Department of Mathematics Education
EMAT 6690

by Jan White

What is a lattice polygon?  Given a lattice of dots all equidistant from each other in the form of a grid, a lattice polygon is a polygon with vertices located at each point on the lattice.

On the lattice below there are three triangles all with the same area, since each has the same base and height.  If the lattice are units then the area of the triangels is 1/2 the area of say a square.

The area of each of the triangles are 1/2

Now finding the area of the polygon below is not difficult, just connect all the points inside the polygon and count the squares.

The area of ABCD is 8.

The area of the polygon below is not too hard either.  Just count up the squares and add the area of the two triangles.

The area of the polygon is 10

Below is a polygon that is a little harder to find the area.  We still count up the squares and add in the half squares but then we must find the areas of P1, P2 and P3. None of the examples are very difficult but you can imagine how tedious this would be on a large lattice with many different vertices such as finding the approximate areas of land on a large scale map.

There are 17 squares, 7 half squares, and  P1=P2=P3=1 (3).
So there is a total area of 23 1/2.

Now there is a better way of finding the area of a lattice polygon developed by Georg Alexander Pick.  He was born around 1859 in Vienna and died in 1943 in the Theresienstadt concentration camp.  Pick's Theorem states that the area of a lattice polygon is:  A(p) = I(p) + B(p)/2 -1, where I = the number of interior points and B = the number of boundary points.  Pick's Theorem interestingly enough does not rely on squares or triangles but on circles. We will now try and derive Pick's Theorem.

Let's define a weighting system for each of the points inside and on the boundary of the polygon.  We need to show that A(p) = W(s).

Let Ap = Angle of visibility from any point "p" in/on "s", a lattice polygon, to "s".

Let  I = the number of interior points in the polygon
n = the number of vertices
b = the number of boundary points not including the vertices
B = n + b

Wp = Ap divided by 2¶

Looking at the polygon below, the interior points have an angle of visibility of 2pi.  The points along the boundary not including the vertices have an angle of visiblility of pi and the vetices have an agle of visibility of (n-2)/pi.

W(s) = Wp= Sum(Wv) + Sum(Wb) + Sum(WI)

We get W(s) = (n-2)/2 + b/2 + I
= (n+b -2)/2 + I
= I + (n+b)/2 - 1
W(s) = I + B/2 - 1  = A(p)     Q.E.D.

Let's look at the example below.
Interior pts      = 31
Vertices          = 13
Boundary pts. = 3

Area = 31 + 3/2 - 1 = 31.5 sq units.  You can check the old fashion way.

References: Lecture by a student, J. Taylor, in Math 3200
www.cut-the-knot.com