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2006 AMC 12A Problems/Problem 16

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The following problem is from both the 2006 AMC 12A #16 and 2006 AMC 10A #23, so both problems redirect to this page.

Problem

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?

\mathrm{(A) \ } 13\qquad\mathrm{(B) \ } \frac{44}{3}\qquad\mathrm{(C) \ } \sqrt{221}\qquad\mathrm{(D) \ } \sqrt{255}\qquad\ma...

Image:2006_AMC12A-16.png

Solution

Image:2006_AMC12A-16a.png

\angle AEC and \angle BED are vertical angles so they are congruent, as are angles \angle ACE and \angle BDE (both are right angles because the radius and tangent line at a point on a circle are always perpendicular). Thus, \triangle ACE \sim \triangle BDE.

By the Pythagorean Theorem, line segment CE = 4. The sides are proportional, so \frac{CE}{AC} = \frac{DE}{BD} \Rightarrow \frac{4}{3} = \frac{DE}{8}. This makes DE = \frac{32}{3} and CD = CE + DE = 4 + \frac{32}{3} = \frac{44}{3} \Longrightarrow \mathrm{B}.

See also

2006 AMC 12A (ProblemsResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
2006 AMC 10A (ProblemsResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
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