Question

圆心为 $A$ 和 $B$ 的两个圆半径分别为 $3$ 和 $8$ ，一条公共内切线分别与两个圆交于 $C$ 和 $D$ ，直线 $AB$ 与 $CD$ 交于 $E$ ，且 $AE = 5$ 。求 $CD$ ？

Circles with centers $A$ and $B$ have radii $3$ and $8,$ respectively. A common internal tangent intersects the circles at $C$ and $D,$ respectively. Lines $AB$ and $CD$ intersect at $E,$ and $AE = 5.$ What is $CD?$

$13$

$\dfrac{44}{3}$

$\sqrt{221}$

$\sqrt{255}$

$\dfrac{55}{3}$

答案：B

知识点：相似切线勾股定理

难度评级：1760

解答：

半径满足 $AC \perp CD$ 和 $BD \perp CD$ 。由勾股定理， $CE = \sqrt{5^2 - 3^2} = 4$ 。

因为 $\triangle ACE \sim \triangle BDE$ ，得 $\tfrac{DE}{CE} = \tfrac{BD}{AC} = \tfrac{8}{3}$ ，所以 $DE = 4 \cdot \tfrac{8}{3} = \tfrac{32}{3}$ 。因此 $\begin{gathered} CD = CE + DE \\ = 4 + \frac{32}{3} \\ = \frac{44}{3}. \end{gathered}$

因此，正确答案是 B。

The radii satisfy $AC \perp CD$ and $BD \perp CD.$ By the Pythagorean theorem, $CE = \sqrt{5^2 - 3^2} = 4.$

Since $\triangle ACE \sim \triangle BDE,$ we get $\tfrac{DE}{CE} = \tfrac{BD}{AC} = \tfrac{8}{3},$ so $DE = 4 \cdot \tfrac{8}{3} = \tfrac{32}{3}.$ Then $\begin{gathered} CD = CE + DE \\ = 4 + \frac{32}{3} \\ = \frac{44}{3}. \end{gathered}$

Thus, the correct answer is B.

2006 AMC 12A 第 16 题

答案：B

难度评级：1760

解答：

其他年份的第 16 题