2007 AMC 10A Problem 11

You do not need to find the actual arrangement. You only need a way to use all eight numbers at once.

1. Add up all the labels. $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36$.
2. Look at two opposite faces, say the top and the bottom. The top has $4$ corners and the bottom has the other $4$. They share no corner, and together they use all $8$ corners.
3. So the two face sums add to $36$. Every face has the same sum, so each one is half of that:
$$\text{face sum} = \frac{36}{2} = 18.$$
4. Such a labeling really exists. For example, top corners $1, 4, 5, 8$ going around, with $6, 7, 2, 3$ directly below them. Every face then adds to $18$.

The answer is $\boxed{\textbf{(C)}\ 18}$.

When several groups must all have the same sum, look for some of those groups that together use every number exactly once. Their sums add to the grand total, which forces the common value without knowing the arrangement. Two opposite faces of a cube are such a pair.

Every corner of a cube touches $3$ faces. If you add the sums of all $6$ faces, each number is counted $3$ times, so the total is $3 \times 36 = 108$. Six equal face sums make $108$, so each is $108 \div 6 = 18$.

Call the common face sum $S$. Take any face and the face opposite it. Those two faces share no vertices, and between them they use all $8$ vertices of the cube.

So the two face sums together equal the sum of all eight labels:
$$2S = 1 + 2 + \cdots + 8 = 36 \quad\Longrightarrow\quad S = 18.$$

(Such a labeling exists, for instance top face $1, 4, 5, 8$ in cyclic order with $6, 7, 2, 3$ directly below them, so the question is meaningful.)

The answer is $\boxed{\textbf{(C)}\ 18}$.