1980 AHSME Problems/Problem 26

Problem

Four balls of radius $1$ are mutually tangent, three resting on the floor and the fourth resting on the others. A tetrahedron, each of whose edges have length $s$ , is circumscribed around the balls. Then $s$ equals

$\text{(A)} \ 4\sqrt 2 \qquad \text{(B)} \ 4\sqrt 3 \qquad \text{(C)} \ 2\sqrt 6 \qquad \text{(D)}\ 1+2\sqrt 6\qquad \text{(E)}\ 2+2\sqrt 6$

Solution

Let $A$ , $B$ , $C$ , and $D$ be the centers of the four spheres with radius $1$ . These points must form a regular tetrahedron of side length $2$ . Let $O$ be the center of $\triangle ABC$ . Then $OA = \frac{2\sqrt{3}}{3}$ . By the Pythagorean theorem, the height $DO$ of this pyramid must be $\sqrt{2^2 - \left(\frac{2\sqrt{3}}{3}\right)^2} = \frac{2\sqrt{6}}{3}$ $\fbox{E}$

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 25	Followed by Problem 27
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1980 AHSME Problems/Problem 26

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