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美国数学邀请赛(AIME Ⅱ 2016)试题与解答殷琦涛;朱兆和【期刊名称】《上海中学数学》【年(卷),期】2016(000)005【总页数】5页(P44-48)【作者】殷琦涛;朱兆和【作者单位】200001 上海市格致中学;200001 上海市格致中学【正文语种】中文1.InitiallyAlex,Betty,andCharliehadatotalof444peanuts.Charliehadthemostp eanuts,andAlexhadtheleast.Thethreenumbersofpeanutsthateachpersonhad formedageometricprogression.Alexeats5ofhispeanuts,Bettyeats9ofherpean uts,andCharlieeats25ofhispeanuts.Nowthethreenumbersofpeanutseachpers onhasformsanarithmeticprogression.FindthenumberofpeanutsAlexhadinitia lly.译：亚历克斯、贝蒂和查利共有444颗花生，其中查利的花生最多，亚历克斯的花生最少，这三个人的花生数构成一个等比数列.亚历克斯吃掉5颗花生，贝蒂吃掉9颗花生，查利吃掉25颗花生，这时三个人的花生数构成一个等差数列.问：亚历克斯开始时有几颗花生？解：设亚历克斯、贝蒂、查利开始时各有a、aq、aq2(a∈N*,q>1)颗花生，根据题意可以得，即可以得，解得或(舍)，代入(1)得a=108.所以亚利克斯开始时有108颗花生.2.There is a 40% chance of rain on Saturday and a 30% chance of rain on Sunday. However, it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is, whereaandbare relatively prime positive integers. Finda+b.译：已知星期六下雨的概率是40%，星期天下雨的概率是30%.当星期六下雨时，星期天也下雨的概率是当星期六不下雨时、星期天下雨的概率的两倍.设周末至少有一天下雨的概率为，其中a,b是互质的正整数，求a+b的值.解：设当星期六不下雨时，星期天下雨的概率为p，则星期六下雨时，星期天也下雨的概率为2P.则有(1-40%)·p+40%·2p=30%，解得.周末两天都不下雨的概率为，所以周末两天至少有一天下雨的概率为.则a+b=107.3.Letx,yand z be real numbers satisfying the system.,,..译：设实数x,y,z满足,,.求的值.解：由题得，三式相加得3xyz+log5xyz=378.因为函数f(t)=3t+log5t是(0,+∞)上的单调递增函数，又f(125)=378，所以xyz=125.代入求得log5x=-90，log5y=-41，log5z=134.得.4.Ana×b×crectangular box is built froma·b·cunit cubes. Each unit cube is colored red, green, or yellow. Each of thealayers of size 1×b×cparallel to the (b×c)-faces of the box contains exactly 9 red cubes, exactly 12 green cubes, and some yellow cubes. Each of theblayers of sizea×1×cparallel to the (a×c)-faces of the box contains exactly 20 green cubes, exactly 25 yellow cubes, and some red cubes. Find the smallest possible volume ofthe box.译：一个a×b×c的盒子由a·b·c个单位正方体组成，每个单位正方体被染成红色、绿色或黄色.平行于盒子面的a层(大小为1×b×c)中，每一层刚好有9个红色单位正方体，12个绿色单位正方体和若干个黄色单位正方体.而平行于盒子面的b层(大小为a×1×c)中，每一层刚好有20个绿色单位正方体，25个黄色单位正方体和若干个红色单位正方体.求盒子体积的最小值.解：设平行于盒子b×c面的每一层中，黄色正方体有x个，平行于盒子a×c面的每一层中，红色正方体有y个.由题意，盒子中绿色正方体为12a或20b个，所以12a=20b，即b.同理,9a=by,ax=25b，所以有x=y=15.那么bc=21+x=36,ac=45+y=60.因a、b、c都是正整数，由知b≥3.V=abc=60b≥180，当b=3，c=12,a=5时等号成立.所以盒子体积的最小值为180.5.TriangleABC0has a right angle atC0. Its side lengths are pairwise relatively prime positive integers, and its perimeter isp. LetC1be the foot of the altitude to, and forn≥2, letCnbe the foot of the altitude toin △Cn-2Cn-1B. The sumCn-1Cn=6p. Findp.译：在△ABC0中，角C0是直角，三边长是两两互质的正整数，周长为p.C1是AB边上的高的垂足，对n≥2，设Cn是△Cn-2Cn-1B的边Cn-2B上高的垂足.已知，求p的值.解：设△ABC0的边长分别为a、b、c(a、b、c是两两互质的正整数)，且a2+b2=c2.如图1，设∠B=θ，则C0C1=asinθ，CnCn+1=Cn-1Cncosθ(n∈N*)，所以数列是以cosθ为公比的等比数列，首项C0C1=asinθ.，所以，所以ab=6(c-a)(a+b+c)，即7ab=6(c2-a2+bc)，又c2-a2=b2，所以7a=6(b+c)(1). 在(1)两边同乘(c-b)得7a(c-b)=6(b+c)(c-b)，即7a(c-b)=6a2，所以6a=7(c-b)(2).由(1)、(2)得a.设a=84k(k∈N*)，则a=84k,b=13k,c=85k，又a,b,c是两两互质的正整数，所以k=1，即a=84,b=13,c=85，所以p=182.6..Then, wheremandnare relatively prime positive integers. Findm+n.译：已知多项式，设，若，其中m,n是互质的正整数，求m+n的值.解(x2-2x+6)(x6-2x3+6)(x10-2x5+6)(x14-2x7+6)(x18-2x9+6),因为在5个小括号中x的偶次幂的系数为正数，奇数幂的系数为负数，在Q(x)的展开式中，x的偶次幂的项是由5个小括号中，在偶数个括号中取奇次幂，剩下的括号中取偶次幂相乘而得，所以展开式中x的偶次幂的系数为正，同理,展开式中x的奇次幂的系数为负.所以，即m+n=275.7.SquaresABCDandEFGHhave a common center and∥. The area ofABCDis 2016, and the area ofEFGHis a smaller positive integer. SquareIJKLis constructed so that each of its vertices lies on a side ofABCDand each vertex ofEFGHlies on a side ofIJKL. Find the difference between the largest and smallest positive integer values for the area ofIJKL.译：如图2，正方形ABCD与正方形EFGH有公共的中心，且AB∥EF.正方形ABCD的面积为2016，正方形EFGH的面积是比2016小的正整数.正方形IJKL的每个顶点在正方形ABCD的边上，且正方形EFGH的每个顶点在正方形IJKL的边上.求正方形IJKL面积的最大整数值与最小整数值的差.解：设AI=a,BI=b，则，当a=b时，.就在此时S正方形EFGH=504满足条件.依条件，正方形ABCD、正方形IJKL、正方形EFGH的面积成等比数列，设公比为，则，所以n2|2016=122×14，推得n|12.要使得S正方形IJKL最大，则m=n-1，所以，此式关于n递增，所以当n=12时，S正方形IJKL的最大值为1848，则所求差为1848-1008=840.8.Find the number of setsof three distinct positive integers with the property that the product ofa,bandcis equal to the product of 11, 21, 31, 41, 51 and 61.译：设集合是由三个不同的正整数组成的集合，满足a,b,c的积等于11、21、31、41、51、61的积，求满足条件的集合个数.解：abc=11×21×31×41×51×61=32×7×11×17×31×41×61.满足此式的有序数组(a,b,c)共有组.但{a,b,c}中的a、b、c无序且互不相等，所以数组(a,b,c)不等于(3,3,7×11×17×31×41×61)与(1,1,32×7×11×17×31×41×61)及其排列.故集合{a,b,c}的个数为.9.The sequences of positive integers 1,a2,a3,… and 1,b2,b3,… are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Letcn=an+bn. There is an integerksuch thatck-1=100andck+1=1000. Findck.译：正整数数列1,a2,a3,…是递增的等差数列，正整数数列1,b2,b3,…是递增的等比数列，设cn=an+bn，若存在整数k，使得ck-1=100，且ck+1=1000，求ck 的值.解：设{an}的公差为d(d∈N*)，{bn}的公比为q(q∈N*,q≥2)，由题意得：，也即，依条件有k≥3.(1)k=3时，，解得q=9,d=90，此时ck=1+2d+q2=262；(2)k=4时，由(1)、(2)得⟹q4-2q2=801⟹(q2-1)2=802，此方程没有正整数解；(3)k≥5时，q5≤qk=999-kd<999，q只能取2或3，①当q=2时，，即(3k-8)d+603=0，又(3k-8)d>0，此式无解，②当q=3时，，即(8k-18)d+108=0，此式也无解.综上可知ck=262.10.TriangleABCis inscribed in circleω.Q. RaysCPandCQmeetωagain atSandT(other thanC), respectively.s. Findm+n.译：如图3，△ABC内接于圆ω，点P,Q在边AB上，且满足AP<AQ，射线CP,CQ分别交圆ω于点S,T(不同于点C).若AP=4,PQ=3,QB=6,BT=5，AS=7，那么，其中m,n是互质的正整数，求m+n的值.解：设∠ACP=α,∠PCQ=β,∠QCB=γ，△ACQ相似于△TBQ，所以，同理.⟹，⟹.又，所以，则，所以，即m+n=43.11.For positive integersNandk, defineNto bek-nice if there exists a positive integerasuch thatakhas exactlyNpositive divisors. Find the number of positive integers less than 1000 that are neither 7-nice nor 8-nice.