Question

一个正二十面体是一个有 $20$ 个面的立体，每个面都是等边三角形，并且每个顶点都有五个三角形相交。下图所示的正二十面体有一个顶点在最上方，一个顶点在最下方，一个由五个顶点组成的上五边形；这五个顶点都与顶点相邻，且都在同一水平平面内；还有一个由五个顶点组成的下五边形；这五个顶点都与底点相邻，且在另一水平平面内。求从顶点到底点的路径数，要求路径的每一段都沿二十面体的边向下或水平移动，并且不重复经过任何顶点。

A regular icosahedron is a $20$ -faced solid where each face is an equilateral triangle and five triangles meet at every vertex. The regular icosahedron shown below has one vertex at the top, one vertex at the bottom, an upper pentagon of five vertices all adjacent to the top vertex and all in the same horizontal plane, and a lower pentagon of five vertices all adjacent to the bottom vertex and all in another horizontal plane. Find the number of paths from the top vertex to the bottom vertex such that each part of a path goes downward or horizontally along an edge of the icosahedron, and no vertex is repeated.

答案

答案：810

知识点：图论乘法原理

难度评级：2230

解答：

上五边形的每个顶点都与顶点、两个上五边形邻点以及两个下五边形顶点相邻；下五边形的每个顶点都与两个上层顶点、两个下五边形邻点以及底点相邻。因此一条不重复顶点且只向下或水平走的路径，必须先下降到上五边形，沿某一个方向绕过其中一部分，再下降到下五边形，沿某一个方向绕过其中一部分，最后到达底点。

第一步向下有 $5$ 种选择。在上五边形上，路径可以走 $0, 1, 2, 3$ ，或 $4$ 条水平边，方向有两种（若转向就会重复顶点），共有 $1 + 2 \cdot 4 = 9$ 种选择。然后有 $2$ 条边下降到下五边形，在那里又有 $9$ 种水平走法，最后再走 $1$ 步向下。

总数为 $5 \cdot 9 \cdot 2 \cdot 9 = 810$ 。

Each vertex of the upper pentagon is adjacent to the top vertex, two upper-pentagon neighbors, and two vertices of the lower pentagon; each vertex of the lower pentagon is adjacent to two upper vertices, two lower-pentagon neighbors, and the bottom vertex. So a downward-or-horizontal path with no repeated vertex must descend to the upper pentagon, circle part of it in one direction, drop to the lower pentagon, circle part of it in one direction, and end at the bottom.

There are $5$ choices for the first step down. On the upper pentagon the path can take $0, 1, 2, 3,$ or $4$ horizontal steps, in either of two directions (a reversal would repeat a vertex), for $1 + 2 \cdot 4 = 9$ options. Then there are $2$ edges down to the lower pentagon, again $9$ horizontal options there, and $1$ final step down.

The total is $5 \cdot 9 \cdot 2 \cdot 9 = 810.$

2016 AIME I 第 3 题

答案：810

难度评级：2230

解答：

其他年份的第 3 题