# University of South Carolina High School Math Contest/1993 Exam/Problem 3

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## Problem

If 3 circles of radius 1 are mutually tangent as shown, what is the area of the gap they enclose?

$\mathrm{(A) \ }\sqrt{3}-\frac{\pi}2 \qquad \mathrm{(B) \ } \frac 16 \qquad \mathrm{(C) \ }\frac 13 \qquad \mathrm{(D) \ } \frac{\sqrt{3}}2 - \frac{\pi}6 \qquad \mathrm{(E) \ } \frac{\pi}6$

## Solution

We connect the centers of the three circles, and we get an equilateral triangle, composed of three congruent sectors and the gap in question. The area of the three gaps is half the area of one of the circles, and is thus $\frac{\pi}{2}$. The area of the whole triangle is $\frac{2^2\sqrt{3}}{4}=\sqrt{3}$, so the area of the gap is $\sqrt{3}-\frac{\pi}{2}$, $\mathrm{(A)}$.