2021 AMC Fall 10B Problem 24

A construction is a choice of $4$ of the $8$ corner positions to be white, up to rotation. Sort by the largest number of white cubes on a single face, and rotate so that face is the top.

Four white on top. The whole top layer is white: $1$ way.

Three white on top. The top has one blue corner $D$; call the corner opposite $D$ on the top face $B$, and the other two $A, C$. The fourth white cube is on the bottom, in one of four positions:
- directly below $D$: an L-shape plus an isolated cube ($1$ way);
- directly below $B$: a "claw" of three cubes around $B$ ($1$ way);
- below $A$ or below $C$: a spiral path of three edges twisting through all three directions. The two are mirror images, and rotations preserve handedness, so they are different ($2$ ways).

Total $4$ ways. (Any other bottom position would create a face with four white cubes.)

At most two white on every face. A face with $0$ or $1$ white cubes has an opposite face with $4$ or $3$, so every face has exactly $2$. If the top whites are adjacent (an edge), the side face containing that edge forbids the two cubes below it, so the bottom whites are the opposite edge: $1$ way. If the top whites are diagonal, each side face already has one white, so the bottom whites are also diagonal; one choice repeats the "two opposite edges" shape and the other gives a regular tetrahedron (no two whites adjacent): $1$ new way.

Total: $1 + 4 + 2 = 7$.

The answer is $\boxed{\textbf{(A)}\ 7}$.

Counting up to rotation is safest by choosing a canonical orientation (the face with the most white cubes on top) and then listing shapes by adjacency structure, since rotations preserve which cubes touch. The one subtlety of 3-D counting is chirality: a shape and its mirror image are distinct unless some rotation maps one to the other. As a check, the orbit sizes $6 + 24 + 8 + 12 + 12 + 6 + 2 = 70 = \binom84$.

Burnside's lemma over the $24$ rotations: the identity fixes $\binom84 = 70$ colorings; the six $90^\circ$ face turns fix $2$ each; the three $180^\circ$ face turns fix $\binom42 = 6$ each; the eight $120^\circ$ vertex turns fix $4$ each; the six $180^\circ$ edge turns fix $6$ each. Total $(70 + 12 + 18 + 32 + 36)/24 = 168/24 = 7$.

• Merging the two mirror-image spiral shapes into one because they look alike; a rotation can never turn a left-handed spiral into a right-handed one.
• Counting the "diagonal top, diagonal bottom" case as two new shapes, when one of them is just the two-opposite-edges shape seen from another angle.