Question

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?

$\dfrac{9}{64}$

$\dfrac{289}{2048}$

$\dfrac{73}{512}$

$\dfrac{147}{1024}$

$\dfrac{589}{4096}$

Solution:

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 $\sum (\text{orbit size})^2 = 1 + 36 + (9 + 144) + (64 + 144) + (9 + 144) + 36 + 1 = 588.$

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

Thus, the correct answer is D.

2020 AMC 12B Problem 20

Solution:

Problem 20 in Other Years