Question

有无限多个全等的彩纸等边三角形。每个三角形都是纯色，并且纸的两面颜色相同。用其中四个纸三角形按图所示拼成一个大的等边三角形。如果不能通过平移、旋转和/或反射把一个大三角形放到另一个上面并使对应的小三角形颜色相同，则认为这两个大三角形可区分。已知可选的三角形有六种不同颜色，可以构造多少个可区分的大等边三角形？

There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles as shown. Two large triangles are considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so that their corresponding small triangles are of the same color. Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles can be constructed?

答案

答案：336

知识点：组合对称性分类讨论

难度评级：2390

解答：

大三角形的旋转和反射能实现三个角上三角形的任意置换，同时固定中心三角形。因此两个大三角形不可区分，当且仅当它们有相同的中心颜色以及相同的三个角上颜色的多重集合。

从六种颜色中数角上颜色的多重集合：三个全相同有 $6$ 种，恰有两个相同有 $6 \cdot 5 = 30$ 种（选择重复颜色和另一个不同颜色），三个全不同有 $\binom{6}{3} = 20$ 种。总共 $6 + 30 + 20 = 56$ 个多重集合。

中心颜色可独立选择 $6$ 种，所以总数为 $6 \cdot 56 = 336$ 。

The rotations and reflections of the large triangle realize every permutation of the three corner triangles while fixing the center triangle. So two large triangles are indistinguishable exactly when they have the same center color and the same multiset of three corner colors.

Count the multisets of corner colors from six colors: all three the same ( $6$ ways), exactly two the same ( $6 \cdot 5 = 30$ ways, choosing the repeated color and then the different one), or all three different ( $\binom{6}{3} = 20$ ways). That is $6 + 30 + 20 = 56$ multisets.

With $6$ independent choices for the center color, the total is $6 \cdot 56 = 336.$

2006 AIME II 第 8 题

答案：336

难度评级：2390

解答：

其他年份的第 8 题