Difference between revisions of "2006 AMC 12A Problems/Problem 16"

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== Solution ==
 
== 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 proportional, 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} \Rightarrow \mathrm{B}</math>.
  
 
== See also ==
 
== See also ==
 
* [[2006 AMC 12A Problems]]
 
* [[2006 AMC 12A Problems]]
*[[2006 AMC 12A Problems/Problem 15|Previous Problem]]
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*[[2006 AMC 12A Problems/Problem 17|Next Problem]]
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{{AMC box|year=2006|n=12A|num-b=15|num-a=17}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 20:26, 31 January 2007

Problem


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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}$$\mathrm{(E) \ }  \frac{55}{3}$

Solution

$\angle AEC$ and $\angle BED$ (vertical angles) are congruent, as are right angles $\angle ACE$ and $\angle BDE$ (since radii intersect tangents at right angles). 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 $CD = CE + DE = 4 + \frac{32}{3} = \frac{44}{3} \Rightarrow \mathrm{B}$.

See also


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Preceded by
Problem 15
AMC 12A
2006
Followed by
Problem 17