2003 AMC 10B Problem 19

Put the large center at $O$ and the small centers at $L$, $O$, $R$ with $LO = OR = 1$. The left and middle small semicircles meet at a point $P$ with $LP = OP = LO = 1$, so triangle $LOP$ is equilateral; similarly the middle and right ones meet at $Q$ with $OQR$ equilateral. The outer two semicircles only touch at $O$, so there are no other overlaps.

Draw $LP, OP, OQ, RQ$. The white region (the union of the three small semicircles) splits into:

• sector of the left semicircle from ray $LP$ (at $60^\circ$) around to $A$: $120^\circ$, area $\frac{1}{3}\pi$;
• sector of the middle semicircle between rays $OP$ and $OQ$: $60^\circ$, area $\frac{1}{6}\pi$;
• sector of the right semicircle from $B$ around to ray $RQ$: $120^\circ$, area $\frac{1}{3}\pi$;
• the two equilateral triangles $LOP$ and $OQR$ of side $1$, area $\frac{\sqrt{3}}{4}$ each.

White area: $\left(\frac{1}{3} + \frac{1}{6} + \frac{1}{3}\right)\pi + \frac{\sqrt{3}}{2} = \frac{5}{6}\pi + \frac{\sqrt{3}}{2}$.

The large semicircle has area $\frac{1}{2}\pi \cdot 2^2 = 2\pi$, so the shaded area is
$$2\pi - \frac{5}{6}\pi - \frac{\sqrt{3}}{2} = \frac{7}{6}\pi - \frac{\sqrt{3}}{2}.$$

The answer is $\boxed{\textbf{(E)}\ \frac{7}{6}\pi - \frac{\sqrt{3}}{2}}$.

Two unit circles whose centers are one unit apart intersect at the third vertex of an equilateral triangle, so all the angles are $60^\circ$ multiples and every piece is a sector or an equilateral triangle. Connecting intersection points to centers is the standard move for overlapping circles: it exposes the radii and turns a curvy region into sectors plus triangles.

Inclusion-exclusion: the three semicircles have total area $\frac{3}{2}\pi$; each of the two overlaps is half a lens of two unit circles at distance $1$, i.e. $\frac{1}{2}\left(\frac{2}{3}\pi - \frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} - \frac{\sqrt{3}}{4}$. White area $= \frac{3}{2}\pi - 2\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right) = \frac{5}{6}\pi + \frac{\sqrt{3}}{2}$, as before.

• Adding the three small semicircle areas ($\frac{3}{2}\pi$) without subtracting their overlaps, which gives $\frac{\pi}{2}$ and no listed answer, or double-counting the overlap.
• Treating the lens between two neighboring semicircles as a full lens ($\frac{2}{3}\pi - \frac{\sqrt3}{2}$) instead of the half that lies above $\overline{AB}$.