美国数学大联盟2015年复赛试题

This Solutions Pamphlet gives at least one solution for each problem on this year’s exam and shows that all the problems can be solved using material normally associated with the mathematics curriculum for students in eighth grade or below. These solutions are by no means the only ones possible, nor are they necessarily superior to others the reader may devise.We hope that teachers will share these solutions with their students. However, the publication, reproduction, or communication of the problems or solutions of the AMC 8 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination at any time via copier, telephone, email, internet or media of any type is a violation of the competition rules.Correspondence about the problems and solutions should be addressed to:Prof. Norbert Kuenzi, AMC 8 Chair934 Nicolet AveOshkosh, WI 54901-1634Orders for prior year exam questions and solutions pamphlets should be addressed to:MAA American Mathematics CompetitionsAttn: PublicationsPO Box 471Annapolis Junction, MD 20701© 2015 Mathematical Association of AmericaWe thank the following donors for their generous support of the MAA American Mathematics Competitions, MOSP and the IMOPatron’s CircleAkamai FoundationSimons FoundationWinner’s CircleAmerican Mathematical SocietyThe D.E. Shaw GroupDropboxMathWorksTwo SigmaTudor Investment CorporationAchiever’s CircleArt of Problem SolvingJane Street CapitalMath for AmericaSustainer’s circleAcademy of Applied ScienceArmy Educational Outreach ProgramCollaborator’s CircleAmerican Statistical AssociationCasualty Actuarial SocietyConference Board of the Mathematical SciencesExpii, Inc.IDEA MATHMu Alpha ThetaNational Council of Teachers of MathematicsSociety for Industrial and Applied MathematicsStar League。

2015 美国数学竞赛(十年级)中图分类号：G424．79 文献标识码iA 文章编号：1005—6416(2015)07—0026—07 1．计算：(2。

一1+5 一0) X5的值为( )．(A)一125 (B)一120 (C)25(D) (E 12．箱子中放有三角形和正方形的瓷砖共25块，共有84条边．则箱子中的正方形瓷砖有( )块．(A)3 (B)5 (C)7 (D)9 (E)113．如图l，安琪儿用18根牙签拼三层楼梯．照这样计算，要想拼五层楼梯，她还需( )根牙签．(A)9 (B)l8 (C)2O(D)22 (E)24图14．巴勃罗、索菲亚和米娅在一次聚会上各分得一些糖果．巴勃罗的糖果数为索菲亚糖果数的3倍，索菲亚的糖果数为米娅糖果数的2倍．巴勃罗决定将自己的糖果分给索菲亚和米娅一部分，这样三个人的糖果数相等．则巴勃罗分给索菲亚的糖果数占自己原来糖果数的( )．(A) (B1(c1(D) (E)寻5．帕特里克先生是l5名学生的数学老师．一次测验后他发现，去掉佩顿的成绩，其余人的平均成绩为80分，加上佩顿的成绩后，全班的平均成绩为81分．则在这次考试中，佩顿的成绩为( )分．(A)81 (B)85 (C)91 (0)94 (E)95注意到，1D，1C =90。

+_1_ C = C．于是，D、，。

、、C四点共圆．则E= DC，2，CI2F= cD，故GEF= EDIl+ DILE= ADI 一DB + DCIz1= ÷厶( ADC+ BcD)一ADB，GFE ：FCI2+ FlC= BCI2一+ ，．1= ÷( Dc+ BcD)一ADB．从而，∞F= G阳，得钮=G又GI3平分AGB，故上F．14．若一共有13个数，则可排列如下：a l，a2，a3，…，a 8，a9，a 2 ，a 3 ，a 4 ，…，a 9 ，a lo ，a 3 ，a 4 ，a 5 ，…，a lo ，a l1，04 ，a5 ，a6，…，0 11，a 12，U 5 ，a 6 ，a 7 ，…，a 12 ，a 13·由已知，其中每行数之和为正，从而，数表中所有数之和为正；另一方面，数表中每列数之和为负，从而，数表中所有数之和为负．矛盾．这表明，满足要求的数串至多有12项．考虑如下的一串数字：一4、一4、一4、15、一4、一4、一4、一4、15、一4、一4、一4，这串数满足题中要求且有12项，故知满足题中要求的，的最大值为12．(张宇鹏提供)2015年第7期276．已知两个正数的和是差的5倍．则较大数与较小数之比为( )．(A5(B3(c9(D)2(E)詈7．等差数列l3，16，19，…，70，73中，共有( )项．(A)20 (B)21 (C)24 (D)60 (E)618．两年前，皮特的年龄为其表弟克莱尔年龄的3倍；四年前，皮特的年龄为克莱尔年龄的4倍．( )年后，皮特和克莱尔的年龄比为2：1．(A)2 (B)4 (C)5 (D)6 (E)89．已知两个圆柱的体积相同，第二个圆柱半径比第一个圆柱半径多10％．下列叙述正确的为( )．(A)第二个圆柱比第一个圆柱低10％(B)第一个圆柱比第二个圆柱高10％(c)第二个圆柱比第一个圆柱低21％(D)第一个圆柱比第二个圆柱高21％(E)第二个圆柱的高是第一个圆柱高的80％10．对于字母排列abed，有( )种不同的重排，使得原来排列中相邻的两个字母重排后不能相邻(如ab、ba在重排后不能出现)．(A)0 (B)1 (C)2 (D)3 (E)411．已知矩形的长与宽之比为4：3，对角线的长为d．若用|l} 表示矩形的面积，则k 的值为( )．(A)号(B)号(c) (D) (E)12．已知A(√7c，a)、B(√7c，b)为曲线Y + =2x Y +I_ L的两不同点．则Ia—bI的值为( )．(A)1 (B)詈(c)2(D) (E)1+13．克劳迪娅有5分和10分的硬币共12枚，用这些硬币中的一部分或者全部硬币恰可以组合成17种不同的币值．则克劳迪娅有( )枚1O分的硬币．(A)3 (B)4 (C)5 (D)6 (E)714．钟表圆盘的半径为2O厘米，它与另一个半径为10厘米的小圆盘在12点位置外切，小圆盘上有一个固定指针，开始时指针竖直指向上方．当小圆盘按顺时针方向沿大表盘外沿滚动，且始终保持相切，直到小圆盘上的指针再一次竖直指向上方时停止．此时，两圆盘外切的切点位于大表盘上的( )点位置．(A)2 (B)3 (C)4 (D)6 (E)815．考虑分数(x,y为两个互素的正整Y数)组成的集合．若分子和分母均增加1，则分数的值增加10％．那么，集合中这样的分数有( )个．(A)0 (B)1 (C)2 (D)3(E)无数多16．若Y+4=( 一2)，+4=(Y一2)，且x#y，则+ 的值为( )．(A)10 (B)15 (C)20 (D)25 (E)3017．一条经过坐标原点的直线，与直线= 1、Y=1+ 围成一个等边三角形．则这个等边三角形的周长为( )．(A)2．／6 (B)2+2,3 (c)6(D)3+2√3 (E)6+18．已知十六进制数是由O一9十个数码和A至F六个字母构成的，其中，A，B，…，F分别代表1O，11，…，15．在前1000个正整数中，找出所有只用数码表示的十六进制数．求出这些数的个数n，此时，凡的各位数码之和为( )．(A)17 (B)18 (C)19 (D)20 (E)2119．在等腰Rt△ABC中，已知BC=AC，28 中等数学C=90。

2015 年美国大学生数学建模竞赛MCM、ICM 试题2015 MCM A: Eradicating EbolaThe world medical association has announced that their new medication could stop Ebola and cure patients whose disease is not advanced. Build a realistic, sensible, and useful model that considers not only the spread of the disease, the quantity of the medicine needed, possible feasible delivery systems, locations of delivery, speed of manufacturing of the vaccine or drug, but also any other critical factors your team considers necessary as part of the model to optimize the eradication of Ebola, or at least its current strain. In addition to your modeling approach for the contest, prepare a 1-2 page non-technical letter for the world medical association to use in their announcement.2015 MCM B: Searching for a lost planeRecall the lost Malaysian flight MH370. Build a generic mathematical model that could assist “searchers” in planning a useful search for a lost plane feared to have crashed in open water such as the Atlantic, Pacific, Indian, Southern, or Arctic Ocean while flying from Point A to Point B. Assume that there are no signals from the downed plane. Your model should recognize that there are many different types of planes for which we might be searching and that there are many different types of search planes, often using different electronics or sensors. Additionally, prepare a 1-2 page non-technical paper for the airlines to use in their press conferences concerning their plan for future searches.。

2015年高中数学竞赛 复赛试题及答案一、选择题（本大题共6小题，每小题6分，共36分．每小题各有四个选择支，仅有一个选择支正确．请把正确选择支号填在答题卡的相应位置．）1．从集合｛1，3，6，8｝中任取两个数相乘，积是偶数的概率是A ．56B ．23C ．12D ．132．若α是第四象限角，且2cos2sin212cos2sinαααα-=-，则2α是A ．第一象限角B ．第二象限角C ．第三象限角D ．第四象限角3. 已知点O A B 、、不在同一条直线上，点P 为该平面上一点，且22+OP OA BA =，则A ．点P 不在直线AB 上 B ．点P 在线段AB 上C ．点P 在线段AB 的延长线上D ．点P 在线段AB 的反向延长线上4．设+∈R n m ,，若直线04)1()1(=-+++y n x m 与圆4)2()2(22=-+-y x 相切，则m n +的取值范围是A .]31,0(+B .),31[+∞+C . ),222[+∞+D .]222,0(+ 5. 已知正方体C 1的棱长为C 1的各个面的中心为顶点的凸多面体记为C 2，以C 2的各个面的中心为顶点的凸多面体记为C 3，则凸多面体C 3的棱长为A ．18B ．29C ．9D ．266. 已知定义在R 上的奇函数)(x f ，满足(3)()f x f x +=-,且在区间]23,0[上是增函数,若方程m x f =)()0(<m 在区间[]6,6-上有四个不同的根1234,,,x x x x ,则1234x x x x +++=A ．6-B ． 6C ．8-D ．8 二、填空题（本大题共6小题，每小题6分，共36分．请把答案填在答题卡相应题的横线上．）7．已知1ln ,0()1,0x xf x x x⎧>⎪⎪=⎨⎪<⎪⎩，则不等式()1f x >-的解集为 ▲ .8．随机抽查某中学高二年级100名学生的视力情况，发现学生的视力全部介于4.3至5.2.现将这些数据分成9组，得其频率分布直方图如下.又知前4组的频数成等比数列，后6组的频数成等差数列，则视力在4.6到5.0之间的学生有 ▲ 人.9．在ABC ∆中，角,,A B C 所对应的边长分别为,,a b c ，若2222a b c +=，则cos C 的最小值为 ▲ . 10．给出下列四个命题：（1）如果平面α与平面β相交，那么平面α内所有的直线都与平面β相交； （2）如果平面α⊥平面β，那么平面α内所有直线都垂直于平面β；（3）如果平面α⊥平面β，那么平面α内与它们的交线不垂直的直线与平面β也不垂直； （4）如果平面α不垂直于平面β，那么平面α内一定不存在直线垂直于平面β．其中真命题．．．的序号是 ▲ ．（写出所有真命题的序号） 11．若动点00(,)M x y 在直线20x y --=上运动，且满足2200(2)(2)x y -++≤8，则2200x y +的取值范围是▲ ．12．设函数()1121++⎪⎭⎫⎝⎛=x x x f x，0A 为坐标原点，n A 为函数()x f y =图象上横坐标为n （n ∈N *）的点，向量∑=-=nk k k n A A a 11，向量)0,1(=i ，设n θ为向量n a 与向量i 的夹角，满足15tan 3n k k θ=<∑的最大整数n 是▲ ．答 题 卡一、选择题（本大题共6小题，每小题6分，共36分．）二、填空题（本大题共6小题，每小题6分，共36分．）7． 8． 9． 10． 11． 12．4.3 4.4 4.5 4.6 4.7 4.8 4.95.0 5.1.5.2三、解答题（本大题共6小题，共78分.解答应写出必要的文字说明、证明过程或演算步骤.） 13.