The vertices of a triangle are lattice points in the plane. Show that the diameter of its circumcircle does not exceed the product of its side lengths.
Let the side lengths be a, b, c and the circumradius R. Let θ be the angle opposite side a. Then the area of the triangle A = 1/2 bc sin θ. The side a subtends an angle 2θ at the centre of the circumcircle, so a = 2R sin θ. Hence 2A = abc/(2R). So we have to show that A ≥ 1/2.
This follows at once from the well-known Pick's theorem: the area of any (non-self-intersecting) polygon whose vertices are lattice points is v/2 + i - 1, where v is the number of lattice points on the perimeter and i is the number of lattice points inside the polygon (so since for a triangle v ≥ 3, and i ≥ 0, we have area at least 3/2 - 1 = 1/2).
32nd Putnam 1971
© John Scholes
7 Feb 2001