Problem:
In β³ABC,AB=10,β A=30β, and β C=45β. Let H,D, and M be points on line BC such that AHβ₯BC,β BAD=β CAD, and BM=CM. Point N is the midpoint of segment HM, and point P is on ray AD such that PNβ₯BC. Then AP2=nmβ, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let Ο be the circumcircle of β³ABC, and let E be the intersection of ray AD and Ο. Because β BAE=β CAE,E is the midpoint of arc BC, and so EMβ₯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, β CBE=β EAB. By the Law of Sines