译：对正整数N,k，若存在一个正整数a，使得ak的正因子个数刚好为N，则称N是k阶好数.在小于1000的正整数中，求既不是7阶好数，又不是8阶好数的数的个数.解：先证明引理,即N是k阶好数的充要条件是N≡1(modk).充分性：若N≡1(modk)，则N=km+1(k∈N)，设a=pm(p为素数)，则ak=pmk的正因子个数为km+1=N，所以N为k阶好数.必要性：设N是k阶好数，则存在a=p1α1·p2α2·…·pnαn，其中pi为两两互不相等的素数，αi∈N*(i=1,2,…n).则ak的正因子个数为，所以N≡1(modk).回到原题：在1，…，999中，7阶好数有个，8阶好数有个，56阶好数有个.所以满足条件的数有999-(143+125-18)=749个.12.The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.译：如图4，一个分成六个部分的环，用四种颜色染色，每个部分染一种颜色，且相邻的部分不能染同一种颜色，求不同的染色方法的种数.解：一个分成n个部分的环用四种颜色染色，相邻部分不同色，记不同的染色方法有an种，则a1=4,a2=4×3=12,a3=4×3×2=24.当n≥4时，将环的各个部分依次编号为1,2,3,…,n.在满足条件的染色方法下，先将第n部分去掉：(1)若第1部分与第n-1部分异色，则得到了一个有n-1部分的环的染色方法；反之，一个有n-1部分的环，在其第1部分与第n-1部分之间插入一段区域作为第n部分，则该部分有2种颜色可选.所以在这种情况下有2an-1种染色方法. (2)若第1部分与第n-1部分同色，则去掉第n部分后得到一个有n-2个部分的环的染色方法；反之，一个有n-2个部分的环，在其第n-2部分中插入一段区域，将第n-2部分分成两段，则新插入部分有三种不同的染色方法，所以在这种情况下有3an-2种染色方法.所以an=2an-1+3an-2(n≥4)，由a2=12,a3=24得a4=84,a5=240,a6=732.即不同的染色方法有732种.13.Beatrix is going to place six rooks on a 6×6 chessboard where both therows and columns are labeled 1 to 6; the rooks are placed so that no two rooks are in the same row or the same column. The value of a square is the sum of its row number and column number. The score of an arrangementof rooks is the least value of any occupied square. The average score over all valid configurations is, wherepandqare relatively prime positive integers. Findp+q.译：比阿特丽克斯在一个6×6的棋盘上放六个棋子，这些棋子中的任何两个既不同行也不同列，把棋盘的行和列用1，…,6编号.棋盘中每一格的“值”是它所在的行与列的编号数之和.在每种满足条件的放法中，每个棋子所占格子的“值”的最小值定义为这种放法的分数.若所有放法的分数的平均值是，其中p,q是互质的正整数，求p+q的值.解：用(a,b)表示棋子所占格子的位置，其中a、b分别表示该格所在的行、列的编号数.在每种满足条件的放法中，六个棋子的位置分别为(i,bi)(i=1,2,3,4,5,6)，其中b1,…,b6是1,…,6的一个排列.因为六个格子的“值”的平均值为=7，所以满足条件的放法的分数可能值为2,3,4,5,6,7.(1)分数为2时，b1=1，满足条件的放法有种.(2)分数为3时：①b1=2时，有种；②b1=3,4,5,6且b2=1时，有种，所以分数为3时的放法有120+96=216种.(3)分数为4时：①b1=3,则b2≥2，此时有种放法；②b1≥4时，又可分为b2=2，有放法，或b2≥3,b3=1，有种放法.所以分数为4的放法共有96+72+54=222种.(4)分数为5时：①b1=4时，则b2≥3且b3≥2，有种放法；②b1≥5，又可分为：b2=3,b3≥2，有种放法，或b2≥4,b3=2，有种放法，或b2≥4,b3≥3,b4=1，有种放法.所以分数为5的放法共有54+36+24+16=130种.(5)分数为6时：①b1=5时，则b2≥4,b3≥3,b4≥2，有种放法；②b1=6又可分为：b2=4,b3≥3,b4≥2，有种放法，或b2=5,b3=3,b4≥2，有种放法，或b2=5,b3=4,b4=2，有种放法，或b2=5,b3=4,b4=3,b5=1，只有1种放法.所以分数为6时的放法共有16+8+4+2+1=31种.(6)分数为7的放法仅有1种.综上=，所以p+q=371.14.Equilateral △ABChas side length 600. PointsPandQlie outside the plane of △ABCand are on opposite sides of the plane. Furthermore,PA=PB=PC, andQA=QB=QC, and the planes of △PABand △QABform a 120° dihedral angle (the angle between the two planes). There is a pointOwhose distance from each ofA,B,C,PandQisd. Findd.译：如图5，等边△ABC边长为600，P,Q两点位于△ABC所在平面异侧，满足PA=PB=PC，QA=QB=QC，二面角P-AB-Q为120°.若存在一点O，点O到A,B,C,P,Q的距离均为d，求d的值.解：设D为△ABC的中心，依条件P,D,Q三点共线，且O在PQ上.设PD=x,QD=y，可设x≥y，设AB中点为E，则，即100(x+y)=xy-3×104(1).又OA=OP，所以⟹xy=12×104，代入(1)得x+y=900，则.15.s. Findm+n.译：对1≤i≤215，设，且.正实数x1,x2,…,x216满足，且.设x2的最大值为，其中m,n是互质的正整数，求m+n的值.解，即,由柯西不等式，所以，由(1)，不等式取等号，所以，则，所以m+n=863.。
AMC竞赛是近年来热度比较高的竞赛，含金量也是非常高的，对于学生个人能力提升也有着很大的帮助!本文我们就详细介绍一下AMC12竞赛详情，一起来看看吧!美国高中数学竞赛由美国数学协会(MAA)组织，每年举办，针对12年级及以下学生(对应国内高一和高二学生)的数学竞赛，第一届AMC12最早追溯到于1950年举办的美国高中生数学考试AHSME，2000年AHSME正式改名为AMC12, 2002年开始AMC12分A赛和B赛，分别于每年的2月初和2月中举行，参赛者任选其中一项参加。
1.考试形式AMC12竞赛考试时间75分钟，一共25道选择题，问题的难度随着考试进行逐渐加大。
满分150分，2008年起，AMC12不再允许使用计算器。
AMC12的评分方式为：答对得6分，答错不得分，不答得1.5分，在2002-2006年间，不答的2.5分，2002年前，不答得2分。
AMC12参赛者的优胜者可受邀参加美国数学邀请赛(AIME)。
2.考试安排AMC12通常在每年11月举办AMC 10/12 A卷▶早期注册截止日期：2021年9月24日▶常规注册截止日期：2021年10月15日▶延迟注册截止日期：2021年10月22日▶比赛日期：2021年11月10日上午8点(美东时间) AMC 10/12 B卷▶早期注册截止日期：2021年10月1日▶常规注册截止日期：2021年10月22日▶延迟注册截止日期：2021 年10 月29 日▶比赛日期：2021年11月16日上午8点(美东时间) 3.考试内容AMC12 的考试内容包括(但不局限于)整数、分数、小数、百分数、比例、数论、日常的几何、面积、体积、概率及统计、逻辑推理等。
不需要任何微积分知识4.AMC12知识点分布在AMC10基础上新增：进阶代数：复杂不等式、调和不等式、轮换不等式、柯西不等式;复杂函数问题，反函数和符合函数，三角函数和差化积、积化和差，万能公式;复数，复平面，欧拉公式，蒂莫夫公式;数学归纳法、复杂数列和极限。
《数学奥林匹克报》 Mathematical Olympiad Express 26th Annual American Invitational Mathematics Examination 2008 第 26 届美国数学邀请赛(AIME1 2008 年 3 月 18 日注意: 1,直到监考老师给出信号方可答题. 2,本卷共 15 道题,答题时间 3 小时.每题的答案都是 000～999 之间的整数.如果某题你的答案是 7, 请在相应位置涂黑 007;如果某题你的答案是 43,请在相应位置涂黑 043.你答对的题数就是你的得分,每题没有部分得分,做错也不倒扣分. 3,可以使用草稿纸,方格纸,直尺,圆规,量角器,橡皮.特别地,不许使用计算器和计算机. 4,AIME 与 AMC10 或AMC12 的总分将用来决定美国数学奥林匹克(USAMO的参赛资格. 2008 第 37 届USAMO 于 4 月 29 日～30 日举行. 5,请把试题答案和相关信息填涂于 AIME 答题卡. 6,答题卡请用 2B 铅笔在相应的圆圈内涂黑,方框处可用签字笔书写. 7,答题卡不许折叠,不许使用涂改液,涂改带.若需修改请用橡皮擦拭干净后再修改. 8,请不要忘记答题卡背面的签名(先英文再中文"Yes"处也请涂黑. , 9,Last Name 姓,First Name名,Gender 性别,Female 女,Male 男简体中文版试题 1,在参加学校聚会的学生中,60%的学生是女生,40%的学生喜欢跳舞.随后又多了 20 名都会跳舞的男生,现在聚会的人中有 58%是女生.那么现在聚会的人中有多少人喜欢跳舞? 2,正方形 AIME 的边长为 10,等腰△ GEM 的底是 EM ,且△ GEM 与正方形 AIME 的公共部分的面积为 80.求△ GEM 的底 EM 上的高. 3,艾德和苏骑自行车的速度相等且恒定,他们慢跑的速度也是相等且恒定,他们游泳的速度也是相等且恒定.艾德在骑车 2 小时,慢跑 3 小时,游泳 4 小时后共行 74 千米,苏在慢跑 2 小时,游泳 3 小时, 骑车 4 小时后共行 91 千米.他们骑车,慢跑,游泳的速度都是以每小时整千米数行进.求艾德骑车, 慢跑,游泳的速度的平方和. 普及数学知识,传播奥林文化,快递竞赛信息.《数学奥林匹克报》 Mathematical Olympiad Express 2 2 4,存在唯一的正整数 x , y 满足方程 x + 84 x + 2008 = y .求 x + y . 5,圆锥的底面圆半径为r ,高为 h .把此圆锥的侧面平放于桌面,它沿着桌面无滑动地滚动.当圆锥再次回到起始位置时正好滚动 17 周,这时圆锥的底座边沿在桌面上形成一个以圆锥顶点为圆心的圆弧. 比值 h 能写成 m n 的形式,其中 m , n 是正整数且 n 不能被任何质数的平方整除.求 m + n . r 6,三角形阵列中的第一行数字按照奇数1,3,5,……,99 的递增顺序书写,在第一行下面的每一行比上一行少一个数字,最底部的一行只有一个数字.每行的相邻两个数字之和写在这两个数字中间的下一行的位置.这个阵列中有多少个数是67 的倍数? 1 4 3 8 5 ……… 12 …… …… … 7,设 Si 是满足条件100i ≤ n < 100 ( i + 1 的所有整数 n 的集合.