Difference between revisions of "2006 AMC 10A Problems/Problem 23"

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== Problem ==
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#redirect [[2006 AMC 12A Problems/Problem 16]]
[[Circle]]s with [[center]]s <math>A</math> and <math>B</math> have [[radius |radii]] 3 and 8, respectively. A [[common internal tangent line | common internal tangent]] [[intersect]]s the circles at <math>C</math> and <math>D</math>, respectively. [[Line]]s <math>AB</math> and <math>CD</math> intersect at <math>E</math>, and <math>AE=5</math>. What is <math>CD</math>?
 
 
 
<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>
 
 
 
[[Image:2006_AMC12A-16.png]]
 
 
 
== Solution ==
 
[[Image:2006_AMC12A-16a.png]]
 
 
 
<math>\angle AEC</math> and <math>\angle BED</math> are [[vertical angles]] so they are [[congruent (geometry) | congruent]], as are [[angle]]s <math>\angle ACE</math> and <math>\angle BDE</math> (both are [[right angle]]s because the radius and [[tangent line]] at a point on a circle are always [[perpendicular]]). Thus, <math>\triangle ACE \sim \triangle BDE</math>.
 
 
 
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>.
 
 
 
== See also ==
 
{{AMC10 box|year=2006|ab=A|num-b=22|num-a=24}}
 
 
 
[[Category:Introductory Geometry Problems]]
 

Latest revision as of 22:30, 1 December 2007