2021 AMC Fall 10A Problem 24

P21-25~10 minPrint

Arrangements with Restrictions Basic Counting Solid Geometry Split into exhaustive cases and handle each Use an invariant, parity, or coloring argument

Each of the $12$ edges of a cube is labeled $0$ or $1$ . Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$ ?

Show solution without answering

Key insight

Exactly six edges are 1s, and top and bottom use four, so exactly 2 upright edges are 1s. Opposite uprights give 12; neighboring uprights give 8.

Solution

How many 1s? Each face adds to $2$ , and there are $6$ faces, so the face totals add to $12$ . Every edge lies on exactly $2$ faces, so each $1$ is counted twice in that $12$ . So there are exactly $12 \div 2 = 6$ edges labeled $1$ .
Split the edges. Stand the cube up. There are $4$ edges around the top square, $4$ around the bottom square, and $4$ upright edges joining them. The top face needs exactly two $1$ s, all on its own edges; the same for the bottom. That uses $4$ of the six $1$ s, so exactly $2$ of the $4$ upright edges are $1$ s.
Name things. Number the upright edges $1, 2, 3, 4$ going around. Each side face contains two neighboring upright edges, one top edge and one bottom edge. Call the side faces $12$ , $23$ , $34$ , $41$ by the uprights they contain.

Case A: the two upright 1s are opposite (uprights $1$ and $3$ , or $2$ and $4$ ). There are $2$ such choices.

Take uprights $1$ and $3$ . Every side face contains exactly one of them. So each side face needs exactly one more $1$ : its top edge or its bottom edge, not both.
The top face needs two $1$ s among its $4$ edges: $\binom42 = 6$ ways. Each side face whose top edge is $0$ must then have its bottom edge be $1$ , so the bottom edges are fixed, and the bottom face also gets exactly two $1$ s. Everything works.
Case A total: $2 \times 6 = 12$ .

Case B: the two upright 1s are neighbors (like uprights $1$ and $2$ ). There are $4$ such choices.

Face $12$ already has two $1$ s, so its top and bottom edges are $0$ .
Face $34$ has no upright $1$ s, so its top and bottom edges must both be $1$ .
The top face now has one $1$ (from face $34$ ) and needs one more, from the top edge of face $23$ or of face $41$ . Say it is face $23$ . Then face $23$ is full, so its bottom edge is $0$ . Face $41$ has top edge $0$ , so its bottom edge is $1$ . Check the bottom face: $1$ (from face $34$ ) $+ 1$ (from face $41$ ) $= 2$ . It works. The other choice (face $41$ ) works the same way.
So there are $2$ ways for each choice of uprights: $4 \times 2 = 8$ .

Total. $12 + 8 = 20$ .

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

Why this works

First a counting fact for the whole cube (six $1$ s, because each edge sits on two faces), then a smart place to start (the four upright edges, which every side face uses). Once the uprights are chosen, every other edge is either forced or a simple two-way choice. Also note the sentence about rotations and reflections: it tells you not to divide by symmetries.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(E)}\ 20}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.5)
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Geometry, Chapter 14: Three-Dimensional Geometry
Introduction to Counting & Probability, Chapter 3: Correcting for Overcounting (§3.2, §3.4)