例如, S 4 = {400, 401, 402, 在集合S0 , S1 , S 2 ,……, S999 中有多少个不包括完全平方数? 97 196 99 , 499} . 8,求出满足条件arctan 1 1 1 1 π + arctan + arctan + arctan = 的正整数 n . 3 4 5 n 4 9,同样规格的木箱的长宽高分别是 3 英尺,4 英尺,6 英尺.第一个木箱放置在地面上,其余九个木箱依次被放置于前一个木箱的上面,且每个木箱的朝向是被随机地选择来放置.设高度正好达 41 英尺的概率,这里 m , n 是互质的正整数.求 m . m 是使得堆放的木箱的 n 10,等腰梯形 ABCD 中, AD ‖ BC ,下底 AD 上的底角是π 3 .对角线长 10 21 . EA = 10 7 , ED = 30 7 .CF ⊥ AD 于 F .线段 EF 的长度可以表示为 m n 的形式,其中 m ,n 是正整数且 n 不能被任何质数的平方整除.求 m + n . 11,考察完全由字母 A 和 B 组成的具有如下性质的序列:每个连写的 A 有偶数长度,每个连写的 B 有奇普及数学知识,传播奥林文化,快递竞赛信息.《数学奥林匹克报》 Mathematical Olympiad Express 数长度.例如 AA , B , AABAA 都是这样的序列,而 BBAB 就不是这样的序列. 在长度为 14 的序列中有多少个具有这种性质? 12,在一段单车道单行线的长直高速公路上,汽车均以相同的速度且都遵循安全规则:每个 15 千米/小时的速度, 前一辆车的车尾到下一辆车的车头的距离是一辆车的长度 (不足 15 千米的按 15 千米/千米算, 因此,当前一辆车每小时行 52 千米时,前一辆车的车尾与后一辆车的车头将相距 4 个车长 .安放在路边的电子眼摄像头用来记录一小时内通过的车辆数目,假定每辆车是 4 米长,且汽车可以以任何速度穿过.设 M 是一小时内电子眼摄像头能够记录的通过的最大整数.求 M 被 10 除所得的商. 13,设 p ( x, y = a0 + a1 x + a2 y + a3 x + a4 xy + a5 y + a6 x + a7 x y + a8 xy + a9 y , 2 2 3 2 2 3 且 p ( 0, 0 = p (1, 0 = p ( 1, 0 = p ( 0,1 = p ( 0, 1 = p (1,1 = p (1, 1 = p ( 2, 2 =0. 对所有这样的多项式存在点 a b a b , 使得 p , =0, c c c c 这里 a , b , c 是正整数,且 a , c 互质, c >1.求 a + b + c . 14,⊙ ω 的直径为 AB ,延长 AB 到 C ,过 C 作 CT 切⊙ ω 于 T , AP ⊥ CT 于 P .设 AB =18, m 表示线段 BP 的最大值.求 m . 15, 正方形纸片的边长为 100, 在其每个角上按如下方式剪下一个楔形: 距离正方形顶点 17 处入刀下剪, 使得两刀口在正方形的对角线相交得 60°角(如图 ,接着把这个顶点处的两条切口折起并胶粘在一起,这样这张纸就做成一个纸盘,这个纸盘的侧面和底面所成的角不是直角.这个纸盘的高度(即纸盘最高处到纸盘底面的垂直距离可以写成 n m 的形式,这里 m ,n 是正整数,其中 m <1000,且 m 不被任何质数的 n 次方整除.求 m + n . 折痕折痕 30° 30°切口 2 切口 17 17 普及数学知识,传播奥林文化,快递竞赛信息.《数学奥林匹克报》 Mathematical Olympiad Express 2008 第26 届美国数学邀请赛(AIME1 American Invitational Mathematics Examination Tuesday,March 18,2008 1,Of the students attending a school party,60% of the students are girls,and 40% of the students like to dance. After these students are joined by 20 more boy students,all of whom like to dance,the party is now 58% girls. How many students now at the party like to dance? 2,Square AIME has sides of length 10 units. Isosceles triangle GEM has base EM , and the area common to triangle GEM and square AIME is 80 square units. Find the length of the altitude to EM in GEM . 3, and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim Ed at equal and constant rates. Ed covers 74 kilometers after biking for 2 hours, jogging for 3 hours, and swimming for 4 hours, while Sue covers 91 kilometers after jogging for 2 hours, swimming for 3 hours, and biking for 4 hours. Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour. Find the sum of the squares of Ed's biking,jogging,and swimming rates. 4,There exist unique positive integers x and y that satisfy the equation x + 84 x + 2008 = y . 2 2 Find x + y . 5, right circular cone has base radius r and height h . The cone lies on its side on a flat table. As the cone A rolls on the surface of the table without slipping, the point where the cone's base meets the table traces a circular arc centered at the point where the vertex touches the table. The cone first returns to its original position on the table after making 17 complete rotations. The value of h can be written in the form r m n , where m and n are positive integers and n is not divisible by the square of any prime. Find m + n . 6, triangular array of numbers has a first row consisting of the odd integers 1, 5, A 3, ……, in increasing order. 99 Each row below the first has one fewer entry than the row above it, and the bottom row has asingle entry. Each entry in any row after the top row equals the sum of the two entries diagonally above it in the row immediately above it. How many entries in the array are multiples of 67? 普及数学知识,传播奥林文化,快递竞赛信息.《数学奥林匹克报》Mathematical Olympiad Express 1 4 3 8 5 ………12 …… …… … 97 99 196 7,Let Si be the set of all integers n such that 100i ≤ n < 100 ( i + 1 . For example, S 4 is the set {400, 401, 402, , 499} . How many of the sets S0 ,S1 ,S2 , ……,S999 do not contain a perfect square? 8,Find the positive integer n such that arctan 1 1 1 1 π + arctan + arctan + arctan = . 3 4 5 n 4 9, identical crates each of dimensions 3 ft×4 ft×6 ft. The first crate is placed flat on the floor. Each of the Ten remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random. Let m be the probability that the stack of crates is exactly 41 ft tall, where m and n n are relatively prime positive integers.Find m . 10, Let ABCD be an isosceles trapezoid with AD ‖ BC whose angle at the longer base AD is π 3 . The diagonals have length 10 21 ,and point E is at dist ances10 7 and 30 7 from vertices A and D , respectively. Let F be the foot of the altitude fromC to AD . The distance EF can be expressed in the form m n , where m and n are positive integers and n is not divisible by the square of any prime. Find m + n . 11, Consider sequences that consist entirely of A's and B's and that have the property that every run of consecutive A's has even length,and every run of consecutive B's has oddlength.Examples of such sequences are AA, B, and AABAA,while BBAB is not such a sequence.How many such sequences have length 14? 