Difference between revisions of "2006 AMC 12A Problems/Problem 16"
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== Solution == | == Solution == | ||
| + | <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]] | ||
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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 19:26, 31 January 2007
Problem
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Circles with centers 
and 
have radii 
and 
, respectively. A common internal tangent intersects the circles at 
and 
, respectively. Lines 
and 
intersect at 
, and 
. What is ?
Solution
and 
(vertical angles) are congruent, as are right angles 
and 
(since radii intersect tangents at right angles). Thus, 
.
By the Pythagorean Theorem, line segment 
. The sides are proportional, so 
. This makes 
and 
.
See also
| {{{header}}} | ||
| Preceded by Problem 15 |
AMC 12A 2006 |
Followed by Problem 17 |
