AMC美国数学竞赛AMCB试题及答案解析
2019年美国数学竞赛(AMC10B)的试题与解答
这时第二个容器里充满了 3 的水. 问小容器与大容器的容积 4
之比是多少?
解: 设第一、二个容器的容积分别为 V1 和 V2, 依题意有
5 6 V1 =
3 4 V2,
从而
V1 V2
=
3 4 5 6
=
9 , 故 (D) 正确. 10
2. Consider the statement, ”If n is not prime, then n − 2 is
√ 由垂线段最短知, m2 + n2
|√am + bn|
=
√ 5.
当且仅当
a2 + b2
OA 与直线 ax + by = 5 垂直即 an = bm 时取最小值.
解法 2 由柯西不等式, (a2 +b2)(m2 +n2) (am+bn)2,
即 5(m2 + n2) 等号成立.
√ 25, m2 + n2
6 ter and the second was empty. She poured all the water from the
first container into the second container, at which point the sec3
ond container was full of water. What is the ratio of the volume 4
解: 反例是指符合命题的条件但不符合命题的结论的例
子, 即满足“若 n 不是素数, 则 n − 2 是素数. ”选项中只有 27
满足, 故 (E) 正确.
评注 运用柯西不等式关键是“配凑”形式, 充分利用已
知条件. 当出现分式和的形式时, 一般构造分式分母和作为
美国数学竞赛AMC题目及答案
2.3.Whatisthevalueof4.5.Hammieisinthegradeandweighs106pounds.Hisquadrupletsistersaretinybabiesandweigh5,5,6,and8pounds.W hichisgreater,theaverage(mean)weightofthesefivechildrenorthemedianweight,andbyhowman ypounds6.Thenumberineachboxbelowistheproductofthenumbersinthetwoboxesthattouchitintherowab ove.Forexample,.Whatisthemissingnumberinthetoprow7.8.Afaircoinistossed3times.Whatistheprobabilityofatleasttwoconsecutiveheads9.TheIncredibleHulkcandoublethedistancehejumpswitheachsucceedingjump.Ifhisfirstjumpis1 meter,thesecondjumpis2meters,thethirdjumpis4meters,andsoon,thenonwhichjumpwillhefirst beabletojumpmorethan1kilometer10.Whatistheratiooftheleastcommonmultipleof180and594tothegreatestcommonfactorof180a nd59411.12.Atthe2013WinnebagoCountyFairavendorisofferinga"fairspecial"onsandals.Ifyoubuyonepai rofsandalsattheregularpriceof$50,yougetasecondpairata40%discount,andathirdpairathalfthe regularprice.Javiertookadvantageofthe"fairspecial"tobuythreepairsofsandals.Whatpercentag eofthe$150regularpricedidhesave13.WhenClaratotaledherscores,sheinadvertentlyreversedtheunitsdigitandthetensdigitofones core.Bywhichofthefollowingmightherincorrectsumhavedifferedfromthecorrectone14.Letthetwodigitsbe and.Thecorrectscorewas.Claramisinterpreteditas.Thedifferencebetweenthetwoiswhichfactorsinto.Therefore,sincethedifferenceisamultipleof9,theonlyanswerchoicethatisamultipleof9 is.15.If,,and,whatistheproductof,,and16.AnumberofstudentsfromFibonacciMiddleSchoolaretakingpartinacommunityserviceproject.Theratioof-gradersto-gradersis,andthetheratioof-gradersto-gradersis .Whatisthesmallestnumberofstudentsthatcouldbeparticipatingintheproject17.Thesumofsixconsecutivepositiveintegersis2013.Whatisthelargestofthesesixintegers18.--Arpanliku16:22,27November2013(EST)Courtesyof19.Bridget,Cassie,andHannaharediscussingtheresultsoftheirlastmathtest.HannahshowsBridg etandCassiehertest,butBridgetandCassiedon'tshowtheirstoanyone.Cassiesays,'Ididn'tgetthel owestscoreinourclass,'andBridgetadds,'Ididn'tgetthehighestscore.'Whatistherankingofthethr eegirlsfromhighesttolowest20.Arectangleisinscribedinasemicirclewithlongersideonthediameter.Whatistheareaofthesemicircle21.22.Toothpicksareusedtomakeagridthatis60toothpickslongand32toothpickswide.Howmanytoo thpicksareusedaltogether23.Angle of isarightangle.Thesidesofarethediametersofsemicirclesasshown.Theareaofthesemicircleon equals,andthearcofthesemicircleon haslength.Whatistheradiusofthesemicircleon24.Squares,,and areequalinarea.Points andarethemidpointsofsides and,respectively.Whatistheratiooftheareaoftheshadedpentagontothesumoftheareasofthethreesquares25.Aballwithdiameter4inchesstartsatpointAtorollalongthetrackshown.Thetrackiscomprisedof3semicirculararcswhoseradiiare inches,inches,and1.2.The50%offpriceofhalfapoundoffishis$3,sothe100%,ortheregularprice,ofahalfpoundoffishis$6.Consequently,ifhalfapoundoffishcosts$6,thenawholepoundoffishis dollars.3.Noticethatwecanpairupeverytwonumberstomakeasumof1:Therefore,theansweris.4.Eachofhersevenfriendspaid tocoverJudi'sportion.Therefore,Judi'sportionmustbe.SinceJudiwassupposedtopay ofthetotalbill,thetotalbillmustbe.5.Themedianhereisobviouslylessthanthemean,sooption(A)and(B)areout. Liningupthenumbers(5,5,6,8,106),weseethatthemedianweightis6pounds.Theaverageweightofthefivekidsis.Therefore,theaverageweightisbigger,by pounds,makingtheanswer.6.Solution1:WorkingBackwardsLetthevalueintheemptyboxinthemiddlerowbe,andthevalueintheemptyboxinthetoprowbe. istheanswerwe'relookingfor.Weseethat,making.Itfollowsthat,so.Solution2:JumpingBacktotheStartAnotherwaytodothisproblemistorealizewhatmakesupthebottommostnumber.Thismethoddoes n'tworkquiteaswellforthisproblem,butinalargertree,itmightbefaster.(Inthiscase,Solution1wou ldbefastersincethere'sonlytwomissingnumbers.)Again,letthevalueintheemptyboxinthemiddlerowbe,andthevalueintheemptyboxinthetoprowbe.istheanswerwe'relookingfor. Wecanwritesomeequations: Nowwecansubstituteintothefirstequationusingthetwoothers:7.IfTreysaw,thenhesaw.2minutesand45secondscanalsobeexpressedas seconds.Trey'srateofseeingcars,,canbemultipliedbyonthetopandbottom(andpreservethesamerate):.Itfollowsthatthemostlikelynumberofcarsis. Solution2minutesand secondsisequalto.SinceTreyprobablycountsaround carsevery seconds,thereare groupsofcarsthatTreymostlikelycounts.Since,theclosestanswerchoiceis.8.First,thereare waystoflipthecoins,inorder. ThewaystogettwoconsecutiveheadsareHHT andTHH. ThewaytogetthreeconsecutiveheadsisHHH.Therefore,theprobabilityofflippingatleasttwoconsecutiveheadsis.9.Thisisageometricsequenceinwhichthecommonratiois2.Tofindthejumpthatwouldbeovera1000meters,wenotethat.However,becausethefirsttermis andnot,thesolutiontotheproblemis10.TofindeithertheLCMortheGCFoftwonumbers,alwaysprimefactorizefirst. Theprimefactorizationof.Theprimefactorizationof.Then,findthegreatestpowerofallthenumbersthereare;ifonenumberisonebutnottheother,useit(t hisis).Multiplyallofthesetoget5940.FortheGCFof180and594,usetheleastpowerofallofthenumbersthatareinbothfactorizationsandmultiply.=18.Thustheanswer==. Westartoffwithasimilarapproachastheoriginalsolution.Fromtheprimefactorizations,theGCFis .Itisawellknownfactthat.Sowehave,.Dividingby yields.Therefore,.11.Weusethatfactthat.Letd=distance,r=rateorspeed,andt=time.Inthiscase,letrepresentthetime.OnMonday,hewasatarateof.So,.ForWednesday,hewalkedatarateof.Therefore,.OnFriday,hewalkedatarateof.So,.Addingupthehoursyields++=.WenowfindtheamountoftimeGrandfatherwouldhavetakenifhewalkedatperday.Setuptheequation,.Tofindtheamountoftimesaved,subtractthetwoamounts:-=.Toconvertthistominutes,wemultiplyby.Thus,thesolutiontothisproblemis12.First,findtheamountofmoneyonewillpayforthreesandalswithoutthediscount.Wehave.Then,findtheamountofmoneyusingthediscount:.Findingthepercentageyields.Tofindthepercentsaved,wehave13.Letthetwodigitsbe and.Thecorrectscorewas.Claramisinterpreteditas.Thedifferencebetweenthetwoiswhichfactorsinto.Therefore,sincethedifferenceisamultipleof9,theonlyanswerchoicethatisamultipleof9is.14.Theprobabilitythatbothshowagreenbeanis.Theprobabilitythatbothshowaredbeanis.Thereforetheprobabilityis15.Therefore,.Therefore,.Tomostpeople,itwouldnotbeimmediatelyevidentthat,sowecanmultiply6'suntilwegetthedesirednumber:,so.Thereforetheansweris.16.Solution1:AlgebraWemultiplythefirstratioby8onbothsides,andthesecondratioby5togetthesamenumberfor8thgra ders,inorderthatwecanputthetworatiostogether:Therefore,theratioof8thgradersto7thgradersto6thgradersis.Sincetheratioisinlowestterms,thesmallestnumberofstudentsparticipatingintheprojectis.Solution2:FakesolvingThenumberof8thgradershastobeamultipleof8and5,soassumeitis40(thesmallestpossibility).Thenthereare6thgradersand7thgraders.Thenumbersofstudentsis17.Solution1Themeanofthesenumbersis.Thereforethenumbersare,sotheanswerisSolution2Letthe numberbe.Thenourdesirednumberis.Ourintegersare,sowehavethat.Solution3Letthefirsttermbe.Ourintegersare.Wehave,18.Solution1Therearecubesonthebaseofthebox.Then,foreachofthe4layersabovethebottom(assinceeachcubeis1foot by1footby1footandtheboxis5feettall,thereare4feetleft),therearecubes.Hence,theansweris.Solution2Wecanjustcalculatethevolumeoftheprismthatwascutoutoftheoriginalbox.Eachinteriorsideofthefortwillbe feetshorterthaneachsideoftheoutside.Sincethefloorisfoot,theheightwillbe feet.Sothevolumeoftheinteriorboxis.Thevolumeoftheoriginalboxis.Therefore,thenumberofblockscontainedinthefortis.19.IfHannahdidbetterthanCassie,therewouldbenowayshecouldknowforsurethatshedidn'tgett helowestscoreintheclass.Therefore,HannahdidworsethanCassie.Similarly,ifHannahdidworseth anBridget,thereisnowayBridgetcouldhaveknownthatshedidn'tgetthehighestintheclass.Therefore,HannahdidbetterthanBridget,soourorderis.20.Asemicirclehassymmetry,sothecenterisexactlyatthemidpointofthe2sideontherectangle,makingtheradius,bythePythagoreanTheorem,.Theareais.21.ThenumberofwaystogetfromSamantha'shousetoCityParkis,andthenumberofwaystogetfromCityParktoschoolis.Sincethere'sonewaytogothroughCityPark(justwalkingstraightthrough),thenumberofdifferentwaystogofromSamantha'shousetoCityParktoschool.22.Thereare verticalcolumnswithalengthof toothpicks,andthereare horizontalrowswithalengthof toothpicks.Aneffectivewaytoverifythisistotryasmallcase,i.e.agridoftoothpicks.Thus,ouransweris.23.Solution1.Solution2WegoasinSolution1,findingthediameterofthecircleonACandAB.Then,anextendedversionofthet heoremsaysthatthesumofthesemicirclesontheleftisequaltothebiggestone,sotheareaofthelargestis,andthemiddleoneis,sotheradiusis.24.Firstlet(where isthesidelengthofthesquares)forsimplicity.Wecanextenduntilithitstheextensionof.Callthispoint.Theareaoftriangle thenis Theareaofrectangle is.Thus,ourdesiredareais.Now,theratiooftheshadedareatothecombinedareaofthethreesquaresis.Solution2Letthesidelengthofeachsquarebe.Lettheintersectionof and be.Since,.Since andareverticalangles,theyarecongruent.Wealsohave bydefinition.Sowehave by congruence.Therefore,.Since and aremidpointsofsides,.Thiscombinedwith yields.Theareaoftrapezoid is.Theareaoftriangle is.Sotheareaofthepentagon is.Theareaofthe squaresis.Therefore,.Solution3Lettheintersectionof and be.Nowwehave and.Becausebothtriangleshasasideoncongruentsquarestherefore.Because and areverticalangles.Alsoboth and arerightanglesso.ThereforebyAAS(Angle,Angle,Side).Thentranslating/rotatingtheshaded intothepositionofSotheshadedareanowcompletelycoversthesquareSettheareaofasquareasTherefore,.25.Solution1Theradiusoftheballis2inches.Ifyouthinkabouttheballrollingordrawapathfortheball(seefigurebelow),youseethatinAandCitloses inches,anditgains inchesonB.So,thedeparturefromthelengthofthetrackmeansthattheansweris.Solution2Thetotallengthofallofthearcsis.Sincewewantthepathfromthecenter,theactualdistancewillbeshorter.Therefore,theonlyanswerchoicelessthan is.Thissolutionmaybeinvalidbecausetheactualdistancecanbelongerifthepaththecent ertravelsisontheoutsideofthecurve,asitisinthemiddlebump.。
