# 1997 AJHSME Problems/Problem 17

## Problem

A cube has eight vertices (corners) and twelve edges. A segment, such as $x$, which joins two vertices not joined by an edge is called a diagonal. Segment $y$ is also a diagonal. How many diagonals does a cube have? $[asy] draw((0,3)--(0,0)--(3,0)--(5.5,1)--(5.5,4)--(3,3)--(0,3)--(2.5,4)--(5.5,4)); draw((3,0)--(3,3)); draw((0,0)--(2.5,1)--(5.5,1)--(0,3)--(5.5,4),dashed); draw((2.5,4)--(2.5,1),dashed); label("x",(2.75,3.5),NNE); label("y",(4.125,1.5),NNE); [/asy]$ $\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$

## Solution 1

On each face, there are $2$ diagonals like $x$. There are $6$ faces on a cube. Thus, there are $2\times 6 = 12$ diagonals that are "x-like".

Every "y-like" diagonal must connect the bottom of the cube to the top of the cube. Thus, for each of the $4$ bottom vertices of the cube, there is a different "y-like" diagonal. So there are $4$ "y-like" diagonals.

This gives a total of $12 + 4 = 16$ diagonals on the cube, which is answer $\boxed{E}$.

## Solution 2

There are $8$ vertices on a cube(A, B, C, D, E, F, G, H). If you pick one vertice, it will have 7 line segments associated with it, so you will have $\frac{8\cdot7}{2} = 28$ segments within the cube. The division by $2$ is necessary because you counted both the segment from $A$ to $B$ and the segment from $B$ to $A$.

But not all $28$ of these segments are diagonals. Some are edges. There are $4$ edges on the top, $4$ edges on the bottom, and $4$ edges that connect the top to the bottom. So there are $12$ edges total, meaning that there are $28 - 12 = 16$ segments that are not edges. All of these segments are diagonals, and thus the answer is $\boxed{E}$.

## Solution 3

Consider picking one point on the corner of the cube. That point has $3$ "x-like" diagonals , and $1$ "y-like" diagonal that ends on the opposite vertex. Thus, each vertex has $3 + 1 = 4$ diagonals associated with it. There are $8$ vertices on the cube, giving a total of $8 \cdot 4 = 32$ diagonals.

However, each diagonal was counted as both a "starting point" and an "ending point". So there are really $\frac{32}{2} = 16$ diagonals, giving an answer of $\boxed{E}$.

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