2006 AMC 12A Problem 16

Attempt Problem 16 of the 2006 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 12A solutions, or check the answer key.

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16.

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?

13

\dfrac{44}{3}

\sqrt{221}

\sqrt{255}

\dfrac{55}{3}

Answer: B

Concepts:similarity tangent line Pythagorean Theorem

Difficulty rating: 1760

Small hint:

Radii

AC

and

BD

are perpendicular to the tangent, so

\triangle ACE \sim \triangle BDE

Big hint:

CE = \sqrt{AE^2 - AC^2} = 4,

and

\tfrac{DE}{CE} = \tfrac{BD}{AC}

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.

Problem 16 in other years