amc试题及答案
amc试题及答案1. 问题：已知函数 \( f(x) = 3x^2 - 2x + 5 \)，求 \( f(2) \) 的值。
答案：将 \( x = 2 \) 代入函数 \( f(x) \) 中，得到 \( f(2) = 3(2)^2 - 2(2) + 5 = 12 - 4 + 5 = 13 \)。
2. 问题：解方程 \( 2x - 5 = 9 \)。
答案：首先将方程两边同时加5，得到 \( 2x = 14 \)，然后除以2，得到 \( x = 7 \)。
3. 问题：如果一个圆的直径是10厘米，那么它的半径是多少？答案：圆的半径是直径的一半，所以半径 \( r = \frac{10}{2} = 5 \) 厘米。
4. 问题：计算 \( \sqrt{49} \)。
答案：\( \sqrt{49} = 7 \)。
5. 问题：一个班级有30名学生，其中20名是男生。
这个班级的男生比例是多少？答案：男生比例是 \( \frac{20}{30} = \frac{2}{3} \)。
6. 问题：如果一个三角形的两个内角分别是 \( 45^\circ \) 和\( 60^\circ \)，那么第三个角是多少度？答案：三角形的内角和为 \( 180^\circ \)，所以第三个角是\( 180^\circ - 45^\circ - 60^\circ = 75^\circ \)。
7. 问题：计算 \( 2^3 \)。
答案：\( 2^3 = 2 \times 2 \times 2 = 8 \)。
8. 问题：一个数的平方根是4，这个数是多少？答案：这个数是 \( 4^2 = 16 \)。
9. 问题：如果一个数的两倍加上3等于15，求这个数。
答案：设这个数为 \( x \)，那么 \( 2x + 3 = 15 \)。
解这个方程，得到 \( 2x = 12 \)，所以 \( x = 6 \)。
10. 问题：一个长方形的长是10厘米，宽是5厘米，求它的周长。
美国数学竞赛AMC题目及答案
2.is the value offriends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $ to cover her portion of the total bill. What was the total billis in the grade and weighs 106 pounds. His quadruplet sisters are tiny babies and weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many poundsnumber in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, . What is the missing number in the top rowand his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the trainfair coin is tossed 3 times. What is the probability of at least two consecutive headsIncredible Hulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will he first be able to jump more than 1 kilometeris the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 59411. Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treadmill. How many minutes less12. At the 2013 Winnebago County Fair a vendor is offering a "fair special" on sandals. If you buy one pair of sandals at the regular price of $50, you get a second pair at a 40% discount, and a third pair at half the regular price. Javier took advantage of the "fair special" to buy three pairs of sandals. What percentage of the $150 regular price did he save13.When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one14. Let the two digits be and .The correct score was . Clara misinterpreted it as . The difference between the two is which factors into . Therefore, since the difference is a multiple of 9, the only answer choice that is a multiple of 9 is .15. If , , and , what is the product of , , and16. A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of -graders to -graders is , and the the ratio of -graders to -graders is . What is the smallest number of students that could be participating in the project17. The sum of six consecutive positive integers is 2013. What is the largest of these six integers18. Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain--Arpanliku 16:22, 27 November 2013 (EST) Courtesy of19. Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score in our class,' and Bridget adds, 'I didn't get the highest score.' What is the ranking of the three girls from highest to lowest20. A rectangle is inscribed in a semicircle with longer side on the diameter. What is the area of the semicircle21.Samantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school. If her route is as short as possible, how many different routes can she take22.Toothpicks are used to make a grid that is 60 toothpicks long and 32 toothpicks wide. How many toothpicks are used altogether23.Angle of is a right angle. The sides of are the diameters of semicircles as shown. The area of the semicircle on equals , and the arc of the semicircle on has length . What isthe radius of the semicircle on24. Squares , , and are equal in area. Points and are the midpoints of sidesand , respectively. What is the ratio of the area of the shaded pentagon to the sum of the areas of the three squares25. A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are inches, inches, and inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of theball travels over the course from A to B1.50% off price of half a pound of fish is $3, so the 100%, or the regular price, of a half pound of fish is $6. Consequently, if half a pound of fish costs $6, then a whole pound of fish is dollars.that we can pair up every two numbers to make a sum of 1:Therefore, the answer is .of her seven friends paid to cover Judi's portion. Therefore, Judi's portion must be . Since Judi was supposed to pay of the total bill, the total bill must be .median here is obviously less than the mean, so option (A) and (B) are out.Lining up the numbers (5, 5, 6, 8, 106), we see that the median weight is 6 pounds.The average weight of the five kids is .Therefore, the average weight is bigger, by pounds, making the answer .1: Working BackwardsLet the value in the empty box in the middle row be , and the value in the empty box in the top row be . is the answer we're looking for.We see that , making .It follows that , so .Solution 2: Jumping Back to the StartAnother way to do this problem is to realize what makes up the bottommost number. This method doesn't work quite as well for this problem, but in a larger tree, it might be faster. (In this case, Solution 1 would be faster since there's only two missing numbers.)Again, let the value in the empty box in the middle row be , and the value in the empty box in the top row be . is the answer we're looking for.We can write some equations:Now we can substitute into the first equation using the two others:Trey saw , then he saw .2 minutes and 45 seconds can also be expressed as seconds.Trey's rate of seeing cars, , can be multiplied by on the top and bottom (and preserve the same rate):. It follows that the most likely number of cars is . Solution 2minutes and seconds is equal to .Since Trey probably counts around cars every seconds, there are groups of cars that Trey most likely counts. Since , the closest answer choice is ., there are ways to flip the coins, in order.The ways to get two consecutive heads are HHT and THH.The way to get three consecutive heads is HHH.Therefore, the probability of flipping at least two consecutive heads is .is a geometric sequence in which the common ratio is 2. To find the jump that would be over a 1000 meters, we note that .However, because the first term is and not , the solution to the problem is10. To find either the LCM or the GCF of two numbers, always prime factorize first.The prime factorization of .The prime factorization of .Then, find the greatest power of all the numbers there are; if one number is one but not the other, use it (this is ). Multiply all of these to get 5940.For the GCF of 180 and 594, use the least power of all of the numbers that are in both factorizations and multiply. = 18.Thus the answer = = .We start off with a similar approach as the original solution. From the prime factorizations, the GCF is .It is a well known fact that . So we have,.Dividing by yields .Therefore, .11.We use that fact that . Let d= distance, r= rate or speed, and t=time. In this case, let represent the time.On Monday, he was at a rate of . So, .For Wednesday, he walked at a rate of . Therefore, .On Friday, he walked at a rate of . So, .Adding up the hours yields + + = .We now find the amount of time Grandfather would have taken if he walked at per day. Set up the equation,.To find the amount of time saved, subtract the two amounts: - = . To convert this to minutes, we multiply by .Thus, the solution to this problem is12. First, find the amount of money one will pay for three sandals without the discount. We have.Then, find the amount of money using the discount: .Finding the percentage yields .To find the percent saved, we have13. Let the two digits be and .The correct score was . Clara misinterpreted it as . The difference between the two is which factors into . Therefore, since the difference is a multiple of 9, the only answer choice that is a multiple of 9 is .14. The probability that both show a green bean is . The probability that both show a red bean is . Therefore the probability is15.Therefore, .Therefore, .To most people, it would not be immediately evident that , so we can multiply 6's until we get the desired number:, so .Therefore the answer is .16. Solution 1: AlgebraWe multiply the first ratio by 8 on both sides, and the second ratio by 5 to get the same number for 8th graders, in order that we can put the two ratios together:Therefore, the ratio of 8th graders to 7th graders to 6th graders is . Since the ratio is in lowest terms, the smallest number of students participating in the project is .Solution 2: FakesolvingThe number of 8th graders has to be a multiple of 8 and 5, so assume it is 40 (the smallest possibility). Then there are 6th graders and 7th graders. The numbers of students is17. Solution 1The mean of these numbers is . Therefore the numbers are , so the answer isSolution 2Let the number be . Then our desired number is .Our integers are , so we have that.Solution 3Let the first term be . Our integers are . We have,18. Solution 1There are cubes on the base of the box. Then, for each of the 4 layers above the bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are 4 