2020年第38届美国数学邀请赛

2017-2018年度美国“数学大联盟杯赛”（中国赛区）题目翻译及解题tips【翻译】：2018与以下哪个数字相加的总和是偶数？The sum of…总和…；the even number偶数【翻译】：约翰和吉尔一共有92美元。

约翰的钱是吉尔的三倍。

问约翰有多少钱？①…has three times（倍数）as many(修饰可数名词)/much(修饰不可数名词)as…A的…是B的几倍②As···as···和什么一样多【翻译】：汤姆是一个篮球热爱者！在他的书中，他写了100次“ILOVENBA”（我爱NBA）。

问他写的第500个字母是什么。

（提示：本题考查周期循环规律题）【翻译】：一个长*宽为8*25的长方形和以下哪个长方形有相同的面积。

【翻译】：前100个正整数（1-100）的和与后50个正整数（51-100）的和之间的差是多少？①Positive difference···与···的差；②positive integers正整数【翻译】：你有一根10英尺长的杆子需要被切成10等份。

若每一份需要10秒去切，完成这份工作一共需要多少秒。

【翻译】：Amy将2018四舍五入约至十位（rounded···to the nearest tens）得到的数字与Ben将2018四舍五入约至百位得到的数字，这两个数字之和是多少？【翻译】：下列哪组数有最小公倍数？【翻译】：Dan每买2支铅笔的同时也会5支钢笔。

如果他买了10支铅笔，那他一共买了几支钢笔？【翻译】：星期四的20天后是星期几？【翻译】：下列哪个角的度数最小？①an obtuse钝角②an acute锐角③a right直角④a stright平角【翻译】：我们班的每位学生都要轮流喊一个整数。

第一个人喊的是1。

《et■较牧拉》2020年第3期美国国际数学奥林匹克国家队的成就、经验与启示*张丽玉1何忆捷2熊斌3(华东师范大学数学科学学院，上海200241 )摘要：美国国际数学奥林匹克（IMO)国家队在历届国际数学奥林匹克中取得了突 出的成就，特别是近几年屡次获得团体第一名。

美国IMO国家队在准备国际数学奥林匹 克的过程中积累了丰富的经验，特别是在人才保障措施、教练团队建设、队员选拔集训以 及竞赛赛题命制等方面形成了不少独具特色的经验。

学习和借鉴美国IMO国家队及其 数学资优教育形成的一系列优良传统和做法对我国IMO国家队以及数学资优教育的发 展大有裨益关键词：国际数学奥林匹克（IMO);美国IMO国家队；资优生教育2〇19年2月25日，罗马尼亚大师杯（Romanian M aster of Mathematics)数学竞赛成绩揭晓，中国国家代表队获得团体第六名，个人成绩最高排名第15名，仅获4银1铜，无 缘金牌，而美国国家代表队则又一次夺得了团体第一名。

111美国队的突出表现再一次引 起了人们的强烈关注。

事实上,近年来美国队在各种国际中学生数学竞赛中取得了极为 突出的成绩。

世界上最具影响力的中学生数学竞赛是国际数学奥林匹克（International MathematiMl Olympiad，简称IM O)，自从1959年第一届IMO举办以来，国际数学奥林匹克 的影响力越来越大，参赛的国家也越来越多。

截至2019年8月，国际数学奥林匹克已经 成功举办了 6〇届，参赛代表队超过100个。

我国自1986年正式参加IMO以来，取得了举 世瞩目的成绩，特别是在2000 _ 2010年间更是9次获得团体第一。

然而，在最近的几届 1M0竞赛中成绩最突出的要数美国IM0国家队。

美国1M0国家队自从1974年参赛以来 也取得了优异的成绩，特别是最近几年,美国队数次在国际数学奥林匹克竞赛中获得团体 第一名。

美国队优异成绩的背后离不开其高质量的数学资优教育体系，尤其是美国形成*基金项目：上海市核心数学与实践重点实验室课题—数学实践（课题编号：18dz()2271000 )〇作者简介：张丽玉，华东师范大学数学科学学院博士研究生；何忆捷，华东师范大 学数学科学学院讲师；熊斌，华东师范大学数学科学学院教授,博士生导师。

AMC8（美国数学邀请赛）AMC8(美国数学邀请赛）AMC8（American Mathematics Competition8）AMC8是美国初中数学竞赛，是针对八年级以下学生的数学科测试，有些小学四～六年级的优秀学生也可以参加，该竞赛开始于1985年，于每年11月中旬的一个星期二举行。

AMC8竞赛内容与美国7、8年级数学大纲相对应，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。

美国数学协会（MAA）组织AMC8竞赛的目的是通过这样一种对学生有吸引力的考试，增加学生在数学方面的兴趣及学习数学的热情，促进学生学习中学数学必修课程之外的数学内容，增强问题解决的能力，该考试给参加者提供了应用初中所学概念处理由易到难，并包含广泛应用的问题的机会，以使他们得到在初中数学课堂中所不能得到的解决问题的经验，获得高分的部分学生将受邀参加美国高中数学竞赛AMC10。

