2018 AMC 8 第 23 题

先试着解答 2018 AMC 8 第 23 题,然后核对你的答案与视频与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2018 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

23.

从一个正八边形中随机选择三个顶点并连接成一个三角形。这个三角形至少有一条边也是八边形的一条边的概率是多少?
From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?
27\dfrac{2}{7}
542\dfrac{5}{42}
1114\dfrac{11}{14}
57\dfrac{5}{7}
67\dfrac{6}{7}
答案:D
知识点:补集计数隔板法
小提示:
使用补集:数没有相邻顶点被选中的三角形
Use complementary counting: count triangles with no adjacent chosen vertices.
大提示:
固定一个顶点后,数三个选中顶点之间的正间隔
After fixing one vertex, count the positive gaps around the octagon between the chosen vertices.
视频讲解:
解答视频缩略图
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文字解答:
用补集计数:计算任意两个所选顶点都不相邻的三角形。从一个所选顶点出发,设三个间隔中未选顶点的正整数个数为 x,y,zx,y,z。那么 x+y+z=5x+y+z=5,共有 (42)=6\binom42=6 个正整数解。
起点有 88 种选择,而每个三角形会从它的 33 个顶点各被计算一次,所以补集中有 8⋅63=16\frac{8\cdot6}{3}=16 个三角形。全部三角形共有 (83)=56\binom83=56 个,因此所求概率是 1−1656=571-\frac{16}{56}=\frac{5}{7}。
所以正确答案是 D。
Count the complement: triangles with no two chosen vertices adjacent. Starting at a chosen vertex, let x,y,zx,y,z be the positive numbers of unchosen vertices in the three gaps. Then x+y+z=5x+y+z=5, which has (42)=6\binom42=6 positive solutions.
There are 88 choices for the starting vertex, and each triangle is counted from each of its 33 vertices, so the complement contains 8⋅63=16\frac{8\cdot6}{3}=16 triangles. Out of (83)=56\binom83=56 total triangles, the desired probability is 1−1656=57.1-\frac{16}{56}=\frac{5}{7}.
Thus, D is the correct answer.
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