12,On a long straight stretch ofone-way single-lane highway, cars all travel at the same speed and all obey the safety rule: the distance from the back of the car ahead to the front of the car behind is exactly one car length for each 15 kilometers per hour of speed or fraction thereof (Thus the front of a car traveling 52 kilometers per hour will be four car lengths behind the back of the car in front of it. A photoelectric eye by the side of the road counts the number of cars that pass in one hour. Assuming that each car is 4 meters long and that the cars can travel at anyspeed, let M be the maximum whole number of cars that can pass the 普及数学知识,传播奥林文化,快递竞赛信息.《数学奥林匹克报》 Mathematical Olympiad Express photoelectric eye in one hour. Find the quotient when M is divided by 10. 13,Let p ( x, y = a0 + a1 x + a2 y + a3 x + a4 xy + a5 y + a6 x + a7 x y + a8 xy + a9 y , 2 2 3 2 2 3 Suppose that p ( 0, 0 = p (1, 0 = p ( 1, 0 = p ( 0,1 = p ( 0, 1 = p (1,1 = p (1, 1 = p ( 2, 2 =0 There is a point a b a b , for which p , =0 for all such polynomials,where a , b ,and c are c c c c positive integers, a and c are relatively prime,and c >1. Find a + b + c . 14, Let AB be a diameter of circle ω . Extend AB through A to C . Point T lies on ω so that line CT is tangent to ω . Point P is the foot of the perpendicular from A to line CT . Suppose AB = 18, and let m denote the maximum possible length of segment BP . Find m . 2 15, square piece of paper has sides of length 100. From each corner a wedge is cut in the following manner: at A each corner, the two cuts for the wedge each start at distance 17 from the corner, and they meet on the diagonal at an angle of 60° (see the figure below. The paper is then folded up along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upper edges, can be written in the form n m , where m and n are positive integers,m <1000 , and m is not divisible by the n-th power of any prime. Find m + n . fold cut 30° 30° cut fold 17 17 普及数学知识,传播奥林文化,快递竞赛信息.。
美国数学竞赛AMC8 – 2008年真题解析(英文解析+中文解析)Problem 1Answer: BSolution:50-12-24=14中文解析：总共花的钱是：12+12*2=36元。
剩余50-36=14元。
答案是BProblem 2Answer: ASolution:We can derive that c=8,L=6, U=7,and E=1. Therefore, the answer is 8671.中文解析：这10个字母的对应关系是： B -0；E-1; S-2; ......K -9. 按照这个对应关系：C-8，L-6，U-7，E-1. 即8671. 答案是A。
Problem 3Answer: ASolution:We can go backwards by days, but we can also backwards by weeks. If we go backwards by weeks, we see that February 6 is a Friday. If we now go backwards by days, February 1 is a Sunday.中文解析：13日是周五，则13-7=6，即6日也是周五，则倒推2月1日是周日。
答案是A。
Problem 4Answer: CSolution:The area outside the small triangle but inside the large triangle is 16-1=15. This is equally distributed between the three trapezoids. Each trapezoid has an area of 15/3=5.中文解析：大三角形的面积等于小的等边三角形的面积加上3个梯形的面积。
据此，三个梯形的面积是16-1=15. 每个梯形的面积是15/3=5. 答案是C。
裘布依公式范文裘布依公式（also known as Juebu Yi formula）是一个用来解决二次多项式最优化问题的公式。
它最早由英国数学家艾伦·喬治·裘布依（Alan George Joubert）在1911年提出，得到了广泛的应用和研究。
裘布依公式在工程、经济学和运筹学等领域都有重要的应用。
x = \frac{-b ± \sqrt{b^2-4ac}}{2a}\]其中，a、b和c都是常数，且$a \neq 0$。
这是一个二次方程的解析解，可以得到方程的根x。
裘布依公式基于这个二次方程的特点，对于给定的a、b和c，可以通过计算出根的值来解决相应的最优化问题。
\underset{x}{\text{minimize}} \quad \frac{1}{2} x^TAx + bx \]其中，A是一个对称矩阵，x是一个向量，b是一个常数。
约束条件可以表示为：\begin{align*}Ax &\geq b \\Cx&=d\\Fx &\leq g\end{align*}\]其中，C和F是矩阵，d和g是向量。
裘布依公式可以用来求解这个二次规划问题的最优解。
具体来说，裘布依公式可以通过将二次规划问题转化为一个等价的线性规划问题来解决。
线性规划问题是一个目标函数为线性函数，约束条件为线性函数的最优化问题。
通过裘布依公式，可以将二次规划问题转化为一个等价的线性规划问题，然后利用线性规划的常规方法来求解最优解。
裘布依公式的应用不仅限于二次规划问题。
在经济学中，裘布依公式可以用来求解最大化利润的方程，最小化成本的方程等。
在工程学中，裘布依公式可以应用于优化问题的设计和优化，如最优化控制问题、最优化参数估计问题等。
总之，裘布依公式是一个重要的数学工具，可以用来解决二次多项式最优化问题。
它的应用广泛，并且不仅限于二次规划问题，还可以推广到更高次的多项式。
裘布依公式为解决最优化问题提供了一种有效的方法和工具。
2013I 1The AIME Triathlon consists of a half-mile swim,a 30-mile bicycle,and an eight-mile run.Tom swims,bicycles,and runs at constant rates.He runs five times as fast as he swims,and he bicycles twice as fast as he runs.Tom completes the AIME Triathlon in four and a quarter hours.How many minutes does he spend bicycling?2Find the number of five-digit positive integers,n ,that satisfy the following conditions:(a)the number n is divisible by 5,(b)the first and last digits of n are equal,and (c)the sum of the digits of n is divisible by 5.3Let ABCD be a square,and let E and F be points on AB and BC ,respectively.The line through E parallel to BC and the line through F parallel to AB divide ABCD into twosquares and two non square rectangles.The sum of the areas of the two squares is 910of thearea of square ABCD .Find AE EB +EB AE .4In the array of 13squares shown below,8squares are colored red,and the remaining 5squares are colored blue.If one of all possible such colorings is chosen at random,the probability that the chosen colored array appears the same when rotated 90◦around the central square is 1n ,where n is a positive integer.Find n .5The real root of the equation 8x 3−3x 2−3x −1=0can be written in the form3√a +3√b +1c,where a ,b ,and c are positive integers.Find a +b +c .6Melinda has three empty boxes and 12textbooks,three of which are mathematics textbooks.One box will hold any three of her textbooks,one will hold any four of her textbooks,and one will hold any five of her textbooks.If Melinda packs her textbooks into these boxes in random order,the probability that all three mathematics textbooks end up in the same box can be written as m n ,where m and n Are