题数︰25题时间︰40分钟题型︰选择题满分︰25分成绩处理︰AMC总部计分方式︰答对一题一分，答错不倒扣美国数学竞赛amc8的常用数学英语单词美国数学竞赛amc8的常用数学英语单词数学mathematics, maths（BrE）, math（AmE）被除数dividend除数divisor 商quotient 等于equals, is equal to, is equivalent to 大于is greater than小于is lesser than大于等于is equal or greater than小于等于is equal or lesser than运算符operator数字digit数number自然数natural number公理axiom定理theorem计算calculation运算operation证明prove假设hypothesis, hypotheses（pl.）命题proposition算术arithmetic加plus（prep.）, add（v.）, addition（n.）被加数augend, summand加数addend和sum减minus（prep.）, subtract（v.）, subtraction（n.）被减数minuend减数subtrahend差remainder乘times（prep.）, multiply（v.）, multiplication（n.）被乘数multiplicand, faciend乘数multiplicator积product除divided by（prep.）, divide（v.）, division（n.）整数integer小数decimal小数点decimal point分数fraction分子numerator分母denominator比ratio正positive负negative零null, zero, nought, nil十进制decimal system二进制binary system十六进制hexadecimal system权weight, significance进位carry截尾truncation四舍五入round下舍入round down上舍入round up有效数字significant digit无效数字insignificant digit代数algebra公式formula, formulae（pl.）单项式monomial多项式polynomial, multinomial系数coefficient未知数unknown, x-factor, y-factor, z-factor 等式，方程式equation一次方程simple equation二次方程quadratic equation三次方程cubic equation四次方程quartic equation不等式inequation阶乘factorial对数logarithm指数，幂exponent乘方power二次方，平方square三次方，立方cube四次方the power of four, the fourth power n次方the power of n, the nth power开方evolution, extraction二次方根，平方根square root三次方根，立方根cube root四次方根the root of four, the fourth root n次方根the root of n, the nth root集合aggregate元素element空集void子集subset交集intersection并集union补集complement映射mapping函数function定义域domain, field of definition值域range常量constant变量variable单调性monotonicity奇偶性parity周期性periodicity图象image数列，级数series微积分calculus微分differential导数derivative极限limit无穷大infinite（a.）infinity（n.）无穷小infinitesimal积分integral定积分definite integral不定积分indefinite integral有理数rational number无理数irrational number实数real number虚数imaginary number复数complex number矩阵matrix行列式determinant几何geometry点point线line面plane体solid线段segment射线radial平行parallel相交intersect角angle角度degree弧度radian锐角acute angle直角right angle钝角obtuse angle平角straight angle周角perigon底base边side高height三角形triangle锐角三角形acute triangle直角三角形right triangle直角边leg斜边hypotenuse勾股定理Pythagorean theorem钝角三角形obtuse triangle不等边三角形scalene triangle等腰三角形isosceles triangle等边三角形equilateral triangle四边形quadrilateral平行四边形parallelogram矩形rectangle长length宽width菱形rhomb, rhombus, rhombi（pl.）, diamond 正方形square 梯形trapezoid直角梯形right trapezoid等腰梯形isosceles trapezoid五边形pentagon六边形hexagon七边形heptagon八边形octagon九边形enneagon十边形decagon十一边形hendecagon十二边形dodecagon多边形polygon正多边形equilateral polygon圆circle圆心centre（BrE）, center（AmE）半径radius 直径diameter圆周率pi弧arc半圆semicircle扇形sector环ring椭圆ellipse圆周circumference周长perimeter面积area轨迹locus, loca（pl.）相似similar全等congruent四面体tetrahedron五面体pentahedron六面体hexahedron平行六面体parallelepiped立方体cube七面体heptahedron八面体octahedron九面体enneahedron十面体decahedron十一面体hendecahedron 十二面体dodecahedron 二十面体icosahedron多面体polyhedron棱锥pyramid棱柱prism棱台frustum of a prism 旋转rotation轴axis圆锥cone圆柱cylinder圆台frustum of a cone球sphere半球hemisphere底面undersurface表面积surface area体积volume空间space坐标系coordinates坐标轴x-axis, y-axis, z-axis 横坐标x-coordinate纵坐标y-coordinate原点origin双曲线hyperbola抛物线parabola三角trigonometry正弦sine余弦cosine正切tangent余切cotangent正割secant余割cosecant反正弦arc sine反余弦arc cosine反正切arc tangent反余切arc cotangent反正割arc secant反余割arc cosecant相位phase周期period振幅amplitude内心incentre（BrE）, incenter（AmE）外心excentre（BrE）, excenter（AmE）旁心escentre（BrE）, escenter（AmE）垂心orthocentre（BrE）, orthocenter（AmE）重心barycentre（BrE）, barycenter（AmE）内切圆inscribed circle 外切圆circumcircle统计statistics平均数average加权平均数weighted average方差variance标准差root-mean-square deviation, standard deviation 比例propotion百分比percent百分点percentage百分位数percentile排列permutation组合combination概率，或然率probability分布distribution正态分布normal distribution非正态分布abnormal distribution 图表graph条形统计图bar graph柱形统计图histogram折线统计图broken line graph曲线统计图curve diagram扇形统计图pie diagram。

首先感谢胡锐博士从美国引进了“数学大联盟杯赛”这么好的一个活动，它让中国的学生得以领略美国最为普及、最受欢迎的数学竞赛的原貌，同时也为数英双优的培训开辟了新的模式，尤其是目前在北京及全国都缺少这样一种培养方式。

古希腊伟大的哲学家亚里士多德曾说：“虽然数学没有明显提到善和美，但善和美也不能和数学完全分开，因为美的形式，就是‘秩序、匀称和确定性’，这些正是数学研究的原则。

”让孩子们年少时就开始接触数学，感受数学之美，从而培养他们对数学的兴趣，这在国外被称为“天才少年培养计划”。

虽然国内也很早就开始了这样的尝试，但像引进美国“数学大联盟杯赛”这种形式的活动还尚属首次。

美国“数学大联盟杯赛”至今已连续举办了35年，它不仅是美国及北美地区影响力最大的中小学数学赛事，也是一项具有世界影响力的中小学数学赛事。

杯赛每年举办一次，分为初赛(包括复赛)和决赛两个阶段。

初赛(包括复赛)试题灵活、生动，富有趣味性，同时也贴近生活。

让学生理解数学、欣赏数学，激励学生创新，更能激发学生学习数学的兴趣，培养学生主动探索的精神。

初赛的目的是普及数学。

决赛的试题由美国“数学大联盟杯赛”和斯坦福大学联合命题，具有很高的挑战性，着力培养学生创造性地解决问题和创新思维的能力(fun and creative problems that promote critical-thinking and problem-solving skills)，是数学超常少年展现才华的绝佳平台。

2012年伊始，在各方的努力之下美国“数学大联盟杯赛”得以成功登陆中国，中国赛区初赛于2012年2月25日举行，首次举办即得到广大中小学生的积极响应及踊跃参与，同时也涌现出大批数学能力与英语水平俱佳的优秀学生。

而随着2011-2012年度美国“数学大联盟杯赛”颁奖会的圆满结束，决赛的大幕正式拉起，我们也马不停蹄的投入到决赛和数学夏令营的组织工作中。

2011-2012年度美国“数学大联盟杯赛”中学组的决赛和数学夏令营是由美国“数学大联盟杯赛”和斯坦福大学联合举办，试题新颖、灵活、生动，富有趣味性和挑战性，培养学生创造性地解决问题和创新思维的能力。

美国数学邀请赛（AIME）
AIME全称American Invitational Mathematics Examination AIME是介于AMC10、AMC12及美国数学奥林匹克竞赛（USAMO）之间的一个数学竞赛，竞赛开始于1983年，在每年的3月底举行。

在AMC12测试中得分在100分以上或成绩在所有参赛者中排名前5％的学生，以及在AMC10测试中成绩在所有参赛者中排名前2.5％的学生可被邀请参加AIME数学竞赛。

AIME竞赛与美国高中数学竞赛及美国数学奥林匹克竞赛一样，考题基本可以使用中学数学方法解决。

AIME的考题相对于AMC10及AMC12而言更具难度，提供了更进一步的挑战和认可，而AIME成绩优异的美国籍学生将再被邀请参加USJMO和USAMO数学竞赛。

因此透过这一国际数学测试，也可让美国地区以外的，在数学方面有优异才能的学生，通过对比对自己的优异得到肯定。

题数︰15题
时间︰3小时
题型︰问答题(应用题)
满分︰15分
成绩处理︰AMC总部。

计分方式︰一题一分，答错不倒扣。

测试日期为每年3月的下旬举行。

2011AIME参赛分数线确定为：
2011年AMC10(A)分数达到117分； 2011年AMC10(B)分数达到117分； 2011年AMC12(A)分数达到93分； 2011年AMC12(B)分数达到97.5分的学生，取得参加AIME的资格； 2012年AMC12(B)分数达到99分的学生，取得参加AIME的资格。

母亲啊，您是我最好的老师——第38届国际数学奥赛金牌得主安金鹏和他的母亲１９９７年７月２８日，天津一中高三学生安金鹏在阿根廷举行的第３８届国际奥林匹克数学竞赛中喜获金牌，这是天津历史上第一个获得国际奥林匹克知识竞赛的冠军。

当我们前往天津武清县农村采访这位１９岁的数学奇才时，这位朴素的农村小伙子几乎是一字一泪地为我们讲述她的母亲哺育他成长的故事。

下面是他的自述咱不能让＂穷＂字把娃的前程耽误了１９９７年９月５日，是我离家去北京大学数学研究院报到的日子。

袅袅的炊烟一大早就在我家那幢破旧的农房上升腾，跛着脚的母亲在为我擀面，这面粉是母亲用５个鸡蛋和邻居换来的，她的脚是前天为给我多筹点学费，推着一平板车蔬菜在去镇里的路上扭伤的。

端着碗，我哭了。

我撂下筷子跪到地上，久久地抚摸着母亲肿得比馒头还高的脚，眼泪一滴滴地滚落在地上……我的家在天津武清县大友岱村，我有一个天下最好的母亲、她今年四十七岁，名叫李艳霞。

我家太穷了，我生下来的时候，奶奶便病倒在炕头上了，４岁那年，爷爷又患了支气管哮喘和半身不遂，家里欠的债一年比一年多。

７岁那年，我也上学了，我的学费是妈妈找人借的，可我发现，自从我上学以后，妈妈反而不爱坐在我身边看我念书了。

时间长了，我便明白了：我总是把同学扔掉的铅笔头捡回来，把它用细线捆在一根小棍上接着用，或者用橡皮把写过字的练习本擦干净，再接着用，急得妈妈有时为了买铅笔和本子的几分钱也要去找人借，我越是懂事，妈妈越是伤心，于是她就再不看我用捆着小棍的铅笔头做作业了。

