2014 AIME II Problems/Problem 11

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 $AE = PM = (CD)/2$ .

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2014 AIME II Problems/Problem 11

Problem 11

Solution