Pick’s Theorem 93

Since this is indeed the area of an n×m rectangle, we have therefore proved Pick’s theorem

in the easy case of a rectangle aligned with the lattice lines. Let us state this as a separate

theorem; we have already just given the proof.

Theorem 64 (Pick’s theorem, rectangle case). For any nondegenerate rectangle in the

integer lattice, whose sides are aligned with the lattice lines, the area is

A = i +

b

2

− 1,

where i is the number of interior lattice points and b is the number of lattice points on the

boundary of the rectangle.

8.3 Pick’s theorem for triangles

Let us now try a somewhat harder case. How

about a triangle? Well, what kind of triangle?

Perhaps we should start with a right triangle,

whose legs are oriented with the lattice lines.

We’ll have to pay attention to whether there

are boundary points on the hypotenuse. For

the 7 × 3 triangle pictured here, we have six

interior vertices and eleven boundary vertices

(none on the hypotenuse), so Pick’s formula

gives 6 + 11/2 − 1 = 10.5, which is indeed the area of this triangle

1

2

· base · height =

1

2

· 7 ·3 = 10.5. Let us try to consider the more general case.

Theorem 65 (Pick’s theorem, right triangle case). For any nondegenerate right triangle

in the integer lattice, whose legs align with the lattice lines, the area is given by Pick’s

formula

A = i +

b

2

− 1,

where i is the number of interior vertices and b is the number of vertices on the boundary.

Proof. Consider such a triangle, with legs of length n and m. The hypotenuse of the

triangle is the diagonal of a corresponding n×m rectangle. The area of the triangle is nm/2,

exactly half of the area of the corresponding rectangle. Let h be the number of boundary

points on the hypotenuse, not counting the vertices of the triangle. The triangle has exactly

half of the interior points of the rectangle, except for the h points on the hypotenuse itself.

In the proof of theorem 64, we calculated the number of interior points of the rectangle

as (n − 1)(m − 1), and so for the triangle, we have i = [(n − 1)(m − 1) − h]/2. For the

boundary, the triangle has exactly half of the boundary vertices of the rectangle, plus the

h extra points on the hypotenuse, plus one extra, since we have one horizontal side, one

vertical side, but we include three corners instead of only two. So for our triangle, we have