Question

两个大小相同但彼此不同的立方体要被涂色，每个面的颜色独立随机选择为黑色或白色。涂色后，这两个立方体可以通过旋转变得外观相同的概率是多少？

Two different cubes of the same size are to be painted, with the color of each face being chosen independently and at random to be either black or white. What is the probability that after they are painted, the cubes can be rotated to be identical in appearance?

$\dfrac{9}{64}$

$\dfrac{289}{2048}$

$\dfrac{73}{512}$

$\dfrac{147}{1024}$

$\dfrac{589}{4096}$

答案：D

知识点：伯恩赛德引理分类讨论

难度评级：2040

解答：

固定第一个立方体后，与它旋转后相同的第二个立方体数等于其旋转轨道的大小。因此所求概率为 $\tfrac{1}{64^2}\sum_{\text{orbits}} (\text{orbit size})^2$ 。

按黑色面数量分类： $0$ 或 $6$ 个黑面时 $\to 1$ $1$ 或 $5$ 个黑面时 $\to 6$ $2$ 或 $4$ 个黑面时，对面情形 $\to 3$ ，相邻情形的轨道大小为 $12$ ； $3$ 个黑面时，共顶点情形 $\to 8$ ，环带情形的轨道大小为 $12$ 。于是 $\begin{gathered} \sum (\text{orbit size})^2 \\ {}= 1 + 36 + (9 + 144) \\ \quad {}+ (64 + 144) + (9 + 144) \\ \quad {}+ 36 + 1 = 588. \end{gathered}$

因此概率为 $\tfrac{588}{4096} = \tfrac{147}{1024}$ 。

所以正确答案是 D。

For a fixed first cube, the number of second cubes matching it (up to rotation) equals the size of its rotation orbit. So the desired probability is $\tfrac{1}{64^2}\sum_{\text{orbits}} (\text{orbit size})^2.$

Grouping by black-face count, the orbit sizes are: $0$ or $6$ black $\to 1;$ $1$ or $5$ black $\to 6;$ $2$ or $4$ black $\to 3$ (opposite) and $12$ (adjacent); $3$ black $\to 8$ (corner) and $12$ (band). Then $\begin{gathered} \sum (\text{orbit size})^2 \\ {}= 1 + 36 + (9 + 144) \\ \quad {}+ (64 + 144) + (9 + 144) \\ \quad {}+ 36 + 1 = 588. \end{gathered}$

The probability is $\tfrac{588}{4096} = \tfrac{147}{1024}.$

Thus, the correct answer is D.

2020 AMC 12B 第 20 题

答案：D

难度评级：2040

解答：

其他年份的第 20 题