不过妈妈也有高兴的时候，学校里不论大考小考，我总能考第一，数学总是满分。

在妈妈的鼓励下，我越学越快乐。

我真的不知道天下还有什么比读书更快乐的事。

我没上小学就学完了四则运算和分数小数：上小学就靠自学弄懂了初中的数理化；上初中就自学完卞高中的数理化课程。

１９９４年５月，天津市举办初中物理竞赛，我是天津市郊五县学生中唯一考进前三名的农村娃。

那年６月，我被着名的天津一中破格录取，我欣喜若狂地跑回家，可我没想到，当我把喜讯告诉家人时，他们的脸上竟会堆满愁云：奶奶去世不到半年，爷爷现在也生命垂危了。

题目：探讨2020中美洲及加勒比数学奥林匹克试解答---1. 引言2020年的中美洲及加勒比数学奥林匹克试解答，是一个备受关注的事件。

在数学领域，这些试题代表着挑战和机遇。

本文将深入探讨这些试题，给您带来一场关于数学奥赛的知识盛宴。

2. 试题概述2020年中美洲及加勒比地区的数学奥林匹克试题，涵盖了代数、几何、概率等多个领域。

每道题目都设计严谨，考察了考生对数学知识的深度和灵活运用能力。

试题内容不仅注重基础知识的考察，更将数学与现实生活相结合，增加了题目的趣味性和挑战性。

3. 题目解析（1）代数题在代数题部分，考生需要熟练运用代数运算、方程式求解等技巧，解决问题。

一道题目要求计算多项式的值，另一道题目则考察了对齐次方程组的解法。

这些题目旨在考察考生对代数知识的掌握情况，并培养其逻辑推理能力。

（2）几何题几何题部分则考察了考生的空间想象能力和几何推理能力。

题目设计涉及了线段、角度、多边形等几何形状，要求考生在解题过程中画出清晰准确的图形，并通过严密的证明、推理解决问题。

（3）概率题概率题部分考察了考生对概率的理解和应用能力。

题目设计涉及了随机事件、概率计算、独立事件等内容，要求考生通过分析情况、计算概率，解决实际问题。

4. 试题意义2020年中美洲及加勒比数学奥林匹克试题的设计，旨在提高学生的数学综合能力，培养他们的逻辑思维和解决问题的能力。

通过解答这些试题，考生不仅能够巩固已学的数学知识，还能够开拓思维，拓展数学视野，提高解决实际问题的能力。

5. 个人观点作为一名数学爱好者，我认为2020年中美洲及加勒比数学奥林匹克试题的设计非常有价值。

这些试题既考察了考生的数学基础知识，又注重了数学在现实生活中的应用。

通过解答这些试题，我深切感受到数学的魅力和实用性，也在实践中不断提高自己的数学素养。

6. 总结2020年中美洲及加勒比数学奥林匹克试题的内容丰富多样，题目设计严谨合理，体现了数学的广度和深度。

解答这些试题不仅有助于考生巩固数学知识，还能够培养他们的逻辑思维和问题解决能力。

2021年第4期252020美国数学竞赛（八年级）中图分类号:G 424.79文献标识码:A文章编号：1005 - 6416(2021 )04 - 0025 - 061. 卢卡在一次学校集资活动中卖自己制作的柠檬水.制作的过程中，他加入的水是糖 浆的4倍，糖浆是柠檬汁的2倍.他使用了 3 杯柠檬汁，则他能制作（ ）杯柠檬水.(A )6 (B )8 (C )12 (D )18 (E )242. 四位好友在周末为邻居整理庭院，分别获得15美元、20美元、25美元和40美元 的报酬.他们决定平分他们的酬劳，则酬劳为 40美元的人一共需要给其他人（）美元•(A )5 (B )10 (C )15(D )20(E )253.凯瑞在6 x 8平方英尺的花园中全部 种植草莓.凯瑞能够在一平方英尺上种植4 株草莓，并且每株草莓平均结10颗草莓.则 预计她能收获（）颗草莓•(A )560 (B )960 (C )1 120(D)l 920(E )3 8404.如图1，有三个边长依次增大的正六 边形.在原有正六边形点阵的基础上增多一 层，形成下一层正六边 _形内的点阵.则第四层 正六边形内有（ ）个点.(A ) 35(B ) 37(C ) 39(D ) 43(E ) 49图l5. —个容器内装有$的菠萝汁,将这些的果汁量是容器容量的百分之（）.(A )5 (B )10 (C )15(D )20(E )256J 、fi、C 、£>、£坐在一辆共有5节车厢 的小火车上，每节车厢只能坐一个人.已知£» 坐在最后一节车厢，4紧挨着£但在£后面， S 坐在4的前面车厢，在C 和fi之间至少间 隔一个人.则（）坐在正中间的位置.(A)/l (B)fi (C )C (D )Z) (E )E 7.在2 020到2 400之间有（ ）个整数，其四位数字顺序是严格递增的，且各不相同（如2 357是满足要求的一个整数）.(A )9 (B )10 (C )15(D )20(E )288•里卡尔多有2 020枚硬币，其中，一些是1分的，剩下的是5分的，且至少有一 枚1分的，也至少有一枚5分的.则里卡尔 多可能拥有的最大金额与最小金额的差是 ()分.(A )8 062 (B )8 068 (C )8 072(D )8 076(E )8 0829.阿卡什的生日蛋糕 是一个4 x 4 x 4的正方 体，如图2.蛋糕的顶部和 四个侧面都涂有糖霜，底 部没有糖霜.若将蛋糕切 成64个1 x 1 x 1的小正方体，则恰有两个面都有 糖霜的小正方体蛋糕有（(A )12 (B )16 (C )18菠萝汁平均倒人5个空杯子.则每个杯子中（D )20 (E)24j . ! = 1 x 2x ，..x n •若 12x ；V !，则iV的值为(C )12(A )6 (D )2012.对于正整数i正整数Af满足5!x 9!:()•(A )10 (B)ll (D )13(E )1413. 贾马尔的一个抽屉里有6只绿色的袜子，18只紫色的袜子和12只橘色的袜子. 现又添加了一些紫色的袜子后，发现此时从 抽屉里随机抽取一只袜子是紫色的概率为 60% .则贾马尔一共添加了（）只袜子.(A )6 (B )9 (C )12 (D )18 (E )2414.牛顿郡有20座城市，这些城市的人口数如图4.所有城市的人口数的平均数是 图4中的水平虚线.则这20座城市的总人数10. 扎拉搜集了 4颗弹珠，分别是玛瑙石的、大黄蜂宝石的、钢的和虎皮石的.她要在 一个架子上将这4颗弹珠排成一排展示，但 是不让钢弹珠和虎皮石弹珠挨在一起.则有 ()种排法.(A )6 (B )8 (C )12 (D )18 (E )2411. 放学后，玛雅和内奥米前往6英里处的一个海滩.玛雅骑自行车去，内奥米坐公交 车去.图3显示了他们行程中的时间与路程 关系.则内奥米和玛雅两人的平均速度的差 值为（）英里/小时•最接近选项（Q-<8 0006 0004 0002 000城市图4(A )65 000 (B )75 000 (C )85 000(D )95 000 (E ) 105 00015. 若;c 的15%等于y 的20%，则；k 是工的百分之（）.(A )5(B )35(C )75(D )133y (E )30016. 如图5，4,5,…，F 六个点代表1,2,…，6这六个不同的数字.五条直线中的每一 条都经过其中的一些点.将每条直线上的点 对应的数相加，可以得到五个和数，且这五个 数之和为47•则点S 对应的数为（）.(A)l (B )2 (C )3 (D )4 (E )517•在2 020的所有约数中，有（）个数的约数多于3个（如12有六个约数，分别为 1、2、3、4、6 和 12).(A )6 (B )7 (C )8 (D )9 (E )1018.如图6,矩形/IfiCD内接于半圆，为半圆的直径，似=16，/^=狀=9.则矩形的面积为（）•)英里/小时.内奥米117111/1」1/11/111 ./5 10 15 20 25时间（分钟）图3(B )12 (C )18(E )2426中等数学(W 梂)被圏2021年第4期27^16FDA E 图6(A )240 (B )248 (C )256(D )264(E )27219. 若一个数是由两个不同数字交替组成，则称此数为“交替数”（如2 020、37 373 都是交替数，而3 883、123 123不是交替数)• 则有（）个五位交替数能被15整除.(A )3 (B )4 (C )5 (D )6 (E )820. —位科学家走在一片森林中，她将一 排的5棵树的高度用整数记录下来.她注意 到每一棵树的高度要么是右边树高度的2 倍，要么是右边树高度的一半.不幸的是由于 下雨，她的记录 本上的一些数据 丢失了，其中，横 线处为丢失的数 据（如表1).基 于她的观察，她 能够重新确定这 些数.则这些树 的平均高度为 ()米.(A )22.2 (B )24.2 (C )33.2(D )35.2 (E )37.221.如图7,一个棋盘上有64个黑白相间的方格，格分别位于最下面一行、最上面一行.一枚棋 子从P 处开始移动，每一次移动都是移动 到所在格的上面一行 相邻的白色格内.则 图7一共有（）种不同的路径，使得恰用七步，棋子从P 移动到<?