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 $AE = PM = (CD)/2$ . (someone incorporate LATEX please) 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 = 1/2 and EO = RO = √3/2, so D(1/2, 0), E(-√3/2, 0), and R(0, √3/2). M = midpoint(D, R) = (1/4, √3/4) and slope(ME) = √3/4 / (1/4 + √3/2) = √3 / (1 + 2√3), so slope(RC) = -(1 + 2√3)/√3. Instead of finding the equation of the line, we use the definition of slope: for every CO = x to the left, we go (1 + 2√3)/√3 * x = √3/2 up. Thus, x = 3/2 / (1 + 2√3) = 3 / (4√3 + 2) = 3(4√3 - 2) / 44 = (6√3 - 3) / 22. DO = 1/2 - x = 1/2 - (6√3 - 3)/22 = (14 - 6√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