Question

如下图，六个半圆位于边长为 $2$ 的正六边形内部，且这些半圆的直径分别与六边形的各边重合。阴影区域在六边形内部、但在所有半圆外部。阴影区域的面积是多少？

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?

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

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

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

$3\sqrt{3} - \pi$

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

答案：D

知识点：正多边形面积分割容斥原理

难度评级：1590

小提示：

六边形面积为 $\dfrac{3\sqrt3}{2}s^2$ ，其中 $s = 2$ 且每个半圆的半径为 $1$ 。

The hexagon area is $\dfrac{3\sqrt3}{2}s^2$ with $s = 2,$ and each semicircle has radius $1$

大提示：

相邻半圆会重叠；减去半圆并集时，要计入六个重叠透镜形区域。

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

解答：

六边形的面积为 $\tfrac{3\sqrt3}{2}\cdot 2^2 = 6\sqrt3$ 。每个半圆的半径为 $1$ ，面积为 $\tfrac{\pi}{2}$ ，所以六个半圆的面积之和为 $3\pi$ 。

相邻半圆的圆心距离为 $\sqrt3$ ，因此每对相邻半圆的重叠透镜形面积为 $2\cos^{-1}\!\left(\tfrac{\sqrt3}{2}\right) - \tfrac{\sqrt3}{2} = \tfrac{\pi}{3} - \tfrac{\sqrt3}{2}$ 。

共有六个这样的透镜形，所以六个半圆的并集面积为 $3\pi - 6\left(\frac{\pi}{3} - \frac{\sqrt3}{2}\right) = \pi + 3\sqrt3\text{。}$ 从六边形面积中减去它，得到阴影面积 $6\sqrt3 - (\pi + 3\sqrt3) = 3\sqrt3 - \pi$ 。

所以正确答案是 D。

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 第 11 题

答案：D

难度评级：1590

小提示：

大提示：

解答：

其他年份的第 11 题