（本小题满分12分）已知函数2()2sincos 222x x xf x =-+． （1）求函数()f x 的单调减区间；（2）该函数的图象可由)(sin R x x y ∈=的图象经过怎样的变换得到?（3）已知2π,63πα⎛⎫∈ ⎪⎝⎭，且6()5f α=，求()6f πα-的值．菱形ABCD 中，)2,1(A ，)0,6(=AB ，点M 是线段AB 的中点，线段CM 与BD 交于点P . （1）若向量)7,3(=AD ，求点C 的坐标； （2）当点D 运动时，求点P 的轨迹.如图，四边形ABCD 是边长为2的正方形，△ABE 为等腰三角形，AE ＝BE ，平面ABCD ⊥平面ABE ，点F 在CE 上，且BF ⊥平面ACE. （1）判断平面ADE 与平面BCE 是否垂直，并说明理由； （2）求点D 到平面ACE 的距离. ABCDEF如图，某化工集团在一条河流的上、下游分别建有甲、乙两家化工厂，其中甲厂每天向河道内排放污水2万m3，每天流过甲厂的河水流量是500万m3（含甲厂排放的污水）；乙厂每天向河道内排放污水1.4万m3，每天流过乙厂的河水流量是700万m3（含乙厂排放的污水）.由于两厂之间有一条支流的作用，使得甲厂排放的污水在流到乙厂时，有20%可自然净化.假设工厂排放的污水能迅速与河水混合，且甲厂上游及支流均无污水排放.根据环保部门的要求，整个河流中污水含量不能超过0.2%，为此，甲、乙两个工厂都必须各自处理一部分污水.（1）设甲、乙两个化工厂每天各自处理的污水分别为x、y万m3，试根据环保部门的要求写出x、y 所满足的所有条件；（2）已知甲厂处理污水的成本是1200元/万m3，乙厂处理污水的成本是1000元/万m3，在满足环保部门要求的条件下，甲、乙两个化工厂每天应分别各自处理污水多少万m3，才能使这两个工厂处理污水的总费用最小？最小总费用是多少元？已知),,(42)(2R c b a c bx ax x f ∈++=.（1）当0≠a 时，若函数)(x f 的图象与直线x y ±=均无公共点，求证：;4142>-b ac （2）43,4==c b 时，对于给定的负数8-≤a ，记使不等式5|)(|≤x f 成立的x 的最大值为)(a M .问a 为何值时，)(a M 最大，并求出这个最大的)(a M ，证明你的结论.2014年高中数学竞赛决赛参考答案11.24一、选择题（本大题共6小题，每小题6分，共36分．）二、填空题（本大题共6小题，每小题6分，共36分．）7．),0()1,(e --∞ 8． 78 9．1210． （3）（4） 11． [2，8] 12． 3三、解答题（本大题共6小题，共78分.解答应写出必要的文字说明、证明过程或演算步骤.） 13.（本小题满分12分）解：（1）2()sin 2sin)2x f x x =+- sin x x =π2sin 3x ⎛⎫=+ ⎪⎝⎭． …………………2分 令πππππk x k 223322+≤+≤+，Z k ∈． 得ππππk x k 26726+≤≤+，Z k ∈． ()f x ∴的单调减区间为]267,26[ππππk k ++，Z k ∈． …………………5分 （2）先把函数)(sin R x x y ∈=的图象向左平移3π个单位，就得到函数))(3sin(R x x y ∈+=π的图象；再把其纵坐标伸长为原来的2倍，横坐标不变，就得到π2sin 3y x ⎛⎫=+ ⎪⎝⎭)(R x ∈的图象．…………7分（3）由56)(=αf 得：π62sin(),35α+=即π3sin(),35α+= …………………8分 因为2π,63πα⎛⎫∈⎪⎝⎭,所以π()(,)32παπ+∈.从而π4cos()35α+==- …………………10分 于是()2sin[()]2[sin()cos cos()sin ]6363636f πππππππαααα-=+-=+-+ 5433]21542353[2+=⨯+⨯=. …………………12分14.（本小题满分12分）解：（1）菱形ABCD 中，)7,9()0,6()7,3(=+=+=AB AD AC ，且)2,1(A ，所以)9,10(C .…4分 （2）设),(y x P ，则)2,7()0,6()2,1(--=---=-=y x y x AB AP BP . …………………5分又因为点M 是线段AB 的中点，线段CM 与BD 交于点P ，即点P 是ABC ∆的重心，从而有MP MC 3=,所以11133()3222AC AM MC AB MP AB AP AB AP AB =+=+=+-=-3(1,2)(6,0)(39,36)x y x y =---=-- …………………7分菱形ABCD 的对角线互相垂直，所以AC BP ⊥， 即 0)63,93()2,7(=--⋅--y x y x ， 亦即0)63)(2()93()7(=--+-⋅-y y x x ，整理得：4)2()5(22=-+-y x （2≠y ）， …………………11分 故P 点的轨迹是以)2,5(为圆心，2为半径的圆，除去与2=y 的交点. …………………12分15．（本题满分13分）解：（1）平面ADE 与平面BCE 垂直. …………………1分证明如下：因为BF ⊥平面ACE ，所以BF ⊥AE. …………………3分 因为平面ABCD ⊥平面ABE ，且ABCD 是正方形，BC ⊥AB ，CD平面ABCD ∩平面ABE ＝AB ，所以BC ⊥平面ABE ，从而BC ⊥AE. …………………6分 于是AE ⊥平面BCE ，故平面ADE ⊥平面BCE. ………………7分 （2）连结BD 交AC 与点M ，则点M 是BD 的中点，所以点D 与点B 到平面ACE 的距离相等. …………………8分 因为BF ⊥平面ACE ，所以BF 为点B 到平面ACE 的距离. …9分 因为AE ⊥平面BCE ，所以AE ⊥BE.又AE ＝BE ，所以△AEB 是等腰直角三角形. …………………10分 因为AB ＝2，所以BE＝2sin 45︒= …………………11分在Rt △CBE 中，CE = 3B C B E BF CE ⨯=== 故点D 到平面ACE 的距离是332. …………………13分16．（本题满分13分）解：（1）据题意，x 、y 所满足的所有条件是()20.25001000.8(2) 1.40.2700100020 1.4x x y x y -⎧≤⎪⎪-+-⎪≤⎨⎪≤≤⎪⎪≤≤⎩， …………………4分即⎪⎩⎪⎨⎧≤≤≤≤≥+4.1021854y x y x . …………………5分 （2）设甲、乙两厂处理污水的总费用为z 元，则目标函数z ＝1200x ＋1000y ＝200(6x ＋5y ).…………7分 作可行域，如图. ……………10分 平移直线l ：6x ＋5y=0，当直线经过点A (1，0.8)时，z 取最大值，此时ABCDEFMGz ＝1200×1＋1000×0.8＝2000（元）. ……………12分故甲、乙两厂每天应分别处理1万m3、0.8万m3污水，才能使两厂处理污水的总费用最小，且最小总费用是2000元. …………………13分17．（本题满分14分）解：(1)由),,(42)(2R c b a c bx ax x f ∈++=与直线x y ±=均无公共点（0≠a ），可知x c bx ax ±=++422无解， ………………1分 由04)12(2=+++c x b ax 无解，得：016)12(2<-+=∆ac b ， 整理得：b b ac +>-4142(1) ………………3分 由04)12(2=+-+c x b ax 无解，得：016)12(2<--=∆ac b ，整理得：b b ac ->-4142(2) ………………5分 由(1),(2)得: 4142>-b ac . ………………6分(2) 由43,4==c b ，所以38)(2++=x ax x f ………………7分因为a a f 163)4(-=-, 由8-≤a 得，5163)4(≤-=-aa f ………………9分 所以()5f x ≤恒成立，故不等式5|)(|≤x f 成立的x 的最大值也就是不等式()5f x ≥-成立的x 的最大值，…………10分 因此)(a M 为方程5382-=++x ax 的较大根，即aaa M 2424)(---=（8-≤a ） ………………11分当8-≤a 时，()M a ==a 的增函数， ………………13分 所以，当8a =-时，)(a M 取得最大值，其最大值为251)(+=a M . ………………14分 18．（本题满分14分）解：（1）由条件可得3n n x =，45n y n =+，根据题意知，23n n c =． …………………1分由k c 为数列{}n y 中的第m 项，则有2345km =+， …………………2分因910m *+∈N ,所以1k c +是数列{}n y 中的第910m +项． …………………5分（2）设在区间[1,2]上存在实数b 使得数列{}n x 和{}n y 有公共项，即存在正整数s ，t 使(1)sa a tb =++，∴1+-=a b a t s ， 因自然数2a ≥,s ，t 为正整数，∴sa b -能被1a +整除． …………………6分 ①当1s =时，1s a b t a -=<+1a a *∉+N ． ②当2s n = (n *∈N )时，若1b =，2222111[1()()()]111()s n nn a b a a a a a a a a ----==-=-+-+-++-++-- 2422(1)[1]n a a a a -*=-+++∈N ，即s a b -能被1a +整除， …………………8分 此时数列{}n x 和{}n y 有公共项组成的数列{}n z ，通项公式为2n n z a =(n *∈N )；若2b =， 显然，222111111s n n a b a a a a a a *---==-∉++++N ，即s a b -不能被1a +整除． ………………9分 ③当21s n =+(n *∈N )时, 2()11n sb a a a b a t a a --==++， …………………10分 若2a >，则2n b a a *-∉N ，又a 与1a +互质，故此时2()1n b a a a t a *-=∉+N ． ………………11分 若2a =，要2n b a a *-∈N ，则要2b =，此时221n n b a a a-=-， …………………12分 由②知，21n a -能被1a +整除， 故2()1n b a a a t a *-=∈+N ，即s a b -能被1a +整除． 当且仅当2b a ==时，b a S -能被1a +整除． …………………13分此时数列{}n x 和{}n y 有公共项组成的数列{}n z ，通项公式为212n n z +=(n *∈N ).综上所述，存在{1,2}b ∈，使得数列{}n x 和{}n y 有公共项组成的数列{}n z ，且当1b =时,数列2n n z a=(n *∈N )；当2b a ==时,数列212n n z +=(n *∈N ). ……………14分18．（本题满分14分）已知数列{}n x 和{}n y 的通项公式分别为n n x a =和()1,n y a n b n N +=++∈.（1）当3,5a b ==时，记2n n c x =，若k c 是{}n y 中的第m 项(,)k m N +∈，试问：1k c +是数列{}n y 中的第几项？请说明理由．（2）对给定自然数2a ≥，试问是否存在{}1,2b ∈，使得数列{}n x 和{}n y 有公共项？若存在，求出b 的值及相应的公共项组成的数列{}n z ，若不存在，请说明理由．。

2015 AMC 12B竞赛真题Problem 1What is the value of ?Problem 2Marie does three equally time-consuming tasks in a row without taking breaks. She begins the first task at 1:00 PM and finishes the second task at 2:40 PM. When does she finish the third task?Problem 3Isaac has written down one integer two times and another integer three times. The sum of the five numbers is 100, and one of the numbers is 28. What is the other number?Problem 4David, Hikmet, Jack, Marta, Rand, and Todd were in a 12-person race with 6 other people. Rand finished 6 places ahead of Hikmet. Marta finished 1 place behind Jack. David finished 2 places behind Hikmet. Jack finished 2 places behind Todd. Todd finished 1 place behind Rand. Marta finished in 6th place. Who finished in 8th place?Problem 5The Tigers beat the Sharks 2 out of the 3 times they played. They then played more times, and the Sharks ended up winning at least 95% of all the games played. What is the minimum possible value for ?Problem 6Back in 1930, Tillie had to memorize her multiplication facts fromto . The multiplication table she was given had rows and columns labeled with the factors, and the products formed the body of the table. To the nearest hundredth, what fraction of the numbers in the body of the table are odd?Problem 7A regular 15-gon has lines of symmetry, and the smallest positive angle for which it has rotational symmetry is degrees. What is ?Problem 8What is the value of ?Problem 9Larry and Julius are playing a game, taking turns throwing a ball at a bottle sitting on a ledge. Larry throws first. The winner is the first person to knock the bottle off the ledge. At each turn the probability that a playerknocks the bottle off the ledge is , independently of what has happened before. What is the probability that Larry wins the game?Problem 10How many noncongruent integer-sided triangles with positive area and perimeter less than 15 are neither equilateral, isosceles, nor right triangles?Problem 11The line forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?Problem 12Let , , and be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation?Problem 13Quadrilateral is inscribed in a