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| − | == Problem ==
| + | #redirect [[2006 AMC 12A Problems/Problem 16]] |
| − | [[Circle]]s with centers A and B have [[radius |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?
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| − | <math>\mathrm{(A) \ } 13\qquad\mathrm{(B) \ } \frac{44}{3}\qquad\mathrm{(C) \ } \sqrt{221}\qquad\mathrm{(D) \ } \sqrt{255}\qquad\mathrm{(E) \ } \frac{55}{3}\qquad</math>
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| − | == Solution ==
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| − | <math>\angle AEC</math> and <math>\angle BED</math> ([[vertical angles]]) are [[congruent]], as are [[right angle]]s <math>\angle ACE</math> and <math>\angle BDE</math> (since radii intersect tangents at right angles). Thus, <math>\triangle ACE \sim \triangle BDE</math>.
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| − | By the [[Pythagorean Theorem]], [[line segment]] <math>CE = 4</math>. The sides are [[proportion]]al, so <math>\frac{CE}{AC} = \frac{DE}{BD} \Rightarrow \frac{4}{3} = \frac{DE}{8}</math>. This makes <math>DE = \frac{32}{3}</math> and <math>CD = CE + DE = 4 + \frac{32}{3} = \frac{44}{3} \Longrightarrow \mathrm{B}</math>.
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| − | == See also ==
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| − | *[[2006 AMC 12A Problems/Problem 16]]
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| − | {{AMC10 box|year=2006|ab=A|num-b=22|num-a=24}}
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| − | [[Category:Introductory Geometry Problems]]
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