1980 AMC 12 Problem 16

Question

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

\sqrt2

\sqrt3

\sqrt{\frac32}

\frac2{\sqrt3}

2

Answer: B

Concepts:cube geometry surface area equilateral triangle

Difficulty rating: 1780

Small hint:

If the cube edge is

s,

each tetrahedron edge is a face diagonal

Big hint:

Compare

6s^2

with four equilateral triangles of side

s\sqrt2

Solution:

Let the cube edge be

s.

Each tetrahedron edge is a cube face diagonal, so it has length

s\sqrt2.

The tetrahedron’s surface area is

4\left(\frac{\sqrt3}{4}(s\sqrt2)^2\right) =2\sqrt3\,s^2.

The cube’s surface area is

6s^2,

so the ratio is

\frac{6}{2\sqrt3}=\sqrt3.

Therefore, the correct answer is B.