2020 AMC 10B Problem 14

Call the center of the hexagon $O$. The hexagon has side $2$, and each semicircle has radius $1$ and is centered at the midpoint of a side.

1. Split the hexagon. Draw lines from $O$ to all six corners. This cuts the hexagon into $6$ equilateral triangles of side $2$. The picture is the same in each one, so study one of them: triangle $OV_1V_2$, where $V_1V_2$ is a side of the hexagon with midpoint $M$.
2. Mark midpoints. Let $P_1$ be the midpoint of $OV_1$ and $P_2$ the midpoint of $OV_2$. Joining $M$, $P_1$ and $P_2$ cuts the big triangle into $4$ small equilateral triangles of side $1$: triangle $OP_1P_2$ at the top, and three small triangles that each have $M$ as a corner.
3. Where the semicircle on $V_1V_2$ reaches. Its center is $M$ and its radius is $1$. The points $V_1$, $P_1$, $P_2$, $V_2$ are all exactly $1$ from $M$, since they are corners of small triangles of side $1$ with a corner at $M$. So the three small triangles touching $M$ are completely covered, and the curved edge of the semicircle passes through $P_1$ and $P_2$.
4. The top small triangle. Between $P_1$ and $P_2$ the arc bulges past the straight segment $P_1P_2$ into triangle $OP_1P_2$. That bulge is a circular "bite". The semicircles on the neighboring sides come in no further than line $OV_1$ or $OV_2$ between the corner and $P_1$ or $P_2$, so they add nothing new here. So the shaded part of this big triangle is triangle $OP_1P_2$ with the bite removed.
5. Area of the bite. Triangle $MP_1P_2$ is equilateral with side $1$, so the arc from $P_1$ to $P_2$ is a $60^\circ$ arc of a circle of radius $1$. The bite is a $60^\circ$ slice of the circle minus triangle $MP_1P_2$:
$$\frac{60}{360}\pi \times 1^2 - \frac{\sqrt3}{4} = \frac{\pi}{6} - \frac{\sqrt3}{4}.$$
6. Shaded part of one big triangle.
$$\frac{\sqrt3}{4} - \left(\frac{\pi}{6} - \frac{\sqrt3}{4}\right) = \frac{\sqrt3}{2} - \frac{\pi}{6}.$$
7. All six.
$$6\left(\frac{\sqrt3}{2} - \frac{\pi}{6}\right) = 3\sqrt3 - \pi.$$

The answer is $\boxed{\textbf{(D)}\ 3\sqrt{3} - \pi}$.

Before you subtract shapes from a region, check whether those shapes overlap. Here neighboring semicircles overlap near each corner, so "hexagon minus six half-circles" removes some area twice. Cutting the hexagon into equal pieces and marking midpoints shows exactly where each arc runs. The shaded region is then just small triangles minus circular bites, and each bite is a slice of a circle minus a triangle.

Subtract first, then add back the overlaps. Hexagon area: $6 \times \frac{\sqrt3}{4} \times 2^2 = 6\sqrt3$. Six half-circles of radius $1$: $6 \times \frac{\pi}{2} = 3\pi$. Neighboring semicircles overlap in a lens shape between the shared corner and the crossing point near the center (the point $P_1$ or $P_2$ above). Each lens is two circular bites like the one in step 5, so its area is $2\left(\frac{\pi}{6} - \frac{\sqrt3}{4}\right) = \frac{\pi}{3} - \frac{\sqrt3}{2}$. That area was removed twice, so add it back once for each of the $6$ lenses:
$$6\sqrt3 - 3\pi + 6\left(\frac{\pi}{3} - \frac{\sqrt3}{2}\right) = 3\sqrt3 - \pi.$$

Let $O$ be the center. Each semicircle has radius $1$ and is centered at the midpoint $M$ of a side, with $OM = \sqrt3$ (the apothem).

Where do two adjacent semicircles meet? Their centers $M_1, M_2$ and the shared vertex $V$ satisfy $M_1V = M_2V = 1$ and $M_1M_2 = \sqrt3$ (triangle $OM_1M_2$ is equilateral). The second intersection point $P$ is the mirror image of $V$ across line $M_1M_2$. Since $O$ is at distance $\tfrac{3}{2}$ from that line and $OV = 2$, the reflection puts $P$ at distance $2 - 2\cdot\tfrac12 = 1$ from $O$.

So the six meeting points lie on a circle of radius $1$ around $O$, spaced $60^\circ$ apart: they form a regular hexagon of side $1$, and the shaded region is that inner hexagon with six circular arcs bulging inward.

Consider one side $P_1P_2$ of the inner hexagon and the semicircle center $M$ on its far side. $MP_1 = MP_2 = P_1P_2 = 1$, so the arc $P_1P_2$ subtends $60^\circ$ at $M$, and the bite it takes out of the inner hexagon is a circular segment of area
$$\frac{\pi}{6} - \frac{\sqrt3}{4}.$$

Shaded area $=$ inner hexagon $-$ six segments:
$$6\cdot\frac{\sqrt3}{4} - 6\left(\frac{\pi}{6} - \frac{\sqrt3}{4}\right) = 3\sqrt3 - \pi.$$

The answer is $\boxed{\textbf{(D)}\ 3\sqrt{3} - \pi}$.