(图中是其中一种移表1第一颗树米第二棵树11米第三颗树米第四棵树米第五棵树米平均高度• 2米动的方式).(A )28(B )30 (C )32(D )33(E )3522.将正整数/V 输人到图8中的一台机 器中，机器按照程序输出一个正整数（如八= 7,则机器输出3x 7+1 =22).N若yv 为偶数/V"2若/V 为奇数3/V +1图8现将输出的结果继续输人到机器中，一 直重复五次，则输出为26,即7 — 22 — 11 — 34 -»■ 1752 — 26.若输人/V 值，进行六次操作，最终得到1，N —> _ —► _ —> _ —> _ —y _ —► 1则所有这些正整数iv的和为（)_(A )73 (B )74 (C )75(D )82(E )8323. 将五个不同的奖品分给三名学生，每名学生至少得到一个奖品•则共有（ ）种不同的分配方法.(A )120 (B )150 (C )180(D )210(E )24024.在一个大正方形的区域上铺有r a 2块灰色方砖，每一块方砖的边长均为s，每块方 砖周围均有宽度为d的边.图9是n = 3的 情况•当n = 24时，576块方砖覆盖了整 个大正方形区域的64%.则此时4的值S为（）.(A )蠢(B )16(C )笞(D )I (E 4□ □□□ □□ □ □□图928中等数学25.矩形、/?2和正方形S,、S2、S3按照图10方式摆放成一个长为3 322、宽为2 020的矩形•则S2的边长为（）•S,r2s2S,(A)651 (D)662m i〇(B)655(E)666(C)656参考答案1.E.由已知条件得7jC:糖楽:柠檬汁=4x2: 2:1.故卢卡一共制作了 3 x8 =24杯柠檬水.2. C.四位朋友一共获得报酬15 +20 +25 +40 = 100(美元），平分后每人获得25美元.因此，酬劳为40美元的人需要拿出15 美元.3. D.由题意，知凯瑞的花园为48平方英尺，能种植48 x4 = 192株草莓•因此，预计收获 192 x10 = 1 920 颗草莓.4. B.设第〃个六边形内有、个点.则"1=1，hn=hn-i +6(n - l) (n€Z+,n^2).故 /i2 = 1 + 6 = 7，/i3 = 7+12= 19,/i4 =19 + 18 =37.5. C.每个杯子里的果汁占容器容量的比例为3_ 1_ 3 _ 15=20=100'6. A.将车厢的顺序记为□□□□□，其中，最左边表示第一节车厢，最右边表示最后一节车厢.由第一个条件，知□□□□/).由第二个条件，知必为如下三种情况之一：□DEAD, \JEAUD,EAUBD.(1)由fi在4前，知排列为和，均不满足最后一个条件；(2) 得排列满足所有条件；(3) 显然不满足fi在4前.因此，唯一满足条件的顺序是B£4C£>，即坐在中间位置的是儿7. C.由于各位数字是严格递增的，于是，第二位数字不能为1和2.又四位数不能大于2 400,则第二位数字不能为4.从而，第二位数字只能为3.故后两位只能选自集合14，5，6，7,8，9丨.6x5因此，共有G1x215个.8. C.显然，里卡尔多拥有的5分最多,则他拥有的金额就最大，此时，有2 019x5+1分.类似地，他拥有的1分最多，则他拥有的金额就最小，此时，有2 019 x 1 +5分.因此，他能拥有的最大金额和最小金额的差为(2 019 x5 +1)-(2 019 xl +5)=2 019 x4 -4=2018 x4 =8 072.9. D.注意到，对于不含底面的小正方体，每一条棱的中间两块均恰有两个面有糖霜，一共有12 -4 =8条这样的棱•于是，有8 x2 = 16个满足要求的小正方体.当含有底面时，4个角的小正方体满足要求.2021年第4期29因此，共有16 +4 =20个小正方体蛋糕恰有两个面有糖霜.10. C.记s、:r分别表示钢的和虎皮石的，先排由题意，知这五个和数分别为4+ fi+ c、A+E + f\C + D + E、B + D、B+F.将其相加得2A +3B+2C+2D+2E+2F = 47s和r.因为s、;r不能相邻，所以，只能是snrn,snnr,nsnr,以及交换S、r的情况，共有3 x 2种方式.另外两个位置只有2种方式.因此，一共有2 x6 = 12种不同的排法.11.E.由图3,知内奥米在这6英里中用了 10分钟，于是，她的速度为+=36英里/小时.6"类似地，玛雅的速度为12英里/小时.故平均速度差为36 -12 =24英里/小时.12. A.注意到，5! =120.则 120 x9! =12 x/V!=> /V! =10 x9! =10!4 /V = 10.=?>2(A+B + C + D+E + F) + B =47.又4到F是1到6这六个不同的数，则A +B +C +D +E +F = 1+2 + •■• +6 =21.故 5=47 -21 x2 =5.17. B.由 2 020 = 22 x 5 x 101，知 2 020 有 12 个 约数，分别为 1、2、4、5、10、20、101、202、404、505、1010、2 020，其中，1、2、4、5、101的约数个数不大于3.故满足要求的约数有12 -5 =7个.18.A.记半圆圆心为〇,其直径为9 + 16+9 =34.于是，OC= 17.由对称性，得〇为A4的中点.贝I J 0/) = 〇4 =8.在Rt A〇况中，由勾股定理得13. B.当贾马尔添加了 x只紫色袜子之后，抽 屉里共有18 + x只紫色袜子，于是，袜子总数 为6+18 +12 + x= 36 + a：.这表明，他任意选取一只袜子是紫色的概率为.36 +x14. D.由图4,知虚线位于4 500和5 000正 中，即4 750处.于是,20座城市的总人数约 为 4 750 x 20 =95 000.15.C.由已知有〇. 15尤15 75：20 =1〇0 '：0. 20y.则LX 16. E.CD =V l72 -82= 15.故矩形的面积为16 x15 =240.19. B.一个数若能被5和3同时整除，则这个 数能被15整除.于是，这个数的个位数为5 或〇,且各位数字之和为3的倍数.若这个数的个位数字为〇，则由于此数为五位数，故首位也为0,显然不可能.从而，个位数必然为5，且这个数为5口5口5.记未知的数字为1则5 +1+5 + i+ 5=0(mod 3)=> 2^=0(mod 3).因为2与3互素，所以，$为3的倍数，即*只能取〇、3、6、9.故只有 50 505、53 535、56 565、59 595 这 四个数满足题意.30中等数学20. B.注意到，每棵树的高度均为整数.由第二棵树的高度为11米，知第一棵与 第三棵树的高度之和必为22米.下面考虑第四棵树的高度.它不可能为11米，否则，第五棵树只能 为22米，于是，五棵树的高度之和的平均值 的末尾不为0.2.从而，第四棵树的高度为44米.此时，前 四棵树高度之和为99,因此，第五棵树的高度不能为88,而只能为22米.故五棵树的平均高度为22^99 = 1|1=24_2(^)_21. A.除了位置P，棋子的每一个位置都是从它下面一行与之相邻的两个位置中的一个移 动过来的.这表明，从P移动到这个位置的方法数，就是移动到它下面一行与之相邻两格的方法数之和.如图11的白格中数值即表示移动到这个格的方法数.图11从而，移动到位置<?共有28种不同方法（即不同路径).22. E.从1开始逆向推导.考虑所有可能的情况.则|1(->|2}-^|4}->|1,8|^|2,16}-> |4,5,32|-^| 1,8,10,64}.例如，对于2，一定是由4得到（因为没有正整数使得3n +1 = 2)，但16可以由32 或5得到（因为^=3 x5 +1 =16).通过构造，最后一个集合就是通过六次操作得到1的所有数组成的集合.因此，所求结果为1 +8 + 10 +64 =83.23. B.若没有“每名学生至少得到一个奖品”的限制，一共有35 =243种不同分配方法.下面考虑其中不满足要求的情况，即至 少有一个人没有奖品的情况.若某名学生没有得到奖品，则有25 =32 种方式.因为是三名学生，所以，共有3 x32 =96 种，其中，恰有两名学生没有得到奖品的情况 被重复计数，而某两名学生没有得到奖品的方法只有1种，三名学生中选定两名学生有3种方法.从而，有1 x3 =3种方法•因此，满足题目要求的方法种数为243 -96 + 3 =150.24. E.由题意，知阴影部分的面积为(2如）2.注意到，没有被覆盖的部分有23 +2 = 25行和25列.于是，整个大正方形的边长为2知+251则(245)2{2As+25d)264100=>245 _ 8 _ 424s+25d= \0=~5=>120s =965+ 100^d67 =25'25. A.记正方形足的边长为~则5, + 52 + s3 =3 322.类似地 h_ s2 + s3= 2 020.两式相减得2s2=1 302 s2 =651.(吴建平提供张广民翻译)。

