Art of Problem Solving

Mock AIME 4 2006-2007 Problems/Problem 11: Difference between revisions

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==Problem==
==Problem==
Let <math>\triangle ABC</math> be an equilateral triangle. Two points <math>D</math> and <math>E</math> are chosen on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, such that <math>AD = CE</math>. Let <math>F</math> be the intersection of <math>\overline{BE}</math> and <math>\overline{CD}</math>. The area of <math>\triangle ABC</math> is 13 and the area of <math>\triangle ACF</math> is 3. If <math>\frac{CE}{EA}=\frac{p+\sqrt{q}}{r}</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are relatively prime positive integers, compute <math>p+q+r</math>.
Let <math>\triangle ABC</math> be an equilateral triangle. Two points <math>D</math> and <math>E</math> are chosen on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, such that <math>AD = CE</math>. Let <math>F</math> be the intersection of <math>\overline{AE}</math> and <math>\overline{CD}</math>. The area of <math>\triangle ABC</math> is 13 and the area of <math>\triangle ACF</math> is 3. If <math>\frac{CE}{EA}=\frac{p+\sqrt{q}}{r}</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are relatively prime positive integers, compute <math>p+q+r</math>.
==Solution==
==Solution==



Revision as of 14:04, 20 January 2007

Problem

Let $\triangle ABC$ be an equilateral triangle. Two points $D$ and $E$ are chosen on $\overline{AB}$ and $\overline{AC}$, respectively, such that $AD = CE$. Let $F$ be the intersection of $\overline{AE}$ and $\overline{CD}$. The area of $\triangle ABC$ is 13 and the area of $\triangle ACF$ is 3. If $\frac{CE}{EA}=\frac{p+\sqrt{q}}{r}$, where $p$, $q$, and $r$ are relatively prime positive integers, compute $p+q+r$.

Solution

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