Difference between revisions of "2006 AMC 10A Problems/Problem 16"

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== Problem ==
 
== Problem ==
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[[Image:2006_AMC10A-16.png]]
  
{{image}}
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A [[circle]] of [[radius]] 1 is tangent to a circle of radius 2. The sides of <math>\triangle ABC</math> are [[tangent]] to the circles as shown, and the sides <math>\overline{AB}</math>  and <math>\overline{AC}</math> are [[congruent]]. What is the area of <math>\triangle ABC</math>?
  
A circle of radius 1 is tangent to a circle of radius 2. The sides of <math>\triangle ABC</math> are tangent to the circles as shown, and the sides <math>\overline{AB}</math>  and <math>\overline{AC}</math>  are congruent. What is the area of <math>\triangle ABC</math>?
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<math>\mathrm{(A) \ } \frac{35}{2}\qquad\mathrm{(B) \ } 15\sqrt{2}\qquad\mathrm{(C) \ } \frac{64}{3}\qquad\mathrm{(D) \ } 16\sqrt{2}\qquad\mathrm{(E) \ } 24\qquad</math>
  
<math>\mathrm{(A) \ } \frac{35}{2}\qquad\mathrm{(B) \ } 15\sqrt{2}\qquad\mathrm{(C) \ } \frac{64}{3}\qquad\mathrm{(D) \ } 16\sqrt{2}\qquad\mathrm{(E) \ } 24\qquad</math>
 
 
== Solution ==
 
== Solution ==
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[[Image:2006_AMC10A-16a.png]]
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Note that <math>\triangle ADO_1 \sim \triangle AEO_2 \sim \triangle AFC \displaystyle</math>. Using the first pair of [[similar triangles]], we write the [[proportion]]:
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<div style="text-align:center;"><math>\frac{AO_1}{AO_2} = \frac{DO_1}{DO_2} \Longrightarrow \frac{AO_1}{AO_1 + 3} = \frac{1}{2} \Longrightarrow AO_1 = 3</math></div>
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By the [[Pythagorean Theorem]] we have that <math>AD = \sqrt{3^2-1^2} = \sqrt{8}</math>.
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Now using <math>\triangle ADO_1 \sim \triangle AFC</math>,
  
{{solution}}
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<div style="text-align:center;"><math>\frac{AD}{AF} = \frac{DO_1}{CF} \Longrightarrow \frac{2\sqrt{2}}{8} = \frac{1}{CF} \Longrightarrow CF = 2\sqrt{2}</math></div>
  
== See Also ==
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The area of the triangle is <math>\frac{1}{2}bh = \frac{1}{2}\left(2\cdot 2\sqrt{2}\right)\left(8\right) = 16\sqrt{2}\ \mathrm{(D)}</math>.
  
* [[2006 AMC 10A Problems]]
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== See also ==
* [[2006 AMC 10A Problems/Problem 15|Previous Problem]]
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{{AMC10 box|year=2006|num-b=15|num-a=17|ab=A}}
* [[2006 AMC 10A Problems/Problem 17|Next Problem]]
 
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 21:11, 15 September 2007

Problem

2006 AMC10A-16.png

A circle of radius 1 is tangent to a circle of radius 2. The sides of $\triangle ABC$ are tangent to the circles as shown, and the sides $\overline{AB}$ and $\overline{AC}$ are congruent. What is the area of $\triangle ABC$?

$\mathrm{(A) \ } \frac{35}{2}\qquad\mathrm{(B) \ } 15\sqrt{2}\qquad\mathrm{(C) \ } \frac{64}{3}\qquad\mathrm{(D) \ } 16\sqrt{2}\qquad\mathrm{(E) \ } 24\qquad$

Solution

2006 AMC10A-16a.png

Note that $\triangle ADO_1 \sim \triangle AEO_2 \sim \triangle AFC \displaystyle$. Using the first pair of similar triangles, we write the proportion:

$\frac{AO_1}{AO_2} = \frac{DO_1}{DO_2} \Longrightarrow \frac{AO_1}{AO_1 + 3} = \frac{1}{2} \Longrightarrow AO_1 = 3$

By the Pythagorean Theorem we have that $AD = \sqrt{3^2-1^2} = \sqrt{8}$.

Now using $\triangle ADO_1 \sim \triangle AFC$,

$\frac{AD}{AF} = \frac{DO_1}{CF} \Longrightarrow \frac{2\sqrt{2}}{8} = \frac{1}{CF} \Longrightarrow CF = 2\sqrt{2}$

The area of the triangle is $\frac{1}{2}bh = \frac{1}{2}\left(2\cdot 2\sqrt{2}\right)\left(8\right) = 16\sqrt{2}\ \mathrm{(D)}$.

See also

2006 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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