**Theorem:** *(Pick)* Let be a simple polygon whose vertices are in . Let (resp. ) denote the cardinality of (resp. of ). Then the area of is given by the formula

.

**Proof.** We first prove the theorem in specific cases.

- Pick’s formula is true for rectangles:

Let be a rectangle whose vertices are in . Then and , hence

.

- Pick’s formula is true for right triangles with a vertical edge:

Let be the rectangle and let denote the cardinality of . Then and hence

.

The same argument applies to other kind of right triangles with a vertical edge (like for example).

- Pick’s formula is true for such triangle:

Let be the rectangle . Let denote the cardinality of ; and are defined in the same way. Then and , hence using the previous cases:

- Pick formula is true for such triangle:

Let be the rectangle . Let denote the cardinality of ; and are defined in the same way. Let (resp. ) be the cardinality of (resp. ). Then and . Exactly like the previous case, we deduce

.

Finally, we deduce that Pick’s formula is true for all triangle (whose vertices are in ).

Now, we show our theorem by induction on the number of vertices of . If , is a triangle and we already proved that Pick’s theorem holds in this case. Suppose that the theorem holds for every polygon with vertices and suppose that has vertices. Using the following lemma:

**Lemma:** Let be a polygon. Then contains a diagonal, that is edge between two vertices lying in the interior of .

to decompose as the union of two polygons and whose intersection is an edge. Let denote the cardinality of . Now and , so using our induction hypothesis:

.

Therefore, our proof by induction is completed.

**Proof of the lemma: **Let be a simple polygon. First, we prove has an angle smaller than .

For , let be the line ; let denote the upper closed half-space determined by . Because is compact, is well-defined and .

If contains a vertex then the angle of at cannot be greater than , otherwise would have a point below . If contains an edge then the angles of at and cannot be greater than , otherwise would have a point below .

Therefore, has an angle smaller than at some vertex . Let and denote the two vertices of adjacent to .

If lies in the interior of , then is a diagonal. Otherwise, the triangle contains a finite number of vertices of . Let be the bisectrix of the angle and, for each , let be the line perpendicular to passing through .

Let be such that . In particular, separates and the , so is a diagonal.

**Corollary 1:** At least one vertex of an equilateral triangle has a non-integer coordinate.

**Proof.** Suppose by contradiction that there exists an equilateral triangle whose vertices are in . Then

.

On the other hand, the length of the edges of is for some , hence

.

Therefore, that is : a contradiction.

**Corollary 2:** An square is randomly tossed onto an integer grid. Then it may never cover more than grid points.

**Proof.** First, notice that where is the perimeter of the square. Then, the square covers

grid points, according to Pick’s theorem.

**Definition:** Let . The sequence obtained by ordering is called the Farey sequence of order .

For example, the beginning of the Farey sequence of order three is

**Corollary 3:** Let be two consecutive terms in a Farey sequence. Then .

This fact has been already noticed and proved by the French mathematician Augustin Cauchy in *Démonstration d’un théorème curieux sur les nombres *(*Proof of a curious theorem about numbers*). Surprisingly, this theorem has a geometric interpretation and can be proved using Pick’s theorem:

**Proof.** Let be two consecutive terms in the Farey sequence of order .

Let the set of lines passing through the origin and meeting . If , we say that if the gradient of is greater than the gradient of . Ordering with respect to , we get a sequence of lines.

For each , let be the closest point of to . Then is the -th term of the Farey sequence of order .

Let , and . Now we compute the area of using Pick’s theorem.

Because the fractions and are irreducible, the open segments and do not meet . If or meeted there would be a line such that , that is and would not be consecutive in the Farey sequence. Therefore, according to Pick’s theorem.

On the other hand, , hence .

To conclude our note, we prove a more general formula when the polygon has “holes”:

**Theorem:** Let and be simple polygons. Let denote the “holed polygon” . Then the area of is given by

,

**Proof. **First, notice that

.

Then, using Pick’s theorem: