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Difference between revisions of "2006 AIME II Problems/Problem 12"

 
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#REDIRECT [[2006 AIME A Problems/Problem 12]]
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== Problem ==
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[[Equilateral triangle | Equilateral]] <math> \triangle ABC </math> is inscribed in a [[circle]] of [[radius]] 2. Extend <math> \overline{AB} </math> through <math> B </math> to point <math> D </math> so that <math> AD=13, </math> and extend <math> \overline{AC} </math> through <math> C </math> to point <math> E </math> so that <math> AE = 11. </math> Through <math> D, </math> draw a line <math> l_1 </math> [[parallel]] to <math> \overline{AE}, </math> and through <math> E, </math> draw a line <math> l_2 </math> parallel to <math> \overline{AD}. </math> Let <math> F </math> be the [[intersection]] of <math> l_1 </math> and <math> l_2. </math> Let <math> G </math> be the point on the circle that is [[collinear]] with <math> A </math> and <math> F </math> and distinct from <math> A. </math> Given that the [[area]] of <math> \triangle CBG </math> can be expressed in the form <math> \frac{p\sqrt{q}}{r}, </math>  where <math> p, q, </math> and <math> r </math> are [[positive integer]]s, <math> p </math> and <math> r</math>  are [[relatively prime]], and <math> q </math> is not [[divisibility | divisible]] by the [[perfect square | square]] of any [[prime]], find <math> p+q+r. </math>
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== Solution ==
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{{solution}}
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== See also ==
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{{AIME box|year=2006|n=II|num-b=11|num-a=13}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 18:13, 25 September 2007

Problem

Equilateral $\triangle ABC$ is inscribed in a circle of radius 2. Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime, and $q$ is not divisible by the square of any prime, find $p+q+r.$


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Solution

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See also

2006 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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