AMC10 2020 B

AMC10 2020 B · Q14

AMC10 2020 B · Q14. It mainly tests Circle theorems, Area & perimeter.

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?

如图所示，六条半圆位于边长为 2 的正六边形内部，其半圆的直径与六边形的边重合。阴影区域（在六边形内但在所有半圆外）的面积是多少？

(A) 6\sqrt{3}-3\pi 6\sqrt{3}-3\pi

(B) \frac{9\sqrt{3}}{2}-2\pi \frac{9\sqrt{3}}{2}-2\pi

(D) 3\sqrt{3}-\pi 3\sqrt{3}-\pi

(E) \frac{9\sqrt{3}}{2}-\pi \frac{9\sqrt{3}}{2}-\pi

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Consider the semicircles on two adjacent sides of the hexagon shown in the figure below, where $F$ and $G$ are the centers of the semicircles and the midpoints of their respective sides. Because $\angle FBG = 120^\circ$, it follows that $\angle FBD = 60^\circ$ and $\triangle FBD$ is equilateral. Thus the segment of the circle centered at $F$ determined by arc $BD$ (that is, the region, shaded in the figure, that lies inside the sector spanned by arc $BD$ but outside $\triangle FBD$) has area $\frac{\pi}{6}-\frac{\sqrt{3}}{4}$. The requested area is then the area of the hexagon, minus the areas of the six semicircles, plus 12 times the area of that segment, which is $$ 6\left(\frac{\sqrt{3}}{4}\cdot 2^2\right)-6\left(\frac{\pi}{2}\right)+12\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right)=3\sqrt{3}-\pi. $$

答案（D）：考虑下图所示六边形相邻两边上的半圆，其中 $F$ 和 $G$ 为半圆的圆心，同时也是各自对应边的中点。由于 $\angle FBG=120^\circ$，可得 $\angle FBD=60^\circ$，且 $\triangle FBD$ 为等边三角形。因此，以 $F$ 为圆心、由弧 $BD$ 所确定的圆弓形（即图中阴影部分：位于弧 $BD$ 张成的扇形内部但在 $\triangle FBD$ 外部的区域）面积为 $\frac{\pi}{6}-\frac{\sqrt{3}}{4}$。所求面积等于六边形面积减去 6 个半圆面积，再加上该圆弓形面积的 12 倍，即 $$ 6\left(\frac{\sqrt{3}}{4}\cdot 2^2\right)-6\left(\frac{\pi}{2}\right)+12\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right)=3\sqrt{3}-\pi. $$

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