relatively prime positive integers.Find m +n .This file was downloaded from the AoPS Math Olympiad Resources PagePage 120137A rectangular box has width 12inches,length 16inches,and height m n inches,where m and n are relatively prime positive integers.Three faces of the box meet at a corner of the box.The center points of those three faces are the vertices of a triangle with an area of 30square inches.Find m +n .8The domain of the function f (x )=arcsin(log m (nx ))is a closed interval of length 12013,wherem and n are positive integers and m >1.Find the remainder when the smallest possible sum m +n is divided by 1000.9A paper equilateral triangle ABC has side length 12.The paper triangle is folded so that vertex A touches a point on side BC a distance 9from point B .The length of the line segment along which the triangle is folded can be written as m √p n ,where m ,n ,and p are positive integers,m and n are relatively prime,and p is not divisible by the square of any prime.Find m +n +p.B CB AC 10There are nonzero integers a ,b ,r ,and s such that the complex number r +si is a zero of thepolynomial P (x )=x 3−ax 2+bx −65.For each possible combination of a and b ,let p a,b be the sum of the zeroes of P (x ).Find the sum of the p a,b ’s for all possible combinations of a and b .11Ms.Math’s kindergarten class has 16registered students.The classroom has a very largenumber,N ,of play blocks which satisfies the conditions:(a)If 16,15,or 14students are present,then in each case all the blocks can be distributed in equal numbers to each student,and (b)There are three integers 0<x <y <z <14such that when x ,y ,or z students are present and the blocks are distributed in equal numbers to each student,there are exactly three blocks left over.Find the sum of the distinct prime divisors of the least possible value of N satisfying the above conditions.201312Let P QR be a triangle with ∠P =75◦and ∠Q =60◦.A regular hexagon ABCDEF withside length 1is drawn inside P QR so that side AB lies on P Q ,side CD lies on QR ,andone of the remaining vertices lies on RP .There are positive integers a ,b ,c ,and d such that the area of P QR can be expressed in the form a +b √cd ,where a and d are relatively primeand c is not divisible by the square of any prime.Find a +b +c +d .13Triangle AB 0C 0has side lengths AB 0=12,B 0C 0=17,and C 0A =25.For each positiveinteger n ,points B n and C n are located on AB n −1and AC n −1,respectively,creating three similar triangles AB n C n ∼ B n −1C n C n −1∼ AB n −1C n −1.The area of the union of all triangles B n −1C n B n for n ≥1can be expressed as pq ,where p and q are relatively primepositive integers.Find q .14For π≤θ<2π,letP =12cos θ−14sin 2θ−18cos 3θ+116sin 4θ+132cos 5θ−164sin 6θ−1128cos 7θ+...and Q =1−12sin θ−14cos 2θ+18sin 3θ+116cos 4θ−132sin 5θ−164cos 6θ+1128sin 7θ+...so that P Q =2√27.Then sin θ=−m n where m and n are relatively prime positive integers.Find m +n .15Let N be the number of ordered triples (A,B,C )of integers satisfying the conditions(a)0≤A <B <C ≤99,(b)there exist integers a ,b ,and c ,and prime p where 0≤b <a <c <p ,(c)p divides A −a ,B −b ,and C −c ,and (d)each ordered triple (A,B,C )and each ordered triple (b,a,c )form arithmetic sequences.Find N .2013II 1Suppose that the measurement of time during the day is converted to the metric system so that each day has 10metric hours,and each metric hour has 100metric minutes.Digital clocks would then be produced that would read 9:99just before midnight,0:00at midnight,1:25at the former 3:00am ,and 7:50at the former 6:00pm .After the conversion,a person who wanted to wake up at the equivalent of the former 6:36am would have to set his new digital alarm clock for A:BC,where A,B,and C are digits.Find 100A +10B +C.2Positive integers a and b satisfy the conditionlog 2(log 2a (log 2b (21000)))=0.Find the sum of all possible values of a +b .3A large candle is 119centimeters tall.It is designed to burn down more quickly when it is first lit and more slowly as it approaches its bottom.Specifically,the candle takes 10seconds to burn down the first centimeter from the top,20seconds to burn down the second centimeter,and 10k seconds to burn down the k -th centimeter.Suppose it takes T seconds for the candle to burn down completely.Then T 2seconds after it is lit,the candle’s height in centimeters will be h .Find 10h .4In the Cartesian plane let A =(1,0)and B = 2,2√3 .Equilateral triangle ABC is con-structed so that C lies in the first quadrant.Let P =(x,y )be the center of ABC .Then x ·y can be written as p √q r ,where p and r are relatively prime positive integers and q is an integer that is not divisible by the square of any prime.Find p +q +r .5In equilateral ABC let points D and E trisect BC .Then sin (∠DAE )can be expressed in the form a √b c ,where a and c are relatively prime positive integers,and b is an integer that is not divisible by the square of any prime.Find a +b +c .6Find the least positive integer N such that the set of 1000consecutive integers beginning with 1000·N contains no square of an integer.7A group of clerks is assigned the task of sorting 1775files.Each clerk sorts at a constant