AMC10 2004 B

AMC10 2004 B · Q25

AMC10 2004 B · Q25. It mainly tests Circle theorems, Area & perimeter.

A circle of radius 1 is internally tangent to two circles of radius 2 at points $A$ and $B$, where $AB$ is a diameter of the smaller circle. What is the area of the region, shaded in the figure, that is outside the smaller circle and inside each of the two larger circles?

一个半径为 1 的圆在两个半径为 2 的圆内切于点 $A$ 和 $B$，其中 $AB$ 是小圆的直径。阴影区域（图中所示）是在两个大圆内部但小圆外部的面积是多少？

(A) \frac{5}{3}\pi - 3\sqrt{2} \frac{5}{3}\pi - 3\sqrt{2}

(B) \frac{5}{3}\pi - 2\sqrt{3} \frac{5}{3}\pi - 2\sqrt{3}

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

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

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) The centers of the two larger circles are at $A$ and $B$. Let $C$ be the center of the smaller circle, and let $D$ be one of the points of intersection of the two larger circles. Then $\triangle ACD$ is a right triangle with $AC=1$ and $AD=2$, so $CD=\sqrt{3}$, $\angle CAD=60^\circ$, and the area of $\triangle ACD$ is $\sqrt{3}/2$. The area of $1/4$ of the shaded region, as shown in the Figure, is the area of sector $BAD$ of the circle centered at $A$, minus the area of $\triangle ACD$, minus the area of $1/4$ of the smaller circle. That area is $$ \frac{2}{3}\pi-\frac{\sqrt{3}}{2}-\frac{1}{4}\pi=\frac{5}{12}\pi-\frac{\sqrt{3}}{2}, $$ so the area of the entire shaded region is $$ 4\left(\frac{5}{12}\pi-\frac{\sqrt{3}}{2}\right)=\frac{5}{3}\pi-2\sqrt{3}. $$

（B）两个较大圆的圆心分别在 $A$ 和 $B$。设 $C$ 为较小圆的圆心，并设 $D$ 为两个较大圆的一个交点。则 $\triangle ACD$ 为直角三角形，且 $AC=1$、$AD=2$，所以 $CD=\sqrt{3}$，$\angle CAD=60^\circ$，并且 $\triangle ACD$ 的面积为 $\sqrt{3}/2$。如图所示，阴影区域的 $\frac{1}{4}$ 的面积等于以 $A$ 为圆心的圆中扇形 $BAD$ 的面积，减去 $\triangle ACD$ 的面积，再减去较小圆的 $\frac{1}{4}$ 的面积。该面积为 $$ \frac{2}{3}\pi-\frac{\sqrt{3}}{2}-\frac{1}{4}\pi=\frac{5}{12}\pi-\frac{\sqrt{3}}{2}, $$ 因此整个阴影区域的面积为 $$ 4\left(\frac{5}{12}\pi-\frac{\sqrt{3}}{2}\right)=\frac{5}{3}\pi-2\sqrt{3}. $$

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