Problem:
Let ABCD be an isosceles trapezoid with AD∥BC whose angle at the longer base AD is 3π. The diagonals have length 1021, and point E is at distances 107 and 307 from vertices A and D, respectively. Let F be the foot of the altitude from C to AD. The distance EF can be expressed in the form mn, where m and n are positive integers and n is not divisible by the square of any prime. Find m+n.
Solution:
Because 307=DE≤DA+AE=DA+107, it follows that DA≥207. Let θ=∠DCA. Applying the Law of Sines to △DCA yields sin(π/3)1021=sinθDA≥sinθ207, which implies sinθ≥1. Then θ must be 2π, DA=207, and point E lies on the extension of side DA. Applying the Pythagorean Theorem to △DCA yields DC=DA2−CA2=107. Then DF=DC⋅cos3π=57. Therefore EF=DE−DF=307−57=257, and m+n=32.