circle withand . What is ?Problem 14A circle of radius 2 is centered at . An equilateral triangle with side4 has a vertex at . What is the difference between the area of the regionthat lies inside the circle but outside the triangle and the area of the region that lies inside the triangle but outside the circle?Problem 15At Rachelle's school an A counts 4 points, a B 3 points, a C 2 points, and a D 1 point. Her GPA on the four classes she is taking is computed as the total sum of points divided by 4. She is certain that she will get As in both Mathematics and Science, and at least a C in each of English and History.She thinks she has a chance of getting an A in English, and a chance of getting a B. In History, she has a chance of getting an A, and a chanceof getting a B, independently of what she gets in English. What is the probability that Rachelle will get a GPA of at least 3.5?Problem 16A regular hexagon with sides of length 6 has an isosceles triangle attached to each side. Each of these triangles has two sides of length 8. The isosceles triangles are folded to make a pyramid with the hexagon as the base of the pyramid. What is the volume of the pyramid?Problem 17An unfair coin lands on heads with a probability of . When tossed times,the probability of exactly two heads is the same as the probability of exactly three heads. What is the value of ?Problem 18For every composite positive integer , define to be the sum of the factors in the prime factorization of . For example, because the prime factorization of is , and . What is the range of the function , ?Problem 19In , and . Squares and are constructed outside of the triangle. The points , , , and lie on a circle. What is the perimeter of the triangle?Problem 20For every positive integer , let be the remainder obtained when is divided by 5. Define a functionrecursively as follows:What is ?Problem 21Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of ?Problem 22Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same chair and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?Problem 23A rectangular box measures , where , , and are integers and. The volume and the surface area of the box are numerically equal. How many ordered triples are possible?Problem 24Four circles, no two of which are congruent, have centers at , , , and , and points and lie on all four circles. The radius of circleis times the radius of circle , and the radius of circle is timesthe radius of circle . Furthermore, and . Let be the midpoint of . What is ?Problem 25A bee starts flying from point . She flies inch due east to point . For , once the bee reaches point , she turns counterclockwise and then flies inches straight to point . When the bee reaches she is exactly inches away from , where , , andare positive integers and and are not divisible by the square of any prime. What is ?2015 AMC 12B竞赛真题答案1.C2.b3.a4.b5.b6.a7.d8.d9.c 10.c 11.e 12.d 13.b 14.d 15.d 16.c 17.d 18.d 19.c 20.b 21.d 22.d 23.b 24.d 25.b。

Copyright © 2016 Art of Problem Solving How many square yards of carpet are required to cover a rectangular floor that is feet long and feetwide? (There are 3 feet in a yard.)First, we multiply to get that you need square feet of carpet you need to cover. Since thereare square feet in a square yard, you divide by to get square yards, so our answer is .Since there are feet in a yard, we divide by to get , andby to get . To find the area of the carpet, we then multiply these two values together to get .2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded byFirst Problem Followed by Problem 21 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21• 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Placement:Easy GeometryRetrieved from "/wiki/index.php?title=2015_AMC_8_Problems/Problem_1&oldid=80323"SolutionSolution 2See AlsoPoint is the center of the regular octagon , and is the midpoint of the side What fraction of the area of the octagon is shaded?Since octagon is a regular octagon, it is split into equal parts, such as triangles, etc. These parts, since they are all equal, are of the octagon each. The shaded region consists of of these equal parts plus half of another, so the fraction of the octagon that is shaded isThe octagon has been divided up into identical triangles (and thus they each have equal area). Since the shaded region occupiesout of the total triangles, the answer is .For starters what I find helpful is to divide the whole octagon up into triangles as shown here:Now it is just a matter of counting the larger triangles remember that and are notfull triangles and are only half for these purposes. We count it up and we get a total ofof the shape shaded. We then simplify it to get our answer of .2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded byProblem 1Followed by Problem 31 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsSolution 3See AlsoCopyright © 2016 Art of Problem Solving Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of miles per hour. Jack walks tothe pool at a constant speed of miles per hour. How many minutes before Jack does Jill arrive?Using , we can set up an equation for when Jill arrives at the swimming pool:Solving for , we get that Jill gets to the pool inof an hour, which is minutes. Doing the same for Jack, we get that Jack arrives at the pool inof an hour, which in turn is minutes. Thus, Jill has to waitminutes for Jack to arrive at the pool.2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded byProblem 2Followed by Problem 41 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2015_AMC_8_Problems/Problem_3&oldid=81064"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingThe Centerville Middle School chess team consists of two boys and three girls. A photographer wants to take a picture of the team to appear in the local newspaper. She decides to have them sit in a row with a boy ateach end and the three girls in the middle. How many such arrangements are possible?There are ways to order the boys on the end, and there are ways to order the girls in the middle.We get the answer to be .2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded byProblem 3Followed by Problem 51 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2015_AMC_8_Problems/Problem_4&oldid=73224"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingBilly's basketball team scored the following points over the course of the first 11 games of the season:If his team scores 40 in the 12th game, which of the following statistics will show an increase?When they score a on the next game, the range increases from to . This means the increased.Because is less than the score of every game they've played so far, the measures of center will neverrise. Only measures of spread, such as the, may increase.2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded byProblem 4Followed by Problem 61 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22• 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2015_AMC_8_Problems/Problem_5&oldid=80825"SolutionSolution 2Copyright © 2016 Art of Problem Solving In , , and . What is