50th Putnam 1989 Problem B1 S is a square side 1. R is the subset of points closer to its center than to any side. Find the area of R. Solution
: (4√2 - 5)/3. Take axes through the center of the square. A point is closer to the center than to the side y = 1/2 if it lies below the parabola y = 1/4 - x 2 . A similar result applies in the other three directions. So the area is 4 times that of the region S bounded by the parabola and the lines y = x, y = - x. The line y = x cuts the parabola at x = 1/√2 - 1/2 (= k for convenience). The area S is the area under the parabola between x = - k and k, less the area under the two diagonals. So the area S is 2 ∫ 0 k (1/4 - x 2 ) dx - k 2 = k/2 - k 2 - 2k 3 /3. We have k 2 = 1/4 - k, so k 3 = 5k/4 - 1/4 and k/2 - k 2 - 2k 3 /3 = 2k/3 - 1/12 = (√2)/3 - 5/12. Thus the required area is (4√2 - 5)/3. 50th Putnam 1989 © John Scholes jscholes@kalva.demon.co.uk 1 Jan 2001