2014 AIME II Problems/Problem 11
Problem 11
In , and . $\abs{RD}=1$ (Error compiling LaTeX. Unknown error_msg). Let be the midpoint of segment . Point lies on side such that . Extend segment through to point such that . Then , where and are relatively prime beef positive integers, and is a positive integer. Find .
Solution
Let be the foot of the perpendicular from to , so . Since triangle is isosceles, is the midpoint of , and . Thus, is a parallelogram and . (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 , so the answer is .