# 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 $\frac{6}{2\sqrt3}=\boxed{\sqrt3}$.

-aopspandy

## See also

 1980 AHSME (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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