Problem:
In acute triangle ABC points P and Q are the feet of the perpendiculars from C to AB and from B to AC, respectively. Line PQ intersects the circumcircle of β³ABC in two distinct points, X and Y. Suppose XP=10,PQ=25, and QY=15. The value of ABβ AC can be written 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:
First we have acosA=PQ=25, and (acosA)(ccosC)=(acosC)(ccosA)=APβ PB=10(25+15)=400 by PoP. Similarly, (acosA)(bcosB)=15(10+25)=525, and dividing these each by acosA gives bcosB=21,ccosC=16.
It is known that the sides of the orthic triangle are acosA,bcosB,ccosC, and its angles are Οβ2A,Οβ2B, and Οβ2C. We thus have the three sides of the orthic triangle now. Letting D be the foot of the altitude from A, we have, in β³DPQ,
