# 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 1

$[asy] unitsize(1cm); draw((0,3sqrt(3))--(3,0)--(12,0)--cycle); draw((3,0)--(84/19,36sqrt(3)/19)); draw((3,0)--(48/19, 4.10223)); draw((3,0)--(120/19,2.46134)); label("A",(0,3sqrt(3)),NNW); label("B",(3,0),SW); label("C",(12,0),ESE); label("P",(48/19,4.10223),NNE); label("Q",(120/19,2.46134),NE); label("H",(84/19,36sqrt(3)/19),NNE); [/asy]$

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 $m+n=54+19=\boxed{073}$.

## Solution 2

Let $\angle A = \alpha$ and $BP = PQ = QB = x$. By the Law of Cosines, $AC = 3\sqrt{19}$. It is easy to see that $\angle APB = 120^\circ$. Since $\angle ABC = 120^\circ$, by AA similarity$\triangle ABC \sim \triangle APB$. From this, we have: $$\frac{AB}{PB} = \frac{AC}{BC}$$ $$\frac{6}{x}=\frac{3\sqrt{19}}{9}$$ Solving, we find that $x = \frac{18}{\sqrt{19}}$, so the perimeter is $3x = \frac{54}{\sqrt{19}}$, and our answer is $m+n=\boxed{73}$