AMC10 2006 A

AMC10 2006 A · Q23

AMC10 2006 A · Q23. It mainly tests Similarity, Circle theorems.

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

圆心为 A 和 B 的圆分别半径 3 和 8。有公内切线触圆于 C 和 D，如图所示。直线 AB 和 CD 交于 E，且 $AE=5$。求 $CD$。

(A) 13 13

(B) $\dfrac{44}{3}$ $\dfrac{44}{3}$

(D) $\sqrt{255}$ $\sqrt{255}$

(E) $\dfrac{55}{3}$ $\dfrac{55}{3}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) Radii $\overline{AC}$ and $\overline{BD}$ are each perpendicular to $\overline{CD}$. By the Pythagorean Theorem, $CE=\sqrt{5^2-3^2}=4.$ Because $\triangle ACE$ and $\triangle BDE$ are similar, $\dfrac{DE}{CE}=\dfrac{BD}{AC},\quad \text{so}\quad DE=CE\cdot\dfrac{BD}{AC}=4\cdot\dfrac{8}{3}=\dfrac{32}{3}.$ Therefore $CD=CE+DE=4+\dfrac{32}{3}=\dfrac{44}{3}.$

（B）半径 $\overline{AC}$ 和 $\overline{BD}$ 都与 $\overline{CD}$ 垂直。由勾股定理， $CE=\sqrt{5^2-3^2}=4.$ 因为 $\triangle ACE$ 与 $\triangle BDE$ 相似， $\dfrac{DE}{CE}=\dfrac{BD}{AC},\quad \text{因此}\quad DE=CE\cdot\dfrac{BD}{AC}=4\cdot\dfrac{8}{3}=\dfrac{32}{3}.$ 所以 $CD=CE+DE=4+\dfrac{32}{3}=\dfrac{44}{3}.$

Topics