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.
Solution
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).
© John Scholes
jscholes@kalva.demon.co.uk
7 Feb 2001