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Mock Geometry AIME 2011 Problems/Problem 3

Revision as of 19:43, 1 January 2012 by Fro116 (talk | contribs) (Created page with "==Problem== In triangle <math>ABC,</math> <math>BC=9.</math> Points <math>P</math> and <math>Q</math> are located on <math>BC</math> such that <math>BP=PQ=2,</math> <math>QC=5.<...")
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Problem

In triangle $ABC,$ $BC=9.$ Points $P$ and $Q$ are located on $BC$ such that $BP=PQ=2,$ $QC=5.$ The circumcircle of $APQ$ cuts $AB,AC$ at $D,E$ respectively. If $BD=CE,$ then the ratio $\frac{AB}{AC}$ can be expressed in the form $\frac{m}{n},$ where $m,n$ are relatively prime positive integers. Find $m+n.$

Solution

By the Power of a Point Theorem on $B$, we have $BD*BA=BP*BQ=2*4=8$. By the Power of a Point on $C$, we have $CE*CA=CQ*CP=5*7=35$. Dividing these two results yields $\frac{BD*BA}{CE*CA}=\frac{8}{35}$. We are given $BD=CE$ and so $\frac{BD}{CE}=1$. Then the previous equation simplifies to $\frac{AB}{AC}=\frac{8}{35}$. Hence $m+n=8+35=\boxed{43}$

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