Question

在一个正八边形中画出全部 $20$ 条对角线。在八边形内部，不在边界上的位置，有多少个不同的点是两条或更多对角线的交点？

All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior of the octagon (not on the boundary) do two or more diagonals intersect?

$49$

$65$

$70$

$96$

$128$

答案：A

知识点：交点计数正多边形组合

难度评级：2180

解答：

若没有三条对角线共点，每选 $4$ 个顶点会确定一个内部交点，共 $\binom84=70$ 个。

通过对顶点的 $4$ 条长对角线都在中心相交。中心原本被数了 $\binom42=6$ 次，但应只数一次，所以减去 $5$ 。

另外还有 $8$ 个对称位置各有 $3$ 条对角线共点。每个位置原本被数 $\binom32=3$ 次，但应只数一次，所以再减去 $8(3-1)=16$ 。

不同内部交点数为 $70-5-16=49$ 。

所以正确答案是 A。

If no three diagonals were concurrent, each choice of $4$ vertices would determine one interior intersection, giving $\binom84=70$ .

The $4$ long diagonals through opposite vertices all meet at the center, so the center was counted $\binom42=6$ times and should be counted once. Subtract $5$ .

There are also $8$ symmetric points where $3$ diagonals meet. Each was counted $\binom32=3$ times and should be counted once, so subtract $8(3-1)=16$ .

The number of distinct interior intersection points is $70-5-16=49$ .

Thus, A is the correct answer.

2013 AMC 10A 第 25 题

答案：A

难度评级：2180

解答：

其他年份的第 25 题