AMC12 2020 B

AMC12 2020 B · Q11

AMC12 2020 B · Q11. 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，六个半圆位于边长为2的正六边形内部，其半圆的直径与六边形的边重合。阴影区域的面积是多少——在六边形内部但在所有半圆外部？

(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

Consider the semicircles on two adjacent sides of the hexagon, where $F$ and $G$ are the centers and midpoints of the sides. Because $\angle FBG = 120^\circ$, $\angle FBD = 60^\circ$ and $\triangle FBD$ is equilateral. The segment of the circle at $F$ by arc $BD$ has area $\frac{\pi}{6} - \frac{\sqrt{3}}{4}$. The area is hexagon area minus six semicircles plus 12 segments: $6 \left( \frac{\sqrt{3}}{4} \cdot 4 \right) - 6 \left( \frac{\pi \cdot 1^2}{2} \right) + 12 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) = 3\sqrt{3} - \pi$.

考虑六边形两条相邻边的半圆，其中$F$和$G$是边的中心和中点。因为$\angle FBG = 120^\circ$，$\angle FBD = 60^\circ$且$\triangle FBD$是等边三角形。由弧$BD$在$F$处的圆段面积为$\frac{\pi}{6} - \frac{\sqrt{3}}{4}$。面积为六边形面积减去六个半圆加上12个圆段： $6 \left( \frac{\sqrt{3}}{4} \cdot 4 \right) - 6 \left( \frac{\pi \cdot 1^2}{2} \right) + 12 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) = 3\sqrt{3} - \pi$。

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