the area of?We know the semi-perimeter of is . Next, we use Heron's Formula to find that the area of the triangle is just .Splitting the isosceles triangle in half, we get a right triangle with hypotenuseand leg . Using the Pythagorean Theorem , we know the height is. Now that we know the height, the area is.2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded byProblem 5Followed by Problem 71 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 •24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2015_AMC_8_Problems/Problem_6&oldid=80483"Solution 1Solution 2See AlsoEach of two boxes contains three chips numbered , , . A chip is drawn randomly from each box and thenumbers on the two chips are multiplied. What is the probability that their product is even?We can instead find the probability that their product is odd, and subtract this from . In order to get an odd product, we have to draw an odd number from each box. We have a probability of drawing an odd numberfrom one box, so there is a probability of having an odd product. Thus, there is a probability of having an even product.You can also make this problem into a spinner problem. You have the first spinner with equally divided sections, andYou make a second spinner that is identical to the first, with equal sections of ,, and . If the first spinner lands on , to be even, it must land on two. You write down the first combination of numbers . Next, if the spinner lands on , it can land on any number on the second spinner. We now have the combinations of . Finally, if the first spinner ends on , we have Since there arepossible combinations, and we have evens, the final answer is.2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded byProblem 6Followed by Problem 81 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2015_AMC_8_Problems/Problem_7&oldid=73737"SolutionSolution 2See AlsoCopyright © 2016 Art of Problem Solving What is the smallest whole number larger than the perimeter of any triangle with a side of length and aside of length ?We know from the triangle inequality that the last side, , fulfills . Adding to both sides of the inequality, we get , and becauseis the perimeter ofour triangle, is our answer.2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded byProblem 7Followed by Problem 91 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23• 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2015_AMC_8_Problems/Problem_8&oldid=78101"SolutionSee AlsoCopyright © 2016 Art of Problem Solving On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previousday. How many widgets in total had Janabel sold after working days?The sum of is The sum is just the sum of the first odd integers, which is2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded byProblem 8Followed by Problem 101 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2015_AMC_8_Problems/Problem_9&oldid=79933"Solution 1Solution 2See AlsoCopyright © 2016 Art of Problem SolvingHow many integers betweenandhave four distinct digits?The question can be rephrased to "How many four-digit positive integers have four distinct digits?",since numbers between and are four-digit integers. There are choices for the first number, since it cannot be , there are only choices left for the second number since it must differ from the first, choices for the third number, since it must differ from the first two, and choices for the fourth number,since it must differ from all three. This means there are integersbetweenandwith four distinct digits.2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded by Problem 9Followed by Problem 111 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22• 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2015_AMC_8_Problems/Problem_10&oldid=81128"Solution 1See AlsoCopyright © 2016 Art of Problem SolvingIn the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and thefourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, whatis the probability that the plate will read "AMC8"?There is one favorable case, which is the license plate says "AMC8". We must now find how many total cases there are. There are choices for the first letter (since it must be a vowel), choices for the second letter (since it must be of consonants), choices for the third letter (since it must differ from the second letter), and choices for the number. This leads to total possible license plates. That means the probability of a license plate saying "AMC8" is.The probability of choosing A as the first letter is . The probability of choosing next is. Theprobability of choosing C as the third letter is(since there areother consonants to choose fromother then M). The probability of having as the last number is . We multiply all these to obtain2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded by Problem 10Followed by Problem 121 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 •24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2015_AMC_8_Problems/Problem_11&oldid=73554"Solution 1Solution 2See AlsoHow many pairs of parallel edges, such asandorand, does a cube have?We first count the number of pairs of parallel lines that are in the same direction as. The pairs ofparallel lines are ,, , , ,and . These are pairs total. We can do the same for the lines in the same direction asand. This means there aretotal pairs of parallel lines.Pick a random edge. Given another edge, the probability that it is parallel to this edge is. Keep in mind we already used one edge. There are edges so pairs. So our answer is.2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded by Problem 11Followed by Problem 131 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsSolution 1Solution 2See AlsoCopyright © 2016 Art of Problem SolvingHow many subsets of two elements can be removed from the set so thatthe mean (average) of the remaining numbers is 6?Since there will be elements after removal, and their mean is , we know their sum is . We also knowthat the sum of the set pre-removal is . Thus, the sum of the elements removed is .There are onlysubsets of elements that sum to:.2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded by Problem 12Followed by Problem 141 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22• 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2015_AMC_8_Problems/Problem_13&oldid=73377"SolutionSee AlsoLet our numbers be , where is odd. Then our sum is . The only answer choice that cannot be written as , where is odd, is .If the four consecutive odd integers are and then the sum is . All the integers are divisible by except .If the four consecutive odd integers are and , the sum is , and divided by gives . This means that must be even. The only integer that does notgive an even integer when divided by is , so the answer is .From Solution 1, we have the sum of the numbers to be equal to . Taking mod 8 gives usfor some residue and for some odd integer . Since , we can express it as the equation for some integer . Multiplying 4 to each side of the equation yields , and taking mod 8 gets us , so . All the answer choicesexcept choice