Question

设 $A_1, A_2, A_3, \ldots, A_{12}$ 是正十二边形的顶点。该十二边形所在平面内，有多少个不同的正方形至少有两个顶点属于集合 $\{A_1, A_2, A_3, \ldots, A_{12}\}$ ？

Let $A_1, A_2, A_3, \ldots, A_{12}$ be the vertices of a regular dodecagon. How many distinct squares in the plane of the dodecagon have at least two vertices in the set $\{A_1, A_2, A_3, \ldots, A_{12}\}?$

答案

答案：183

知识点：数对计数正多边形图形中的形状计数

难度评级：2480

解答：

$\binom{12}{2} = 66$ 对顶点中的每一对都恰好确定三个正方形：两个以这对点为边（线段两侧各一个），一个以这对点为对角线。这样共计 $3 \cdot 66 = 198$ 个正方形。

只有当一个正方形有多于两个顶点属于 $A_i$ 时才会重复计算。若一个正方形的三个顶点在该外接圆上，则正方形自己的外接圆与它共有三点，因此两圆重合；而内接正方形的顶点相隔 $90^\circ$ ，也就是十二边形的三步，所以第四个顶点也必为某个 $A_i$ 。完全内接的正方形正好是 $A_1A_4A_7A_{10}$ 、 $A_2A_5A_8A_{11}$ 、 $A_3A_6A_9A_{12}$ 。每个都由 $\binom{4}{2} = 6$ 对顶点产生，所以每个被数了 $6$ 次而不是一次。

不同正方形的个数为 $198 - 3 \cdot 5 = 183$ 。

Each of the $\binom{12}{2} = 66$ pairs of vertices determines exactly three squares: two having the pair as a side (one on each side of the segment) and one having it as a diagonal. That counts $3 \cdot 66 = 198$ squares.

A square is overcounted only if it has more than two vertices among the $A_i.$ If three vertices of a square lie on the circumcircle, the square's own circumcircle shares three points with it and hence coincides with it, and an inscribed square's vertices are spaced $90^\circ$ apart — three steps of the dodecagon — so the fourth vertex is also an $A_i.$ The fully inscribed squares are exactly $A_1A_4A_7A_{10},$ $A_2A_5A_8A_{11},$ and $A_3A_6A_9A_{12},$ and each is generated by all $\binom{4}{2} = 6$ of its vertex pairs, so each is counted $6$ times instead of once.

The number of distinct squares is $198 - 3 \cdot 5 = 183.$

2002 AIME I 第 5 题

答案：183

难度评级：2480

解答：

其他年份的第 5 题