Question

平面上给定十个点，任意三点不共线。随机选择四条由这些点中的两点连接而成的不同线段，所有这样的线段集合等可能。这些线段中有三条能组成一个以这十个给定点中的点为顶点的三角形的概率为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质的正整数。求 $m + n$ 。

Ten points in the plane are given, with no three collinear. Four distinct segments joining pairs of these points are chosen at random, all such segments being equally likely. The probability that some three of the segments form a triangle whose vertices are among the ten given points is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：489

知识点：基本概率组合

难度评级：2480

解答：

共有 $\binom{10}{2} = 45$ 条线段，所以等可能选择数为 $\binom{45}{4} = 148995$ 。两个不同三角形至多共用一条边，所以合起来至少使用 $5$ 条线段；因此一组 $4$ 条线段至多包含一个三角形。有利集合可以通过先选一个三角形、再选第四条线段来恰好计数一次： $\begin{aligned} &\binom{10}{3} \cdot (45 - 3) = 120 \cdot 42 \\ &= 5040. \end{aligned}$

概率为 $\frac{5040}{148995} = \frac{16}{473}$ ，已经是最简分数（ $473 = 11 \cdot 43$ ），所以 $m + n = 16 + 473 = 489$ 。

There are $\binom{10}{2} = 45$ segments, so $\binom{45}{4} = 148995$ equally likely choices. Two distinct triangles share at most one edge, so together they use at least $5$ segments; hence a set of $4$ segments contains at most one triangle, and the favorable sets are counted exactly once by choosing a triangle and then a fourth segment: $\begin{aligned} &\binom{10}{3} \cdot (45 - 3) = 120 \cdot 42 \\ &= 5040. \end{aligned}$

The probability is $\frac{5040}{148995} = \frac{16}{473},$ already in lowest terms ( $473 = 11 \cdot 43$ ), so $m + n = 16 + 473 = 489.$

1999 AIME 第 10 题

答案：489

难度评级：2480

解答：

其他年份的第 10 题