D is a multiple of 8, and since 100 satisfies all the conditions the answer is .The problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Copyright © 2016 Art of Problem SolvingAt Euler Middle School,students voted on two issues in a school referendum with the following results:voted in favor of the first issue and voted in favor of the second issue. If there were exactlystudents who voted against both issues, how many students voted in favor of both issues?We can see that this is a Venn Diagram Problem.First, we analyze the information given. There are students. Let's use A as the first issue and B asthe second issue.students were for the A, and students were for B. There were also students against both A andB.Solving this without a Venn Diagram, we subtract away from the total,. Out of the remaining,we havepeople for A andpeople for B. We add this up to get. Since that is more than what we need, we subtract fromto getThere are 198 people. We know that 29 people voted against both the first issue and the second issue. That leaves us with 169 people that voted for at least one of them. If 119 people voted for both of them, then that would leave 20 people out of the vote, because 149 is less than 198 people. 198-149 is 20, so to make it even, we have to take 20 away from the 119 people, which leaves us with2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded by Problem 14Followed by Problem 161 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22• 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2015_AMC_8_Problems/Problem_15&oldid=80999"Solution 1Solution 2See AlsoCopyright © 2016 Art of Problem SolvingIn a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If of all the ninth graders are paired withof all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?Let the number of sixth graders be , and the number of ninth graders be . Thus, , whichsimplifies to. Since we are trying to find the value of, we can just substitute forinto the equation. We then get a value ofWe see that the minimum number of ninth graders is , because if there arethen there is ninth grader with a buddy, which would mean sixth graders with a buddy, and that's impossible. With ninth graders, of them are in the buddy program, so theresixth graders total, two of whom have a buddy. Thus,the desired fraction is .2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded by Problem 15Followed by Problem 171 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22• 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2015_AMC_8_Problems/Problem_16&oldid=73512"Solution 1Solution 2See AlsoSo and .This gives , which gives , which then givesSolution 2, is obviously constantso , plug into the first one and it's miles to schoolWe set up an equation in terms of the distance and the speed In miles per hour. We have SoAn arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, is an arithmetic sequence with five terms, in which thefirst term is and the constant added is . Each row and each column in this array is an arithmetic sequence with five terms. What is the value of ?We begin filling in the table. The top row has a first term and a fifth term , so we have the common difference is . This means we can fill in the first row of the table:The fifth row has a first term of and a fifth term of , so the common difference is. We can fill in the fifth row of the table as shown:Copyright © 2016 Art of Problem SolvingWe must find the third term of the arithmetic sequence with a first term of and a fifth term of . The common difference of this sequence is, so the third term is.The middle term of the first row is, since the middle number is just the average in anarithmetic sequence. Similarly, the middle of the bottom row is . Applying this again forthe middle column, the answer is.The value ofis simply the average of the average values of both diagonals that contain. This is2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded by Problem 17Followed by Problem 191 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24• 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2015_AMC_8_Problems/Problem_18&oldid=73460"Solution 2Solution 3See AlsoA triangle with vertices as , , and is plotted on a grid. What fraction of the grid is covered by the triangle?The area of is equal to half the product of its base and height. By the Pythagorean Theorem, we find its height is , and its base is . We multiply these and divideby to find the of the triangle is . Since the grid has an area of , the fraction of the grid covered by the triangle is .Note angle is right, thus the area is thus the fraction of the total isCopyright © 2016 Art of Problem SolvingBy the Shoelace theorem, the area of.This means the fraction of the total area isThe smallest rectangle that follows the grid lines and completely encloses has an area of,where splits the rectangle into four triangles. The area ofis therefore . That means thattakes upof the grid.Using Pick's Theorem, the area of the triangle is. Therefore, the triangle takes upof the grid.2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded by Problem 18Followed by Problem 201 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22• 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2015_AMC_8_Problems/Problem_19&oldid=75879"Solution 4Solution 5See AlsoRalph went to the store and bought 12 pairs of socks for a total of $24. Some of the socks he bought cost $1a pair, some of the socks he bought cost $3 a pair, and some of the socks he bought cost $4 a pair. If hebought at least one pair of each type, how many pairs of $1 socks did Ralph buy?So let there be pairs of socks, pairs ofsocks, pairs of socks.We have,, and.Now we subtract to find , and . It follows that is a multiple of and is amultiple of , so since , we must have .Therefore, , and it follows that. Now, asdesired.Since the total cost of the socks was and Ralph boughtpairs, the average cost of each pair ofsocks is.There are two ways to make packages of socks that average to . You can have:Twopairs and one pair (package adds up to )Onepair and onepair (package adds up to)So now we need to solvewhere is the number of packages and is the number of packages. We see our only solution (thathas at least one of each pair of sock) is , which yields the answer of.2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded by Problem 19Followed by Problem 211 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19• 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?2015 AMC 8 Problems/Problem 20Solution 1Solution 2See AlsoIn the given figure hexagon is equiangular, andare squares with areasandrespectively, is equilateral and . What is the area of ?.Clearly, since is a side of a square with area , . Now, since , we have .Now, is a side of a square with area, so . Since is equilateral, . Lastly, is a right triangle. We see that, so is a right triangle with legs and . Now, its area is .2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded byProblem 20Followed byProblem 221 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 •21•22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2015_AMC_8_Problems/Problem_21&oldid=73361"2015 AMC 8 Problems/Problem 21SolutionSee AlsoCopyright © 2016 Art of Problem SolvingCopyright © 2016 Art of Problem SolvingOn June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This processcontinues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in thegroup?As we read through this text, we find that the given information means that the number of students in thegroup has factors, since each arrangement is a factor. The smallest integer with factors is.2015 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2015))Preceded by Problem 21Followed by Problem 231 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 •25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2015_AMC_8_Problems/Problem_22&oldid=73359"SolutionSee Also。

