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

(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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==Solution==
 
==Solution==
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>AB*BC*\sin{B}</math>, and <math>\frac{BH*AC}{2}</math>.
 
  
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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>.
  
By the Law of Cosines, <math>AC^2=9^2+6^2-2*9*6\cos{120}=171</math> and so <math>AC=3\sqrt{19}</math>.  Therefore, <math>[ABC]=6*9\sin{120}=\frac{3\sqrt{19}BH}{2}</math>. Solving for <math>BH</math> yields <math>BH=\frac{18\sqrt{3}}{\sqrt{19}}</math>.
 
  
Let <math>s</math> be the side length of <math>BPQ</math>. The height of an equilateral triangle is given by the formula <math>\frac{s\sqrt3}{2}</math>. Then <math>BH=\frac{s\sqrt{3}}{2}=\frac{18\sqrt{3}}{\sqrt{19}}</math>. Solving for <math>s</math> yields <math>s=\frac{36}{\sqrt{19}}</math>. Then the perimeter of the triangle is <math>3s=\frac{108}{\sqrt{19}}</math> and <math>m+n=108+19=\boxed{127}</math>.
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By the Law of Cosines, <math>AC^2=9^2+6^2-2*9*6\cos{120}=171</math> and so <math>AC=3\sqrt{19}</math>.  Therefore, <math>[ABC]=\frac{1}{2}*6*9\sin{120}=\frac{3\sqrt{19}BH}{2}</math>. Solving for <math>BH</math> yields <math>BH=\frac{9\sqrt{3}}{\sqrt{19}}</math>.
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Let <math>s</math> be the side length of <math>BPQ</math>. The height of an equilateral triangle is given by the formula <math>\frac{s\sqrt3}{2}</math>. Then <math>BH=\frac{s\sqrt{3}}{2}=\frac{9\sqrt{3}}{\sqrt{19}}</math>. Solving for <math>s</math> yields <math>s=\frac{18}{\sqrt{19}}</math>. Then the perimeter of the triangle is <math>3s=\frac{54}{\sqrt{19}}</math> and <math>m+n=54+19=\boxed{073}</math>.

Revision as of 22:57, 1 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

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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