# 1996 AHSME Problems/Problem 10

## Problem

How many line segments have both their endpoints located at the vertices of a given cube? $\text{(A)}\ 12\qquad\text{(B)}\ 15\qquad\text{(C)}\ 24\qquad\text{(D)}\ 28\qquad\text{(E)}\ 56$

## Contents

### Solution 1

There are $8$ choices for the first endpoint of the line segment, and $7$ choices for the second endpoint, giving a total of $8\cdot 7 = 56$ segments. However, both $\overline{AB}$ and $\overline{BA}$ were counted, while they really are the same line segment. Every segment got double counted in a similar manner, so there are really $\frac{56}{2} = 28$ line segments, and the answer is $\boxed{D}$.

In shorthand notation, we're choosing $2$ endpoints from a set of $8$ endpoints, and the answer is $\binom{8}{2} = \frac{8!}{6!2!} = 28$.

### Solution 2

Each segment is either an edge, a facial diagonal, or a long/main/spacial diagonal.

A cube has $12$ edges: Four on the top face, four on the bottom face, and four that connect the top face to the bottom face.

A cube has $6$ square faces, each of which has $2$ facial diagonals, for a total of $6\cdot 2 = 12$.

A cube has $4$ spacial diagonals: each diagonal goes from one of the bottom vertices to the "opposite" top vertex.

Thus, there are $12 + 12 + 4 = 28$ segments, and the answer is $\boxed{D}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 