A convex pentagon has all its vertices lattice points in the plane (and no three collinear). Prove that its area is at least 5/2.
Solution
We need Pick's theorem: given a polygon (whose sides are not self-intersecting, but which is not necessarily convex) whose vertices are lattice points, its area is A + B/2 - 1, where A is the number of interior lattice points and B is the number of lattice points on the perimeter.
So we have to show that a convex pentagon either has an interior lattice point or has at least two lattice points on its sides (apart from its vertices).
We can divide the coordinates into four parity classes: (odd, odd), (odd, even), (even, odd) and (even, even). At least two of the vertices must belong to the same class. But then their midpoint must be a lattice point. If the vertices are not adjacent, then the midpoint is an interior lattice point and we are done. So suppose the two vertices are A and B, adjacent and with midpoint M, also a lattice point. Let the other vertices be C, D, and E. A, C, D and E must all have different parity classes, otherwise we get another lattice point and we are done. But now M must have the same parity class as one of A, C, D and E, which gives us another lattice point.
Note that the result is the best possible. For example, (0, 0), (1, 0), (2, 2), (1, 2), (0, 1) has area exactly 5/2.
© John Scholes
jscholes@kalva.demon.co.uk
1 Jan 2001