2021 AMC 10B Fall Problem 11

Question

A regular hexagon of side length

1

is inscribed in a circle. Each minor arc of the circle determined by a side of the hexagon is reflected over that side. What is the area of the region bounded by these

6

reflected arcs?

\frac{5\sqrt{3}}{2} - \pi

3\sqrt{3}-\pi

4\sqrt{3}-\frac{3\pi}{2}

\pi - \frac{\sqrt{3}}{2}

\frac{\pi + \sqrt{3}}{2}

Answer: B

Concepts:area decomposition regular polygon circle area

Difficulty rating: 1630

Small hint:

Compare the reflected-arc region with the original circle

Big hint:

The hexagon is the average of the two regions

Solution:

The original circle is made from the regular hexagon plus

6

equal circular segments. Reflecting each minor arc over its side puts those same

6

segments inside the hexagon instead.

Therefore the average of the circle’s area and the reflected-arc region’s area is the area of the regular hexagon. The hexagon has area

6\cdot\frac{\sqrt3}{4}=\frac{3\sqrt3}{2},

and the circle has radius

1,

so its area is

\pi.

If the desired area is

A,

then

\frac{A+\pi}{2}=\frac{3\sqrt3}{2},

so

A=3\sqrt3-\pi.

Thus, the answer is B .