2006 AMC 10A Problems/Problem 23

Problem

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?

$\mathrm{(A) \ } 13\qquad\mathrm{(B) \ } \frac{44}{3}\qquad\mathrm{(C) \ } \sqrt{221}\qquad\mathrm{(D) \ } \sqrt{255}\qquad\mathrm{(E) \ } \frac{55}{3}\qquad$

Solution

$\angle AEC$ and $\angle BED$ (vertical angles) are congruent, as are right angles $\angle ACE$ and $\angle BDE$ (since radii intersect tangents at right angles). Thus, $\triangle ACE \sim \triangle BDE$ .

By the Pythagorean Theorem, line segment $CE = 4$ . The sides are proportional, so $\frac{CE}{AC} = \frac{DE}{BD} \Rightarrow \frac{4}{3} = \frac{DE}{8}$ . This makes $DE = \frac{32}{3}$ and $CD = CE + DE = 4 + \frac{32}{3} = \frac{44}{3} \Longrightarrow \mathrm{B}$ .

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2006 AMC 10A Problems/Problem 23

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