………………………………………………装……………………………………………订……………………………………………线…………………………………………………………… ……………………………………装…………………订…………………线…………………内…………………不…………………答…………………题………………………………………2014-2015 年度美国“数学大联盟杯赛”(中国赛区)初赛(五年级)(初赛时间：2015 年 1 月 3 日，考试时间 90 分钟，总分 200 分)学生诚信协议：考试期间，我确定没有就所涉及的问题或结论，与任何人、用任何方式交流或讨论， 我确定以下的答案均为我个人独立完成的成果，否则愿接受本次成绩无效的处罚。

一、选择题(每小题 5 分，答对加 5 分，答错不扣分，共 175 分，请将正确答案 A 、B 、C 或者 D 写在每题后面的圆括号内。

)正确答案填写示例如下：20 − 5 × 2 = 2 × ? ( A )A) 5 B) 15 C) 25D) 301. Three cases of 24 cans each is the same number of cans as twelve boxes of ? cans each. ( )A) 6 B) 15 C) 21 D) 96 2. A trapezoid has ? sides. ( )11. Thok has a simple plan. He will spend 50% of the day in the cave, 25% of the rest of the day on the hunt, and the remainder of the day watching films outside. How many hours will Thok spend watching films? ( ) A) 3 B) 6 C) 9 D) 25 12. 2 × 3 × 6 × 36 × 2 × 3 × 6 × 36 = ( )A) 65 B) 66C) 67D) 6813. I have 5 pennies, 4 nickels, 3 quarters, 2 half-dollars, and 1 dollar. The average value of one of my coins is ( )14.Wyatt O’Vine’s sheep weighs twice as much as Wyatt, and Wyatt weights twice as much as his hat. If Wyatt, his sheep, and his hat have a combined weight of 210 kg, how much does Wyatt weigh? ( ) A) 30 kg B) 35 kg C) 60 kg D) 70 kg 15. (12 + 34) × (56 + 78) = 12 × (56 + 78) + ? × (56 + 78) ()A) 12 B) 34 C) 56 D) 78 16. If 2 flocks equal 5 flecks, then 500 flocks equal ? flecks. ( )3. Abby put an X through 7 of the 28 days in this month. Abby put anX through ? of the days. ()17.2= ( )4. 60 ÷ 4 = 3 × ? ( )A) 20 B) 12 C) 6 D) 5 5. 30 × 40 × 50 = 120 × ? ( )A) 50 B) 200 C) 500 D) 6006. If 3 out of 4 of the students in my class have brown eyes, and there are 24 students in my class, then how many have brown eyes? ( ) A) 6 B) 8 C)18 D)217. Four identical squares, each with a perimeter of 36, are put together as shown to make one big square. What is the perimeter of the big square? ()8. 80 + (160 + 240) ÷ 4 = 40 + 80 + (120 ÷ ? ) ( )A) 4 B) 2 C) 1 D) 0 9. In which of the following divisions is the remainder greatest? ()A) 1111 ÷ 8 B) 2222 ÷ 7 C) 3333 ÷ 6 D) 4444 ÷ 5 10. Of the following, which is a factor of 20 × 14 × 20 × 15? ( )A) 13 B) 11 C) 9 D) 718.If the sum of seven consecutive even numbers is 182, the smallest of the seven numbers is ( ) A) 20 B) 23 C) 26 D) 3219. While he was standing on his head, Flip decided to count backward from 777 by eights. Which of the following numbers did Flip count? ( A) 123 B) 125 C) 127 D) 129 20. The cost of five apples is the same as the cost of six pears. If one apple costs 15 cents more than one pear, then what is the total cost of 5 apples and 6 pears? ( ) A) $3 B) $6 C) $9 D) $1821. What is the difference between 27 and the product of all its whole-number factors? ( ) A) 2 B) 27 C) 2 × 27 D) 26 × 27 22. The prime factorization of a whole number less than 100 is the product of at most ? primes (not necessarily different). ( )A) 3 B) 4 C) 5 D) 6 23. A rectangle with sides of integer length is divided into a square regionand a shaded rectangular region as shown. If the area of the shaded)城市区 学校年级性别姓名（正楷）A) 3 B) 4 C) 5 D) 10 A) 0.07 B) 0.25C) 0.28D) 0.35A) 36B) 48C) 72D) 144A) $0.20 B) $0.60 C) $1.50 D) $3.00A) 200 B) 250 C) 1000 D) 1250 A) 16B) 64C) 128D) 256)2 ………………………………………………装……………………………………………订……………………………………………线…………………………………………………………… ……………………………………装…………………订…………………线…………………内…………………不…………………答…………………题………………………………………region is 8, the area of the entire figure is at most ( )A) 24 B) 64 C) 72 D) 8124. Gabriel wrote his name 100 times in the sand. What was the 100th letter hewrote? ( ) A) a B) b C) r D) i 25. I put $100 in a savings account. At the end of each year, I earned 10% interest on the balance of my account at that point. With no deposits or withdrawals, after 5 years I had ? . (Round to the nearest dollar.) ( ) A) $162 B) $161 C) $160 D) $150 26. Mrs. Andrews enjoys feeding the birds her special mix of birdseed. She mixes 1 part sunflower seed with 3 parts sesame seed to make a total of 200 6-kg bags. If sunflower seed comes in 12-kg bags, how many such bags does she need? ( ) A) 25 B) 34 C) 50 D) 6737. There are several apples in a basket. Tony removes half of the apples in the basket and then places 15 additional apples in the basket. Sophie then removes half of the apples in the basket, and then places 10 additional apples in the basket. The number of apples Tony removes from the basket is the same as the number of apples Sophie removes. How many apples are now in the basket?Answer: .38. In this problem as shown on the right, each square represents a digit from 0 to 9. What is the final product? Answer: .39. In triangle ABC on the right, BD :DC = 1:2. If the area of triangle ABP is 7 and the area of triangle ACP is 5, what is the area of triangle ADP ?Answer: .40. 甲、乙、丙三位同学参加运动会，他们的号码都 27. The product of 100 one hundreds is the same as the sum of ? one hundreds. ()A) 100100 B) 10099 C) 10010 D) 1002 不一样。

This Pamphlet gives at least one solution for each problem on this year’s contest and shows that all problems can be solved without the use of a calculator. When more than one solution is provided, this is done to illustrate a significant contrast in methods, e.g., algebraic vs geometric, computational vs conceptual, elementary vs advanced. These solutions are by no means the only ones possible, nor are they superior to others the reader may devise.We hope that teachers will inform their students about these solutions, both as illustrations of the kinds of ingenuity needed to solve nonroutine problems and as examples of good mathematical exposition. However, the publication, reproduction or communication of the problems or solutions of the AMC 10 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination via copier, telephone, email, internet or media of any type during this period is a violation of the competition rules.After the contest period, permission to make copies of problems in paper or electronic form including posting on web-pages for educational use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear the copyright notice. Correspondence about the problems/solutions for this AMC 10 and orders for any publications should be addressed to:MAA American Mathematics CompetitionsAttn: Publications, PO Box 471, Annapolis Junction, MD 20701Phone 800.527.3690 | Fax 240.396.5647 | amcinfo@The problems and solutions for this AMC 10 were prepared by the MAA’s Committee on theAMC 10 and AMC 12 under the direction of AMC 10 Subcommittee Chair:Silvia Fernandez© 2015 Mathematical Association of AmericaTheMAA American Mathematics Competitionsare supported byAcademy of Applied ScienceAkamai FoundationAmerican Mathematical SocietyAmerican Statistical AssociationArt of Problem Solving, Inc.Casualty Actuarial Society Conference Board of the Mathematical SciencesThe D.E. Shaw GroupGoogleIDEA MATH, LLCJane Street CapitalMath for America Inc.Mu Alpha ThetaNational Council of Teachers of MathematicsSimons FoundationSociety for Industrial & Applied Mathematics。

