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

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* [[2006 AMC 12A Problems]]
 
* [[2006 AMC 12A Problems]]
  
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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 19:21, 2 February 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

2006 AMC 12A (ProblemsAnswer KeyResources)
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
All AMC 12 Problems and Solutions