### 8th Putnam 1948

Problem B5

Find the area of the region { (x, y) : |x + yt + t2| ≤ 1 for all t ∈ [0, 1] }.

Solution

We cannot have |x| > 1, for then the inequality is not satisfied at t = 0. We cannot have y > 1 for then yt + t2 > 2 at t = 1 and hence x + yt + t2 > 1. Similarly, we cannot have y < -3 for then yt + t2 < -2 at t = 1 and hence x + yt + t2 < -1.

We can write x + yt + t2 = x - y2/4 + (t + y/2)2. So for -3 ≤ y ≤ -2, the maximum of (t + y/2)2 occurs at t = 0 with value y2/4, and its minimum is (1 + y/2)2 at t = 1. Hence the minimum of x + yt + t2 is 1 + x + y and the maximum is x. So the region is bounded by x ≤ 1 and x + y ≥ -2.

For -2 ≤ y ≤ -1, the maximum of (t + y/2)2 is y2/4 at t = 0, but the minimum is 0 at t = -y/2. Hence the maximum of x + yt + t2 is x and the minimum is x - y2/4. So the region is bounded by x ≤ 1 and x - y2/4 ≥ -1.

For -1 ≤ y ≤ 0, the maximum of (t + y/2)2 is (1 + y/2)2 and the minimum is 0 at t = -y/2. Hence the maximum of x + yt + t2 is 1 + x + y and the minimum is x - y2/4. So the region is bounded by x + y ≤ 0 and x - y2/4 ≥ -1.

For 0 ≤ y ≤ 1, the maximum of (t + y/2)2 is (1 + y/2)2 and the minimum is y2/4. Hence the maximum of x + yt + t2 is 1 + x + y and the minimum is x. So the region is bounded by x + y ≤ 0 and x ≥ -1.

Thus the region is the rhombus with vertices at (1, -1), (1, -3), (-1, -1), (-1, 1), except that the vertex (-1, -1) is cut off by the curve x = y2/4 - 1, which forms the boundary between (0, -2) and (-1, 0). The rhombus has area 4 (base 2 and height 2). The area between the parabola and the lines x = -1 and y = -2 is ∫02 y2/4 dy = 8/12. The area between the line y = -2 -x and the lines x = -1 and y = -2 is 1/2, so the area cut off the rhombus by the parabola is 2/3 - 1/2 = 1/6. Hence the area required is 4 - 1/6 = 3 5/6.