AMC10 2008 A

AMC10 2008 A · Q16

AMC10 2008 A · Q16. It mainly tests Triangles (properties), Circle theorems.

Points $A$ and $B$ lie on a circle centered at $O$, and $\angle AOB = 60^\circ$. A second circle is internally tangent to the first and tangent to both $\overline{OA}$ and $\overline{OB}$. What is the ratio of the area of the smaller circle to that of the larger circle?

点 $A$ 和 $B$ 位于以 $O$ 为圆心的大圆上，且 $\angle AOB = 60^\circ$。有一个小圆内切于大圆并同时与 $\overline{OA}$ 和 $\overline{OB}$ 相切。小圆与大圆面积之比是多少？

(A) $\frac{1}{16}$ $\frac{1}{16}$

(B) $\frac{1}{9}$ $\frac{1}{9}$

(D) $\frac{1}{6}$ $\frac{1}{6}$

(E) $\frac{1}{4}$ $\frac{1}{4}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Let $r$ and $R$ be the radii of the smaller and larger circles, respectively. Let $E$ be the center of the smaller circle, let $\overline{OC}$ be the radius of the larger circle that contains $E$, and let $D$ be the point of tangency of the smaller circle to $\overline{OA}$. Then $OE=R-r$, and because $\triangle EDO$ is a $30-60-90^\circ$ triangle, $OE=2DE=2r$. Thus $2r=R-r$, so $\frac{r}{R}=\frac{1}{3}$. The ratio of the areas is $\left(\frac{1}{3}\right)^2=\frac{1}{9}$.

答案（B）：设较小圆和较大圆的半径分别为 $r$ 和 $R$。设 $E$ 为小圆的圆心，$\overline{OC}$ 为包含点 $E$ 的大圆半径，$D$ 为小圆与 $\overline{OA}$ 的切点。则 $OE=R-r$，且因为 $\triangle EDO$ 是一个 $30-60-90^\circ$ 三角形，所以 $OE=2DE=2r$。因此 $2r=R-r$，从而 $\frac{r}{R}=\frac{1}{3}$。面积之比为 $\left(\frac{1}{3}\right)^2=\frac{1}{9}$。

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