2006 AMC 10A Problem 23

Below is the professionally curated solution for Problem 23 of the 2006 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 10A solutions, or check the answer key.

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Concepts:tangent line similarity Pythagorean Theorem

Difficulty rating: 1720

23.

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 $CD?$

$13$

$\dfrac{44}{3}$

$\sqrt{221}$

$\sqrt{255}$

$\dfrac{55}{3}$

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.

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