AMC10 2017 A

AMC10 2017 A · Q22

AMC10 2017 A · Q22. It mainly tests Triangles (properties), Circle theorems.

Sides AB and AC of equilateral triangle ABC are tangent to a circle at points B and C, respectively. What fraction of the area of $\triangle ABC$ lies outside the circle?

正三角形 ABC 的边 AB 和 AC 分别在点 B 和 C 处与一个圆相切。该圆外部的 $\triangle ABC$ 面积占 $\triangle ABC$ 总面积的几分之几？

(A) $\frac{4\sqrt{3\pi}}{27} - \frac{1}{3}$ $\frac{4\sqrt{3\pi}}{27} - \frac{1}{3}$

(B) $\frac{\sqrt{3}}{2} - \frac{\pi}{8}$ $\frac{\sqrt{3}}{2} - \frac{\pi}{8}$

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

(E) $\frac{4}{3} - \frac{4\sqrt{3\pi}}{27}$ $\frac{4}{3} - \frac{4\sqrt{3\pi}}{27}$

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Let $O$ be the center of the circle, and without loss of generality, assume that radius $OB=1$. Because $\triangle ABO$ is a $30-60-90^\circ$ right triangle, $AO=2$ and $AB=BC=\sqrt{3}$. Kite $ABOC$ has diagonals of lengths $2$ and $\sqrt{3}$, so its area is $\sqrt{3}$. Because $\angle BOC=120^\circ$, the area of the sector cut off by $\angle BOC$ is $\frac{1}{3}\pi$. The area of the portion of $\triangle ABC$ lying outside the circle (shaded in the figure) is therefore $\sqrt{3}-\frac{1}{3}\pi$. The area of $\triangle ABC$ is $\frac{1}{4}\sqrt{3}(\sqrt{3})^2=\frac{3}{4}\sqrt{3}$, so the requested fraction is \[ \frac{\sqrt{3}-\frac{1}{3}\pi}{\frac{3}{4}\sqrt{3}}=\frac{4}{3}-\frac{4\sqrt{3}\pi}{27}. \]

答案（E）：设 $O$ 为圆心，不失一般性，令半径 $OB=1$。由于 $\triangle ABO$ 是一个 $30-60-90^\circ$ 的直角三角形，$AO=2$ 且 $AB=BC=\sqrt{3}$。风筝形 $ABOC$ 的两条对角线长度分别为 $2$ 和 $\sqrt{3}$，因此其面积为 $\sqrt{3}$。因为 $\angle BOC=120^\circ$，由 $\angle BOC$ 截得的扇形面积为 $\frac{1}{3}\pi$。因此，$\triangle ABC$ 在圆外的部分（图中阴影部分）的面积为 $\sqrt{3}-\frac{1}{3}\pi$。$\triangle ABC$ 的面积为 $\frac{1}{4}\sqrt{3}(\sqrt{3})^2=\frac{3}{4}\sqrt{3}$，所以所求比例为 \[ \frac{\sqrt{3}-\frac{1}{3}\pi}{\frac{3}{4}\sqrt{3}}=\frac{4}{3}-\frac{4\sqrt{3}\pi}{27}. \]

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