feet left), there are cubes. Hence, the answer is .Solution 2We can just calculate the volume of the prism that was cut out of the original box. Each interior side of the fort will be feet shorter than each side of the outside. Since the floor is foot, the height will be feet. So the volume of the interior box is .The volume of the original box is . Therefore, the number of blocks contained in the fort is .19. If Hannah did better than Cassie, there would be no way she could know for sure that she didn't get the lowest score in the class. Therefore, Hannah did worse than Cassie. Similarly, if Hannah did worse than Bridget, there is no way Bridget could have known that she didn't get the highest in the class. Therefore, Hannah did better than Bridget, so our order is .20.A semicircle has symmetry, so the center is exactly at the midpoint of the 2 side on the rectangle, making the radius, by the Pythagorean Theorem, . The area is .21.The number of ways to get from Samantha's house to City Park is , and the number of ways to get from City Park to school is . Since there's one way to go through City Park (just walking straight through), the number of different ways to go from Samantha's house to City Park to school .22.There are vertical columns with a length of toothpicks, and there are horizontal rows with a length of toothpicks. An effective way to verify this is to try a small case, . a grid of toothpicks. Thus, our answer is .23. Solution 1If the semicircle on AB were a full circle, the area would be 16pi. Therefore the diameter of the first circle is 8. The arc of the largest semicircle would normally have a complete diameter of 17. The Pythagorean theorem says that the other side has length 15, so the radius is .Solution 2We go as in Solution 1, finding the diameter of the circle on AC and AB. Then, an extended version of the theorem says that the sum of the semicircles on the left is equal to the biggest one, so the area of the largestis , and the middle one is , so the radius is .24.First let (where is the side length of the squares) for simplicity. We can extend until it hits the extension of . Call this point . The area of triangle then is The area of rectangle is . Thus, our desired area is . Now, the ratio of the shaded area to the combinedarea of the three squares is .Solution 2Let the side length of each square be .Let the intersection of and be .Since , . Since and are vertical angles, they are congruent. We also have by definition.So we have by congruence. Therefore, .Since and are midpoints of sides, . This combined with yields.The area of trapezoid is .The area of triangle is .So the area of the pentagon is .The area of the squares is .Therefore, .Solution 3Let the intersection of and be .Now we have and .Because both triangles has a side on congruent squares therefore .Because and are vertical angles .Also both and are right angles so .Therefore by AAS(Angle, Angle, Side) .Then translating/rotating the shaded into the position ofSo the shaded area now completely covers the squareSet the area of a square asTherefore, .25. Solution 1The radius of the ball is 2 inches. If you think about the ball rolling or draw a path for the ball (see figure below), you see that in A and C it loses inches, and it gains inches on B.So, the departure from the length of the track means that the answer is . Solution 2The total length of all of the arcs is . Since we want the path from the center, the actual distance will be shorter. Therefore, the only answer choice less than is . This solution may be invalid because the actual distance can be longer if the path the center travels is on the outside of the curve, as it is in the middle bump.。
AMC美国数学竞赛AMC.B 试题及答案解析
2003A M C10 B 1、Which of the following is the same asSolution2、Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs more than a pink pill, and Al’s pills cost a total of for the two weeks. How much does one green pill costSolution3、The sum of 5 consecutive even integers is less than the sum of the rst consecutive odd counting numbers. What is the smallest of the even integersSolution4、Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the gure. She plants one flower per square foot in each region. Asters cost 1 each, begonias 1.50 each, cannas 2 each, dahlias 2.50 each, and Easter lilies 3 each. What is the least possible cost, in dollars, for her gardenSolution5、Moe uses a mower to cut his rectangular -foot by -foot lawn. The swath he cuts is inches wide, but he overlaps each cut by inches tomake sure that no grass is missed. He walks at the rate of feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow his lawnSolution.6、Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is . The horizontal length of a “-inch” television screen is closest, in inches, to which of the followingSolution7、The symbolism denotes the largest integer not exceeding . For example. , and . ComputeSolution.8、The second and fourth terms of a geometric sequence are and . Which of the following is a possible first termSolution9、Find the value of that satisfies the equationSolution10、Nebraska, the home of the AMC, changed its license plate scheme. Each old license plate consisted of a letter followed by four digits. Each new license plate consists of three letters followed by three digits. By how many times is the number of possible license plates increasedSolution11、A line with slope intersects a line with slope at the point . What is the distance between the -intercepts of these two linesSolution12、Al, Betty, and Clare split among them to be invested in different ways. Each begins with a different amount. At the end of one year they have a total of . Betty and Clare have both doubled their money, whereas Al has managed to lose . What was Al’s original portionSolution.13、Let denote the sum of the digits of the positive integer . For example, and . For how many two-digit values of isSolution14、Given that , where both and are positive integers, find the smallest possible value for .Solution15、There are players in a singles tennis tournament. The tournament is single elimination, meaning that a player who loses a match is eliminated. In the first round, the strongest players are given a bye, and the remaining players are paired off to play. After each round, the remaining players play in the next round. The match continues until only one player remains unbeaten. The total number of matches played isSolution16、A restaurant offers three desserts, and exactly twice as many appetizers as main courses. A dinner consists of an appetizer, a main course, and a dessert. What is the least number of main courses that the restaurant should offer so that a customer could have a different dinner each night in the yearSolution.17、An ice cream cone consists of a sphere of vanilla ice cream and a right circular cone that has the same diameter as the sphere. If the ice cream melts, it will exactly ll the cone. Assume that the melted ice cream occupies of the volume of the frozen ice cream. What is the ratio of the cone’s height to its radiusSolution18、What is the largest integer that is a divisor offor all positive even integersSolution19、Three semicircles of radius are constructed on diameter of a semicircle of radius . The centers of the small semicircles divide into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicirclesSolution20、In rectangle , and . Points and are on so that and . Lines and intersect at . Find the area of .Solution21、A bag contains two red beads and two green beads. You reach into the bag and pull out a bead, replacing it with a red bead regardless of the color you pulled out. What is the probability that all beads in the bag are red after three such replacementsSolution22、A clock chimes once at minutes past each hour and chimes on the hour according to the hour. For example, at 1 PM there is one chime and at noon and midnight there are twelve chimes. Starting at 11:15 AM on February , , on what date will the chime occurSolution23、A regular octagon has an area of one square unit. What is the area of the rectangleSolution24、The rst four terms in an arithmetic sequence are , , , and, in that order. What is the fth termSolution25、How many distinct four-digit numbers are divisible by and have as their last two digitsSolution。
amc 美国数学竞赛 amc 10b 试题及答案解析
2004 AMC 10BProblem 1Each row of the Misty Moon Amphitheater has 33 seats. Rows 12 through 22 are reserved for a youth club. How many seats are reserved for this club?There are rows of seats, giving seats.Problem 2How many two-digit positive integers have at least one 7 as a digit?Ten numbers () have as the tens digit. Nine numbers () have it as the ones digit. Number is in both sets.Thus the result is .Problem 3At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made48 free throws. How many free throws did she make at the first practice?At the fourth practice she made throws, at the third one it was , then we get throws for the second practice, and finally throws at the first one.Problem 4A standard six-sided die is rolled, and P is the product of the five numbers that are visible. What is the largest number that is certain to divide P?Solution 1The product of all six numbers is . The products of numbers that can be visible are , , ..., . The answer to this problem is their greatest common divisor -- which is , where is the least common multiple of . Clearly and the answer is .Solution 2Clearly, can not have a prime factor other than , and .We can not guarantee that the product will be divisible by , as the number can end on the bottom.We can guarantee that the product will be divisible by (one of and will always be visible), but not by .Finally, there are three even numbers, hence two of them are always visible and thus the product is divisible by . This is the most we can guarantee, as