AMC10 2025 B

AMC10 2025 B · Q21

AMC10 2025 B · Q21. It mainly tests Casework, Counting with symmetry / Burnside (rare).

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 considered the same. How many different colorings are possible?

一个 $3 \times 3$ 网格中的 $9$ 个方格将被涂成红色、蓝色或黄色，使得每个红色方格与至少一个蓝色方格共享一条边，每个蓝色方格与至少一个黄色方格共享一条边，每个黄色方格与至少一个红色方格共享一条边。通过旋转和/或反射可以从彼此获得的涂色被视为相同的。有多少种不同的涂色可能？

(A) 3 3

(B) 9 9

(D) 18 18

(E) 27 27

Answer

Correct choice: (C)

正确答案：（C）

Solution

Denote $1=\text{red}$, $2=\text{blue}$, $3=\text{yellow}$. We need $1\to 2\to 3\to 1$. WLOG place $1$ in the center $(0,0)$, $2$ on the left edge $(-1,0)$, $3$ on the top-left corner $(-1,1)$. \[\begin{bmatrix} 3 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\] $3$ must see $1$, so the top edge $(0,1)$ must also have $1$. Then, $1$ must see $2$, so the top-right corner $(1,1)$ becomes $2$, which must see $3$, so the right edge $(1,0)$ must have $3$. \[\begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 3 \\ 0 & 0 & 0 \end{bmatrix}\] Now this bottom-right corner can vary either as $1$, $2$ or $3$. Cases on $(1,-1)$: If $1$: \[\begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 1 \end{bmatrix}\] but the 3 needs a 1 and does not have it, so there are $0$ cases. If $2$: \[\begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 3 \\ b & a & 2 \end{bmatrix}\] if $a=1$, $b$ can be $1$ or $3$. If $a=2$, $b=3$, but the 3 needs a 1 and can't get it. If $a=3$, $b=1,2$, so there are $4$ cases in total. If $3$: \[\begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 3 \\ 2 & 1 & 3 \end{bmatrix}\] Since the 2 in the bottom left corner does not have a 3 nearby, there are $0$ cases. WLOG the center fixes a factor of $3$, so the answer is $4\cdot 3=\boxed{\textbf{(C) } 12}$.

记 $1=\text{红色}$，$2=\text{蓝色}$，$3=\text{黄色}$。我们需要 $1\to 2\to 3\to 1$。不失一般性，将 $1$ 放在中心 $(0,0)$，$2$ 放在左边 $(-1,0)$，$3$ 放在左上角 $(-1,1)$。 \[\begin{bmatrix} 3 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\] $3$ 必须与 $1$ 相邻，因此上边 $(0,1)$ 必须是 $1$。然后，$1$ 必须与 $2$ 相邻，因此右上角 $(1,1)$ 变为 $2$，它必须与 $3$ 相邻，因此右边 $(1,0)$ 必须是 $3$。 \[\begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 3 \\ 0 & 0 & 0 \end{bmatrix}\] 现在右下角可以是 $1$、$2$ 或 $3$。对 $(1,-1)$ 分类讨论：如果 $1$： \[\begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 1 \end{bmatrix}\] 但 $3$ 需要一个 $1$ 却没有，因此 $0$ 种。如果 $2$： \[\begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 3 \\ b & a & 2 \end{bmatrix}\] 若 $a=1$，$b$ 可以是 $1$ 或 $3$。若 $a=2$，$b=3$，但 $3$ 需要 $1$ 却得不到。若 $a=3$，$b=1,2$，总共 $4$ 种。如果 $3$： \[\begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 3 \\ 2 & 1 & 3 \end{bmatrix}\] 左下角的 $2$ 附近没有 $3$，因此 $0$ 种。不失一般性，中心固定带来 $3$ 倍因子，因此答案是 $4\cdot 3=\boxed{\textbf{(C) } 12}$。

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