2006 AMC 10A Problem 23

Problem 23 of 25HarderGeometry

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?

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Solution

Since

AC \perp CD,

we have

CE = \sqrt{AE^2 - AC^2}

= \sqrt{25 - 9}

= 4.

Because

\triangle ACE \sim \triangle BDE,

\frac{DE}{CE} = \frac{BD}{AC},

so

DE = 4 \cdot \frac{8}{3} = \frac{32}{3}.

Then

CD = CE + DE

= 4 + \frac{32}{3}

= \frac{44}{3}.

Thus, the correct answer is B.

More practice

Concepts: tangent line · similarity · Pythagorean Theorem

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