AMC12 2006 A

AMC12 2006 A · Q16

AMC12 2006 A · Q16. It mainly tests Triangles (properties), Pythagorean theorem.

Circles with centers $A$ and $B$ have radius 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 $CD$?

中心分别为 $A$ 和 $B$ 的圆分别有半径 3 和 8。一个公共内切线分别与圆相交于 $C$ 和 $D$。直线 $AB$ 和 $CD$ 相交于 $E$，且 $AE=5$。求 $CD$ 的长。

(A) 13 13

(B) \frac{44}{3} \frac{44}{3}

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

(E) \frac{55}{3} \frac{55}{3}

Answer

Correct choice: (B)

正确答案：（B）

Solution

Error creating thumbnail: Unable to save thumbnail to destination $\angle AED$ and $\angle BEC$ are vertical angles so they are congruent, as are angles $\angle ADE$ and $\angle BCE$ (both are right angles because the radius and tangent line at a point on a circle are always perpendicular). Thus, $\triangle ACE \sim \triangle BDE$. By the Pythagorean Theorem, line segment $DE = 4$. The sides are proportional, so $\frac{DE}{AD} = \frac{CE}{BC} \Rightarrow \frac{4}{3} = \frac{CE}{8}$. This makes $CE = \frac{32}{3}$ and $CD = CE + DE = 4 + \frac{32}{3} = \boxed{\textbf{(B) }\frac{44}{3}}$.

$\angle AED$ 和 $\angle BEC$ 是对顶角，所以它们全等；同样地，$\angle ADE$ 和 $\angle BCE$ 也全等（两者都是直角，因为圆上一点处的半径与该点处的切线总是垂直）。因此，$\triangle ACE \sim \triangle BDE$。由勾股定理，线段 $DE = 4$。对应边成比例，所以 $\frac{DE}{AD} = \frac{CE}{BC} \Rightarrow \frac{4}{3} = \frac{CE}{8}$。因此 $CE = \frac{32}{3}$，从而 $CD = CE + DE = 4 + \frac{32}{3} = \boxed{\textbf{(B) }\frac{44}{3}}$。

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