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 $AE = PM = \frac{CD}{2}$ . We can then use coordinates. Let $O$ be the foot of altitude $RO$ and set $O$ as the origin. Now we notice special right triangles! In particular, $DO = \frac{1}{2}$ and $EO = RO = \frac{\sqrt{3}}{2}$ , so $D(\frac{1}{2}, 0)$ , $E(-\frac{\sqrt{3}}{2}, 0)$ , and $R(0, \frac{\sqrt{3}}{2}).$ $M =$ midpoint $(D, R) = (\frac{1}{4}, \frac{\sqrt{3}}{4})$ and the slope of $ME = \frac{\frac{\sqrt{3}}{4}}{\frac{1}{4} + \frac{\sqrt{3}}{2}} = \frac{\sqrt{3}}{1 + \frac{2}{\sqrt{3}}}$ , so the slope of $RC = -\frac{1 + 2\sqrt{3}}{\sqrt{3}}.$ Instead of finding the equation of the line, we use the definition of slope: for every $CO = x$ to the left, we go $\frac{1 + 2\sqrt{3}}{x\sqrt{3}} = \frac{\sqrt{3}}{2}$ up. Thus, $x = \frac{\frac{3}{2}}{1 + 2\sqrt{3}} = \frac{3}{4\sqrt{3} + 2} = \frac{3(4\sqrt{3} - 2)}{44} = \frac{6\sqrt{3} - 3}{22}.$ $DO = \frac{1}{2} - x = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{14 - 6\sqrt{3}}{22}$ , and $AE = \frac{7 - \sqrt{27}}{22}$ , so the answer is $\boxed{056}$ .

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2014 AIME II Problems/Problem 11

Problem 11

Solution