when the is on the bottom side, the two visible even numbers are and , and their product is not divisible by .Hence .Problem 5In the expression , the values of , , , and are , , , and , although not necessarily in that order. What is the maximum possible value of the result?If or , the expression evaluates to .If , the expression evaluates to .Case remains.In that case, we want to maximize where .Trying out the six possibilities we get that the best one is, where .Problem 6Which of the following numbers is a perfect square?Using the fact that , we can write:▪▪▪▪▪Clearly is a square, and as , , and are primes, none of the other four are squares.Problem 7On a trip from the United States to Canada, Isabella took U.S. dollars. At the border she exchanged them all, receiving Canadian dollars for every U.S. dollars. After spending Canadian dollars, she had Canadian dollars left. What is the sum of the digits of ?Solution 1Isabella had Canadian dollars. Setting up an equation we get, which solves to , and the sum of digits of isSolution 2Each time Isabelle exchanges U.S. dollars, she gets Canadian dollars and Canadian dollars extra. Isabelle received a total of Canadian dollars extra, therefore she exchanged U.S. dollars times. Thus .Problem 8Minneapolis-St. Paul International Airport is 8 miles southwest of downtown St. Paul and 10 miles southeast of downtown Minneapolis. Which of the following is closest to the number of miles between downtown St. Paul and downtown Minneapolis?The directions "southwest" and "southeast" are orthogonal. Thus the described situation is a right triangle with legs 8 miles and 10 mileslong. The hypotenuse length is , and thus the answer is .Without a calculator one can note that . Problem 9A square has sides of length 10, and a circle centered at one of its vertices has radius 10. What is the area of the union of the regions enclosed by the square and the circle?The area of the circle is , the area of the square is . Exactly of the circle lies inside the square. Thus the total area is.Problem 10A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains cans, how many rows does it contain?The sum of the first odd numbers is . As in our case , we have .Problem 11Two eight-sided dice each have faces numbered 1 through 8. When the dice are rolled, each face has an equal probability of appearing on the top. What is the probability that the product of the two top numbers is greater than their sum?Solution 1We have , hence if at least one of the numbers is , the sum is larger. There such possibilities.We have .For we already have , hence all other cases are good.Out of the possible cases, we found that in the sum is greater than or equal to the product, hence in it is smaller. Therefore the answer is .Solution 2Let the two rolls be , and .From the restriction:Since and are non-negative integers between and , either , , orif and only if or .There are ordered pairs with , ordered pairs with , and ordered pair with and . So, there areordered pairs such that .if and only if and or equivalently and . This gives ordered pair .So, there are a total of ordered pairs with .Since there are a total of ordered pairs , there are ordered pairs with .Thus, the desired probability is .Problem 12An annulus is the region between two concentric circles. The concentric circles in the figure have radii and , with . Let be a radius of the larger circle, let be tangent to the smaller circle at , and let be the radius of the larger circle that contains . Let , , and . What is the area of the annulus?The area of the large circle is , the area of the small one is , hence the shaded area is .From the for the right triangle we have , hence and thus the shaded area is .Problem 13In the United States, coins have the following thicknesses: penny, mm; nickel, mm; dime, mm; quarter, mm. If a stack of these coins is exactly mm high, how many coins are in the stack?All numbers in this solution will be in hundreds of a millimeter.The thinnest coin is the dime, with thickness . A stack of dimes has height .The other three coin types have thicknesses , , and . By replacing some of the dimes in our stack by other, thicker coins, we can clearly create exactly all heights in the set.If we take an odd , then all the possible heights will be odd, and thus none of them will be . Hence is even.If the stack will be too low and if it will be too high. Thus we are left with cases and .If the possible stack heights are , with the remaining ones exceeding .Therefore there are coins in the stack.Using the above observation we can easily construct such a stack. A stack of dimes would have height , thus we need to add . This can be done for example by replacing five dimes by nickels (for ), and one dime by a penny (for ).Problem 14A bag initially contains red marbles and blue marbles only, with moreblue than red. Red marbles are added to the bag until only of the marbles in the bag are blue. Then yellow marbles are added to the baguntil only of the marbles in the bag are blue. Finally, the number of blue marbles in the bag is doubled. What fraction of the marbles now in the bag are blue?We can ignore most of the problem statement. The only important information is that immediately before the last step blue marbles formed of the marbles in the bag. This means that there were blue and other marbles, for some . When we double the number of blue marbles, there will be blue and other marbles, hence bluemarbles now form of all marbles in the bag.Problem 15Patty has coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have cents more. How much are her coins worth?Solution 1She has nickels and dimes. Their total cost iscents. If the dimes were nickels and vice versa, she would havecents. This value should be cents more than the previous one. We get , which solves to . Her coins are worth .Solution 2Changing a nickel into a dime increases the sum by cents, and changing a dime into a nickel decreases it by the same amount. As the sum increased by cents, there are more nickels than dimes. As the total count is , this means that there are nickels and dimes.Problem 16Three circles of radius are externally tangent to each other and internally tangent to a larger circle. What is the radius of the large circle?The situation in shown in the picture below. The radius we seek is . Clearly . The point is clearly the center of the equilateral triangle , thus is of the altitude of this triangle. We get that . Therefore the radius we seek is.WARNING. Note that the answer does not correspond to any of the five options. Most probably there is a typo in option D.Problem 17The two digits in Jack's age are the same as the digits in Bill's age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?Solution 1If Jack's current age is , then Bill's current age is .In five years, Jack's age will be and Bill's age will be .We are given that . Thus .For we get . For and the value is not an integer, and for it is more than . Thus the only solution is , and the difference in ages is .Solution 2Age difference does not change in time. Thus in five years Bill's age will be equal to their age difference.The age difference is , hence it is a multiple of . Thus Bill's current age modulo must be .Thus Bill's age is in the set .As Jack is older, we only need to consider the cases where the tens digit of Bill's age is smaller than the ones digit. This leaves us with the options .Checking each of them, we see that only works, and gives the solution .Problem 18In the right triangle , we have , , and . Points , , and are located on , , and , respectively, so that , , and . What is the ratio of the area of to that of ?First of all, note that , and therefore.Draw the height from onto as in the picture below:Now consider the area of . Clearly the triangles and are similar, as they have all angles equal. Their ratio is ,hence . Now the area of can be computed as= . Similarly we can find that as well.Hence , and the answer is .Problem 19In the sequence , , , , each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is. What is the term in this sequence?Solution 1We already know that , , , and . Let's compute the next few terms to get the idea how the sequence behaves. We get ,, , and so on.We can now discover the following pattern: and . This is easily proved by induction. It follows that.Solution 2Note that the recurrence can be rewritten as.Hence we get that and alsoFrom the values given in the problem statement we see that .From we get that .From we get that .Following this pattern, we get.Problem 20In points and lie on and , respectively. If and intersect at so that and , what is ?Solution (Triangle Areas)We use the square bracket notation to denote area.Without loss of generality, we can assume . Then , and . We have , so we need to find the area of quadrilateral .Draw the line segment to form the two triangles and . Let , and . By considering trianglesand , we obtain , and by considering triangles and , we obtain . Solving, we get , , so the area of quadrilateral is .ThereforeSolution (Mass points)The presence of only ratios in the problem essentially cries out for mass points.As per the problem, we assign a mass of to point , and a mass of to . Then, to balance and on , has a mass of .Now, were we to assign a mass of to and a mass of to , we'd have . Scaling this down by (to get , which puts and in terms of the masses of and ), we assign a mass of to and a mass of to .Now, to balance and on , we must give a mass of . Finally, the ratio of to is given by the ratio of the mass of tothe mass of , which is .Solution (Coordinates)Affine transformations preserve ratios of distances, and for any pair of triangles there is an affine transformation that maps the first one onto the second one. This is why the answer is the same for any , and we just need