Question

从所有以正 $12$ 边形顶点为端点的线段中随机选取三条不同的线段。这三条线段的长度能作为一个面积为正的三角形的三条边长的概率是多少？

Three distinct segments are chosen at random among the segments whose endpoints are the vertices of a regular $12$ -gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?

$\dfrac{553}{715}$

$\dfrac{443}{572}$

$\dfrac{111}{143}$

$\dfrac{81}{104}$

$\dfrac{223}{286}$

答案：E

知识点：正多边形三角不等式补集计数

难度评级：2650

小提示：

共有 $6$ 种可能长度 $d_k = 2\sin(15k^\circ)$ ；统计每种长度有多少条线段

There are $6$ possible lengths $d_k = 2\sin(15k^\circ);$ count how many segments have each length

大提示：

用补集计数：减去最长边长度大于或等于另外两边长度之和的三元组

Use complementary counting: subtract the triples whose longest length is at least the sum of the other two

解答：

将正 $12$ 边形内接于单位圆。线段长度为 $d_k = 2\sin(15k^\circ)$ ，其中 $1 \le k \le 6$ ，长度 $d_1, \ldots, d_5$ 各有 $12$ 条，长度 $d_6$ 有 $6$ 条。

比较各和，满足 $d_a \le d_b \le d_c$ 且 $d_c \ge d_a + d_b$ 的禁用指标三元组 $(a, b, c)$ 为 $\begin{gathered} (1,1,3),(1,1,4),(1,1,5), \\ (1,1,6),(1,2,4),(1,2,5), \\ (1,2,6),(1,3,5),(1,3,6), \\ (2,2,6) \end{gathered}\text{。}$

前三个以 $3,4,5$ 结尾的三元组贡献 $3\binom{12}{2}12$ ；两个以 $6$ 结尾且有重复长度的三元组贡献 $2\binom{12}{2}6$ ；三个长度互异且不含直径的三元组贡献 $3\cdot12^3$ ；剩下两个含直径的三元组贡献 $2\cdot12^2\cdot6$ 。将总和除以 $\binom{66}{3}$ ，得到失败概率 $\frac{63}{286}$ ，所以答案是 $1-\frac{63}{286}=\frac{223}{286}$ 。

因此，正确答案是 E。

Inscribe the $12$ -gon in a unit circle. The segment lengths are $d_k = 2\sin(15k^\circ)$ for $1 \le k \le 6,$ with $12$ segments of each length $d_1, \ldots, d_5$ and $6$ of length $d_6.$

Comparing sums, the forbidden index triples $(a, b, c)$ with $d_a \le d_b \le d_c$ and $d_c \ge d_a + d_b$ are $\begin{gathered} (1,1,3),(1,1,4),(1,1,5), \\ (1,1,6),(1,2,4),(1,2,5), \\ (1,2,6),(1,3,5),(1,3,6), \\ (2,2,6). \end{gathered}$

The first three triples ending in $3,4,5$ contribute $3\binom{12}{2}12;$ the two repeated-length triples ending in $6$ contribute $2\binom{12}{2}6;$ the three triples of distinct non-diameter lengths contribute $3\cdot12^3;$ and the two remaining diameter triples contribute $2\cdot12^2\cdot6.$ Dividing their sum by $\binom{66}{3}$ gives failure probability $\frac{63}{286},$ so the answer is $1-\frac{63}{286}=\frac{223}{286}.$

Thus, the correct answer is E.

2013 AMC 12A 第 24 题

答案：E

难度评级：2650

小提示：

大提示：

解答：

其他年份的第 24 题