In , and . . 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 positive integers, and is a positive integer. Find .
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Let P be the foot of the perpendicular from A to C R , so A P ∥ E M . Since triangle A R C is isosceles, P is the midpoint of C R , and P M ∥ C D . Thus, A P M E is a parallelogram and A E = P M = 2 C D . We can then use coordinates. Let O be the foot of altitude R O and set O as the origin. Now we notice special right triangles! In particular, D O = 2 1 and E O = R O = 2 3 , so D ( 2 1 , 0 ) , E ( − 2 3 , 0 ) , and R ( 0 , 2 3 ) . M = midpoint ( D , R ) = ( 4 1 , 4 3 ) and the slope of M E = 4 1 + 2 3 4 3 = 1 + 2 3 3 , so the slope of R C = − 3 1 + 2 3 . Instead of finding the equation of the line, we use the definition of slope: for every C O = x to the left, we go 3 x ( 1 + 2 3 ) = 2 3 up. Thus, x = 1 + 2 3 2 3 = 4 3 + 2 3 = 4 4 3 ( 4 3 − 2 ) = 2 2 6 3 − 3 . D C = 2 1 − x = 2 1 − 2 2 6 3 − 3 = 2 2 1 4 − 6 3 , and A E = 2 2 7 − 2 7 , so the answer is 5 6 .