to compute it for any single triangle.We can choose the points , , and . This way we will have , and . The situation is shown in the picture below:The point is the intersection of the lines and . The points on the first line have the form , the points on the second line have the form . Solving for we get , hence.The ratio can now be computed simply by observing the coordinates of , , and :Problem 21Let ; ; and ; ; be two arithmetic progressions. The set is the union of the first terms of each sequence. How many distinct numbers are in ?The two sets of terms are and.Now . We can compute. We will now find .Consider the numbers in . We want to find out how many of them lie in . In other words, we need to find out the number of valid values of for which .The fact "" can be rewritten as ", and ".The first condition gives , the second one gives .Thus the good values of are , and their count is .Therefore , and thus .Problem 22A triangle with sides of 5, 12, and 13 has both an inscribed and a circumscribed circle. What is the distance between the centers of those circles?This is obviously a right triangle. Pick a coordinate system so that the right angle is at and the other two vertices are at and .As this is a right triangle, the center of the circumcircle is in the middle of the hypotenuse, at .The radius of the inscribed circle can be computed using the well-known identity , where is the area of the triangle and its perimeter. In our case, and , thus . As the inscribed circle touches both legs, its center must be at .The distance of these two points is then.Problem 23Each face of a cube is painted either red or blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?Label the six sides of the cube by numbers to as on a classic dice. Then the "four vertical faces" can be: , , or .Let be the set of colorings where are all of the same color, similarly let and be the sets of good colorings for the other two sets of faces.There are possible colorings, and there are goodcolorings. Thus the result is . We need to compute .Using the we can writeClearly , as we have two possibilities for the common color of the four vertical faces, and two possibilities for each of the horizontal faces.What is ? The faces must have the same color, and at the same time faces must have the same color. It turns out thatthe set containing just the two cubes where all six faces have the same color.Therefore , and the result is .Problem 24In we have , , and . Point is on the circumscribed circle of the triangle so that bisects . What is the value of ?Problem 25A circle of radius is internally tangent to two circles of radius at points and , where is a diameter of the smaller circle. What is the area of the region, shaded in the picture, that is outside the smaller circle and inside each of the two larger circles?The area of the small circle is . We can add it to the shaded region, compute the area of the new region, and then subtract the area of the small circle from the result.Let and be the intersections of the two large circles. Connect them to and to get the picture below:Now obviously the triangles and are equilateral with side .Take a look at the bottom circle. The angle is , hence the sector is of the circle. The same is true for the sector of the bottom circle, and sectors and of the top circle.If we now sum the areas of these four sectors, we will almost get the area of the new shaded region - except that each of the two equilateral triangles will be counted twice.Hence the area of the new shaded region is, and the area of the original shared region is .。
amc 美国数学竞赛 00 amc 0b 试题及答案解析
2003 AMC 10B1、Which of the following is the same asSolution2、Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs more than a pinkpill, and Al’s pills cost a total of for the two weeks. How much doesone green pill cost?Solution3、The sum of 5 consecutive even integers is less than the sum of thefirst consecutive odd counting numbers. What is the smallest of theeven integers?Solution4、Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the figure. She plants one flower per square foot in each region. Asters cost 1 each,begonias 1.50 each, cannas 2 each, dahlias 2.50 each, and Easterlilies 3 each. What is the least possible cost, in dollars, for hergarden?Solution5、Moe uses a mower to cut his rectangular -foot by -foot lawn.The swath he cuts is inches wide, but he overlaps each cut byinches to make sure that no grass is missed. He walks at the rate of feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow his lawn?Solution.6、Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is . The horizontal length ofa “-inch” television screen is closest, in inches, to which of thefollowing?Solution7、The symbolism denotes the largest integer not exceeding . Forexample. , and . ComputeSolution.8、The second and fourth terms of a geometric sequence are and .Which of the following is a possible first term?Solution9、Find the value of that satisfies the equationSolution10、Nebraska, the home of the AMC, changed its license plate scheme. Each old license plate consisted of a letter followed by four digits. Each new license plate consists of three letters followed by three digits. By how many times is the number of possible license plates increased?Solution11、A line with slope intersects a line with slope at the point .What is the distance between the -intercepts of these two lines?Solution12、Al, Betty, and Clare split among them to be invested indifferent ways. Each begins with a different amount. At the end of one year they have a total of . Betty and Clare have both doubledtheir money, whereas Al has managed to lose . What was Al’soriginal portion?Solution.13、Let denote the sum of the digits of the positive integer . Forexample, and . For how many two-digitvalues of is ?Solution14、Given that , where both and are positive integers,find the smallest possible value for .Solution15、There are players in a singles tennis tournament. Thetournament is single elimination, meaning that a player who loses a match is eliminated. In the first round, the strongest players aregiven a bye, and the remaining players are paired off to play. Aftereach round, the remaining players play in the next round. The match continues until only one player remains unbeaten. The total number of matches played isSolution16、A restaurant offers three desserts, and exactly twice as many appetizers as main courses. A dinner consists of an appetizer, a main course, and a dessert. What is the least number of main courses that the restaurant should offer so that a customer could have a different dinner each night in the year ?Solution.17、An ice cream cone consists of a sphere of vanilla ice cream and a right circular cone that has the same diameter as the sphere. If the ice cream melts, it will exactly fill the cone. Assume that the melted ice cream occupies of the volume of the frozen ice cream. What is theratio of the cone’s height to its radius?Solution18、What is the largest integer that is a divisor offor all positive even integers ?Solution19、Three semicircles of radius are constructed on diameter of asemicircle of radius . The centers of the small semicircles divideinto four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside thesmaller semicircles?Solution20、In rectangle , and . Points and are onso that and . Lines and intersect at . Findthe area of .Solution21、A bag contains two red beads and two green beads. You reach into the bag and pull out a bead, replacing it with a red bead regardless of the color you pulled out. What is the probability that all beads in the bag are red after three such replacements?22、A clock chimes once at minutes past each hour and chimes onthe hour according to the hour. For example, at 1 PM there is one chime and at noon and midnight there are twelve chimes. Starting at11:15 AM on February , , on what date will the chimeoccur?Solution23、A regular octagon has an area of one square unit.What is the area of the rectangle ?Solution24、The first four terms in an arithmetic sequence are , , ,and , in that order. What is the fifth term?Solution25、How many distinct four-digit numbers are divisible by and haveas their last two digits?Solution。
2019年AMC10B美国数学竞赛(真题加详解)
2019 AMC 10B Problems/Problem 1The following problem is from both the 2019 AMC 10B #1 and 2019 AMC 12B #1, so both problems redirect to this page.ProblemAlicia had two containers. The first was full of water and the second was empty. She poured all the water from the first container into the secondcontainer, at which point the second container was full of water. What is the ratio of the volume of the first container to the volume of the secondcontainer?Solution 1Let the first jar's volume be and the second's be . It is giventhat . We find thatWe already know that this is the ratio of the smaller to the larger volumebecause it is less thanSolution 2We can set up a ratio to solve this problem. If is the volume of the firstcontainer, and is the volume of the second container, then:Cross-multiplying allows us to get . Thus the ratio of the volume of the first container to the second containeris .~IronicNinjaSolution 3An alternate solution is to plug in some maximum volume for the firstcontainer - let's say , so there was a volume of in the first container, and then the second container also has a volume of , so youget . Thus the answer is .2019 AMC 10B Problems/Problem 2The following problem is from both the 2019 AMC 10B #2 and 2019 AMC 12B #2, so both problems redirect to this page.ProblemConsider the statement, "If is not prime, then is prime." Which of the following values of is a counterexample to this statement?SolutionSince a counterexample must be value of which is not prime, must be composite, so we eliminate and . Now we subtract from the remaining answer choices, and we see that the only time is not prime iswhen .2019 AMC 10B Problems/Problem 3 ProblemIn a high school with students, of the seniors play a musical instrument, while of the non-seniors do not play a musical instrument. Inall, of the students do not play a musical instrument. How many non-seniors play a musical instrument?Solution 1of seniors do not play a musical instrument. If we denote as the numberof seniors, thenThus there are non-seniors. Since 70% of the non-seniorsplay a musical instrument, .