AMC12 2020 B

AMC12 2020 B · Q20

AMC12 2020 B · Q20. It mainly tests Counting with symmetry / Burnside (rare), Counting & probability misc.

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?

两个相同大小的不同立方体将被涂漆，每个面的颜色独立随机选择为黑色或白色。涂漆后，这两个立方体能通过旋转变得外观相同的概率是多少？

(A) \frac{9}{64} \frac{9}{64}

(B) \frac{289}{2048} \frac{289}{2048}

(D) \frac{147}{1024} \frac{147}{1024}

(E) \frac{589}{4096} \frac{589}{4096}

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): There are $2^6=64$ ways to color each cube, for a total of $2^{12}=4096$ colorings of the two cubes. The 64 colorings may be grouped as follows: • There is 1 coloring with no white faces, and the same number with no black faces. • There are 6 colorings with exactly one white face, and the same number with exactly one black face. • There are 3 colorings with only two white faces that are opposite each other, and the same number with only two black faces that are opposite each other. • There are 12 colorings with only two white faces that share an edge, and the same number with only two black faces that share an edge. • There are 8 colorings with exactly three white faces that share a vertex. • There are 12 colorings with exactly three white faces that include a pair of opposite faces. The two cubes can be rotated to be identical in appearance if and only if their colorings are in the same group. Therefore the requested probability is $\dfrac{2\cdot(1^2+6^2+3^2+12^2)+8^2+12^2}{4096}=\dfrac{588}{4096}=\dfrac{147}{1024}.$

答案（D）：每个立方体有 $2^6=64$ 种着色方式，因此两个立方体共有 $2^{12}=4096$ 种着色。64 种着色可分组如下： • 恰好没有白色面的着色有 1 种；同样，恰好没有黑色面的也有 1 种。 • 恰好只有 1 个白色面的着色有 6 种；同样，恰好只有 1 个黑色面的也有 6 种。 • 恰好只有 2 个白色面且二者互为对面的着色有 3 种；同样，恰好只有 2 个黑色面且二者互为对面的也有 3 种。 • 恰好只有 2 个白色面且二者共边的着色有 12 种；同样，恰好只有 2 个黑色面且二者共边的也有 12 种。 • 恰好有 3 个白色面且三者共享一个顶点的着色有 8 种。 • 恰好有 3 个白色面且包含一对白色对面的着色有 12 种。当且仅当两个立方体的着色落在同一组时，才能通过旋转使它们外观相同。因此所求概率为 $\dfrac{2\cdot(1^2+6^2+3^2+12^2)+8^2+12^2}{4096}=\dfrac{588}{4096}=\dfrac{147}{1024}.$

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