2020 AMC 12B 第 11 题

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：

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