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Difference between revisions of "2014 AIME II Problems/Problem 11"

(Created page with "==Problem 11== In <math>\triangle RED</math>, <math>\measuredangle DRE=75^{\circ}</math> and <math>\measuredangle RED=45^{\circ}</math>. <math>\abs{RD}=1</math>. Let <math>M</mat...")
 
(→‎Solution)
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==Solution==
 
==Solution==
 
Let <math>P</math> be the foot of the perpendicular from <math>A</math> to <math>\overline{CR}</math>, so <math>\overline{AP}\parallel\overline{EM}</math>. Since triangle <math>ARC</math> is isosceles, <math>P</math> is the midpoint of <math>\overline{CR}</math>, and <math>\overline{PM}\parallel\overline{CD}</math>. Thus, <math>APME</math> is a parallelogram and <math>AE = PM = (CD)/2</math>.
 
Let <math>P</math> be the foot of the perpendicular from <math>A</math> to <math>\overline{CR}</math>, so <math>\overline{AP}\parallel\overline{EM}</math>. Since triangle <math>ARC</math> is isosceles, <math>P</math> is the midpoint of <math>\overline{CR}</math>, and <math>\overline{PM}\parallel\overline{CD}</math>. Thus, <math>APME</math> is a parallelogram and <math>AE = PM = (CD)/2</math>.
 +
 +
We can then use coordinates to find that <math>AE = \frac{7 - \sqrt{27}}{22}</math>, so the answer is <math>\boxed{056}</math>.

Revision as of 12:23, 10 April 2014

Problem 11

In $\triangle RED$, $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$. $\abs{RD}=1$ (Error compiling LaTeX. Unknown error_msg). Let $M$ be the midpoint of segment $\overline{RD}$. Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$. Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$. Then $AE=\frac{a-\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer. Find $a+b+c$.

Solution

Let $P$ be the foot of the perpendicular from $A$ to $\overline{CR}$, so $\overline{AP}\parallel\overline{EM}$. Since triangle $ARC$ is isosceles, $P$ is the midpoint of $\overline{CR}$, and $\overline{PM}\parallel\overline{CD}$. Thus, $APME$ is a parallelogram and $AE = PM = (CD)/2$.

We can then use coordinates to find that $AE = \frac{7 - \sqrt{27}}{22}$, so the answer is $\boxed{056}$.

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