# Difference between revisions of "1980 AHSME Problems/Problem 16"

## Problem

Four of the eight vertices of a cube are the vertices of a regular tetrahedron. Find the ratio of the surface area of the cube to the surface area of the tetrahedron.

$\text{(A)} \ \sqrt 2 \qquad \text{(B)} \ \sqrt 3 \qquad \text{(C)} \ \sqrt{\frac{3}{2}} \qquad \text{(D)} \ \frac{2}{\sqrt{3}} \qquad \text{(E)} \ 2$

## Solution

We assume the side length of the cube is $1$. The side length of the tetrahedron is $\sqrt2$, so the surface area is $4\times\frac{2\sqrt3}{4}=2\sqrt3$. The surface area of the cube is $6\times1\times1=6$, so the ratio of the surface area of the cube to the surface area of the tetrahedron is $\frac{6}{2\sqrt3}=\boxed{\sqrt3}$