### 40th Putnam 1979

Problem B5

A convex set S in the plane contains (0, 0) but no other lattice points. The intersections of S with each of the four quadrants have the same area. Show that the area of S is at most 4.

Solution

Let A, B, C, D be the points (1, 0), (0, 1), (-1, 0), (0, -1). These all lie outside S, so we may take support lines SA, SB, SC, SD through each of them. If SA meets the x-axis in the interval [-1, 1], then we are home, because the part of S in the first quadrant lies inside a triangle area < 1. So assume each support line is either parallel to the other axis or meets it a distance more than 1 from the origin. Hence SA and SB meet at W in the first quadrant, SB and SC meet at X in the second quadrant, SC and SD meet at Y in the third quadrant, and SD and SA meet at Z in the fourth quadrant. S lies entirely inside WXYZ. At least one of the angles of this quadrilateral must be ≤ 90o. Suppose it is W.

Now we show that the area of the quadrilateral OAWB (where O is the origin) is at most 1. The required result then follows immediately. Area OAWB = area OAB + area WAB, and area OAB = 1/2. So we need to show that area WAB <= 1/2. But that is almost immediate. The locus of W given angle W is an arc through A, W, B. We maximise the area of the triangle by taking W the midpoint of the arc. But this triangle lies entirely inside the triangle AW'B with AW' = W'B and angle AW'B = 90o, which has area 1/2.