2006 AMC 12A Problem 16

Problem 16 of 25IntermediateGeometry

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?

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Solution

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.

More practice

Concepts: similarity · tangent line · Pythagorean Theorem

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.