2020 AMC 12B Problem 11

Question

As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length

2

so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region—inside the hexagon but outside all of the semicircles?

6\sqrt{3} - 3\pi

\dfrac{9\sqrt{3}}{2} - 2\pi

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

3\sqrt{3} - \pi

\dfrac{9\sqrt{3}}{2} - \pi

Answer: D

Concepts:regular polygon area decomposition inclusion-exclusion

Difficulty rating: 1590

Small hint:

The hexagon area is

\dfrac{3\sqrt3}{2}s^2

with

s = 2,

and each semicircle has radius

1

Big hint:

Adjacent semicircles overlap; subtract the union, accounting for the six overlapping lenses

Solution:

The hexagon has area

\tfrac{3\sqrt3}{2}\cdot 2^2 = 6\sqrt3.

Each semicircle has radius

1

and area

\tfrac{\pi}{2},

totaling

3\pi.

Adjacent semicircle centers (side midpoints) are a distance

\sqrt3

apart, so each adjacent pair overlaps in a lens of area

2\cos^{-1}\!\left(\tfrac{\sqrt3}{2}\right) - \tfrac{\sqrt3}{2} = \tfrac{\pi}{3} - \tfrac{\sqrt3}{2}.

There are six such lenses.

The union of the semicircles is

3\pi - 6\left(\frac{\pi}{3} - \frac{\sqrt3}{2}\right) = \pi + 3\sqrt3.

Subtracting from the hexagon gives the shaded area

6\sqrt3 - (\pi + 3\sqrt3) = 3\sqrt3 - \pi.

Thus, the correct answer is D.