Difference between revisions of "1980 AHSME Problems/Problem 26"

Revision as of 04:40, 25 June 2025

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
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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

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Difference between revisions of "1980 AHSME Problems/Problem 26"

Revision as of 04:40, 25 June 2025

Problem

Solution

See also