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?

$6\sqrt{3} - 3\pi$

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

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

$3\sqrt{3} - \pi$

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

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.

2020 AMC 12B Problem 11

Solution:

Problem 11 in Other Years