rate of 30files per hour.At the end of the first hour,some of the clerks are reassigned to another task;at the end of the second hour,the same number of the remaining clerks are also reassigned to another task,and a similar reassignment occurs at the end of the third hour.The group finishes the sorting in 3hours and 10minutes.Find the number of files sorted during the first one and a half hours of sorting.20138A hexagon that is inscribed in a circle has side lengths 22,22,20,22,22,and 20in that order.The radius of the circle can be written as p +√q ,where p and q are positive integers.Find p +q .9A 7×1board is completely covered by m ×1tiles without overlap;each tile may cover any number of consecutive squares,and each tile lies completely on the board.Each tile is either red,blue,or green.Let N be the number of tilings of the 7×1board in which all three colors are used at least once.For example,a 1×1red tile followed by a 2×1green tile,a 1×1green tile,a 2×1blue tile,and a 1×1green tile is a valid tiling.Note that if the 2×1blue tile is replaced by two 1×1blue tiles,this results in a different tiling.Find the remainder when N is divided by 1000.10Given a circle of radius √13,let A be a point at a distance 4+√13from the center O of thecircle.Let B be the point on the circle nearest to point A .A line passing through the point A intersects the circle at points K and L .The maximum possible area for BKL can bewritten in the form a −b √c d ,where a ,b ,c ,and d are positive integers,a and d are relativelyprime,and c is not divisible by the square of any prime.Find a +b +c +d .11Let A ={1,2,3,4,5,6,7}and let N be the number of functions f from set A to set A suchthat f (f (x ))is a constant function.Find the remainder when N is divided by 1000.12Let S be the set of all polynomials of the form z 3+az 2+bz +c ,where a ,b ,and c are integers.Find the number of polynomials in S such that each of its roots z satisfies either |z |=20or |z |=13.13In ABC ,AC =BC ,and point D is on BC so that CD =3·BD .Let E be the midpoint of AD .Given that CE =√7and BE =3,the area of ABC can be expressed in the form m √n ,where m and n are positive integers and n is not divisible by the square of any prime.Find m +n .14For positive integers n and k ,let f (n,k )be the remainder when n is divided by k ,and forn >1let F (n )=max 1≤k ≤n 2f (n,k ).Find the remainder when 100 n =20F (n )is divided by 1000.15Let A,B,C be angles of an acute triangle withcos 2A +cos 2B +2sin A sin B cos C =158and cos 2B +cos 2C +2sin B sin C cos A =149.There are positive integers p ,q ,r ,and s for which cos 2C +cos 2A +2sin C sin A cos B =p −q √r s ,2013where p+q and s are relatively prime and r is not divisible by the square of any prime.Find p+q+r+s.。
天道2014 美国加州理工学院录取分析加州理工学院（California Institute of Technology, 缩写为Caltech）是美国的一所久负盛名的大学，位于加利福尼亚州的帕萨蒂纳(Pasadena)，创建于1891 年，最初的校名为Throop University，1920 年更名为California Institute ofTechnology。
下设的学院包括：生物与生物工程学院、工程与应用科学学院、化学与化学工程学院、地质学与行星科学学院、物理学、数学与天文科学学院、人文与社会科学学院。
专业设置方面，涵盖了航空学、天体物理学、天文学、材料科学、机械工程学、化学工程学、土木工程学、计算与神经系统学、计算机科学、电机工程学、地震学、地震工程学、生物学、分子生物学、神经科学、地质学、行星科学、化学、物理学、数学、应用物理学、控制与动态系统、管理学、经济学、社会学、人文与社会科学、人类学、历史、语言与文学、地球物理等。
科学院学生总数997 人（2013）。
能够在这里留下来的学生们，都是出了名的“工作狂”。
他们平均每个星期学习50 小时，每个人在毕业前都必须修满486 小时的课程，也就是说，平均每个学期有5 门课，1 年3 个学期共修15 门课——大部分人甚至比这更多。
另外，所有的学生都必须上5 门物理课，2 门化学、2 门数学、1 门生物，1 门天文或是地质，还要上两门实验课。
在这里，几乎人人抱着一种“不管教授们给我们布置什么难题都能完成”的豁出去了的态度。
在加州理工学院，一个人想完成所有功课是根本不可能的，因此，“集体做题”不但常见而且非常必要，整夜不睡地做题也是家常便饭。
因此有学生说，我最恨的事就是解题才解到一半，就发现太阳已经升起来了。
然而，学校对于每一年的新生，还是应该算是“仁慈”的。
新生们第一年上的课都不给以字母表示的成绩，只给打“通过”或“不通过”，这无疑是校方尽力想给学生们缓解点压力的结果。
一类四边形问题的解题策略美国数学邀请赛aime)试题新颖别致、内容广泛、灵活性强，是公认的既适应大众学生，又不失为较好区分度考查学生能力的好题.正因为aime的这种优越性，其中尤以平几题的这种命题风格，正在影响着我国的中考、高考选学内容平几题的命题，本文以aime 中有关特殊四边形形式出现的题目为例，试谈这类题的解题策略，以求对现实的教学有一定的借鉴意义.一、要重视利用特殊四边形的性质，尤其是与题中已知与所求直接关联的元素所拥有的性质例1 正方形aime的边长为10，等腰三角形gem的底是em，若△gem与正方形aime的公共部分的面积为80，求△gem的底边em 上的高.（2008年第26届aime图 1分析如图1所示，设等腰△gem底边上的高为h，ge和gm 分别交ai于点b，c，有△gem与正方形aime的公共部分的面积为80，又注意到s bcme=s gem-s gbc及ai∥em，得到△gem∽△gbc，故有h-10[]h=bc[]10，bc=10h-100[]h，则80=1[]2h-1[]210h-100[]h(h-10),化简，求得h=25.图 2例2 如图3所示，六边形abcdef被分成5个菱形p,q,r,s,t.菱形p,q,r,s是全等的，面积都为2006,令k为菱形t的面积，已知k是正整数，试求k的所有可能值的个数.（2006年第24届aime分析在本题中，首先想到的是要用已知四个全等的菱形面积数据去求另一菱形的面积，考虑到菱形面积的求法：一是一边与其边上的高乘积的一半，二是其对角线的乘积的一半，据此探索.如图所示，作菱形t的对角线，设z是其交点，x，y是菱形p 和菱形t的公共顶点，y在ab上，设yz=x,xy=z,则2006=fx·yz=zx,故z=2006[]x.则k=1[]2·(2yz)·(2xz)=1[]2·2x·(2z2-x2)=2x2006[]x2-x2=8024-4x 4.因为8024=89，所以有89个正的x使得8024-4x4是一个正整数的平方，因此k一共有89个可能的值.在本题求解中，充分利用了菱形各边相等与菱形对角线互相垂直平分等性质，如菱形afxy中，fx=xy；如菱形xywg中，yg⊥xw，且xz=zw，yz=zg.二、要善于运用特殊四边形的性质，构建起关系式，并且能注意到元素间的替代与转换，实现解题的顺畅进行例3 如图3，正方形abcd的边长为1，点e，f分别在bc，cd 上，且△aef是等边三角形，另有一小正方形以b为顶点，各边分别与abce的各边平行，且有一顶点在线段ae上，若小正方形的边长为a-b[]c，其中a，b，c均为正整数，且b不能被任何素数的平方整除，试求a+b+c的值.（2006年第24届aime分析此题是依赖于大正方形的边长求小正方形的边长，目标找出两者的联系.若设小正方形的边长为x，则利用已知构建起含有x的等式，尝试解之.图 3如图所示，设小正方形为bqpo，点q在线段ab上.设bq=x，则qp=x，aq=x tan75.因此1=aq+qb=x（tan 75+1）=（3+3）x，故x=1[]3+3=3-3[]6.所以a+b+c=12.由于，我国现行教材关于tan75tan75借助相似三角形的比例式与勾股定理关系式，建立方程求解.或者，为避开tan75aef改为顶角为30ae，af是∠dab 的三等分线），这时，小正方形的边长a-b[]c形式变为a-b[]c,试求a+b+c的值.当然，本题也可演变为以下形式：等边三角形abc边长为2，等腰直角三角形def的顶点d在bc的中点上，边ef平行bc，另有一小正方形ghjk的一边与ef重合，另两顶点k，j分别在ab，ac边上，试求小正方形的边长.例4 如图4，在长方形abcd中，ab=12，bc=10.