Question

八个全等的等边三角形，每个颜色都不同，被用来构造一个正八面体。构造这个八面体有多少种可区分的方法？（如果两个着色八面体不能通过旋转变得完全一样，则称它们可区分。）

Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.)

$210$

$560$

$840$

$1260$

$1680$

答案：E

知识点：伯恩赛德引理排列对称性

难度评级：2440

解答：

将八种不同颜色分配给八个面共有 $8!$ 种方式。两个分配表示同一个八面体，当且仅当其中一个可以旋转成另一个。

正八面体的旋转群有 $24$ 个元素。因为八种颜色全都不同，没有非平凡旋转会固定一种着色，所以每个可区分的八面体恰好对应 $24$ 种分配。

因此可区分的八面体数量为 $\frac{8!}{24} = \frac{40320}{24} = 1680.$

因此，正确答案是 E。

There are $8!$ ways to assign the eight distinct colors to the eight faces. Two assignments give the same octahedron exactly when one is a rotation of the other.

The rotation group of a regular octahedron has $24$ elements. Because all eight colors are different, no nontrivial rotation fixes a coloring, so each distinguishable octahedron corresponds to exactly $24$ assignments.

Therefore the number of distinguishable octahedrons is $\frac{8!}{24} = \frac{40320}{24} = 1680.$

Thus, the correct answer is E.

2000 AMC 12 第 25 题

答案：E

难度评级：2440

解答：

其他年份的第 25 题