AMC12 2012 A

AMC12 2012 A · Q14

AMC12 2012 A · Q14. It mainly tests Circle theorems, Polygons.

The closed curve in the figure is made up of 9 congruent circular arcs each of length $\frac{2\pi}{3}$, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?

图中的闭合曲线由 9 段全等的圆弧组成，每段圆弧的长度为 $\frac{2\pi}{3}$，且这些圆弧所对应圆的圆心都位于边长为 2 的正六边形的顶点中。求该曲线所围成的面积。

(A) $2\pi + 6$ $2\pi + 6$

(B) $2\pi + 4\sqrt{3}$ $2\pi + 4\sqrt{3}$

(D) $2\pi + 3\sqrt{3} + 2$ $2\pi + 3\sqrt{3} + 2$

(E) $\pi + 6\sqrt{3}$ $\pi + 6\sqrt{3}$

Answer

Correct choice: (E)

正确答案：（E）

Solution

We can draw the hexagon between the centers of the circles, and compute its area. The hexagon is made of $6$ equilateral triangles each with length $2$, so the area is: \[\frac{\sqrt{3}}{4} \cdot 2^2 \cdot 6=6 \sqrt{3}.\] Then, we add the areas of the three sectors outside the hexagon: \[\frac 23 \pi \cdot 3=2\pi.\] We now subtract the areas of the three sectors inside the hexagon but outside the figure (which is $\pi$) to get the area enclosed in the curved figure: \[6 \sqrt{3}+2\pi-\pi=\pi+6\sqrt{3}.\] Hence, our answer is $\boxed{\textbf{(E)}\ \pi+6\sqrt{3}},$ and we are done. \[\] (Minor edits, made by dbnl.)

我们可以在这些圆的圆心之间画出正六边形，并计算其面积。该六边形由 $6$ 个边长为 $2$ 的等边三角形组成，因此面积为： \[\frac{\sqrt{3}}{4} \cdot 2^2 \cdot 6=6 \sqrt{3}.\] 然后，加上六边形外侧的三个扇形面积： \[\frac 23 \pi \cdot 3=2\pi.\] 再减去六边形内部但在图形外的三个扇形面积（为 $\pi$），得到曲线图形所围面积： \[6 \sqrt{3}+2\pi-\pi=\pi+6\sqrt{3}.\] 因此答案为 $\boxed{\textbf{(E)}\ \pi+6\sqrt{3}},$ 结束。 \[\] (由 dbnl 做了少量修改。)

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