~IronicNinjaSolution 2Let be the number of seniors, and be the number of non-seniors.ThenMultiplying both sides by gives usAlso, because there are 500 students in total.Solving these system of equations give us , .Since of the non-seniors play a musical instrument, the answer issimply of , which gives us .Solution 3 (using the answer choices)We can clearly deduce that of the non-seniors do play an instrument, but,since the total percentage of instrument players is , the non-senior population is quite low. By intuition, we can therefore see that the answer isaround or . Testing both of these gives us the answer . 2019 AMC 10B Problems/Problem 4 ProblemAll lines with equation such that form an arithmeticprogression pass through a common point. What are the coordinates of that point?Solution 1If all lines satisfy the condition, then we can just plug in values for , ,and that form an arithmetic progression. Let's use , , ,and , , . Then the two lines we get are:Use elimination to deduce and plug this into one of the previous line equations. We get Thus the common point is .~IronicNinjaSolution 2We know that , , and form an arithmetic progression, so if the commondifference is , we can say Now wehave , and expandinggives Factoringgives . Since this must always be true (regardless of the values of and ), we musthave and , so and the common point is .2019 AMC 10B Problems/Problem 5 ProblemTriangle lies in the first quadrant. Points , , and are reflected across the line to points , , and , respectively. Assume that none of the vertices of the triangle lie on the line . Which of the following statements is not always true?Triangle lies in the first quadrant.Triangles and have the same area.The slope of line is .The slopes of lines and are the same.Lines and are perpendicular to each other.SolutionLet's analyze all of the options separately.: Clearly is true, because a point in the first quadrant will have non-negative - and -coordinates, and so its reflection, with the coordinates swapped, will also have non-negative - and -coordinates.: The triangles have the same area,since and are the same triangle (congruent). More formally, we can say that area is invariant under reflection.: If point has coordinates , then will have coordinates .The gradient is thus , so this is true. (We know since the question states that none of the points , , or lies on the line , so there is no risk of division by zero).: Repeating the argument for , we see that both lines have slope , so this is also true.: By process of elimination, this must now be the answer. Indeed, ifpoint has coordinates and point has coordinates ,then and will, respectively, have coordinates and . The product of the gradientsof and is , so in fact these lines are never perpendicular to each other (using the "negative reciprocal" condition for perpendicularity).Thus the answer is .CounterexamplesIf and , then the slopeof , , is , while the slope of , ,is . is the reciprocal of , but it is not the negative reciprocal of . To generalize, let denote thecoordinates of point , let denote the coordinates of point ,let denote the slope of segment , and let denote the slope of segment . Then, the coordinates of are , andof are . Then, ,and .If and , , and in these cases, the condition is false.2019 AMC 10B Problems/Problem 6The following problem is from both the 2019 AMC 10B #6 and 2019 AMC 12B #4, so both problems redirect to this page.ProblemThere is a real such that .What is the sum of the digits of ?Solution 1Solving by the quadraticformula,(since clearly ). The answer is therefore .~IronicNinjaSolution 2Dividing both sidesby givesSince is non-negative, . The answer is .Solution 3Dividing both sides by as beforegives . Now factorout , giving . By considering the prime factorization of , a bit of experimentation givesus and , so , so the answeris .2019 AMC 10B Problems/Problem 7The following problem is from both the 2019 AMC 10B #7 and 2019 AMC 12B #5, so both problems redirect to this page.ProblemEach piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either pieces of red candy, pieces of green candy, pieces of blue candy, or pieces of purple candy. A piece of purple candy costs cents. What is the smallest possible value of ?Solution 1If he has enough money to buy pieces of red candy, pieces of green candy, and pieces of blue candy, then the smallest amount of money hecould have is cents. Since a piece of purple candy costs cents, the smallest possible valueof is .~IronicNinjaSolution 2We simply need to find a value of that is divisible by , , and .Observe that is divisible by and , but not . is divisible by , , and , meaning that we have exact change (in this case, cents) to buy each type of candy, so the minimum valueof is .2019 AMC 10B Problems/Problem 8ProblemThe figure below shows a square and four equilateral triangles, with each triangle having a side lying on a side of the square, such that each triangle has side length and the third vertices of the triangles meet at the center of the square. The region inside the square but outside the triangles is shaded. What is the area of the shaded region?Solution 1We notice that the square can be split into congruent smaller squares, with the altitude of the equilateral triangle being the side of this smaller square. Therefore, the area of each shaded part that resides within a square is the total area of the square subtracted from each triangle (which has already been split in half). When we split an equilateral triangle in half, we gettwo triangles. Therefore, the altitude, which is also the side length of one of the smaller squares, is . We can then compute the areaof the two triangles as .The area of the each small squares is the square of the side length,i.e. . Therefore, the area of the shaded region in each of the four squares is .Since there are of these squares, we multiply this by toget as our answer.Solution 2We can see that the side length of the square is by considering thealtitude of the equilateral triangle as in Solution 1. Using the Pythagorean Theorem, the diagonal of the square isthus . Because of this, the height of one ofthe four shaded kites is . Now, we just need to find the length of that kite. By the Pythagorean Theorem again, this lengthis . Nowusing , the area of one of the four kitesis . 2019 AMC 10B Problems/Problem 9 ProblemThe function is defined by for all real numbers , where denotes the greatest integer less than or equal to the real number . What is the range of ?Solution 1There are four cases we need to consider here.Case 1: is a positive integer. Without loss of generality, assume . Then .Case 2: is a positive fraction. Without loss of generality, assume .Then .Case 3: is a negative integer. Without loss of generality, assume . Then .Case 4: is a negative fraction. Without loss of generality, assume . Then .Thus the range of the function is .~IronicNinja, edited by someone elseSolution 2It is easily verified that when is an integer, is zero. We therefore need only to consider the case when is not an integer.When is positive, , soWhen is negative, let be composed of integer part and fractional part (both ):Thus, the range of f is .Note: One could solve the case of as a negative non-integer in thisway:2019 AMC 10B Problems/Problem 10The following problem is from both the 2019 AMC 10B #10 and 2019 AMC 12B #6, so both problems redirect to this page.ProblemIn a given plane, points and are units apart. How manypoints are there in the plane such that the perimeterof is units and the area of is square units?Solution 1Notice that whatever point we pick for , will be the base of thetriangle. Without loss of generality, letpoints and be and , since for any other combination of points, we can just rotate the plane to makethem and under a new coordinate system. When we pickpoint , we have to make sure that its -coordinate is , because that's the only way the area of the triangle can be .Now when the perimeter is minimized, by symmetry, we put in the middle, at . We can easily see that and will bothbe . The perimeter of this minimal triangleis , which is larger than . Since the minimum perimeter is greater than , there is no triangle that satisfies the condition, givingus .~IronicNinjaSolution 2Without loss of generality, let be a horizontal segment of length .Now realize that has to lie on one of the lines parallel to andvertically units away from it. But is already 50, andthis doesn't form a triangle. Otherwise, without loss ofgenerality, . Dropping altitude , we have a righttriangle with hypotenuse and leg , which is clearly impossible, again giving the answer as .2019 AMC 10B Problems/Problem 11 ProblemTwo jars each contain the same number of marbles, and every marble is either blue or green. In Jar the ratio of blue to green marbles is , and the ratio of blue to green marbles in Jar is . There are green marbles in all. How many more blue marbles are in Jar than in Jar ?SolutionCall the number of marbles in each jar (because the problem specifies that they each contain the same number). Thus, is the number of green marbles in Jar , and is the number of green marbles in Jar .Since , we have , so thereare marbles in each jar.Because is the number of blue marbles in Jar , and is the number of blue marbles in Jar , there are more marbles in Jar than Jar . This means the answer is .2019 AMC 10B Problems/Problem 12 ProblemWhat is the greatest possible sum of the digits in the base-seven representation of a positive integer less than ?Solution 1Observe that . To maximize the sum of the digits, we want as many s as possible (since is the highest value in base ), and this will occur with either of the numbers or . Thus, the answeris .