AMC 8 2017 Problem 25

AMC8 2017 年第 25 题

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In the figure shown, US‾\overline{US} and UT‾\overline{UT} are line segments each of length 2, and m∠TUS=60∘m\angle TUS = 60^\circ. Arcs TR⌢\overset{\frown}{TR} and SR⌢\overset{\frown}{SR} are each one-sixth of a circle with radius 2. What is the area of the region shown?
如下图所示，US‾\overline{US} 和 UT‾\overline{UT} 是两条长度都为 2 的线段，且 m∠TUS=60∘m\angle TUS = 60^\circ。弧 TR⌢\overset{\frown}{TR} 和 SR⌢\overset{\frown}{SR} 各是半径为 2 的圆的六分之一。那么图中所示区域的面积是多少？
AMC 8 2017 Problem 25 figure
  1. A.33−π3\sqrt{3}-\pi
  2. B.43−4π3\displaystyle 4\sqrt{3}-\frac{4\pi}{3}
  3. C.232\sqrt{3}
  4. D.43−2π3\displaystyle 4\sqrt{3}-\frac{2\pi}{3}
  5. E.4+4π3\displaystyle 4+\frac{4\pi}{3}

正确答案 · Correct Answer

B

解析 · Solution

解法1

常规解法 · 补全等边三角形再扣 图形 = 边长 4 的等边三角形挖去两个「六分之一圆（半径 2）」。等边面积 34⋅42=43\displaystyle \frac{\sqrt3}{4}\cdot4^2=4\sqrt3，每个扇形 16π⋅22=2π3\displaystyle \frac16\pi\cdot2^2=\frac{2\pi}{3}。阴影 =43−2⋅2π3=43−4π3\displaystyle =4\sqrt3-2\cdot\frac{2\pi}{3}=4\sqrt3-\frac{4\pi}{3}。

Solution 1

Elementary · complete the equilateral triangle, then subtract The region is an equilateral triangle of side 4 with two 60∘60^\circ sectors of radius 2 removed. The triangle has area 34⋅42=43\displaystyle \frac{\sqrt3}{4}\cdot4^2=4\sqrt3, and each sector has area 16π⋅22=2π3\displaystyle \frac16\pi\cdot2^2=\frac{2\pi}{3}. So the shaded area is 43−2⋅2π3=43−4π3\displaystyle 4\sqrt3-2\cdot\frac{2\pi}{3}=4\sqrt3-\frac{4\pi}{3}.

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