点e，f在长方形abcd内，且满足be=9,df=8，be∥df，ef∥ab，直线be与线段ad相交，线段ef的长度可表示为mn-p的形式，其中m,n,p都是正整数，且n不能被任何质数的平方整除.求m+n+p.（2011年第29届aime图 4分析如图4所示，设be与ad交于g，df与bc交于h，ef分别交ad于m，交bc于n，过e作ek⊥ab于k，过f作fj⊥cd 于j.因为bg∥df，dm∥bn，有△dfm∽△gem∽△ben，于是，df[]be=dm[]bn，8[]9=dm[]bn，8+9[]9=dm+bn[]bn=cn+bn[]bn=bc[]dn，bn=9[]17bc=90[]17.在rt ebk中，bk=be2-ek2=be2-bn2=92-90[]172=27[]1721.同理可得，dj=24[]1721.所以，ef=ne+mf-mn=bk+dj-ab=27[]1721+24[]1721-12=321-12，故有m+n+p=36.三、要充分挖掘特殊四边形隐含的相关性质，其中不添置适当的辅助线，将其性质显性化后，搭建起已知与所求之间的桥梁，帮助解题例5 如图5所示，将一张长方形纸abcd中顶点为b的一角折起来，使得顶点b与ad边上的点b′重合，设折痕为ef，其中e点在ab边上，f点在cd边上，已知ab=8，be=17，cf=3，长方形abcd的周长为n[]m，其中m,n为互素的正整数，试求m+n的值.（2004年第22届aime图 5分析注意到折叠产生了以ef为对称轴的轴对称图形，有b′e=be=17.在rt eab′中，由勾股定理求得ab=15；连接bb′，有ef⊥bb′，作fg∥cb交ab于g，显然有ge=17-3=14，△egf∽△b′ab.故fg[]ba=ge[]ab′，解得fg=70[]3.所以，长方形abcd的周长为70[]3+25×2=290[]3，则m+n=293.关于图形的折叠与旋转是近年中考题中命题概率高的一个趋向，它将静态的平面图形置于一定的运动变化中，在不同的对称中，蕴含了相等、全等、垂直、平行、相似等众多关系，考查学生的观察、分析能力.本题的关键是利用轴对称的关系，找到△egf∽△b′ab.本题的另一解法是：设长方形对边ad=bc=x,b′d2+df2=b′f2=b′c′2+c′f2=bc2+cf2,即有(x-15)2+(25-3)2=x2+32,求得x=70[]3.。
1.你擦一下窗好吗？（mind）Do you mind／Would you mind cleaning the window？2．有困难请向警察求助。
（turn to）Turn to the police for help when (you are )in trouble3．他一到上海就和我们取得了联系。
（contact）He contacted us as soon as he arrived in Shanghai4．演出还没有结東，孩子们就睡着了。
（before）The children had fallen asleep/gone to sleep before the performance ended/was over.5．导演得知自己的影片获奖，感到无比自豪。
（award）Learning that he was awarded /won an award for his film, the director felt very proudof himself/was full of pride6．这些十八世纪的油画保存得这样好，使参观者大为惊以。
（so...that)These eighteenth-century oil paintings have been/are preserved so well that the visitors are greatly amazed/ impressed.2008上海高考1．我们打篮球的时间到了。
（time）It’s time for us to play basketball2．他设法把游客及时送到了机场。
（manage）He managed to send the tourists to the airport in time3．你今晚能米参加我的生日聚会？（possible）Is it possible for you to attend my birthday party this evening?4．应该鼓励年轻人按照自己的特长选择职业。
《数学奥林匹克报》 Mathematical Olympiad Express 26th Annual American Invitational Mathematics Examination 2008 第 26 届美国数学邀请赛(AIME1 2008 年 3 月 18 日注意: 1,直到监考老师给出信号方可答题. 2,本卷共 15 道题,答题时间 3 小时.每题的答案都是 000～999 之间的整数.如果某题你的答案是 7, 请在相应位置涂黑 007;如果某题你的答案是 43,请在相应位置涂黑 043.你答对的题数就是你的得分,每题没有部分得分,做错也不倒扣分. 3,可以使用草稿纸,方格纸,直尺,圆规,量角器,橡皮.特别地,不许使用计算器和计算机. 4,AIME 与 AMC10 或AMC12 的总分将用来决定美国数学奥林匹克(USAMO的参赛资格. 2008 第 37 届USAMO 于 4 月 29 日～30 日举行. 5,请把试题答案和相关信息填涂于 AIME 答题卡. 6,答题卡请用 2B 铅笔在相应的圆圈内涂黑,方框处可用签字笔书写. 7,答题卡不许折叠,不许使用涂改液,涂改带.若需修改请用橡皮擦拭干净后再修改. 8,请不要忘记答题卡背面的签名(先英文再中文"Yes"处也请涂黑. , 9,Last Name 姓,First Name名,Gender 性别,Female 女,Male 男简体中文版试题 1,在参加学校聚会的学生中,60%的学生是女生,40%的学生喜欢跳舞.随后又多了 20 名都会跳舞的男生,现在聚会的人中有 58%是女生.那么现在聚会的人中有多少人喜欢跳舞? 2,正方形 AIME 的边长为 10,等腰△ GEM 的底是 EM ,且△ GEM 与正方形 AIME 的公共部分的面积为 80.求△ GEM 的底 EM 上的高. 3,艾德和苏骑自行车的速度相等且恒定,他们慢跑的速度也是相等且恒定,他们游泳的速度也是相等且恒定.艾德在骑车 2 小时,慢跑 3 小时,游泳 4 小时后共行 74 千米,苏在慢跑 2 小时,游泳 3 小时, 骑车 4 小时后共行 91 千米.他们骑车,慢跑,游泳的速度都是以每小时整千米数行进.求艾德骑车, 慢跑,游泳的速度的平方和. 普及数学知识,传播奥林文化,快递竞赛信息.《数学奥林匹克报》 Mathematical Olympiad Express 2 2 4,存在唯一的正整数 x , y 满足方程 x + 84 x + 2008 = y .求 x + y . 5,圆锥的底面圆半径为r ,高为 h .把此圆锥的侧面平放于桌面,它沿着桌面无滑动地滚动.当圆锥再次回到起始位置时正好滚动 17 周,这时圆锥的底座边沿在桌面上形成一个以圆锥顶点为圆心的圆弧. 比值 h 能写成 m n 的形式,其中 m , n 是正整数且 n 不能被任何质数的平方整除.求 m + n . r 6,三角形阵列中的第一行数字按照奇数1,3,5,……,99 的递增顺序书写,在第一行下面的每一行比上一行少一个数字,最底部的一行只有一个数字.每行的相邻两个数字之和写在这两个数字中间的下一行的位置.这个阵列中有多少个数是67 的倍数? 1 4 3 8 5 ……… 12 …… …… … 7,设 Si 是满足条件100i ≤ n < 100 ( i + 1 的所有整数 n 的集合.例如, S 4 = {400, 401, 402, 在集合S0 , S1 , S 2 ,……, S999 中有多少个不包括完全平方数? 97 196 99 , 499} . 8,求出满足条件arctan 1 1 1 1 π + arctan + arctan + arctan = 的正整数 n . 3 4 5 n 4 9,同样规格的木箱的长宽高分别是 3 英尺,4 英尺,6 英尺.第一个木箱放置在地面上,其余九个木箱依次被放置于前一个木箱的上面,且每个木箱的朝向是被随机地选择来放置.设高度正好达 41 英尺的概率,这里 m , n 是互质的正整数.求 m . m 是使得堆放的木箱的 n 10,等腰梯形 ABCD 中, AD ‖ BC ,下底 AD 上的底角是π 3 .对角线长 10 21 . EA = 10 7 , ED = 30 7 .CF ⊥ AD 于 F .线段 EF 的长度可以表示为 m n 的形式,其中 m ,n 是正整数且 n 不能被任何质数的平方整除.求 m + n . 11,考察完全由字母 A 和 B 组成的具有如下性质的序列:每个连写的 A 有偶数长度,每个连写的 B 有奇普及数学知识,传播奥林文化,快递竞赛信息.《数学奥林匹克报》 Mathematical Olympiad Express 数长度.例如 AA , B , AABAA 都是这样的序列,而 BBAB 就不是这样的序列. 在长度为 14 的序列中有多少个具有这种性质? 12,在一段单车道单行线的长直高速公路上,汽车均以相同的速度且都遵循安全规则:每个 15 千米/小时的速度, 前一辆车的车尾到下一辆车的车头的距离是一辆车的长度 (不足 15 千米的按 15 千米/千米算, 因此,当前一辆车每小时行 52 千米时,前一辆车的车尾与后一辆车的车头将相距 4 个车长 .安放在路边的电子眼摄像头用来记录一小时内通过的车辆数目,假定每辆车是 4 米长,且汽车可以以任何速度穿过.设 M 是一小时内电子眼摄像头能够记录的通过的最大整数.求 M 被 10 除所得的商. 13,设 p ( x, y = a0 + a1 x + a2 y + a3 x + a4 xy + a5 y + a6 x + a7 x y + a8 xy + a9 y , 2 2 3 2 2 3 且 p ( 0, 0 = p (1, 0 = p ( 1, 0 = p ( 0,1 = p ( 0, 1 = p (1,1 = p (1, 1 = p ( 2, 2 =0. 对所有这样的多项式存在点 a b a b , 使得 p , =0, c c c c 这里 a , b , c 是正整数,且 a , c 互质, c >1.求 a + b + c . 14,⊙ ω 的直径为 AB ,延长 AB 到 C ,过 C 作 CT 切⊙ ω 于 T , AP ⊥ CT 于 P .设 AB =18, m 表示线段 BP 的最大值.求 m . 15, 正方形纸片的边长为 100, 在其每个角上按如下方式剪下一个楔形: 距离正方形顶点 17 处入刀下剪, 使得两刀口在正方形的对角线相交得 60°角(如图 ,接着把这个顶点处的两条切口折起并胶粘在一起,这样这张纸就做成一个纸盘,这个纸盘的侧面和底面所成的角不是直角.这个纸盘的高度(即纸盘最高处到纸盘底面的垂直距离可以写成 n m 的形式,这里 m ,n 是正整数,其中 m <1000,且 m 不被任何质数的 n 次方整除.求 m + n . 折痕折痕 30° 30°切口 2 切口 17 17 普及数学知识,传播奥林文化,快递竞赛信息.《数学奥林匹克报》 Mathematical Olympiad Express 2008 第26 届美国数学邀请赛(AIME1 American Invitational Mathematics Examination Tuesday,March 18,2008 1,Of the students attending a school party,60% of the students are girls,and 40% of the students like to dance. After these students are joined by 20 more boy students,all of whom like to dance,the party is now 58% girls. How many students now at the party like to dance? 