Question

令集合 $S = \{P_1, P_2, \ldots, P_{12}\}$ 由一个正 $12$ 边形的十二个顶点组成。若 $S$ 的一个子集 $Q$ 满足：存在一个圆，使得 $Q$ 中所有点都在圆内，而 $S$ 中不属于 $Q$ 的所有点都在圆外，则称这样的子集为可聚子集。有多少个可聚子集？（注意空集也是可聚子集。）

Let the set $S = \{P_1, P_2, \ldots, P_{12}\}$ consist of the twelve vertices of a regular $12$ -gon. A subset $Q$ of $S$ is called communal if there is a circle such that all points of $Q$ are inside the circle, and all points of $S$ not in $Q$ are outside of the circle. How many communal subsets are there? (Note that the empty set is a communal subset.)

答案

答案：134

知识点：子集正多边形基本计数

难度评级：2390

解答：

子集 $Q$ 是可聚子集的充要条件是它的顶点在 $12$ 边形周围连续。确实，一个分离圆与这个 $12$ 边形的外接圆最多相交于两点，所以在它内部的顶点形成一段连续弧。反过来，任意一段连续顶点都可以用一条直线与剩余顶点分开，再在这条直线适当一侧取一个足够大的圆，就能恰好包含这段顶点。

对于每个满足 $1 \le k \le 11$ 的大小 $k$ ，都有 $12$ 段由 $k$ 个连续顶点组成的子集（每个顶点都可作为起点），共得到 $12 \cdot 11 = 132$ 个子集；此外空集和整个 $S$ 也都是可聚子集。总数为 $132 + 2 = 134$ 。

A subset $Q$ is communal exactly when its vertices are consecutive around the $12$ -gon. Indeed, a separating circle meets the circumcircle of the $12$ -gon in at most two points, so the vertices inside it form a contiguous arc. Conversely, any run of consecutive vertices can be separated from the remaining vertices by a line, and a sufficiently large circle on the proper side of that line contains exactly that run.

For each size $k$ with $1 \le k \le 11$ there are $12$ runs of $k$ consecutive vertices (one starting at each vertex), giving $12 \cdot 11 = 132$ subsets, and the empty set and all of $S$ are also communal. The total is $132 + 2 = 134.$

2014 AIME I 第 5 题

答案：134

难度评级：2390

解答：

其他年份的第 5 题