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AMC10每日一题（2001年真题#13）

2018-06-04
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　　AMC不但是美国顶尖数学人才的人才库，更为学校提供了解申请入学者在数学科目上的学习成就与表现评估。AMC成功地为许多学生因测验成绩优良而进入理想学校。藉由设计严谨的试题，达到激发应试者解决问题的能力，培养对数学的兴趣。试题由简至难兼具，使任何程度的学生都能感受到挑战，还可以筛选出特有天赋者。AMC10是针对高中一年级及初中三年级学生的数学测验，25题选择题、考试时间75分钟;包含演算概念理解的数学题型。AMC10的测验允许使用计算器(工程用计算器除外)。AMC10的主要目的是在刺激学生对数学的兴趣并且透过以选择题方式来开发学生对数学的才能;测验题型范围由容易到困难。参予AMC10的学生应该不难发现测验的问题都很具挑战性，但测验的题型都不会超过学生的学习范围。这项测验希望每个考生能从竞赛中享受数学。今天课窝小编为大家整理了AMC10真题练习，希望考生们认真阅读，能够对你的考试有所帮助。

2001 AMC 10 竞赛试题/第13题

Problem

Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?

$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 18$

Solution with Stars and Bars

Let the donuts be represented by $O$ s. We wish to find all possible combinations of glazed, chocolate, and powdered donuts that give us $4$ in all. The four donuts we want can be represented as $OOOO$ . Notice that we can add two "dividers" to divide the group of donuts into three different kinds; the first will be glazed, second will be chocolate, and the third will be powdered. For example, $O|OO|O$ represents one glazed, two chocolate, and one powdered. We have six objects in all, and we wish to turn two into dividers, which can be done in $\binom{6}{2}=15$ ways. Our answer is hence $\boxed{\textbf{(D)}\ 15}$ . Notice that this can be generalized to get the balls and urn (stars and bars) identity.

2001 AMC 10 竞赛试题/第14题

Problem

A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length $2000$ . What is the length of each side of the octagon?

$\textbf{(A)} \frac{1}{3}(2000) \qquad \textbf{(B)} {2000(\sqrt{2}-1)} \qquad \textbf{(C)} {2000(2-\sqrt{2})} \qquad \textbf{(D)} {1000} \qquad \textbf{(E)} {1000\sqrt{2}}$

Solution

$[asy] draw((0,0)--(0,10)--(10,10)--(10,0)--cycle); draw((0,7)--(3,10)); draw((7,10)--(10,7)); draw((10,3)--(7,0)); draw((3,0)--(0,3)); label("$x$",(0,1),W); label("$x\sqrt{2}$",(1.5,1.5),NE); label("$2000-2x$",(5,0),S);[/asy]$