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 |
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