Problem:
The circumcircle of acute △ABC has center O. The line passing through point O perpendicular to OB intersects lines AB and BC at P and Q, respectively. Also AB=5,BC=4,BQ=4.5, and BP=nm​, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let line PQ intersect the circumcircle at points M and N as shown in the figure. Because MN is a diameter and OB is perpendicular to MN, it follows that BN=BM. Thus ∠QPB=2BM+AN​=2BN+AN​=2ANB​=∠ACB. Hence △ABC∼△QBP, and BCBP​=ABQ​. It follows that BP=54(4.5)​=518​. The requested sum is 18+5=23​.
OR
Let M and N be defined as above, and let x=BP, so PA=5−x. The Power ofa Point Theorem applied to point P shows x(5−x)=BP⋅PA=PM⋅PN=BO2−OP2, and applied to point Q shows 21​⋅29​=QC⋅QB=QM⋅QN=QO2−BO2. Then 49​+5x−x2=QO2−OP2=(BQ2−BO2)−(BP2−BO2)=BQ2−BP2=(29​)2−x2. Thus 5x=481​−49​=18 and x=518​.
