Pick's theorem


Given a simple polygon constructed on a grid of equal-distanced points such that all the polygon's vertices are grid points, Pick's theorem provides a simple formula for calculating the area of this polygon in terms of the number of lattice points in the interior located in the polygon and the number of lattice points on the boundary placed on the polygon's perimeter:
In the example shown, we have interior points and boundary points, so the area is = 7 + − 1 = 7 + 4 − 1 = 10 square units.
The theorem as stated above is only valid for simple polygons, i.e., ones that consist of a single, non-self-intersecting boundary. For a general polygon, Pick's formula generalizes to
where is the number of vertices both in and on the boundary of the polygon, is the number of lattice edges on the boundary of the polygon, and is the number of holes in the polygon.
As an example, consider the "polygon" made by connecting the points. It has 3 vertices, 0 holes, and 0 area. To make the formula work, it must have 4 edges. Thus, one just has to count each edge twice, "once on each side".
The result was first described by Georg Alexander Pick in 1899. The Reeve tetrahedron shows that there is no analogue of Pick's theorem in three dimensions that expresses the volume of a polytope by counting its interior and boundary points. However, there is a generalization in higher dimensions via Ehrhart polynomials.

Proof

Consider a polygon and a triangle, with one edge in common with. Assume Pick's theorem is true for both and separately; we want to show that it is also true for the polygon obtained by adding to. Since and share an edge, all the boundary points along the edge in common are merged to interior points, except for the two endpoints of the edge, which are merged to boundary points. So, calling the number of boundary points in common, we have
and
From the above follows
and
Since we are assuming the theorem for and for separately,
Therefore, if the theorem is true for polygons constructed from triangles, the theorem is also true for polygons constructed from triangles. For general polytopes, it is well known that they can always be triangulated. That this is true in dimension 2 is an easy fact. To finish the proof by mathematical induction, it remains to show that the theorem is true for triangles. The verification for this case can be done in these short steps:
The last step uses the fact that if the theorem is true for the polygon and for the triangle, then it's also true for ; this can be seen by a calculation very much similar to the one shown above.

Inequality for convex sets

Let be a bounded, convex region in , not necessarily closed. Then
where is the set of lattice points in, and is the number of them. Equality holds iff is a closed lattice polygon.
The proof is by taking the convex hull of, which should be thought of as a lattice approximation of, then apply Pick's theorem to it.
where is the number of boundary points of, which is equal to the number of edges of it, and since each edge is at least length 1,. And the step uses the property that, between two nested, convex, closed curves, the inner one is shorter, which is an application of Crofton formula.
This still works in the degenerate case when is on the same line. One just have to count each edge twice, "once on each side".