Problem:
Let ABCD be an isosceles trapezoid with ADβ₯BC whose angle at the longer base AD is 3Οβ. The diagonals have length 1021β, and point E is at distances 107β and 307β 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 mnβ, where m and n are positive integers and n is not divisible by the square of any prime. Find m+n.
Solution:
Because 307β=DEβ€DA+AE=DA+107β, it follows that DAβ₯207β. Let ΞΈ=β DCA. Applying the Law of Sines to β³DCA yields sin(Ο/3)1021ββ=sinΞΈDAββ₯sinΞΈ207ββ, which implies sinΞΈβ₯1. Then ΞΈ must be 2Οβ, DA=207β, and point E lies on the extension of side DA. Applying the Pythagorean Theorem to β³DCA yields DC=DA2βCA2β=107β. Then DF=DCβ cos3Οβ=57β. Therefore EF=DEβDF=307ββ57β=257β, and m+n=32β.