2025 AMC 12B 第 17 题

Question

一个

3 \times 3

方格中的

9

个小正方形要被涂成红、蓝、黄三色，要求每个红色小方格至少与一个蓝色小方格共边，每个蓝色小方格至少与一个黄色小方格共边，每个黄色小方格至少与一个红色小方格共边。可以通过旋转和/或反射互相得到的涂色视为相同。共有多少种不同的涂色？

Each of the

9

squares in a

3 \times 3

grid is to be colored red, blue, or yellow in such a way that each red square shares an edge with at least one blue square, each blue square shares an edge with at least one yellow square, and each yellow square shares an edge with at least one red square. Colorings that can be obtained from one another by rotations and/or reflections are to be considered the same. How many different colorings are possible?

3

9

12

18

27

答案：C

知识点：伯恩赛德引理分类讨论

难度评级：1980

小提示：

每种颜色都必须出现，而且红

\to

蓝

\to

黄

\to

红的循环条件会强烈限制布局。

Every color must appear, and the cyclic conditions red

\to

blue

\to

yellow

\to

red tightly constrain the layout

大提示：

先数固定带标号的方格上所有合法涂色，再用 Burnside 引理按

8

个旋转和反射对称化简。

Count all valid colorings of the fixed (labeled) grid, then reduce by the

8

rotations and reflections using Burnside’s lemma

解答：

先在方格位置带标号的情形下数涂色数。依次检查

27

种可能的行，一旦某个方格的邻格都已确定却缺少所需的下一种颜色就立即舍弃该行，这样按红、蓝、黄方格的个数可得如下完整计数：

\begin{array}{c|c} (\#R,\#B,\#Y)&\text{合法涂色数}\\ \hline (2,4,3)&16\\ (3,2,4)&16\\ (3,3,3)&36\\ (4,3,2)&16 \end{array}

因此恒等对称固定了

16+16+36+16=84

种涂色。

对于其余的对称，被固定的涂色数为：每个非恒等旋转

0

种，关于水平轴或竖直轴的每个反射

6

种，每个对角线反射

0

种。（对于关于轴的反射，直接检查三个回文行即可得到这

6

种可能。）因此由 Burnside 引理得

\frac{84+2\cdot6}{8}=12\text{。}

因此，正确答案是 C。

First count colorings of the grid with its positions labeled. Checking the

27

possible rows in succession and rejecting a row as soon as a square whose neighbors are now known lacks its required next color gives the following complete count by the numbers of red, blue, and yellow squares:

\begin{array}{c|c} (\#R,\#B,\#Y)&\text{valid colorings}\\ \hline (2,4,3)&16\\ (3,2,4)&16\\ (3,3,3)&36\\ (4,3,2)&16 \end{array}

Thus the identity symmetry fixes

16+16+36+16=84

colorings.

For the other symmetries, the fixed-coloring counts are

0

for each nontrivial rotation,

6

for each reflection across a horizontal or vertical axis, and

0

for each diagonal reflection. (For an axis reflection, a direct check of the three palindromic rows gives the

6

possibilities.) Therefore Burnside’s lemma gives

\frac{84+2\cdot6}{8}=12.

Thus, the correct answer is C.

2025 AMC 12B 第 17 题

答案：C

难度评级：1980

小提示：

大提示：

解答：

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