Line Integral Arguments for Pick’s Theorem
Student researcher: Claire Gianelle
Advisor: Dr. Randall Helmstutler, Department of Mathematics
Over the past fall semester, Dr. Helmstutler and I attempted to write a new proof for
Pick’s Theorem using integrals. The first step was getting a firm grasp on Pick’s Theorem and
it’s original proof. Pick’s Theorem basically states that if a simple polygon has all of its vertices
located on lattice points, coordinates (a, b) where both a and b are integers, then the area of that
polygon can be found using the formula:
½ p + q – 1,
where p is the number of lattice points on the boundary of the polygon, and q is the number of
lattice points inside of the polygon. The original proof of this theorem is shown by first proving
that the formula is additive, which it is, and then by proving it for rectangles, right triangles, any
triangles, and then any polygon, respectively. Sure enough, the theorem holds true for all
polygons, which is significant considering the formula is based mostly off addition of points,
when normally area is found using some form of multiplication.
Knowing that Pick’s Theorem works for certain, we started looking at how integrals
could possibly play into the proof. Our first thought was to look at the vector field
F = -y, x
because when F is integrated over a curve c, it calculates the area of the curve. So, we plugged
in generic numbers (a, b) and (c, d) for starting and ending points respectively to see if there was
a simpler formula we could use to calculate the integral. After solving the integral using brute
force over the linear path from (a, b) to (c, d), we saw that there was an easier formula, and it
happened to be the determinant, or [ad-bc]. Yes, simpler, but not any help in relating the integral
to Pick’s Theorem.
Our next idea was to look at Pick’s Theorem when the polygons had no interior points
because we knew that the formula was additive, hence any polygon could be broken down into
polygons with no interiors and then the areas could simply be added. First, we looked at
rectangles with no interiors, which gave us a better understanding as to why Pick’s Theorem
makes sense, but again it did not really help us when attempting to relate it back to integrals.
And, although this step did not connect to integrals, I thought it was rather important because we
saw a more logical explanation for the Pick’s Theorem proof.
Then, our last and probably most successful idea was to look at the amount of area
around each lattice point by calculating the angle of “vision” each lattice point in the polygon
had. After reading about another proof that looked at angles (Bogomolny), but had a few holes
in its explanation, we thought that we could use the same concept but calculate the angles by
integrating over the vector field,
F=
1 -y, x
x2 + y2
because when this vector field is integrated over a curve, it calculates the angle of the curve, no
matter what the curve is doing otherwise. This method was for the most part successful because
we were able to show that once all of the angles of “vision” were calculated and then scaled
down by 2, the addition of the angles equaled the area of the polygon. The problem we faced
with this idea was that although it did equal area, we had no proof that when calculating the
angles, area was being seen. We attempted to remedy this problem by looking at a box with area
1 unit around the lattice points, but ran into problems showing the area of those boxes with
integrals.
So, long story short, we were unable to find a new proof for Pick’s Theorem via the
theory of integration, although we are close. With more time we may be able to link the angle
and area calculations with the “correct” integrals, which would complete the proof. We have
agreed to work on this in our spare time next semester.
References
A. Bogomolny, Pick's Theorem: Proof from Interactive Mathematics Miscellany and
Puzzles
http://www.cut-the-knot.org/ctk/Pick_proof.shtml, Accessed 15 December 2011