2012 AMC 10B Problem 16

The three circles together have area $3 \cdot \pi \cdot 2^2 = 12\pi$. What remains is the curved "gap" between them.

Connect the three centers. Tangent circles of radius $2$ have centers $4$ apart, so the centers form an equilateral triangle of side $4$ and area $\frac{\sqrt3}{4} \cdot 16 = 4\sqrt3$.

This triangle consists of the gap plus three sectors, one at each center. Each sector has the triangle's $60^\circ$ angle and radius $2$, so together they make $\frac{3 \cdot 60}{360} = \frac12$ of a circle: area $\frac12 \cdot 4\pi = 2\pi$. Hence the gap has area $4\sqrt3 - 2\pi$.

Total shaded area:
$$12\pi + (4\sqrt3 - 2\pi) = 10\pi + 4\sqrt3.$$

The answer is $\boxed{\textbf{(A)}\ 10\pi + 4\sqrt3}$.

Regions bounded by arcs of tangent circles are handled by connecting the centers: the polygon of centers decomposes into the curvy region plus sectors, and the sector angles are exactly the polygon's angles. Three $60^\circ$ sectors of equal radius always total half a circle, a fact worth remembering for any three mutually tangent equal circles.

Sanity check with the choices: the shaded region is three circles ($12\pi \approx 37.7$) plus a small gap, so the answer is slightly more than $37.7$. Only (A), $10\pi + 4\sqrt3 \approx 38.3$, is a bit above $12\pi$ while (B), (C), (E) are far above and (D) is below.

• Adding the whole triangle to the three circles (12π + 4√3) without subtracting the three sectors that are already inside the circles.
• Using side $2$ (the radius) instead of $4$ for the triangle of centers, which shrinks the triangle's area to $\sqrt3$.