# Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 7"

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+ | As above, use Power of a Point to compute <math>CF=2</math> and <math>FB=1</math>. Since triangles <math>AFE</math> and <math>CEF</math> share the same height, <math>\frac{[AEF]}{[CFE]}=\frac{CE}{EA}=5</math>. Similarly, <math>\frac{[CFE]}{[CFD]}=\frac{EF}{FD}</math>. Using Menelaus's Theorem on points <math>E, F, D</math> on the sides of triangle <math>ABC</math>, we see that <cmath>\frac{AE}{EC}\cdot\frac{CF}{FB}\cdot\frac{BD}{DA}=1\implies \frac{AB}{BD}=9.</cmath>Let <math>X=AF\cap CD</math>. Using Ceva's Theorem on points <math>X, E, B</math> lying on the sides of triangle <math>ACD</math>, we find that <cmath>\frac{AE}{EC}\cdot\frac{CX}{XD}\cdot\frac{DB}{BA}=1\implies \frac{CX}{XD}=\frac{9}{5}.</cmath>Then, using Menelaus's Theorem on the points <math>A, F, X</math> on the sides of triangle <math>ECD</math>, we see that <cmath>\frac{EA}{AC}\cdot\frac{CX}{XD}\cdot\frac{DF}{FE}=1\implies \frac{DF}{FE}=\frac{2}{3}.</cmath> Thus, <math>\frac{[CFE]}{[CFD]}=\frac{3}{2},</math> so that <cmath>\frac{[AEF]}{[CFD]}=\frac{[AEF]}{[CFE]}\cdot\frac{[CFE]}{[CFD]}=5\cdot\frac{3}{2}=\frac{15}{2}=\frac{m}{n}\implies m+n=\boxed{17}.</cmath> | ||

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*[[Mock AIME 1 2006-2007 Problems/Problem 6 | Previous Problem]] | *[[Mock AIME 1 2006-2007 Problems/Problem 6 | Previous Problem]] | ||

## Latest revision as of 16:54, 25 November 2015

## Problem

Let have and . Point is such that and . Construct point on segment such that . and are extended to meet at . If where and are positive integers, find (note: denotes the area of ).

## Solution

We can immediately see that quadrilateral is cyclic, since . We then have, from Power of a Point, that . In other words, . is then 2, and is 1. We can now use Menelaus on line with respect to triangle :

This shows that .

Now let , for some real . Therefore , and . Similarly, and . The desired ratio is then

Therefore .

## Alternate Solution

As above, use Power of a Point to compute and . Since triangles and share the same height, . Similarly, . Using Menelaus's Theorem on points on the sides of triangle , we see that Let . Using Ceva's Theorem on points lying on the sides of triangle , we find that Then, using Menelaus's Theorem on the points on the sides of triangle , we see that Thus, so that