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

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== See also ==
 
== See also ==
 
{{AMC12 box|year=2006|ab=A|num-b=15|num-a=17}}
 
{{AMC12 box|year=2006|ab=A|num-b=15|num-a=17}}
 +
 +
{{AMC10 box|year=2006|ab=A|num-b=22|num-a=24}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 21:25, 1 December 2007

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 $CD$?

$\mathrm{(A) \ } 13\qquad\mathrm{(B) \ } \frac{44}{3}\qquad\mathrm{(C) \ } \sqrt{221}\qquad\mathrm{(D) \ } \sqrt{255}\qquad\mathrm{(E) \ } \frac{55}{3}\qquad$

2006 AMC12A-16.png

Solution

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 (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
2006 AMC 10A (ProblemsAnswer KeyResources)
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
All AMC 10 Problems and Solutions