Problem:
In △ABC,AB=10,∠A=30∘, and ∠C=45∘. Let H,D, and M be points on line BC such that AH⊥BC,∠BAD=∠CAD, and BM=CM. Point N is the midpoint of segment HM, and point P is on ray AD such that PN⊥BC. Then AP2=nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let ω be the circumcircle of △ABC, and let E be the intersection of ray AD and ω. Because ∠BAE=∠CAE,E is the midpoint of arc BC, and so EM⊥BC. The projection of three collinear points A,P, and E on line BC are H,N, and M, respectively, with N the midpoint of segment HM. Thus P is the midpoint of segment AE. Because they subtend equal arcs, ∠CBE=∠EAB. By the Law of Sines