2003 AMC 10B Problem 19

1. Set up the picture. $AB = 4$, and the small centers split it into four pieces of length $1$. Call the small centers $L$, $O$, $R$ from left to right; $O$ is also the center of the big semicircle.
2. The shaded part is the big semicircle minus the white part (everything covered by the three small semicircles). The big semicircle has area $\frac{1}{2}\pi \cdot 2^2 = 2\pi$. So we need the white area.
3. The left and middle small semicircles cross at a point $P$. Since $P$ is on both, $LP = 1$ and $OP = 1$. Also $LO = 1$. So triangle $LOP$ has all sides $1$: it is equilateral, and all its angles are $60^\circ$. In the same way the middle and right semicircles cross at $Q$, and triangle $OQR$ is equilateral.
4. The left and right semicircles only touch at the point $O$, so there is no other overlap.
5. Draw the segments $LP$, $OP$, $OQ$, $RQ$. They cut the white region into pieces that are easy to measure:
- Left semicircle, the slice from $P$ around to $A$: the angle at $L$ is $180^\circ - 60^\circ = 120^\circ$, which is $\frac{1}{3}$ of a full circle, so area $\frac{\pi}{3}$.
- Middle semicircle, the slice between $P$ and $Q$: the angle at $O$ is $180^\circ - 60^\circ - 60^\circ = 60^\circ$, which is $\frac{1}{6}$ of a circle, so area $\frac{\pi}{6}$.
- Right semicircle, the slice from $Q$ around to $B$: again $120^\circ$, area $\frac{\pi}{3}$.
- Two equilateral triangles of side $1$, each with area $\frac{\sqrt{3}}{4}$.
6. White area:
$$\frac{\pi}{3} + \frac{\pi}{6} + \frac{\pi}{3} + 2 \cdot \frac{\sqrt{3}}{4} = \frac{5}{6}\pi + \frac{\sqrt{3}}{2}.$$
7. Shaded area:
$$2\pi - \frac{5}{6}\pi - \frac{\sqrt{3}}{2} = \frac{7}{6}\pi - \frac{\sqrt{3}}{2}.$$

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

When two circles of the same radius have centers one radius apart, the crossing point forms an equilateral triangle with the two centers, so every angle is a multiple of $60^\circ$. Drawing lines from crossing points to the centers is the standard move for overlapping circles: it cuts a curvy region into pie slices and triangles, which you can measure.

Add the three small semicircles and take away the overlaps. Together they have area $3 \cdot \frac{\pi}{2} = \frac{3}{2}\pi$. Each overlap (between $L$'s and $O$'s semicircles, and between $O$'s and $R$'s) is a $60^\circ$ slice from each circle minus the equilateral triangle they share: $\frac{\pi}{6} + \frac{\pi}{6} - \frac{\sqrt{3}}{4} = \frac{\pi}{3} - \frac{\sqrt{3}}{4}$. So the white area is $\frac{3}{2}\pi - 2\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right) = \frac{5}{6}\pi + \frac{\sqrt{3}}{2}$, the same as before.