2015-2016 Math League Contests, Grades 9 – 12Second-Round, Jan - Feb 2016Instructions:1.This second-round contest consists of two parts. Part 1 is math questions. Part 2 is essay.2.For all the questions below, login to your account at / , and enter youranswers. Answers written on this document will NOT be credited.3.In Part 1, you are asked to read one math subject, The Mathematics of Normal Distributions, and thesupplementary materials if necessary, Descriptive Statistics. Then you have 28 questions to work on.You will need to give precise, unambiguous answers to Questions 1-19 (The Mathematics of Normal Distributions), and Questions 21-26 (Descriptive Statistics).4.Question 20 (The Mathematics of Normal Distributions) and 27-28 (Descriptive Statistics) areProjects and Papers, which means you need to do your research and write a paper for each question.There is no word limit on each of your papers, but it doesn’t necessary mean the more words thebetter. The best paper is precise and succinct. Please don’t feel frustrated at all if you can’t write apaper, as the topics, Confidence Intervals (Question 20); Lies, Damn Lies, and Statistics (Question27); and Data in Your Daily Life (Question 28), are very hard for a high school student, even for anadult. Please don’t feel frustrated even if you can’t finish all of Questions 1-19, 21-26, as they are not trivial questions and it requires a lot reading and thinking. Students who can work out a few questions should be commended.5.For the subject of Normal Distributions, if you really understand what it is and how it works,then the questions are fairly easy to solve. So our recommendation is don’t rush to solve theproblems. Instead please take your time to read and understand the subject thoroughly. Once you understand how Normal Distributions works, you can solve most of the problems without much difficulty. So this is really a test of your research and analytical skill, your patience, and perseverance.6.In order to understand Normal Distributions, you need to be familiar with basic statistics termssuch as mean, median, standard deviation, percentile, quartile, and etc. It is a good idea to read the subject Descriptive Statistics, the supplementary materials, thoroughly to refresh yourmemory.7.The more questions you answered correctly, the more credit you will get. The total credit, or perfectscore, of Part 1 is 180. The total credit, or perfect score, of Part 2, is 90. The problems are ordered by content, NOT DIFFICULTY. It is to your advantage to attempt problems from throughout the test. 8.You can seek help by reading books, searching the Internet, asking an expert, and etc. But you can’tdelegate this to someone else and turn in whatever he/she wrote for you. To make it clear, the purpose of the second-round contest is to test your ability to read and research. You need to be the one who understand the topics and solve the problems. You will be caught if it is not the case during theinterview.9.For Part 1, you can write in either English or Chinese.10.In Part 2, you are asked to write an essay. You have to write in English in Part 2.11.If you have any questions regarding the contest, please contact us at once atINFO@12.This document contains 16 pages in total, including this page.13.Submission of your answers:a)For all the questions below, login to your account at / , and enter youranswers. Answers written on this document will NOT be credited.b)You need to submit your answers no later than 12:00AM, Feb 7, 2016, Beijing Time. Latersubmission will not be accepted.Part 1 – The Mathematics of Normal DistributionThe following is an excerpt from some math book.The Mathematics of Normal Distributions, (see separate document).Understanding the above Chapter on “Normal Distributions” requires familiarity with basic statistics terms such as mean, median, standard deviation, percentile, quartile, and etc. It helps to read the following Chapter on “Descriptive Statistics”, the supplementary materials, if you need to refresh your memory of these terms.Descriptive Statistics, (the supplementary materials, see separate document).For all the questions below, login to your account at / , and enter your answers. Answers written on this document will NOT be credited.Question 1: (credit: 4)Hint:1Q is the first quartile.3Q is the third quartile. For the definition of 1Q and 3Q , see Section 14.3, of Chapter 14, “Descriptive Statistics.”Question 2: (credit: 4)Estimate the value of the standard deviation (rounded to the nearest inch) of a normal distribution with 81.2μ=inch and 394.7Q =inch.Hint:3Q is the third quartile. For the definition of 3Q , see Section 14.3, of Chapter 14, “Descriptive Statistics.”Question 3: (credit: 4)Note:1Q is the first quartile.3Q is the third quartile. For the definition of 1Q and 3Q , see Section 14.3, of Chapter 14, “Descriptive Statistics.”Note: For this question, please write your answer on file “high-school-answersheet.doc ”, downloadable together with this document at , and submit file “high-school-answersheet.doc ” at after you are done.Question 4: (credit: 4)Question 5: (credit: 4)Question 6: (credit: 4): Question 7: (credit: 4)Question 8: (credit: 4)Question 9: (credit: 4)Questions 10 & 11 refer to the following:Note: For the definition of percentile, see Section 14.3, of Chapter 14, “Descriptive Statistics.”Question 10: (credit: 6)Question 11: (credit: 6)Hint:(c) Rounded to the nearest pound.Question 12: (credit: 6)Hint:(b)Rounded to nearest integer.(c)Enter your answer as a decimal between 0 and 1, rounded to the nearest hundredth. Question 13: (credit: 6)Hint:(b)Rounded to nearest thousandth.(c)Rounded to the nearest hundredth.Question 14: (credit: 4)Questions 15-18:In Questions 15-18, you should use the table above to make your estimates.Note: For the definition of percentile, s ee Section 14.3, of Chapter 14, “Descriptive Statistics.”Question 15: (credit: 4)Question 16: (credit: 4)Consider again the distribution of weights of six-month-old baby boys in Question 15.Question 17: (credit: 6)Question 18: (credit: 8)Question 19: (credit: 4)Question 20: (credit: 20, note: paper with exceptional quality can get up to 40 credits)Note: You can write in either Engish or Chinese.Note: For this question, please write your answer on file “high-school-answersheet.doc”, downloadable together with this document at , and submit file “high-school-answersheet.doc” at after you are done.Question 21: (credit: 4)Question 22: (credit: 4)The two histograms below summarize the team payrolls in Major League Baseball (2008).Using the information in the figure, where did the median payroll of 2008 baseball teams fall?