A polygon without self-intersections is called lattice if all its vertices have integer coordinates in some 2D grid. Pick’s theorem provides a way to compute the area of this polygon through the number of vertices that are lying on the boundary and the number of vertices that lie strictly inside the polygon.

## 1. Formula

Given a certain lattice polygon with non-zero area.

We denote its area by $S$, the number of points with integer coordinates lying strictly inside the polygon by $I$ and the number of points lying on polygon sides by $B$.

Then, the **Pick’s formula** states:

$$S=I+\frac{B}{2}-1$$

In particular, if the values of $I$ and $B$ for a polygon are given, the area can be calculated in $O(1)$ without even knowing the vertices.

This formula was discovered and proven by Austrian mathematician Georg Alexander Pick in 1899.

## 2. Proof

The proof is carried out in many stages: from simple polygons to arbitrary ones:

- A single square: $S=1, I=0, B=4$, which satisfies the formula.
- An arbitrary non-degenerate rectangle with sides parallel to coordinate axes: Assume $a$ and $b$ be the length of the sides of rectangle. Then, $S=ab, I=(a-1)(b-1), B=2(a+b)$. On substituting, we see that formula is true.
- A right angle with legs parallel to the axes: To prove this, note that any such triangle can be obtained by cutting off a rectangle by a diagonal. Denoting the number of integral points lying on diagonal by $c$, it can be shown that Pick’s formula holds for this triangle regardless of $c$.
- An arbitrary triangle: Note that any such triangle can be turned into a rectangle by attaching it to sides of right-angled triangles with legs parallel to the axes (you will not need more than 3 such triangles). From here, we can get correct formula for any triangle.
- An arbitrary polygon: To prove this, triangulate it, ie, divide into triangles with integral coordinates. Further, it is possible to prove that Pick’s theorem retains its validity when a polygon is added to a triangle. Thus, we have proven Pick’s formula for arbitrary polygon.

## 3. Generalization to higher dimensions

Unfortunately, this simple and beautiful formula cannot be generalized to higher dimensions.

John Reeve demonstrated this by proposing a tetrahedron (**Reeve tetrahedron**) with following vertices in 1957:

$$A=(0,0,0), B=(1,0,0), C=(0,1,0), D=(1,1,k),$$

where $k$ can be any natural number. Then for any $k$, the tetrahedron $ABCD$ does not contain integer point inside it and has only $4$ points on its borders, $A, B, C, D$. Thus, the volume and surface area may vary in spite of unchanged number of points within and on boundary. Therefore, Pick’s theorem doesn’t allow generalizations.

However, higher dimensions still has a generalization using **Ehrhart polynomials** but they are quite complex and depends not only on points inside but also on the boundary of polytype.

## 4. Extra Resources

A few simple examples and a simple proof of Pick’s theorem can be found here.