# 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>. | ||

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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>. | 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>. | ||

+ | |||

+ | ===Solution 2=== | ||

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

## Latest revision as of 17:33, 15 June 2017

## 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 .

### Solution 2

Let and . By the Law of Cosines, . It is easy to see that . Since , by AA similarity. From this, we have: Solving, we find that , so the perimeter is , and our answer is