background

Pick's theorem

Theorem 1   Given a simple closed polygon whose vertices have integer coordinates, then

# interior lattice points# boundary lattice points

Clearly we can reduce Pick's Theorem to the case of triangles because

where area, # interior lattice points, and # boundary points. These all imply

We may further assume that there are no other lattice points other than its vertices. Now the idea is to embed the special triangles into rectangles.
Thus, all the problem reduces to is to study and show that Pick's theorem is true for
1. Rectangles.
2. Right angle triangles with integer vertices and no lattice points in the hypotenuse.

For a triangle of case 1

,

By thinking of the triangle as a rectangle,

# of interior points

It's crucial that there are no lattice points in the hypotenuse. QED.

Going back to the case of an arbitrary polygon, there is an alogrithm that can be used to count the number of lattice points inside of it:

1. Triangulate polygon.
2. Count lattice points in each triangle (there are of them). Call this number .
3. Count lattice points on each diagonal. Call this number .
4.    Result .

Definition 2   Let be a triangle whose vertices are integral. We say is primitive if it has no other lattice opints in its boundary or interior.

Lemma 3   Every lattice triangle can be divided into primitive triangles.

Lemma 4   A primitive triangle has area .

Lemma 5   Euler's formula: In a triangulation of a polygon,

where # vertices in primitive triangulation , # edge segments in prmitive triangulation , and # primitive triangles .

Corollary 6   The number of primitive triangles from the first lemma is

where # interior points and # boundary points .

The proof of the third lemma can be done using the corollary and the first lemma:

 (area of primitive triangles) # primitive triangles

The proof of the corollary can be done using the second lemma.

1. I wish to write in terms of
 (1)

2. We have and
3. Substituting equation (1) into the equations from number 2, we get

4. DONE.

The proof of the second lemma goes as follows:

1. Any triangle can be sandwiched into a rectangle with sides parallel to the axis. The proof should be done for all cases, but we will show the hardest case and the simpler proofs can follow.
2. From the above picture we get the following equations:
which all imply

3. So, if we can show , then we are done.
4. To help us show the above we need to define a new notation
 # of lattice points inside but not on the boundary of polygon M # of lattice points lying on the segment PQ except its ends

5. Now we look at two ways to represent :

where the ending comes from the point .
6. Then write the above in terms of to get

7. After simplification we get,

8. DONE!

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