Problem:
In △ABC with side lengths AB=13,BC=14, and CA=15, let M be the midpoint of BC. Let P be the point on the circumcircle of △ABC such that M is on AP. There exists a unique point Q on segment AM such that ∠PBQ=∠PCQ. Then AQ can be written as nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
We notice that BCPQ is a parallelogram because of the angle conditions. this is because ∠PBQ=∠PCQ and BC is bisected by M. By Stewart's Theorem AM=237, and then we have
AM⋅PM=CM⋅BM→PM=14849
Because of this parallelogram condition, notice that QM=PM and it follows that AQ=AM−PM=23799→99+148=247.