Problem:
In right triangle ABC with the right angle at C,β BAC<45β and AB=4. Point P on AB has the properties that β APC=2β ACP and CP=1. The ratio BPAPβ can be represented in the form p+qrβ, where p,q and r are positive integers and r is not divisible by the square of any prime. Find p+q+r.
Solution:
Let the circumcircle of β³ABC have center at O and radius r, and let β ACP=Ξ±. Extend CP to intersect the circle at the point D. Because β AOD=β DPB=2Ξ±, it follows that DO=DP=r. Because inscribed angles subtended by the same arc are equal, it follows that β³APD and β³CPB are similar. Therefore BPCPβ=DPAPβ and APCPβ=DPBPβ. Thus BPCPβ+APCPβ=DPAPβ+DPBPβ=DPABβ=r2rβ=2. Observe that β BAC<45β implies that AP>BP. Because CP=1, the previous equation takes the form 4βAP1β+AP1β=2, giving 2+2β=AP. It follows that BP=2β2β, and so BPAPβ=2β2β2+2ββ=3+22β. Hence p+q+r=7β.
Note: The existence of such a triangle can be shown by using Stewart's Theorem.