2020 AMC 10B Solution Problem1What is the value ofSolutionWe know that when we subtract negative numbers, .The equation becomesProblem2Carl has cubes each having side length , and Kate has cubes each having side length . What is the total volume of these cubes?SolutionA cube with side length has volume , so of these will have a total volume of .A cube with side length has volume , so of these will have a total volume of .~quacker88Problem 3The ratio of to is , the ratio of to is , and the ratioof to is . What is the ratio of toSolution 1WLOG, let and .Since the ratio of to is , we can substitute in the value of toget .The ratio of to is , so .The ratio of to is then so our answeris ~quacker88Solution 2We need to somehow link all three of the ratios together. We can start by connecting the last two ratios together by multiplying the last ratio by two., and since , we can link themtogether to get .Finally, since , we can link this again to get: ,so ~quacker88Problem4The acute angles of a right triangle are and , where andboth and are prime numbers. What is the least possible value of ?SolutionSince the three angles of a triangle add up to and one of the anglesis because it's a right triangle, .The greatest prime number less than is . If ,then , which is not prime.The next greatest prime number less than is . If ,then , which IS prime, so we have our answer ~quacker88 Solution 2Looking at the answer choices, only and are coprime to . Testing , the smaller angle, makes the other angle which is prime, therefore our answerisProblem5How many distinguishable arrangements are there of 1 brown tile, 1 purple tile, 2 green tiles, and 3 yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)SolutionLet's first find how many possibilities there would be if they were all distinguishable, then divide out the ones we overcounted.There are ways to order objects. However, since there's ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and ways to order the green tiles, we have to divide out these possibilities.~quacker88SolutionWe can repeat chooses extensively to find the answer. Thereare choose ways to arrange the brown tiles which is . Then from the remaining tiles there are choose ways to arrange the red tiles. And now from the remaining two tiles and two slots we can see there are two ways to arrange the purple and brown tiles, giving us an answerofProblem6Driving along a highway, Megan noticed that her odometershowed (miles). This number is a palindrome-it reads the same forward and backward. Then hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this -hour period?SolutionIn order to get the smallest palindrome greater than , we need to raise the middle digit. If we were to raise any of the digits after the middle, we would be forced to also raise a digit before the middle to keep it a palindrome, making it unnecessarily larger.So we raise to the next largest value, , but obviously, that's not how place value works, so we're in the s now. To keep this a palindrome, our number is now .So Megan drove miles. Since this happened over hours, she drove at mph. ~quacker88 Problem7How many positive even multiples of less than are perfect squares?SolutionAny even multiple of is a multiple of , so we need to find multiples of that are perfect squares and less than . Any solution that we want will be in theform , where is a positive integer. The smallest possible value isat , and the largest is at (where the expression equals ). Therefore, there are a total of possible numbers.-PCChess Problem8Points and lie in a plane with . How many locations forpoint in this plane are there such that the triangle with vertices , ,and is a right triangle with area square units?Solution 1There are options here:1. is the right angle.It's clear that there are points that fit this, one that's directly to the rightof and one that's directly to the left. We don't need to find the length, we just need to know that it is possible, which it is.2. is the right angle.Using the exact same reasoning, there are also solutions for this one.3. The new point is the right angle.(Diagram temporarily removed due to asymptote error)The diagram looks something like this. We know that the altitude tobase must be since the area is . From here, we must see if there are valid triangles that satisfy the necessary requirements.First of all, because of the area.Next, from the Pythagorean Theorem.From here, we must look to see if there are valid solutions. There are multiple ways to do this:We know that the minimum value of iswhen . In this case, the equationbecomes , which is LESSthan . . The equationbecomes , which is obviously greater than . We canconclude that there are values for and in between that satisfy the Pythagorean Theorem.And since , the triangle is not isoceles, meaning we could reflectit over and/or the line perpendicular to for a total of triangles this case.Solution 2Note that line segment can either be the shorter leg, longer leg or thehypotenuse. If it is the shorter leg, there are two possible points for that cansatisfy the requirements - that being above or below . As such, thereare ways for this case. Similarly, one can find that there are also ways for point to lie if is the longer leg. If it is a hypotenuse, then thereare possible points because the arrangement of the shorter and longer legs can be switched, and can be either above or below the line segment. Therefore, the answer is .Problem9How many ordered pairs of integers satisfy theequationSolutionRearranging the terms and and completing the square for yields theresult . Then, notice that can onlybe , and because any value of that is greater than 1 will causethe term to be less than , which is impossible as must be real. Therefore, plugging in the above values for gives the orderedpairs , , , and gives a totalof ordered pairs.Solution 2Bringing all of the terms to the LHS, we see a quadraticequation in terms of . Applying the quadratic formula, weget In order for to be real, which it must be given the stipulation that we are seekingintegral answers, we know that the discriminant, must benonnegative. Therefore, Here, we see that we must split the inequality into a compound, resultingin .The only integers that satisfy this are . Plugging thesevalues back into the quadratic equation, we see that both produce a discriminant of , meaning that there is only 1 solution for .If , then the discriminant is nonzero, therefore resulting in two solutions for .Thus, the answer is .~TiblisSolution 3, x firstSet it up as a quadratic in terms of y:Then the discriminant is This will clearly only yield real solutionswhen , because it is always positive. Then . Checking each one: and are the same when raised to the 2020th power:This has only has solutions , so are solutions. Next, if :Which has 2 solutions, so andThese are the only 4 solutions, soSolution 4, y firstMove the term to the other side toget . Because for all , then . If or , the right side is and therefore . When , the right side become , therefore . Our solutions are , , , . There are solutions, so the answer is - wwt7535Problem 10A three-quarter sector of a circle of radius inches together with its interior can be rolled up to form the lateral surface area of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubicinches?SolutionNotice that when the cone is created, the radius of the circle will become the slant height of the cone and the intact circumference of the circle will become the circumference of the base of the cone.We can calculate that the intact circumference of the circle is . Since that is also equal to the circumference of the cone, the radius of the cone is . We also have that the slant height of the cone is . Therefore, we use the Pythagorean Theorem to calculate that the height of the coneis . The volume of the coneis -PCChessSolution 2 (Last Resort/Cheap)Using a ruler, measure a circle of radius 4 and cut out the circle and then the quarter missing. Then, fold it into a cone and measure the diameter to be 6cm . You can form a right triangle with sides 3, 4, and then through the Pythagorean theorem the height is found tobe . The volume of a cone is . Plugging in we findProblem11Ms. Carr asks her students to read any 5 of the 10 books on a reading list. Harold randomly selects 5 books from this list, and Betty does the same. What is the probability that there are exactly 2 books that they both select?SolutionWe don't care about which books Harold selects. We just care that Bettypicks books from Harold's list and that aren't on Harold's list.The total amount of combinations of books that Betty can selectis .There are ways for Betty to choose of the books that are on Harold's list.From the remaining books that aren't on Harold's list, thereare ways to choose of them.