~IronicNinja, edited by some peopleNote: the number can also be , which will also give the answer of . Solution 2Note that all base numbers with or more digits are in fact greaterthan . Since the first answer that is possible using a digit number is , we start with the smallest base number that whose digits sum to ,namely . But this is greater than , so we continue bytrying , which is less than 2019. So the answer is .2019 AMC 10B Problems/Problem 13The following problem is from both the 2019 AMC 10B #13 and 2019 AMC 12B #7, so both problems redirect to this page.ProblemWhat is the sum of all real numbers for which the median of thenumbers and is equal to the mean of those five numbers?SolutionThe mean is .There are three possibilities for the median: it is either , , or .Let's start with .has solution , and the sequenceis , which does have median , so this is a valid solution.Now let the median be .gives , so the sequence is , which has median , so this is not valid.Finally we let the median be ., and the sequence is , which has median . This case is therefore again not valid.Hence the only possible value of is2019 AMC 10B Problems/Problem 14 ProblemThe base-ten representationfor is , where , ,and denote digits that are not given. What is ?Solution 1We can figure out by noticing that will end with zeroes, as there are three s in its prime factorization. Next, we use the fact that is a multiple of both and . Their divisibility rules (see Solution 2) tell usthat and that . By inspection, we see that is a valid solution. Therefore the answer is .Solution 2 (similar to Solution 1)We know that and are both factors of . Furthermore, we knowthat , because ends in three zeroes (see Solution 1). We can simply use the divisibility rules for and for this problem to find and . For to be divisible by , the sum of digits must simply be divisible by .Summing the digits, we get that must be divisible by . Thisleaves either or as our answer choice. Now we test for divisibility by . For a number to be divisible by , the alternating sum must be divisibleby (for example, with the number , ,so is divisible by ). Applying the alternating sum test to this problem, we see that must be divisible by 11. By inspection, we can see that this holds if and . The sumis .2019 AMC 10B Problems/Problem 15ProblemRight triangles and , have areas of 1 and 2, respectively. A side of iscongruent to a side of , and a different side of is congruent to a differentside of . What is the square of the product of the lengths of the other (third)side of and ?Solution 1First of all, let the two sides which are congruent be and , where . The only way that the conditions of the problem can be satisfied is if is the shorter leg of and the longer leg of , and is the longer leg of and thehypotenuse of .Notice that this means the value we are looking for is the squareof , which is just . The area conditions give us twoequations: and .This means that and that .Taking the second equation, we get , sosince , .Since , we get .The value we are looking for is just so the answer is .Solution bySolution 2Like in Solution 1, we have and .Squaring both equations yields and .Let and . Then ,and , so .We are looking for the value of , so the answeris .Solution 3Firstly, let the right triangles be and ,with being the smaller triangle. As in Solution 1,let and . Additionally,let and .We are given that and , sousing , we have and . Dividing the twoequations, we get = , so .Thus is a right triangle, meaningthat . Now by the Pythagorean Theoremin ,.The problem requires the square of the product of the third side lengths of each triangle, which is . By substitution, we seethat = . We alsoknow.Since we want , multiplying both sides by getsus . Now squaringgives .2019 AMC 10B Problems/Problem 16ProblemIn with a right angle at , point lies in the interior of andpoint lies in the interior of so that and the ratio . What is the ratioSolution 1Without loss of generality, let and . Let and .As and areisosceles, and .Then , so isa triangle with .Then , and is a triangle.In isosceles triangles and , drop altitudesfrom and onto ; denote the feet of these altitudesby and respectively. Then by AAA similarity, so we get that ,and . Similarly we get ,and .Solution 2Let , and . (For thissolution, is above , and is to the right of ). Also let ,so , whichimplies . Similarly, , whichimplies . This further impliesthat .Now we seethat. Thus is a right triangle, with side lengths of , , and (by the Pythagorean Theorem, or simply the Pythagorean triple ).Therefore (by definition), ,and . Hence (by thedouble angle formula), giving .By the Law of Cosines in , if , wehaveNow . Thus theanswer is .~IronicNinjaSolution 3Draw a nice big diagram and measure. The answers to this problem are not very close, so it is quite easy to get to the correct answer by simply drawing a diagram. (Note: this strategy should only be used as a last resort!)2019 AMC 12B Problems/Problem 13(Redirected from 2019 AMC 10B Problems/Problem 17)The following problem is from both the 2019 AMC 10B #17 and 2019 AMC 12B #13, so both problems redirect to this page.ProblemA red ball and a green ball are randomly and independently tossed into binsnumbered with the positive integers so that for each ball, the probability that it is tossed into bin is for What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?Solution 1By symmetry, the probability of the red ball landing in a higher-numbered bin is the same as the probability of the green ball landing in a higher-numbered bin. Clearly, the probability of both landing in the same binis (by the geometric series sum formula). Therefore the other two probabilities have to bothbe .Solution 2Suppose the green ball goes in bin , for some . The probability of this occurring is . Given that this occurs, the probability that the red ball goesin a higher-numbered bin is (by thegeometric series sum formula). Thus the probability that the green ball goesin bin , and the red ball goes in a bin greater than , is . Summing from to infinity, we getwhere we again used the geometric series sum formula. (Alternatively, if this sum equals , then by writing out the terms andmultiplying both sides by , we see , which gives .) Solution 3The probability that the two balls will go into adjacent binsisby the geometric series sum formula. Similarly, the probability that the two balls will go into bins that have a distance of from each otheris(again recognizing a geometric series). We can see that each time we add a bin between the two balls, the probability halves. Thus, our answeris , which, by the geometric series sum formula,is .-fidgetboss_4000Solution 4 (quick, conceptual)Define a win as a ball appearing in higher numbered box.Start from the first box.There are possible results in the box: Red, Green, Red and Green, ornone, with an equal probability of for each. If none of the balls is in the first box, the game restarts at the second box with the same kind of probability distribution, so if is the probability that Red wins, we canwrite : there is a probability that "Red" wins immediately,a probability in the cases "Green" or "Red and Green", and in the "None"case (which occurs with probability), we then start again, giving the sameprobability . Hence, solving the equation, we get . Solution 5Write out the infinite geometric series as , . To find the probablilty that red goes in a higher-numbered bin than green, we can simply remove all odd-index terms (i.e term , term , etc.), and then sum theremaining terms - this is in fact precisely equivalent to the method of Solution2. Writing this out as another infinite geometric sequence, we are leftwith . Summing, we get2019 AMC 10B Problems/Problem 18 ProblemHenry decides one morning to do a workout, and he walks of the way from his home to his gym. The gym is kilometers away from Henry's home. At thatpoint, he changes his mind and walks of the way from where he is back toward home. When he reaches that point, he changes his mind again and walks ofthe distance from there back toward the gym. If Henry keeps changing his mindwhen he has walked of the distance toward either the gym or home from the point where he last changed his mind, he will get very close to walking back and forth between a point kilometers from home and a point kilometers fromhome. What is ?Solution 1Let the two points that Henry walks in between be and , with being closer to home. As given in the problem statement, the distances of thepoints and from his home are and respectively. By symmetry, the distance of point from the gym is the same as the distance from home to point . Thus, . In addition, when he walks from point to home, he walks of the distance, ending at point . Therefore, we knowthat . By substituting, we get .Adding these equations now gives . Multiplying by , we get ,so .Solution 2 (not rigorous)We assume that Henry is walking back and forth exactly betweenpoints and , with closer to Henry's home than . Denote Henry's home as a point and the gym as a point .Then and ,so .Therefore,. 