Question

从正 $n$ 边形的顶点中随机选取三个不同顶点，它们确定钝角三角形的概率为 $\frac{93}{125}$ 。求所有可能的 $n$ 值之和。

The probability that a set of three distinct vertices chosen at random from among the vertices of a regular $n$ -gon determine an obtuse triangle is $\frac{93}{125}.$ Find the sum of all possible values of $n.$

答案

答案：503

知识点：圆周角基本概率分类讨论

难度评级：2990

解答：

由圆周角定理，圆内接三角形是钝角三角形当且仅当它的三个顶点严格位于某个半圆内。按“第一个”顶点计数钝角三角形，也就是从该顶点出发，另两个顶点沿顺时针方向在半个圆内可到达。若 $n = 2k$ ，某顶点顺时针方向的开半圆内有 $k - 1$ 个顶点，得到 $n\binom{k-1}{2}$ 个钝角三角形；若 $n = 2k + 1$ ，开半圆内有 $k$ 个顶点，得到 $n\binom{k}{2}$ 个。

当 $n = 2k$ 时，概率为所以 $375(k - 2) = 186(2k - 1)$ ，得 $3k = 564$ ， $k = 188$ ， $n = 376$ 。 $\frac{2k\binom{k-1}{2}}{\binom{2k}{3}} = \frac{3(k - 2)}{2(2k - 1)} = \frac{93}{125},$

当 $n = 2k + 1$ 时，概率为 $\frac{3(k - 1)}{2(2k - 1)} = \frac{93}{125}$ ，所以 $375(k - 1) = 186(2k - 1)$ ，得 $3k = 189$ ， $k = 63$ ， $n = 127$ 。所有可能值之和为 $376 + 127 = 503$ 。

By the inscribed angle theorem, an inscribed triangle is obtuse exactly when its three vertices lie strictly within some semicircle. Count obtuse triangles by their "first" vertex, the vertex from which the other two are reached going clockwise within half the circle. If $n = 2k,$ the open semicircle clockwise of a vertex contains $k - 1$ vertices, giving $n\binom{k-1}{2}$ obtuse triangles; if $n = 2k + 1,$ it contains $k$ vertices, giving $n\binom{k}{2}.$

For $n = 2k$ the probability is $\frac{2k\binom{k-1}{2}}{\binom{2k}{3}} = \frac{3(k - 2)}{2(2k - 1)} = \frac{93}{125},$ so $375(k - 2) = 186(2k - 1),$ giving $3k = 564,$ $k = 188,$ and $n = 376.$

For $n = 2k + 1$ the probability is $\frac{3(k - 1)}{2(2k - 1)} = \frac{93}{125},$ so $375(k - 1) = 186(2k - 1),$ giving $3k = 189,$ $k = 63,$ and $n = 127.$ The sum of all possible values is $376 + 127 = 503.$

2011 AIME I 第 10 题

答案：503

难度评级：2990

解答：

其他年份的第 10 题