Problem:
Let P be a point on the circle circumscribing square ABCD that satisfies PA⋅PC=56 and PB⋅PD=90. Find the area of ABCD.
Solution:
Answer (106):
Let Q and R be the projections of P onto AC and BD, respectively, and let O be the center of square ABCD. Also, let r denote the radius of the circle circumscribing ABCD, so that AC=BD=2r.
Observe that
56=PA⋅PC=2⋅Area(△PAC)=PQ⋅AC=PQ⋅2r and 90=PB⋅PD=2⋅Area(△PBD)=PR⋅BD=PR⋅2r.
Furthermore, because ORPQ is a rectangle, r=OP=QR=PQ2+PR2. Squaring both of the above equations and adding yields