Problem:
In right triangle ABC with the right angle at C,∠BAC<45∘ and AB=4. Point P on AB has the properties that ∠APC=2∠ACP and CP=1. The ratio BPAP​ can be represented in the form p+qr​, 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 △ABC have center at O and radius r, and let ∠ACP=α. Extend CP to intersect the circle at the point D. Because ∠AOD=∠DPB=2α, it follows that DO=DP=r. Because inscribed angles subtended by the same arc are equal, it follows that △APD and △CPB are similar. Therefore BPCP​=DPAP​ and APCP​=DPBP​. Thus BPCP​+APCP​=DPAP​+DPBP​=DPAB​=r2r​=2. Observe that ∠BAC<45∘ implies that AP>BP. Because CP=1, the previous equation takes the form 4−AP1​+AP1​=2, giving 2+2​=AP. It follows that BP=2−2​, and so BPAP​=2−2​2+2​​=3+22​. Hence p+q+r=7​.
Note: The existence of such a triangle can be shown by using Stewart's Theorem.
