Solutions to the 76 th William Lowell Putnam Mathematical Competition Saturday , December 5 , 2015
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Second solution: For any λ > 0, the map (x,y) 7→ (λx,λ−1y) preserves both areas and the hyperbola xy = 1. We may thus rescale the picture so that A,B are symmetric across the line y = x, with A above the line. As P moves from A to B, the area of APB increases until P passes through the point (1,1), then decreases. Consequently, P = (1,1) achieves the maximum area, and the desired equality is obvious by symmetry. Alternatively, since the hyperbola is convex, the maximum is uniquely achieved at the point where the tangent line is parallel to AB, and by symmetry that point is P.
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