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Mock AIME 1 2006-2007 Problems/Problem 7

Revision as of 18:39, 22 August 2006 by JBL (talk | contribs)

Problem

Let $\triangle ABC$ have $AC=6$ and $BC=3$. Point $E$ is such that $CE=1$ and $AE=5$. Construct point $F$ on segment $BC$ such that $\angle AEB=\angle AFB$. $EF$ and $AB$ are extended to meet at $D$. If $\frac{[AEF]}{[CFD]}=\frac{m}{n}$ where $m$ and $n$ are positive integers, find $m+n$ (note: $[ABC]$ denotes the area of $\triangle ABC$).

Solution

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