Mock AIME 1 2006-2007 Problems/Problem 7

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

We can immediately see that quadrilateral $AEFB$ is cyclic, since $\angle AEB=\angle AFB$. We then have, from Power of a Point, that $CE\cdot CA=CF\cdot CB$. In other words, $1\cdot 6 = CF\cdot 3$. $CF$ is then 2, and $BF$ is 1. We can now use Menelaus on line $DF$ with respect to triangle $ABC$:

\[\frac{AE}{EC}\cdot \frac{CF}{FB}\cdot \frac{BD}{DA}=1\]

\[\frac{5}{1}\cdot \frac{2}{1}\cdot \frac{BD}{DA}=1\]

\[\frac{BD}{DA}=\frac{1}{10}\]

This shows that $\frac{BA}{BD}=9$.

Now let $[ABC]=x$, for some real $x$. Therefore $[CFA]=\frac{CF}{CB}\cdot [ABC]=\frac{2x}{3}$, and $[AEF]=\frac{AE}{AC}\cdot [CFA]=\frac{5}{6}\cdot \frac{2x}{3}=\frac{5x}{9}$. Similarly, $[CBD]=\frac{DB}{AB}\cdot [ABC]=\frac{x}{9}$ and $[CFD]=\frac{CF}{CB}\cdot [CBD]=\frac{2}{3}\cdot \frac{x}{9}=\frac{2x}{27}$. The desired ratio is then

\[\frac{[AEF]}{[CFD]}=\frac{\frac{5x}{9}}{\frac{2x}{27}}=\frac{5\cdot 27}{9\cdot 2}=\frac{15}{2}\]

Therefore $m+n=\boxed{017}$.


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