Difference between revisions of "2006 AMC 10A Problems/Problem 23"
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| − | + | == Problem == | |
| + | [[Circle]]s with [[center]]s <math>A</math> and <math>B</math> have [[radius |radii]] 3 and 8, respectively. A [[common internal tangent line | common internal tangent]] [[intersect]]s the circles at <math>C</math> and <math>D</math>, respectively. [[Line]]s <math>AB</math> and <math>CD</math> intersect at <math>E</math>, and <math>AE=5</math>. What is <math>CD</math>? | ||
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| + | <math>\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</math> | ||
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| + | [[Image:2006_AMC12A-16.png]] | ||
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| + | == Solution == | ||
| + | [[Image:2006_AMC12A-16a.png]] | ||
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| + | <math>\angle AEC</math> and <math>\angle BED</math> are [[vertical angles]] so they are [[congruent (geometry) | congruent]], as are [[angle]]s <math>\angle ACE</math> and <math>\angle BDE</math> (both are [[right angle]]s because the radius and [[tangent line]] at a point on a circle are always [[perpendicular]]). 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 [[proportion]]al, 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} \Longrightarrow \mathrm{B}</math>. | ||
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| + | == See also == | ||
| + | {{AMC10 box|year=2006|ab=A|num-b=22|num-a=24}} | ||
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| + | [[Category:Introductory Geometry Problems]] | ||
Revision as of 18:18, 1 December 2007
Problem
Circles with centers 
and 
have radii 3 and 8, respectively. A common internal tangent intersects the circles at 
and 
, respectively. Lines 
and 
intersect at 
, and 
. What is ?
Solution
and 
are vertical angles so they are congruent, as are angles 
and 
(both are right angles because the radius and tangent line at a point on a circle are always perpendicular). Thus, 
.
By the Pythagorean Theorem, line segment 
. The sides are proportional, so 
. This makes 
and 
.
See also
| 2006 AMC 10A (Problems • Answer Key • Resources) | ||
| 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 | ||
