Mock Geometry AIME 2011 Problems/Problem 4

Problem

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 $m+n.$

Solution

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: $\frac{1}{2}*AB*BC*\sin{B}$ , 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, $[ABC]=\frac{1}{2}*6*9\sin{120}=\frac{3\sqrt{19}BH}{2}$ . Solving for $BH$ yields $BH=\frac{9\sqrt{3}}{\sqrt{19}}$ .

Let $s$ be the side length of $BPQ$ . The height of an equilateral triangle is given by the formula $\frac{s\sqrt3}{2}$ . Then $BH=\frac{s\sqrt{3}}{2}=\frac{9\sqrt{3}}{\sqrt{19}}$ . Solving for $s$ yields $s=\frac{18}{\sqrt{19}}$ . Then the perimeter of the triangle is $3s=\frac{54}{\sqrt{19}}$ and $m+n=54+19=\boxed{073}$ .

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Mock Geometry AIME 2011 Problems/Problem 4

Problem

Solution