2021 AMC Fall 10B Problem 11

1. Size of the circle. A regular hexagon splits into $6$ equilateral triangles meeting at the center. Each has side $1$, so the circle's radius is $1$. Each side of the hexagon cuts off $\frac{360^\circ}{6} = 60^\circ$ of the circle.
2. The sliver outside one side. Between one side of the hexagon and the circle is a thin sliver. To find its area, take the $60^\circ$ pie slice of the circle and remove the equilateral triangle inside it:
- Pie slice: $\frac{60}{360} \times \pi \times 1^2 = \frac{\pi}{6}$.
- Equilateral triangle with side $1$: $\frac{\sqrt3}{4}$.
- Sliver: $\frac{\pi}{6} - \frac{\sqrt3}{4}$.
3. What reflecting does. Flipping the arc over its side flips the sliver to the inside of the hexagon, with the same area. The six flipped arcs meet at the hexagon's corners and bend inward, so the region they enclose is the hexagon with six slivers cut away.
4. Hexagon area. Six equilateral triangles: $6 \times \frac{\sqrt3}{4} = \frac{3\sqrt3}{2}$.
5. Subtract.
$$\frac{3\sqrt3}{2} - 6\left(\frac{\pi}{6} - \frac{\sqrt3}{4}\right) = \frac{3\sqrt3}{2} - \pi + \frac{3\sqrt3}{2} = 3\sqrt3 - \pi.$$

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

Flipping a shape does not change its area, so each flipped arc cuts off exactly as much from the inside of the hexagon as the original arc added outside it. For any region with curved edges, break it into pie slices and triangles. "Sliver = pie slice minus triangle" is the piece you will use again and again.

Estimate to check. The hexagon's area is about $6 \times 0.433 = 2.6$. Each sliver is about $0.524 - 0.433 = 0.09$, so six of them are about $0.55$. The region should be about $2.6 - 0.55 = 2.05$. Choice (B) is about $5.20 - 3.14 = 2.06$, and no other choice is close.

A regular hexagon with side $1$ inscribed in a circle has circumradius $1$, and each side subtends a $60^\circ$ arc.

The region between a side and its minor arc is a circular segment. Its area is a $60^\circ$ sector minus an equilateral triangle of side $1$:
$$\frac{60}{360}\pi \cdot 1^2 - \frac{\sqrt3}{4} = \frac{\pi}{6} - \frac{\sqrt3}{4}.$$

Reflecting the arc over the side flips this segment to the inside of the hexagon. The region bounded by the six reflected arcs is therefore the hexagon with six such segments removed.

The hexagon consists of six equilateral triangles of side $1$, area $6 \cdot \frac{\sqrt3}{4} = \frac{3\sqrt3}{2}$. So the required area is
$$\frac{3\sqrt3}{2} - 6\left(\frac{\pi}{6} - \frac{\sqrt3}{4}\right) = \frac{3\sqrt3}{2} + \frac{3\sqrt3}{2} - \pi = 3\sqrt3 - \pi.$$

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