2020 AMC 10B Problem 14

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

Before subtracting shapes from a region, check whether they overlap; here neighboring semicircles share a lens near each vertex, so "hexagon minus six semicircles" is wrong. Describing the shaded region directly by its boundary (arcs between intersection points) leads to an honest decomposition: a polygon minus circular segments, each being sector minus triangle.

Inclusion-exclusion: hexagon $6\sqrt3$ minus six semicircles $3\pi$ plus the six lenses. Two unit circles with centers $\sqrt3$ apart form a lens of area $2\cdot\frac{\pi}{6} - \frac{\sqrt3}{2} = \frac{\pi}{3} - \frac{\sqrt3}{2}$, and each lens lies entirely inside both semicircles. Total: $6\sqrt3 - 3\pi + 6\left(\frac{\pi}{3} - \frac{\sqrt3}{2}\right) = 3\sqrt3 - \pi$.

• Computing hexagon area minus six semicircles, 6 sqrt(3) - 3 pi, which double-subtracts the overlaps near each vertex.
• Assuming the semicircles reach the center or are tangent to each other; the apothem $\sqrt3 > 1$ and the center distance $\sqrt3 < 2$ show neither is true.