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

 
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7. Let <math>\triangle ABC</math> have <math>AC=6</math> and <math>BC=3</math>. Point <math>E</math> is such that <math>CE=1</math> and <math>AE=5</math>. Construct point <math>F</math> on segment <math>BC</math> such that <math>\angle AEB=\angle AFB</math>. <math>EF</math> and <math>AB</math> are extended to meet at <math>D</math>. If <math>\frac{[AEF]}{[CFD]}=\frac{m}{n}</math> where <math>m</math> and <math>n</math> are positive integers, find <math>m+n</math> (note: <math>[ABC]</math> denotes the area of <math>\triangle ABC</math>).
 
7. Let <math>\triangle ABC</math> have <math>AC=6</math> and <math>BC=3</math>. Point <math>E</math> is such that <math>CE=1</math> and <math>AE=5</math>. Construct point <math>F</math> on segment <math>BC</math> such that <math>\angle AEB=\angle AFB</math>. <math>EF</math> and <math>AB</math> are extended to meet at <math>D</math>. If <math>\frac{[AEF]}{[CFD]}=\frac{m}{n}</math> where <math>m</math> and <math>n</math> are positive integers, find <math>m+n</math> (note: <math>[ABC]</math> denotes the area of <math>\triangle ABC</math>).
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[[Mock AIME 1 2006-2007]]

Revision as of 14:57, 24 July 2006

7. 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 $\triangle ABC$).

Mock AIME 1 2006-2007

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