2019 AMC 10B Problem 20

The figure already contains two useful straight lines: the tangent line $EG$ and the diameter line $AD$. Cut the big circle along both and handle each part separately.

Set up coordinates with $F$ at $(0, 0)$. The semicircles have radius $1$ and touch line $EG$, so $EG$ is the line $y = 0$ and $AD$ is the line $y = -1$. The semicircle centers (the midpoints of $AB$, $BC$, $CD$) are $(-2, -1)$, $(0, -1)$, $(2, -1)$, so $E = (-2, 0)$ and $G = (2, 0)$. The big circle has center $(0,0)$ and radius $2$.

Part 1: above line $EG$. The semicircles all stay below $EG$, so the whole top half of the big circle is shaded. Area: $\frac12 \cdot \pi \cdot 2^2 = 2\pi$.

Part 2: between $EG$ and $AD$.

1. In this strip the semicircles cover almost everything. What is left inside the big circle are two curved gaps, one between each pair of neighboring semicircles.
2. The left gap sits inside the rectangle with corners $(-2, 0)$, $(0, 0)$, $(0, -1)$, $(-2, -1)$. That rectangle is $2$ by $1$, area $2$.
3. The two semicircles each cover a quarter-circle of radius $1$ inside that rectangle, total $2 \cdot \frac{\pi}{4} = \frac{\pi}{2}$.
4. So one gap has area $2 - \frac{\pi}{2}$, and two gaps have area $4 - \pi$.
5. The parts of the outer semicircles beyond $E$ and $G$ (where $x$ is past $-2$ or $2$) are outside the big circle anyway, so nothing else in this strip is shaded.

Part 3: below line $AD$. This is the slice of the big circle cut off by the chord along $y = -1$.

1. The chord is $1$ unit from the center, and the radius is $2$. So its endpoints are at $x = \pm\sqrt{4 - 1} = \pm\sqrt3$, and the chord has length $2\sqrt3$.
2. Each half of the triangle from the center to the chord has sides $1$, $\sqrt3$, $2$: a $30$-$60$-$90$ triangle. So the angle at the center is $2 \cdot 60^\circ = 120^\circ$.
3. The pie slice with a $120^\circ$ angle has area $\frac{120}{360} \cdot 4\pi = \frac{4\pi}{3}$.
4. Remove the triangle (base $2\sqrt3$, height $1$, area $\sqrt3$). The slice below the chord has area $\frac{4\pi}{3} - \sqrt3$.

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 $\boxed{\textbf{(E)}\ 17}$.

A strange shaded region becomes easy when you cut it along lines the figure already gives you. Above one line and below the other, the pieces are standard shapes: a half-circle and a circle slice (pie slice minus triangle). Only the middle strip needs care, and there the leftovers are "rectangle minus two quarter-circles."