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Difference between revisions of "2006 AMC 12A Problems/Problem 16"

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== Problem ==
 
== Problem ==
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Circles with centers <math>A</math> and <math>B</math> have radii <math>3</math> and <math>8</math>, respectively. A common internal tangent intersects the circles at <math>C</math> and <math>D</math>, respectively. Lines <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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[[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}</math><math>\mathrm{(E) \ }  \frac{55}{3}</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>
 
== Solution ==
 
== Solution ==
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<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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<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 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>.
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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>.
  
 
== See also ==
 
== See also ==
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* [[2006 AMC 12A Problems]]
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* [[2006 AMC 10A Problems/Problem 23]]
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{{AMC12 box|year=2006|ab=A|num-b=15|num-a=17}}
 
{{AMC12 box|year=2006|ab=A|num-b=15|num-a=17}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 11:19, 17 February 2007

Problem

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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 $CD$?

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

Solution

$\angle AEC$ and $\angle BED$ are vertical angles so they are congruent, as are angles $\angle ACE$ and $\angle BDE$ (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 $CE = 4$. The sides are proportional, so $\frac{CE}{AC} = \frac{DE}{BD} \Rightarrow \frac{4}{3} = \frac{DE}{8}$. This makes $DE = \frac{32}{3}$ and $CD = CE + DE = 4 + \frac{32}{3} = \frac{44}{3} \Longrightarrow \mathrm{B}$.

See also

2006 AMC 12A (Problems • Answer Key • Resources)
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
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