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 AA and BB have radii 33 and 8,8, respectively. A common internal tangent intersects the circles at CC and D,D, respectively. Lines ABAB and CDCD intersect at E,E, and AE=5.AE = 5. What is CD?CD?

1313

443\dfrac{44}{3}

221\sqrt{221}

255\sqrt{255}

553\dfrac{55}{3}

Answer: B
Concepts:similaritytangent linePythagorean Theorem
Difficulty rating: 1760
Solution:

The radii satisfy ACCDAC \perp CD and BDCD.BD \perp CD. By the Pythagorean theorem, CE=5232=4.CE = \sqrt{5^2 - 3^2} = 4.

Since ACEBDE,\triangle ACE \sim \triangle BDE, we get DECE=BDAC=83,\tfrac{DE}{CE} = \tfrac{BD}{AC} = \tfrac{8}{3}, so DE=483=323.DE = 4 \cdot \tfrac{8}{3} = \tfrac{32}{3}. Then CD=CE+DE=4+323=443. \begin{gathered} CD = CE + DE \\ = 4 + \frac{32}{3} \\ = \frac{44}{3}. \end{gathered}

Thus, the correct answer is B.

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