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Difference between revisions of "2006 AMC 12A Problems/Problem 16"
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{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #16]] and [[2006 AMC 10A Problems/Problem 23|2006 AMC 10A #23]]}}
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== Problem ==
 
== Problem ==
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Circles with centers <math>A</math> and <math>B</math> have radius 3 and 8, respectively. A [[common internal tangent line | 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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<asy>
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unitsize(2.5mm);
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defaultpen(fontsize(10pt)+linewidth(.8pt));
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dotfactor=3;
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pair A=(0,0), Ep=(5,0), B=(5+40/3,0);
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pair M=midpoint(A--Ep);
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pair C=intersectionpoints(Circle(M,2.5),Circle(A,3))[1];
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pair D=B+8*dir(180+degrees(C));
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dot(A);
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dot(C);
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dot(B);
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dot(D);
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draw(C--D);
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draw(A--B);
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draw(Circle(A,3));
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draw(Circle(B,8));
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label("$A$",A,W);
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label("$B$",B,E);
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label("$C$",C,SE);
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label("$E$",Ep,SSE);
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label("$D$",D,NW);
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</asy>
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<math>\textbf{(A) } 13\qquad\textbf{(B) } \frac{44}{3}\qquad\textbf{(C) } \sqrt{221}\qquad\textbf{(D) } \sqrt{255}\qquad\textbf{(E) } \frac{55}{3}\qquad</math>
  
 
== Solution ==
 
== Solution ==
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[[Image:2006_AMC12A-16a.png]]
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<math>\angle AED</math> and <math>\angle BEC</math> are vertical angles so they are congruent, as are angles <math>\angle ADE</math> and <math>\angle BCE</math> (both are right angles 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>DE = 4</math>.  The sides are proportional, so <math>\frac{DE}{AD} = \frac{CE}{BC} \Rightarrow \frac{4}{3} = \frac{CE}{8}</math>. This makes <math>CE = \frac{32}{3}</math> and <math>CD = CE + DE = 4 + \frac{32}{3} = \boxed{\textbf{(B) }\frac{44}{3}}</math>.
  
 
== See also ==
 
== See also ==
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* [[2006 AMC 12A Problems]]
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{{AMC12 box|year=2006|ab=A|num-b=15|num-a=17}}
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{{AMC10 box|year=2006|ab=A|num-b=22|num-a=24}}
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{{MAA Notice}}
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[[Category:Introductory Geometry Problems]]

Latest revision as of 09:16, 19 December 2021

The following problem is from both the 2006 AMC 12A #16 and 2006 AMC 10A #23, so both problems redirect to this page.

Problem

Circles with centers $A$ and $B$ have radius 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 $CD$?

[asy] unitsize(2.5mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair A=(0,0), Ep=(5,0), B=(5+40/3,0); pair M=midpoint(A--Ep); pair C=intersectionpoints(Circle(M,2.5),Circle(A,3))[1]; pair D=B+8*dir(180+degrees(C)); dot(A); dot(C); dot(B); dot(D); draw(C--D); draw(A--B); draw(Circle(A,3)); draw(Circle(B,8)); label("$A$",A,W); label("$B$",B,E); label("$C$",C,SE); label("$E$",Ep,SSE); label("$D$",D,NW); [/asy]

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

Solution

2006 AMC12A-16a.png

$\angle AED$ and $\angle BEC$ are vertical angles so they are congruent, as are angles $\angle ADE$ and $\angle BCE$ (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 $DE = 4$. The sides are proportional, so $\frac{DE}{AD} = \frac{CE}{BC} \Rightarrow \frac{4}{3} = \frac{CE}{8}$. This makes $CE = \frac{32}{3}$ and $CD = CE + DE = 4 + \frac{32}{3} = \boxed{\textbf{(B) }\frac{44}{3}}$.

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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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