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

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== Solution ==
 
== Solution ==
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Notice that <math>\angle{E} = \angle{BGC} = 120^\circ</math> because <math>\angle{A} = 60^\circ</math>. Also, <math>\angle{GBC} = \angle{GAC} = \angle{FAE}</math> because they both correspond to arc <math>{GC}</math>. So <math>\Delta{GBC}</math>~<math>\Delta{EAF}</math>.
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<math>[EAF] = \frac12\cdot11\cdot13\cdot\sin{120^\circ} = \frac {143\sqrt3}4</math>.
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<math>[GBC] = \frac {BC^2}{AF^2}\cdot[EAF] = \frac {12}{11^2 + 13^2 - 2\cdot11\cdot13\cdot\cos120^\circ}\cdot\frac {143\sqrt3}4 = \frac {429\sqrt3}{433}</math>.
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<math>429+433+3=\boxed{865}</math>
  
 
== See also ==
 
== See also ==

Revision as of 08:50, 3 December 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 $p+q+r.$

Aime2006-2-11.JPG

Solution

Notice that $\angle{E} = \angle{BGC} = 120^\circ$ because $\angle{A} = 60^\circ$. Also, $\angle{GBC} = \angle{GAC} = \angle{FAE}$ because they both correspond to arc ${GC}$. So $\Delta{GBC}$~$\Delta{EAF}$.

$[EAF] = \frac12\cdot11\cdot13\cdot\sin{120^\circ} = \frac {143\sqrt3}4$.

$[GBC] = \frac {BC^2}{AF^2}\cdot[EAF] = \frac {12}{11^2 + 13^2 - 2\cdot11\cdot13\cdot\cos120^\circ}\cdot\frac {143\sqrt3}4 = \frac {429\sqrt3}{433}$.

$429+433+3=\boxed{865}$

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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