# Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 3"

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− | By the Power of a Point Theorem on <math>B</math>, we have <math>BD | + | By the Power of a Point Theorem on <math>B</math>, we have <math>BD \times BA=BP \times BQ=2 \times 4=8</math>. By the Power of a Point on <math>C</math>, we have <math>CE \times CA=CQ \times CP=5 \times 7=35</math>. Dividing these two results yields <math>\frac{BD \times BA}{CE \times CA}=\frac{8}{35}</math>. We are given <math>BD=CE</math> and so <math>\frac{BD}{CE}=1</math>. Then the previous equation simplifies to <math>\frac{AB}{AC}=\frac{8}{35}</math>. Hence <math>m+n=8+35=\boxed{043}</math> |

## Latest revision as of 10:49, 9 March 2014

## Problem

In triangle Points and are located on such that The circumcircle of cuts at respectively. If then the ratio can be expressed in the form where are relatively prime positive integers. Find

## Solution

By the Power of a Point Theorem on , we have . By the Power of a Point on , we have . Dividing these two results yields . We are given and so . Then the previous equation simplifies to . Hence