Difference between revisions of "2006 AMC 10A Problems/Problem 23"
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== Problem == | == Problem == | ||
| − | [[Circle]]s with centers A and B have [[radius |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? | + | [[Circle]]s with centers A and B have [[radius |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? |
<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> | <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> | ||
== 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 [[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>. | |
| − | *[[2006 AMC | + | == See also == |
| + | *[[2006 AMC 12A Problems/Problem 16]] | ||
| + | {{AMC10 box|year=2006|ab=A|num-b=22|num-a=24}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
Revision as of 22:50, 16 February 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?
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
| 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 | ||
