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Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 4"

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==Solution==
 
==Solution==
 +
 +
<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 <math>H</math> be the midpoint of <math>PQ</math>. It follows that <math>BH</math> is perpendicular to <math>PQ</math> and to <math>AC</math>. The area of <math>\Delta ABC</math> can then be calculated two different ways: <math>\frac{1}{2}*AB*BC*\sin{B}</math>, and <math>\frac{BH*AC}{2}</math>.
 
Let <math>H</math> be the midpoint of <math>PQ</math>. It follows that <math>BH</math> is perpendicular to <math>PQ</math> and to <math>AC</math>. The area of <math>\Delta ABC</math> can then be calculated two different ways: <math>\frac{1}{2}*AB*BC*\sin{B}</math>, and <math>\frac{BH*AC}{2}</math>.

Revision as of 11:26, 2 January 2012

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

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

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