2019 AMC 10B Problem 20

Place $F$ at the origin, so line $EG$ is $y = 0$, line $AD$ is $y = -1$, and the semicircle centers are $(-2,-1), (0,-1), (2,-1)$. Cut the big disk into three horizontal zones.

Above $EG$. The semicircles lie below $EG$ (touching it at $E, F, G$), so the whole upper half-disk is shaded: area $\frac12 \cdot 4\pi = 2\pi$.

Between $EG$ and $AD$. Here the semicircles cover everything except the two gaps between neighbors, plus points beyond the outer semicircles; those have $|x| > 2$ and are outside the big circle. Each gap sits in the $2 \times 1$ rectangle with corners $E$, $F$ and the two adjacent semicircle centers, minus two quarter-disks of radius $1$: area $2 - \frac{\pi}{2}$. The gaps lie inside the big circle (their outer arcs run from $E$ or $G$, on the circle, inward to $B$ or $C$ at distance $\sqrt2$). Two gaps: $4 - \pi$.

Below $AD$. The chord along $y = -1$ is at distance $1$ from the center of a radius-$2$ circle, so it subtends a $120^\circ$ central angle (half-angle with $\cos\theta = \frac12$). The segment below it has area
$$\frac{120}{360}\cdot 4\pi - \frac12 \cdot 2 \cdot 2 \cdot \sin 120^\circ = \frac{4\pi}{3} - \sqrt3.$$

Total.
$$2\pi + (4 - \pi) + \left(\frac{4\pi}{3} - \sqrt3\right) = \frac{7\pi}{3} - \sqrt3 + 4.$$
So $a = 7$, $b = 3$, $c = 3$, $d = 4$, and $a + b + c + d = 17$.

The answer is $\boxed{\textbf{(E)}\ 17}$.

The tangent line $EG$ and the diameter line $AD$ are natural cut lines: above one and below the other the shaded region is a plain half-disk and a plain circular segment. The awkward middle strip becomes simple once you notice that the outer semicircles poke outside the big circle exactly where the big circle leaves the strip, so only the two cusp-shaped gaps remain. Decompose along the lines the figure already gives you.

• Assuming all three semicircles lie inside the big circle and computing 4 pi minus 3 pi / 2, or forgetting the shaded segment below line AD.
• Using a $60^\circ$ central angle for the chord at distance $1$ (confusing the half-angle with the full angle), which gives a segment area of $\frac{2\pi}{3} - \sqrt3$.