2019 AMC 10B Problems/Problem 19The following problem is from both the 2019 AMC 10B #19 and 2019 AMC 12B #14, so both problems redirect to this page.ProblemLet be the set of all positive integer divisors of How many numbers are the product of two distinct elements ofSolutionThe prime factorization of is . Thus, we choose twonumbers and where and, whose product is ,where and .Notice that this is analogous to choosing a divisorof , whichhas divisors. However, some of thedivisors of cannot be written as a product of two distinct divisorsof , namely: , , , and . The last twocannot be so written because the maximum factor of containingonly s or s (and not both) is only or . Since the factors chosen must be distinct, the last two numbers cannot be so written because they would require or . This gives candidatenumbers. It is not too hard to show that every number of the form ,where , and are not both or , can be written asa product of two distinct elements in . Hence the answer is . 2019 AMC 10B Problems/Problem 20The following problem is from both the 2019 AMC 10B #20 and 2019 AMC 12B #15, so both problems redirect to this page.ProblemAs shown in the figure, line segment is trisected bypoints and so that Three semicirclesof radius and have their diameters on and are tangent to line at and respectively. A circle ofradius has its center on The area of the region inside the circle butoutside the three semicircles, shaded in the figure, can be expressed in the form where and are positive integersand and are relatively prime. What is ?SolutionDivide the circle into four parts: the top semicircle (); the bottom sector (), whose arc angle is because the large circle's radius is and the short length (the radius of the smaller semicircles) is , givinga triangle; the triangle formed by the radii of and the chord (), and the four parts which are the corners of a circle inscribedin a square (). Then the area is (in , wefind the area of the shaded region above the semicircles but below the diameter, and in we find the area of the bottom shaded region).The area of is .The area of is .For the area of , the radius of , and the distance of (the smaller semicircles' radius) to , creates two triangles,so 's area is .The area of is .Hence, finding , the desired areais , so the answeris2019 AMC 10B Problems/Problem 21 ProblemDebra flips a fair coin repeatedly, keeping track of how many heads and how many tails she has seen in total, until she gets either two heads in a row or two tails in a row, at which point she stops flipping. What is the probability that she gets two heads in a row but she sees a second tail before she sees a second head?SolutionWe firstly want to find out which sequences of coin flips satisfy the given condition. For Debra to see the second tail before the seecond head, her first flip can't be heads, as that would mean she would either end with double tails before seeing the second head, or would see two heads before she sees two tails. Therefore, her first flip must be tails. The shortest sequence of flips by which she can get two heads in a row and see the second tail before she sees the secondhead is , which has a probability of . Furthermore, she can prolong her coin flipping by adding an extra , which itself has aprobability of . Since she can do this indefinitely, this gives an infinite geometric series, which means the answer (by the geometric series sum formula)is .Solution 2 (Easier)Note that the sequence must start in THT, which happens with probability. Now, let be the probability that Debra will get two heads in a row after flippingTHT. Either Debra flips two heads in a row immediately (probability ), or flips a head and then a tail and reverts back to the "original position" (probability ). Therefore, , so , so our final answeris . -Stormersyle get rect2019 AMC 10B Problems/Problem 22The following problem is from both the 2019 AMC 10B #22 and 2019 AMC 12B #19, so both problems redirect to this page.ProblemRaashan, Sylvia, and Ted play the following game. Each starts with . A bell rings every seconds, at which time each of the players who currently have money simultaneously chooses one of the other two playersindependently and at random and gives to that player. What is theprobability that after the bell has rung times, each player willhave ? (For example, Raashan and Ted may each decide to give to Sylvia, and Sylvia may decide to give her her dollar to Ted, at which pointRaashan will have , Sylvia will have , and Ted will have , and that is the end of the first round of play. In the second round Rashaan has no money to give, but Sylvia and Ted might choose each other to give their to, and the holdings will be the same at the end of the second round.)SolutionOn the first turn, each player starts off with . Each turn after that, there are only two possibilities: either everyone stays at , which we will writeas , or the distribution of moneybecomes in some order, which we writeas . We will consider these two states separately.In the state, each person has two choices for whom to give their dollar to, meaning there are possible ways that the money canbe rearranged. Note that there are only two ways that we canreach again:1. Raashan gives his money to Sylvia, who gives her money to Ted, who gives his money to Raashan.2. Raashan gives his money to Ted, who gives his money to Sylvia, who gives her money to Raashan.Thus, the probability of staying in the state is , while theprobability of going to the state is (we can check that the 6 other possibilities lead to )In the state, we will label the person with as person A, the person with as person B, and the person with as person C.Person A has two options for whom to give money to, and person B has 2 options for whom to give money to, meaning there are。
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2003 AMC 10B1、Which of the following is the same as2、Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs more than a pink pill, and Al’s pills cost a total of for the two weeks. How much does one green pill cost3、The sum of 5 consecutive even integers is less than the sum of the rst consecutive odd counting numbers. What is the smallest of the even integers4、Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the gure. She plants one flower per square foot in each region. Asters cost 1 each, begonias each, cannas 2 each, dahlias each, and Easter lilies 3 each. What is the least possible cost, in dollars, for her garden5、Moe uses a mower to cut his rectangular -foot by -foot lawn. The swath he cuts is inches wide, but he overlaps each cut by inches to make sure that no grass is missed. He walks at the rate of feet perhour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow his lawn.6、Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is . The horizontal length of a “-inch” television screen is closest, in inches, to which of the following7、The symbolism denotes the largest integer not exceeding . For example. , and . Compute.8、The second and fourth terms of a geometric sequence are and . Which of the following is a possible first term9、Find the value of that satisfies the equation10、Nebraska, the home of the AMC, changed its license plate scheme. Each old license plate consisted of a letter followed by four digits. Each new license plate consists of three letters followed by three digits. By how many times is the number of possible license plates increased11、A line with slope intersects a line with slope at the point .What is the distance between the -intercepts of these two lines12、Al, Betty, and Clare split among them to be invested in different ways. Each begins with a different amount. At the end of one year they have a total of . Betty and Clare have both doubled their money, whereas Al has managed to lose . What was Al’s original portion.13、Let denote the sum of the digits of the positive integer . For example, and . For how many two-digit values of is14、Given that , where both and are positive integers, find the smallest possible value for .15、There are players in a singles tennis tournament. The tournament is single elimination, meaning that a player who loses a match is eliminated. In the first round, the strongest players are given a bye, and the remaining players are paired off to play. After each round, the remaining players play in the next round. The match continues until only one player remains unbeaten. The total number of matches played is16、A restaurant offers three desserts, and exactly twice as many appetizers as main courses. A dinner consists of an appetizer, a main course, and a dessert. What is the least number of main courses that the restaurant should offer so that a customer could have a different dinner each night in the year.17、An ice cream cone consists of a sphere of vanilla ice cream and a right circular cone that has the same diameter as the sphere. If the ice cream melts, it will exactly ll the cone. Assume that the melted ice cream occupies of the volume of the frozen ice cream. What is the ratio of the cone’s height to its radius18、What is the largest integer that is a divisor offor all positive even integers19、Three semicircles of radius are constructed on diameter of a semicircle of radius . The centers of the small semicircles divideinto four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles20、In rectangle , and . Points and are onso that and . Lines and intersect at . Find the area of .21、A bag contains two red beads and two green beads. You reach into the bag and pull out a bead, replacing it with a red bead regardless of the color you pulled out. What is the probability that all beads in the bag are red after three such replacements22、A clock chimes once at minutes past each hour and chimes on the hour according to the hour. For example, at 1 PM there is one chime and at noon and midnight there are twelve chimes. Starting at 11:15 AM on February , , on what date will the chime occur23、A regular octagon has an area of one square unit. What is the area of the rectangle24、The rst four terms in an arithmetic sequence are , , , and , in that order. What is the fth term25、How many distinct four-digit numbers are divisible by and haveas their last two digits。