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2016 AMC 10A Problem 18

2016 AMC 10A Problem 18AMC 10 Step by Step

Name: Date: September 13, 2026

Arrangements with Restrictions · difficulty 5 of 5 · about 10 min · position P16-20

Each vertex of a cube is to be labeled with an integer $1$ through $8$ , with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?

A)
$1$
B)
$3$
C)
$6$
D)
$12$
E)
$24$

Answer

Key insight

Equal face sums force each antipodal pair to differ by one constant d, and 1 through 8 split into equal-difference pairs only for d = 1, 2, 4.

Solution

Each vertex lies on three faces, so the six face sums total $3 \cdot 36 = 108$ and each face sums to $18$ .

Structure. Two adjacent faces share an edge $e$ ; each equals $e$ plus the edge parallel to $e$ on that face, so those two parallel edges have equal vertex sums. Let $a$ be a vertex with antipode $g$ , and let $b$ be a neighbor of $a$ with antipode $h$ . Edges $ab$ and $gh$ are parallel and share no face, so $a + b = g + h$ , i.e. $a - g = h - b$ . Repeating with the other two neighbors of $a$ shows every antipodal pair has the same difference $d$ , with the larger label on the tetrahedron $T_1$ of vertices at even distance from $a$ .

Converse. Each face holds two $T_1$ vertices and two others whose antipodes are the remaining two $T_1$ vertices, so every face sums to $S_1 - 2d$ ; any such labeling works.

Count. Split $\{1,\dots,8\}$ into four pairs with common difference $d$ : $d = 1$ gives $\{1,2\},\{3,4\},\{5,6\},\{7,8\}$ ; $d = 2$ gives $\{1,3\},\{2,4\},\{5,7\},\{6,8\}$ ; $d = 4$ gives $\{1,5\},\{2,6\},\{3,7\},\{4,8\}$ ; no other $d$ works. For each $d$ , rotate $8$ to a fixed vertex; its partner is opposite, and the other three pairs fill the remaining axes in $3! = 6$ ways, which the three rotations about $8$ 's diagonal collapse to $2$ .

Total: $3 \cdot 2 = 6$ . The answer is $\boxed{\textbf{(C)}\ 6}$ .

Why this works

Subtracting the equations of two overlapping faces cancels the shared edge and exposes a hidden equality; iterating that idea turns a global condition into a local one (constant antipodal difference). Once the structure is forced, counting reduces to a small number-theory question (equal-difference pairings) times an orbit count under rotations. Whenever "all sums equal" appears, difference adjacent equations.

The trap

Trying to place numbers by trial without deriving a structural constraint, or dividing by the wrong symmetry count (24 rotations, not 48 symmetries).

Common mistakes

Trying to place numbers by trial without deriving a structural constraint, or dividing by the wrong symmetry count (24 rotations, not 48 symmetries).
Accepting $d = 3$ (pairs $\{1,4\},\{2,5\},\{3,6\}$ leave $7$ and $8$ unmatched) and answering $12$ or more.
Forgetting the mirror-image arrangements around vertex $8$ are not rotations of each other, which undercounts to $3$ .

Techniques

Split into exhaustive cases and handle each · Exploit symmetry to reduce work or pair up objects