2020 AMC 12B Problem 20

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}

Answer: D

Concepts:Burnside’s Lemma casework

Difficulty rating: 2040

Small hint:

The probability equals

\dfrac{1}{64^2}\sum_{\mathcal O} |\mathcal O|^2

, where the sum is over rotation-equivalence classes

\mathcal O

of colorings

Big hint:

Group colorings by number of black faces and find each class’s orbit size (e.g. two black faces: opposite gives size

3,

adjacent gives size

12

)

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_{\mathcal O} |\mathcal O|^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_{\mathcal O} |\mathcal O|^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 Problem 20

Answer: D

Difficulty rating: 2040

Small hint:

Big hint:

Solution:

Problem 20 in other years