2021 AMC Fall 10A Problem 24

How many ones? Each edge lies on exactly two faces, so adding the six face totals counts every edge twice: $6 \cdot 2 = 2 \cdot (\text{number of ones})$, so exactly $6$ edges are labeled $1$.

Pivot on the vertical edges. The top and bottom faces each contain $2$ ones among their horizontal edges, accounting for $4$ ones, so exactly $2$ of the $4$ vertical edges are $1$. Number the top corners $1, 2, 3, 4$ in order; vertical edge $v_i$ hangs from corner $i$, and side face $F_{i,i+1}$ contains $v_i$, $v_{i+1}$, top edge $t_{i,i+1}$ and bottom edge $b_{i,i+1}$.

Case A: diagonal pair ($v_1, v_3$; $2$ such pairs). Each side face has one vertical $1$ and needs exactly one of its $t, b$: $t_{i,i+1} + b_{i,i+1} = 1$ on all four sides. The top face needs two of the four $t$'s equal to $1$: $\binom42 = 6$ ways. The $b$'s are then forced and the bottom face automatically totals $2$. Total $2 \cdot 6 = 12$.

Case B: adjacent pair ($v_1, v_2$; $4$ such pairs). Face $F_{12}$ is full, so $t_{12} = b_{12} = 0$; face $F_{34}$ has no vertical ones, so $t_{34} = b_{34} = 1$. Faces $F_{23}$ and $F_{41}$ each need exactly one of $t, b$, while the top face needs one more $1$ from $t_{23}, t_{41}$ and the bottom one more from $b_{23}, b_{41}$. Choosing $t_{23} = 1$ forces $b_{23} = 0$, $b_{41} = 1$, $t_{41} = 0$; the mirror choice also works. So $2$ completions, total $4 \cdot 2 = 8$.

Altogether $12 + 8 = 20$.

The answer is $\boxed{\textbf{(E)}\ 20}$.

A global count (six ones, via double counting) plus a well-chosen pivot (the four vertical edges) turns a twelve-variable problem into two small cases where every remaining edge is either forced or a free binary choice. Choosing the pivot set so that every face contains some of it is what makes the rest propagate. Note the wording: labelings related by symmetry are counted separately, so no division by $48$ or $24$.

Structural view of the $20$ labelings: $4$ are "two opposite corners with all three edges at each" (one per antipodal pair), $4$ are skew hexagons (Petrie polygons, one per antipodal pair of omitted vertices), and $12$ are pairs of disjoint three-edge paths. Recognizing these shapes is elegant but easy to get wrong under time pressure; the casework above is safer.

• Dividing by symmetries because of the sentence about rotations and reflections, which actually says NOT to identify symmetric labelings.
• Assuming the two ones on a face must be opposite edges (or must be adjacent); both configurations occur, and only careful casework catches all $20$.