Mock Geometry AIME 2011 Problems/Problem 4

Revision as of 19:33, 1 January 2012 by Fro116 (talk | contribs) (Created page with "==Problem== In triangle <math>ABC,</math> <math>AB=6, BC=9, \angle ABC=120^{\circ}</math> Let <math>P</math> and <math>Q</math> be points on <math>AC</math> such that <math>BPQ<...")
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In triangle $ABC,$ $AB=6, BC=9, \angle ABC=120^{\circ}$ Let $P$ and $Q$ be points on $AC$ such that $BPQ$ is equilateral. The perimeter of $BPQ$ can be expressed in the form $\frac{m} {\sqrt{n}},$ where $m,n$ are relatively prime positive integers. Find $m+n.$


Let $H$ be the midpoint of $PQ$. It follows that $BH$ is perpendicular to $PQ$ and to $AC$. The area of $\Delta ABC$ can then be calculated two different ways: $AB*BC*\sin{B}$, and $\frac{BH*AC}{2}$.

By the Law of Cosines, $AC^2=9^2+6^2-2*9*6\cos{120}=171$ and so $AC=3\sqrt{19}$. Therefore, $[ABC]=6*9\sin{120}=\frac{3\sqrt{19}BH}{2}$. Solving for $BH$ yields $BH=\frac{18\sqrt{3}}{\sqrt{19}}$.

Let $s$ be the side length of $BPQ$. The height of an equilateral triangle is given by the formula $\frac{s\sqrt3}{2}$. Then $BH=\frac{s\sqrt{3}}{2}=\frac{18\sqrt{3}}{\sqrt{19}}$. Solving for $s$ yields $s=\frac{36}{\sqrt{19}}$. Then the perimeter of the triangle is $3s=\frac{108}{\sqrt{19}}$ and $m+n=108+19=\boxed{127}$.

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