(a)Somewhere between $50 million and $80 million(b)Somewhere between $70 million and $80 million(c)Somewhere between $70 million and $100 million(d)Somewhere between $80 million and $100 millionQuestion 23: (credit: 6)This question refers to histograms with unequal class intervals.Question 24: (credit: 8)Note: For (c) and (d), please write your answer on file “high-school-answersheet.doc ”, downloadable together with this document at , and submit file “high-school-answersheet.doc ” at after you are done.Question 25: (credit: 4)Let A denote the mean of data set 12{,,...,}N x x x . Let B denote the mean of dataset 12{,,...,}N x c x c x c +++.(a) Find the relationship between A and B .a) A B <b) A B c =-c) A B c =+d) Nondeterministic(b) Find the mean of 12{,,...,}N x A x A x A ---.Question 26: (credit: 4)Let 1R and 1σdenote the range and standard deviation of data set 12{,,...,}N x x x , respectively. Let 2R and 2σdenote the range and standard deviation of data set 12{,,...,}N x c x c x c +++, respectively. (a) Find the relationship between 1R and 2R .a) Nondeterministicb) 1R = 2Rc) 1R = 2R + cd) 1R = 2R - c(b) Find the relationship between 1σand 2σ.a) Nondeterministicb) 1σ = 2σc) 1σ = 2σ+ cd) 1σ = 2σ- cQuestion 27: (credit: 20, note: paper with exceptional quality can get up to 40 credits)Note: You can write in either Engish or Chinese.Note: For this question, please write your answer on fi le “high -school-answersheet .doc”, downloadable together with this document at , and submit file“high-school-answersheet.doc” at after you are done.Question 28: (credit: 20, note: paper with exceptional quality can get up to 40 credits)Note: You can write in either Engish or Chinese.Note: For this question, please write your answer on fi le “high-school-answersheet.do c”, downloadable together with this document at , and submit file “high-school-answersheet.doc” at after you are done.Part 2 Essay (Credit: 90)Pros and Cons of Math Competitionsby Richard RusczykMathematics competitions such as MATH LEAGUE, MATHCOUNTS, and the American Mathematics Competitions are probably the extracurricular math programs with the widest participation. The most immediate value of these math contests is obvious –they pique students’ interest in mathematics and encourage them to value intellectual pursuits. Kids love games, and many will turn just about any activity into a contest, or in other words, something to get good at. Math contests thus inspire them to become good at mathematics just like sports encourage physical fitness. Eventually, students put aside the games. By then, hopefully an interest in the underlying activity has developed.Beyond encouraging an interest in mathematics, contests help prepare students for competition. For better or worse, much of life is competition, be it for jobs or resources or whatever. Competition of any sort trains students to deal with success and failure, and teaches them that effective performance requires practice. Moreover, nearly every interesting and worthwhile venture in life comes with some element of pressure; competition teaches students how to handle it.Despite all the benefits of math contests, they are not an unmitigated good. First of all, not all contests are designed well. Students shouldn’t take too seriously contests that greatly emphasize speed or memorization. Curricular contests (particularly calculus contests for high school students) can also be misleading, as they deepen the misconception that there is no more to math than what is in the classroom. Such contests run the risk of encouraging students to overvalue skills that aren’t nearly as valuable as the one asset a contest should help them develop – the ability to think about and solve complex problems.A second danger of contests is extending kids beyond their ability. Students should certainly be challenged with problems they can’t do from time to time, but if this happens consistently, the experien ce goes from humbling and challenging to humiliating and discouraging.A third potential pitfall, burnout, often comes on the heels of the first two. Participants in math contests are just as much at risk of burnout as musicians or athletes. Parents, teachers, and the students themselves should be on the lookout for signs of decreased interest, and they must be willing to back off and allow the student to rediscover an interest in mathematics on his or her own. Burnout is particularly pernicious because th e end result often isn’t a backlash against competition, but against math in general. Indeed, even students not involved in contests have to watch out for burnout, though the pressure of contests tends to encourage burnout more quickly than the classroom.These possible perils are usually more than offset not only by the values we’ve already mentioned, but also by the greatest asset of math contests - cooperation. These competitions bring together students of like interests and abilities, allowing them to form their own community in which they will find friendship, inspiration, and encouragement to a far greater degree than most of these students can find in the typical classroom. Whereas a student may be one of only three or four in her school who pursues math the way others play basketball, she will not find herself so lonely at a math contest, where she’ll find many kindred spirits.In summary, math contests are a tremendous social and intellectual opportunity for students, but exposingstudents to contests must be done wisely, else they become counterproductive to the goal of encouraging a lifelong interest in mathematics and other intellectual pursuits.Direction (Note: For this question, please write your answer on fi le “high-school-answersheet.doc”, downloadable together with this document at , and submit file“high-school-answersheet.doc” at after you are done.):In his article “Pros and Cons of Math Competitions”, Richard Rusczyk lists and explains the advantages and the disadvantages of math competitions. Are math contests beneficial or harmful to students? Everything always has two sides: the good and the bad. This proverb probably also applies to math competitions. Write a response in which you discuss whether students should be encouraged to participate in math competitions. You may draw examples from your reading, studies, experience, observations, and etc.Hint: Here are some questions to think about. You do not have to answer them, but they will help you to craft your response.1. What is the purpose of math contests?2. What are the good and the bad sides of math contests?3. How should you value your scores on math contests?4. What are the most important qualities that a successful mathematician should have?5. To a student who is highly interested in mathematics, what are much more important than math contests?。