# 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 Let and be points on such that is equilateral. The perimeter of can be expressed in the form where are relatively prime positive integers. Find

## Solution

Let be the midpoint of . It follows that is perpendicular to and to . The area of can then be calculated two different ways: , and .

By the Law of Cosines, and so . Therefore, . Solving for yields .

Let be the side length of . The height of an equilateral triangle is given by the formula . Then . Solving for yields . Then the perimeter of the triangle is and .