Main Index Number Theory Geometric Number Theory
  Subject Index
comment on the page

Pick’s theorem

In 1899 G.A.Pick  [1] , [2] found a simple method how to compute the area of  a (not necessarily convex) polygon whose vertices coincide with the points of a plane parallelogrammic lattice (grid) , the so called lattice polygon :  ¹

Theorem. If is the number of points of the lattice inside the lattice polygon whose sides do not cross one another, and is the number of lattice points on its boundary, including the vertices, then the area of the polygon (measured in area units given by the area of the fundamental parallelogram of the lattice) is

In the picture for the green polygon we have  , so its area equals   areas of a fundamental parallelogram, while the blue fundamental parallelogram has clearly the area  . The area of the red (primitive) triangle is of the area of a fundamental parallelogram.

A polygon whose sides do not cross one another is called simple. Fundamental to all proofs of Pick’s theorem is the notion of a primitive triangle, theses are lattice triangle with no interior lattice point and which vertices are the only three lattice points on the its boundary.  

For the proof firstly note that the expression on the right hand side is additive when two polygons are juxtaposed.  If (clearly ) is the number of points on a common side of two such polygons then their contribution to the inside points in the juxtaposed polygon is , while the number of boundary points of the juxtaposition increases by 1.  

The next step of the proof of is based on the fact that every simple polygon can be dissected into triangles. Joining a vertex of a triangle with the lattice points on the opposite side, we can suppose that all triangulating triangles  are primitive.

The proof of Pick’s theorem now follows from the following two results:

Lemma 1. The number of primitive triangles in a decomposition of a simple lattice polygon into primitive triangles is .

To prove [3]  this result suppose that we have such a triangulation (actually in what follows we shall only need  that each region of the decomposition is bounded by three arcs and that no arcs cross). Take an identical copy of the decomposition and superimpose it on the top of the original triangulation and glue them along the matched boundary edges of both polygons. Suppose that the new object is inflated to a sphere. Since the interior vertices of the triangulation occur in both copies, while the boundary vertices are glued together in pairs, the total number of vertices on the sphere is . The number of edges is , where is the number of primitive triangles in our decomposition.  Now Euler’s polyhedron theorem implies that

and Lemma 1 follows.

Lemma 2. The area of a primitive triangle is of the area of a fundamental parallelogram.

Actually, all three results (Pick’s theorem, Lemma 1 and 2) are mutually equivalent.

Notes

References

Cite this web-page as: