2006 AMC 12A Problems/Problem 16

The following problem is from both the 2006 AMC 12A #16 and 2006 AMC 10A #23, so both problems redirect to this page.

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 $CD$?

2006 AMC12A-16.png

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

Solution

2006 AMC12A-16a.png

$\angle AED$ and $\angle BEC$ are vertical angles so they are congruent, as are angles $\angle ADE$ and $\angle BCE$ (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 $DE = 4$. The sides are proportional, so $\frac{DE}{AD} = \frac{CE}{BC} \Rightarrow \frac{4}{3} = \frac{CE}{8}$. This makes $CE = \frac{32}{3}$ 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

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