~quacker88Problem12The decimal representation of consists of a string of zeros after the decimal point, followed by a and then several more digits. How many zeros are in that initial string of zeros after the decimal point?Solution 1Now we do some estimation. Notice that , which meansthat is a little more than . Multiplying itwith , we get that the denominator is about . Notice that whenwe divide by an digit number, there are zeros before the first nonzero digit. This means that when we divide by the digitinteger , there are zeros in the initial string after the decimal point. -PCChessSolution 2First rewrite as . Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to findthe number of digits in .and memming (alternatively use the factthat ),digits.Our answer is .Solution 3 (Brute Force)Just as in Solution we rewrite as We thencalculate entirely by hand, first doing then multiplying that product by itself, resulting in Because this is digits,after dividing this number by fourteen times, the decimal point is beforethe Dividing the number again by twenty-six more times allows a stringof zeroes to be formed. -OreoChocolateSolution 4 (Smarter Brute Force)Just as in Solutions and we rewrite as We can then look at the number of digits in powersof . , , , , ,, and so on. We notice after a few iterations that every power of five with an exponent of , the number of digits doesn't increase. This means should have digits since thereare numbers which are from to , or digits total. This means our expression can be written as , where is in therange . Canceling gives , or zeroes before the since the number should start on where the one would be in . ~aop2014 Solution 5 (Logarithms)Problem13Andy the Ant lives on a coordinate plane and is currently at facingeast (that is, in the positive -direction). Andy moves unit and thenturns degrees left. From there, Andy moves units (north) and thenturns degrees left. He then moves units (west) and againturns degrees left. Andy continues his progress, increasing his distance each time by unit and always turning left. What is the location of the point at which Andy makes the th leftturn?Solution 1You can find that every four moves both coordinates decrease by 2. Therefore, both coordinates need to decrease by two 505 times. You subtract, giving you theanswer of ~happykeeperProblem14As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region — inside the hexagon but outside all of the semicircles?Solution 1Let point A be a vertex of the regular hexagon, let point B be the midpoint of the line connecting point A and a neighboring vertex, and let point C be the second intersection of the two semicircles that pass through point A. Then, , since B is the center of the semicircle with radius 1 that C lies on, , since B is the center of the semicircle with radius 1 that A lies on,and , as a regular hexagon has angles of 120,and is half of any angle in this hexagon. Now, using the sinelaw, , so . Since the angles in a triangle sum to 180, is also 60. Therefore, is an equilateral triangle with side lengths of 1.Since the area of a regular hexagon can be found with the formula , where is the side length of the hexagon, the area of this hexagonis . Since the area of an equilateral triangle can be foundwith the formula , where is the side length of the equilateral triangle,the area of an equilateral triangle with side lengths of 1 is . Since the area of a circle can be found with the formula , the area of a sixthof a circle with radius 1 is . In each sixth of the hexagon, thereare two equilateral triangles colored white, each with an area of , and onesixth of a circle with radius 1 colored white, with an area of . The rest of the sixth is colored gray. Therefore, the total area that is colored white in each sixthof the hexagon is , which equals , and the total areacolored white is , which equals . Since the area colored gray equals the total area of the hexagon minus the area colored white,the area colored gray is , whichequals .Solution 2First, subdivide the hexagon into 24 equilateral triangles with side length1:Now note that the entire shadedregion is just 6 times this part:The entire rhombus is just 2 equilatrial triangles with side lengths of 1, so it has an area of:The arc that is not included has an area of:Hence, the area ofthe shaded region in that section is For a final areaof:Problem15Steve wrote the digits , , , , and in order repeatedly from left to right, forming a list of digits, beginning He thenerased every third digit from his list (that is, the rd, th, th, digits from the left), then erased every fourth digit from the resulting list (that is, the th, th, th, digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions ?Solution 1After erasing every third digit, the list becomes repeated. After erasing every fourth digit from this list, the listbecomes repeated. Finally, after erasing every fifth digit from this list, the list becomes repeated. Since this list repeats every digits andsince are respectively in we have that the th, th, and st digits are the rd, th, and thdigits respectively. It follows that the answer is~dolphin7Problem16Bela and Jenn play the following game on the closed interval of the real number line, where is a fixed integer greater than . They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in theinterval . Thereafter, the player whose turn it is chooses a real numberthat is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?SolutionNotice that to use the optimal strategy to win the game, Bela must select themiddle number in the range and then mirror whatever number Jennselects. Therefore, if Jenn can select a number within the range, so can Bela. Jenn will always be the first person to run out of a number to choose, so theanswer is .Solution 2 (Guessing)First of all, realize that the value of should have no effect on the strategy at all. This is because they can choose real numbers, not integers, so even if is odd, for example, they can still go halfway. Similarly, there is no reason the strategy would change when .So we are left with (A) and (B). From here it is best to try out random numbers and try to find the strategy that will let Bela win, but if you can't find it, realize thatit is more likely the answer is since Bela has the first move and thus has more control.Problem17There are people standing equally spaced around a circle. Each person knows exactly of the other people: the people standing next to her or him, as well as the person directly across the circle. How many ways are there forthe people to split up into pairs so that the members of each pair know each other?SolutionLet us use casework on the number of diagonals.Case 1: diagonals There are ways: either pairs with , pairs with , and so on or pairs with , pairs with , etc.Case 2: diagonal There are possible diagonals to draw (everyone else pairs with the person next to them.Note that there cannot be 2 diagonals.Case 3: diagonalsNote that there cannot be a case with 4 diagonals because then there would have to be 5 diagonals for the two remaining people, thus a contradiction.Case 4: diagonals There is way to do this.Thus, in total there are possible ways. Problem18An urn contains one red ball and one blue ball. A box of extra red and blue balls lie nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?SolutionLet denote that George selects a red ball and that he selects a blue one. Now, in order to get balls of each color, he needs more of both and .There are 6cases:(wecan confirm that there are only since ). However we canclump , ,and together since they are equivalent by symmetry.andLet's find the probability that he picks the balls in the order of .The probability that the first ball he picks is red is .Now there are reds and blue in the urn. The probability that he picks red again is now .There are reds and blue now. The probability that he picks a blue is .Finally, there are reds and blues. The probability that he picks a blue is . So the probability that the case happensis . However, since the case is the exactsame by symmetry, case 1 has a probability of chance of happening.andLet's find the probability that he picks the balls in the order of .The probability that the first ball he picks is red is .Now there