2,Square AIME has sides of length 10 units. Isosceles triangle GEM has base EM , and the area common to triangle GEM and square AIME is 80 square units. Find the length of the altitude to EM in GEM . 3, and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim Ed at equal and constant rates. Ed covers 74 kilometers after biking for 2 hours, jogging for 3 hours, and swimming for 4 hours, while Sue covers 91 kilometers after jogging for 2 hours, swimming for 3 hours, and biking for 4 hours. Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour. Find the sum of the squares of Ed's biking,jogging,and swimming rates. 4,There exist unique positive integers x and y that satisfy the equation x + 84 x + 2008 = y . 2 2 Find x + y . 5, right circular cone has base radius r and height h . The cone lies on its side on a flat table. As the cone A rolls on the surface of the table without slipping, the point where the cone's base meets the table traces a circular arc centered at the point where the vertex touches the table. The cone first returns to its original position on the table after making 17 complete rotations. The value of h can be written in the form r m n , where m and n are positive integers and n is not divisible by the square of any prime. Find m + n . 6, triangular array of numbers has a first row consisting of the odd integers 1, 5, A 3, ……, in increasing order. 99 Each row below the first has one fewer entry than the row above it, and the bottom row has asingle entry. Each entry in any row after the top row equals the sum of the two entries diagonally above it in the row immediately above it. How many entries in the array are multiples of 67? 普及数学知识,传播奥林文化,快递竞赛信息.《数学奥林匹克报》Mathematical Olympiad Express 1 4 3 8 5 ………12 …… …… … 97 99 196 7,Let Si be the set of all integers n such that 100i ≤ n < 100 ( i + 1 . For example, S 4 is the set {400, 401, 402, , 499} . How many of the sets S0 ,S1 ,S2 , ……,S999 do not contain a perfect square? 8,Find the positive integer n such that arctan 1 1 1 1 π + arctan + arctan + arctan = . 3 4 5 n 4 9, identical crates each of dimensions 3 ft×4 ft×6 ft. The first crate is placed flat on the floor. Each of the Ten remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random. Let m be the probability that the stack of crates is exactly 41 ft tall, where m and n n are relatively prime positive integers.Find m . 10, Let ABCD be an isosceles trapezoid with AD ‖ BC whose angle at the longer base AD is π 3 . The diagonals have length 10 21 ,and point E is at dist ances10 7 and 30 7 from vertices A and D , respectively. Let F be the foot of the altitude fromC to AD . The distance EF can be expressed in the form m n , where m and n are positive integers and n is not divisible by the square of any prime. Find m + n . 11, Consider sequences that consist entirely of A's and B's and that have the property that every run of consecutive A's has even length,and every run of consecutive B's has oddlength.Examples of such sequences are AA, B, and AABAA,while BBAB is not such a sequence.How many such sequences have length 14? 12,On a long straight stretch ofone-way single-lane highway, cars all travel at the same speed and all obey the safety rule: the distance from the back of the car ahead to the front of the car behind is exactly one car length for each 15 kilometers per hour of speed or fraction thereof (Thus the front of a car traveling 52 kilometers per hour will be four car lengths behind the back of the car in front of it. A photoelectric eye by the side of the road counts the number of cars that pass in one hour. Assuming that each car is 4 meters long and that the cars can travel at anyspeed, let M be the maximum whole number of cars that can pass the 普及数学知识,传播奥林文化,快递竞赛信息.《数学奥林匹克报》 Mathematical Olympiad Express photoelectric eye in one hour. Find the quotient when M is divided by 10. 13,Let p ( x, y = a0 + a1 x + a2 y + a3 x + a4 xy + a5 y + a6 x + a7 x y + a8 xy + a9 y , 2 2 3 2 2 3 Suppose that p ( 0, 0 = p (1, 0 = p ( 1, 0 = p ( 0,1 = p ( 0, 1 = p (1,1 = p (1, 1 = p ( 2, 2 =0 There is a point a b a b , for which p , =0 for all such polynomials,where a , b ,and c are c c c c positive integers, a and c are relatively prime,and c >1. Find a + b + c . 14, Let AB be a diameter of circle ω . Extend AB through A to C . Point T lies on ω so that line CT is tangent to ω . Point P is the foot of the perpendicular from A to line CT . Suppose AB = 18, and let m denote the maximum possible length of segment BP . Find m . 2 15, square piece of paper has sides of length 100. From each corner a wedge is cut in the following manner: at A each corner, the two cuts for the wedge each start at distance 17 from the corner, and they meet on the diagonal at an angle of 60° (see the figure below. The paper is then folded up along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upper edges, can be written in the form n m , where m and n are positive integers,m <1000 , and m is not divisible by the n-th power of any prime. Find m + n . fold cut 30° 30° cut fold 17 17 普及数学知识,传播奥林文化,快递竞赛信息.。
