AMC12 2008 A

AMC12 2008 A · Q13

AMC12 2008 A · Q13. It mainly tests Angle chasing, 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 $O$ 为圆心的圆上，且 $\angle AOB = 60^\circ$。第二个圆内切于第一个圆，并且与 $\overline{OA}$ 和 $\overline{OB}$ 都相切。较小圆的面积与较大圆面积的比是多少？

(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

Let $P$ be the center of the small circle with radius $r$, and let $Q$ be the point where the small circle is tangent to $OA$. Also, let $C$ be the point where the small circle is tangent to the big circle with radius $R$. Then $PQO$ is a right triangle. Angle $POQ$ is $30$ degrees because line $OP$ bisects angle $AOB$ (this can be proved by dropping a perpendicular line from $P$ to line $OB$, letting their intersection be point $S$, and proving triangles $PQO$ and $PSO$ congruent), meaning that $PQO$ is a $30-60-90$ triangle. Therefore, $OP=2PQ$. Since $OP=OC-PC=OC-r=R-r$, we have $R-r=2PQ$, or $R-r=2r$, or $\frac{1}{3}=\frac{r}{R}$. Ratio of areas of circles is ratio of radii squared, so the answer is $\left(\frac{1}{3}\right)^2 = \frac{1}{9} \Rightarrow \boxed{B}$

设 $P$ 为小圆圆心，半径为 $r$，设 $Q$ 为小圆与 $OA$ 相切的切点。再设 $C$ 为小圆与大圆（半径为 $R$）相切的切点。则 $PQO$ 为直角三角形。由于直线 $OP$ 平分角 $AOB$（可通过从 $P$ 向直线 $OB$ 作垂线，垂足为 $S$，并证明三角形 $PQO$ 与 $PSO$ 全等来证明），所以 $\angle POQ$ 为 $30$ 度，这意味着 $PQO$ 是一个 $30-60-90$ 三角形。因此，$OP=2PQ$。由于 $OP=OC-PC=OC-r=R-r$，我们有 $R-r=2PQ$，即 $R-r=2r$，从而 $\frac{1}{3}=\frac{r}{R}$。圆面积之比等于半径平方之比，所以答案为 $\left(\frac{1}{3}\right)^2 = \frac{1}{9} \Rightarrow \boxed{B}$。

Topics