### 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**

Answer: (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