Question

正四面体的每个面涂成红、白、蓝三色之一。若两个涂色的全等四面体可以通过旋转使外观完全相同，则认为这两种涂色不可区分。共有多少种可区分的涂色？

Each face of a regular tetrahedron is painted either red, white, or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical. How many distinguishable colorings are possible?

$15$

$18$

$27$

$54$

$81$

答案：A

知识点：伯恩赛德引理对称性

难度评级：2000

解答：

四面体旋转群有 $12$ 个元素：恒等、 $8$ 个绕顶点-对面轴的 $3$ 阶旋转，以及 $3$ 个绕对边中点轴的 $2$ 阶旋转。

恒等固定全部 $3^4=81$ 种涂色。每个顶点旋转固定 $3^2=9$ 种；每个边旋转也固定 $3^2=9$ 种。

由 Burnside 引理，可区分涂色数为 $\dfrac{81+8\cdot9+3\cdot9}{12}=\dfrac{180}{12}=15.$

所以正确答案是 A。

The rotation group of the tetrahedron has $12$ elements: the identity, $8$ rotations of order $3$ about a vertex-face axis, and $3$ rotations of order $2$ about an edge-midpoint axis.

The identity fixes all $3^4=81$ colorings. Each vertex rotation fixes one face and cycles the other three, so it fixes $3^2=9$ colorings; likewise each edge rotation swaps two pairs of faces and fixes $3^2=9.$

By Burnside's lemma the number of distinguishable colorings is $\dfrac{81+8\cdot9+3\cdot9}{12}=\dfrac{180}{12}=15.$

Thus, the correct answer is A.

2007 AMC 12B 第 16 题

答案：A

难度评级：2000

解答：

其他年份的第 16 题