InJar1theratioofbluetogreenmarblesis9:1andtheratioofbluetogreenmarblesinJar2is8:
1.Thereare95greenmarblesinall.HowmanymorebluemarblesareinJar1thaninJar2A5B10C25D45E50SolutionCalltheamountofmarblesineachjartbecausetheyareequivalent.ThusJ,/11istheamountofgreenmarblesinIandtheamountofgreenmarblesin
2.工/9+x/10=19j/90j19x/90=95andx=45lmarblesineachjar.Becausethe9/istheamountofbluemarblesinjar1and8/9istheamountofbluemarblesinjar991/1018//9=//Wsotheremustbe5moremarblesinjarIthanjar
2.Theansweris川2019AMC10BProblem12Whatisthegreatestpossiblesumofthedigitsinthebase-sevenrepresentationofapositiveintegerlessthan211flA11B14C22D23E27Solution1Convert201Otobase
7.Thiswillgetyou56137whichwillbetheupperbound.TomaximizethesumofthedigitswewantasmanyGsaspossiblewhichisthehighestvalueinbase7andthiswouldbethenumber
46667.Thus5theansweris4+6+6+6=l^Note:thenumbercanalsobe55667whichwillalsogivetheanswerof22Solution2Notethatallbase7numberswith5digitsormoreisgreaterthan
2019.Sincethefirstanswerthatispossibleusinga4digitnumberis23westartwiththesmallestbase7numberthatdigitsaddsupto
235666.5666inbase10isgreaterthan2017sowecontinuewithtrying4666whichislessthan
2019.C22Sotheansweris_Whatisthesumofallrealnumbers.rforwhichthemedianofthenumbers468Iand.risequaltothemeanofthosefivenumbers、1535A-5B0C5D-E-SolutionThereare3cases:GisthemedianSisthemedianand.risthemedian.Inallx+h.7+fcasesthemeanis
5.Forcase1r——
3.Thisallows6tobethemedianbecausethesetis-
54.
6817.Forcase2r—
5.Thisisimpossiblebecausethesetis
5.6S
17.3535unX=TU--•L.Iu....468—17Forcase
34.Thisisimpossiblebecausethesetis
4.Onlycase1yieldsasolution1=-5sotheanswerisA-
5.2019AMC10BProblem14Thebase-tenrepresentationfor19!is1216T510040Af832tHOOwhereTA/andWdenotedigitsthatarenotgiven.Whatis74-A/4-HA3B8C12D14E17SolutionWecanfigureoutIIObynoticingthat19!willendwithJzeroesastherearethree5sinitsprimefactorization.NextweusethefactthatIOisamu用pieofboth1land
9.SincetheirdivisibilityrulesgivesusthatT4-congruentto3mod9andthatT3/iscongruentto7mod
11.Byinspectionweseethat丁hV=Sjsavalidsolution.Thereforetheansweris4+8+0=12whichisC.TworighttrianglesTandThaveareasof1and2respectively.Onesidelengthofonetriangleiscongruenttoadifferentsidelengthintheotherandanothersidelengthofthefirsttriangleiscongruenttoyetanothersidelengthintheother.Whatistheproductofthethirdsidelengthsof7iandT283234A-B10C-D-E1215jSolutionFirstofallnamethetwosideswhicharecongruenttobe.randwhereyJ.TheonlywaythattheconditionsoftheproblemcanbesatisfiedisifxwastheshorterlegofTLandthelongerlegofandVisthelongerlegofT-iandthehypotenuseofT\.Noticethatthismeansthevaluewearelookingforisthesquareof+*Vy2-l2=vVTwhichisjustyl一工y02铲一炉=/Wehavetwoequations:2and2Thismeansthatrandthatx
21.Takingthesecondequationweget//一=4sosince工y=4/=
12.4425664U=-u==—Sinceweget
123.64-3628u—z=二—Thevaluewearelookingforisjust33sotheansweris回.In△/“witharightangleat「pointDiesintheinteriorofABandpointFliesintheinteriorofRCsothat4c::CDDEEZ,andtheratio4C:DE=4:3WhatistheratioAD:03A2:3B2:\/5Cl:lD3:5/5E3:2SolutionWithoutlossofgeneralityletAC-CD=landDE-EH=
3.Let/A=nand/力90°-Q.AsA4f7JandADEZiareisosceles/.ADC—aand£BDE=0Then/.CDE—180°—c—B=90;so△CDEisa3-4-5trianglewithCE=
5.ThenCB=54-3=8andAABtisa1-2-v5triangle.OnisoscelestrianglesA.1C7and△££/dropaltitudesfromCandEonto1R;denotethefeetofthesealtitudesbyPtandP卜respectively.ThenAPc—PcD——p△ACH、~A/inCbyAAAsimilaritysowegetthatv5AD=2x-r5BD=2xandv5Similarlywegetv5andA2:3。