are reds and blue in the urn. The probability that he picks blue is .There are reds and blues now. The probability that he picks a red is . Finally, there are reds and blues. The probability that he picks a blue is .So the probability that the case happensis . However, since the case is the exactsame by symmetry, case 2 has a probability of chance of happening.andLet's find the probability that he picks the balls in the order of .The probability that the first ball he picks is red is .Now there are reds and blue in the urn. The probability that he picks blueis .There are reds and blues now. The probability that he picks a blue is .Finally, there are reds and blues. The probability that he picks a red is .So the probability that the case happensis . However, since the case is the exactsame by symmetry, case 3 has a probability of chance of happening.Adding up the cases, we have ~quacker88 Solution 2We know that we need to find the probability of adding 2 red and 2 blue balls insome order. There are 6 ways to do this, since there are ways to arrange in some order. We will show that the probability for each of these 6 ways is the same.We first note that the denominators should be counted by the same number. This number is . This is because 2, 3, 4, and 5 represent how many choices there are for the four steps. No matter what the stepinvolves numbers to choose from.The numerators are the number of successful operations. No matter the order, the first time a red is added will come from 1 choice and the second time will come from 2 choices, since that is how many reds are in the urn originally. Thesame goes for the blue ones. The numerator must equal . Therefore, the probability for each of the orderingsof is . There are 6 of these, so the total probabilityis .Solution 3First, notice that when George chooses a ball he just adds another ball of the same color. On George's first move, he either chooses the red or the blue witha chance each. We can assume he chooses Red(chance ), and then multiply the final answer by two for symmetry. Now, there are two red balls andone blue ball in the urn. Then, he can either choose another Red(chance ), in which case he must choose two blues to get three of each, withprobability or a blue for two blue and two red in the urn, withchance . If he chooses blue next, he can either choose a red then a blue, or ablue then a red. Each of these has a for total of . The total probability that he ends up with three red and three blueis . ~aop2014 Solution 4Let the probability that the urn ends up with more red balls be denoted . Since this is equal to the probability there are more blue balls, the probabilitythere are equal amounts is . the probability no more blues are chosen plus the probability only 1 more blue is chosen. The firstcase, .The second case, . Thus,the answer is .~JHawk0224Solution 5By conditional probability after 4 rounds we have 5 cases: RRRBBB, RRRRBB,RRBBBB, RRRRRB and RBBBBB. Thus the probability is . Put .~FANYUCHEN20020715Edited by KinglogicSolution 6Here X stands for R or B, and Y for the remaining color. After 3 rounds one can either have a 4+1 configuration (XXXXY), or 3+2 configuration (XXXYY). Theprobability of getting to XXXYYY from XXXYY is . Observe that the probability of arriving to 4+1 configuration is ( to get from XXY toXXXY, to get from XXXY to XXXXY). Thus the probability of arriving to 3+2configuration is also , and the answer isSolution 7We can try to use dynamic programming to solve this problem. (Informatics Olympiad hahaha)We let be the probability that we end up with red balls and blue balls. Notice that there are only two ways that we can end up with red balls and blue balls: one is by fetching a red ball from the urn when wehave red balls and blue balls and the other is by fetching a blue ball from the urn when we have red balls and blue balls.Then wehaveThen we start can with and try to compute .The answer is .Problem19In a certain card game, a player is dealt a hand of cards from a deckof distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as . What is the digit ?Solution 1We're looking for the amount of ways we can get cards from a deck of ,which is represented by .We need to get rid of the multiples of , which will subsequently get rid of the multiples of (if we didn't, the zeroes would mess with the equation since you can't divide by 0), , leaves us with 17.Converting these into, we have~quacker88 Solution 2Since this number is divisible by but not , the last digits must be divisible by but the last digits cannot be divisible by . This narrows the options down to and .Also, the number cannot be divisible by . Adding up the digits, weget . If , then the expression equals , a multiple of . This would mean that the entire number would be divisible by , which is not what we want. Therefore, the only option is -PCChessSolution 3It is not hard to check that divides thenumber,As , using wehave .Thus , implying so the answer is .Solution 4As mentioned above,We can divide both sidesof by 10 to obtain which means is simply the units digit of the left-hand side. This valueisProblem20Let be a right rectangular prism (box) with edges lengths and ,together with its interior. For real , let be the set of points in -dimensional space that lie within a distance of some point . The volumeof can be expressed as ,where and are positive real numbers. What isSolutionSplit into 4 regions:1. The rectangular prism itself2. The extensions of the faces of3. The quarter cylinders at each edge of4. The one-eighth spheres at each corner ofRegion 1: The volume of is 12, soRegion 2: The volume is equal to the surface area of times . The surfacearea can be computed to be ,so .Region 3: The volume of each quarter cylinder is equal to . The sum of all such cylinders must equal times the sum of the edge lengths. This can be computed as , so the sum of the volumes of the quarter cylinders is , soRegion 4: There is an eighth of a sphere of radius at each corner. Since there are 8 corners, these add up to one full sphere of radius . The volume of thissphere is , so .Using these values,Problem21In square , points and lie on and , respectively, so that Points and lie on and , respectively, and points and lie on so that and . See the figure below. Triangle , quadrilateral ,quadrilateral , and pentagon each has area Whatis ?SolutionSince the total area is , the side length of square is . We see that since triangle is a right isosceles triangle with area 1, we can determinesides and both to be . Now, considerextending and until they intersect. Let the point of intersection be .We note that is also a right isosceles triangle with side and find it's area to be . Now, we notice that is also a rightisosceles triangle and find it's area to be . This is also equalto or . Since we are looking for , we want two times this. That gives .~TLiuSolution 2Since this is a geometry problem involving sides, and we know that is , we can use our ruler and find the ratio between and . Measuring(on the booklet), we get that is about inches and isabout inches. Thus, we can then multiply the length of by the ratioof of which we then get We take the square of that andget and the closest answer to that is . ~Celloboy (Note that this is just a strategy I happened to use that worked. Do not press your luck with this strategy, for it was a lucky guess)Solution 3Draw the auxiliary line . Denote by the point it intersects with , and by the point it intersects with . Last, denote by the segment , and by the segment . We will find two equations for and , and then solve for .Since the overall area of is ,and . In addition, the areaof .The two equations for and are then:Lengthof :Area of CMIF: .Substituting the first into the second,yieldsSolving for gives ~DrBSolution 4Plot a point such that and are collinear and extend line topoint such that forms a square. Extend line to meetline and point is the intersection of the two. The area of this square is equivalent to . We see that the area of square is , meaning each side is of length 2. The area of the pentagon is .Length , thus . Triangle is isosceles, and the area of this triangleis . Adding these two areas, we get . --OGBooger Solution 5 (HARD Calculation)We can easily observe that the area of square is 4 and its side length is 2 since all four regions that build up the square has area 1. Extend and let the intersection with be . Connect , and let the intersectionof and be . Notice that since the area of triangle is 1and , ,therefore . Let ,。