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2003 AMC 12B 第 16 题

先试着解答 2003 AMC 12B 第 16 题，然后核对你的答案与精心整理的解答，解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2003 AMC 12B 解答，或核对答案。

所有题目均经美国数学协会（MAA）官方合法授权使用。

16.

在半径为 22 的半圆的直径 AB‾\overline{AB} 上作三个半径为 11 的半圆。小半圆的圆心把 AB‾\overline{AB} 分成四段相等的线段，如图所示。位于大半圆内且在小半圆外的阴影区域面积是多少？

Three semicircles of radius 11 are constructed on diameter AB‾\overline{AB} of a semicircle of radius 2.2. The centers of the small semicircles divide AB‾\overline{AB} into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?

π−3\pi - \sqrt{3}

π−2\pi - \sqrt{2}

π+22\dfrac{\pi + \sqrt{2}}{2}

π+32\dfrac{\pi + \sqrt{3}}{2}

76π−32\dfrac{7}{6}\pi - \dfrac{\sqrt{3}}{2}

答案：E
知识点：扇形等边三角形面积分割
难度评级：1680
小提示：

大半圆面积为 12π(2)2=2π\dfrac{1}{2}\pi(2)^2 = 2\pi

The large semicircle has area 12π(2)2=2π\dfrac{1}{2}\pi(2)^2 = 2\pi

大提示：

用容斥原理：从三个小半圆的总面积中减去两块重叠部分

Use inclusion-exclusion: subtract the two overlaps from the total area of the three small semicircles

解答：

大半圆面积为 12π(2)2=2π\dfrac{1}{2}\pi(2)^2 = 2\pi。

在计入重叠之前，三个小半圆的总面积为 3π2\dfrac{3\pi}{2}。每相邻两个小半圆的圆心相距 11，所以它们的公共部分由两个 60∘60^\circ 扇形和一个等边三角形围成。因此两块重叠部分的面积各为 π3−34\dfrac{\pi}{3} - \dfrac{\sqrt{3}}{4}。

阴影面积为 2π−[3π2−2(π3−34)]=76π−32。 \begin{aligned} &2\pi - \left[\frac{3\pi}{2} - 2\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right)\right] \\ &= \frac{7}{6}\pi - \frac{\sqrt{3}}{2} \end{aligned}\text{。}

因此，正确答案是 E。

The large semicircle has area 12π(2)2=2π.\dfrac{1}{2}\pi(2)^2 = 2\pi.

The three small semicircles have total area 3π2\dfrac{3\pi}{2} before their overlaps are accounted for. The centers of each adjacent pair are 11 unit apart, so their intersection is bounded by two 60∘60^\circ sectors and an equilateral triangle. Each of the two overlaps therefore has area π3−34.\dfrac{\pi}{3} - \dfrac{\sqrt{3}}{4}.

The shaded area is 2π−[3π2−2(π3−34)]=76π−32. \begin{aligned} &2\pi - \left[\frac{3\pi}{2} - 2\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right)\right] \\ &= \frac{7}{6}\pi - \frac{\sqrt{3}}{2}. \end{aligned}

Thus, the correct answer is E.

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