An Investigation of Pick's Theorem
By Kyle Schultz
What is Pick's Theorem?
Attributed to Georg Pick, this theorem is a useful method for finding the area of any polygon whose vertices are points on a lattice, a regularly spaced array of points. While lattices may have points in different arrangements, this essay will use a square lattice to demonstrate the qualities of Pick's Theorem.
Examples of polygons whose area can be computed by Pick's Theorem are shown below.
While the area of each of these figures can be calculated by using other methods, including being broken into smaller pieces or using a surrounding rectangle , Pick's theorem provides a relatively simple alternative. In order to use it, two definitions must be stated:
Boundary Point (B): a lattice point that lies on the edge of a polygon (including its vertices)
Interior Point (I): a lattice point that lies in the interior of the polygon
Pick's Theorem uses these definitions to state the area of a polygon whose vertices are lattice points:
For example, to find the area of the polygon shaded in yellow above, it is first necessary to count the number of border points (5) and interior points (5). Once this is accomplished, the formula is implemented:
As an exercise, calculate the area of the green and blue figures above using Pick's Theorem. Click Here to view the solutions.
I have used Pick's Theorem in a classroom setting for several years, but did not have a firm grasp of the mathematics behind it. As a result, I began to wonder how it could be proved. My work presented below is an attempt to answer this question.
Building a Proof: Rectangles
To begin to understand how Pick's Theorem works, I began with very simple polygons – rectangles with vertical and horizontal edges. Examining these rectangles now will also help in my analysis of triangles below. Looking at rectangles in this orientation with various sizes and shapes, I noticed several relationships:
For a j x k rectangle, the number of boundary points is 2j + 2k.
For a j x k rectangle, the number of interior points is (j – 1)(k – 1).
Using these relationships, the area of the rectangle can be calculated accurately using Pick's Theorem.
Consider the 5 x 3 rectangle below. There are 16 boundary points and 8 interior points. These calculations agree with the formulas above, as well as the area of 15.
B = 2(5) + 2(3) = 16
I = (5 – 1)(3 – 1) = 8
Using the fact that Pick's Theorem is true for all rectangles of this orientation, I then examined triangles.
Triangles with a Vertical and a Horizontal Leg
There are a lot of similarities between the rectangles I examined above and right triangles that have one vertical and one horizontal leg. Consider the figure below:
Notice that the green triangles are each half the area of a j x k rectangle. Like the rectangles above, relationships exist to describe the number of boundary and interior points:
The number of boundary points is j
+ k + 1 – h,
is the number of lattice points intersected by the hypotenuse of the triangle
that are not endpoints of the hypotenuse . The value of h can be calculated
for any triangle that is half of a j x k rectangle by finding the greatest
common factor of j and k and subtracting 1 (h = gcf(h, k) - 1). For example
the larger green triangle depicted above is half of a 4 x 8 rectangle. The
greatest common factor of 4 and 8 is four, so h = 3. This is verified by
. The value of h can be calculated for any triangle that is half of a j x k rectangle by finding the greatest common factor of j and k and subtracting 1 (h = gcf(h, k) - 1). For example the larger green triangle depicted above is half of a 4 x 8 rectangle. The greatest common factor of 4 and 8 is four, so h = 3. This is verified by the figure.
The number of interior points for any triangle with a
vertical and a horizontal leg can also be calculated. Recall that the number
of interior points for a j x k rectangle is I = (j - 1)(k - 1).When the rectangle
is partitioned into two triangles, some of its interior points may become
boundary points if they coincide with the hypotenuse. This quantity has already
been defined as h. Thus, the number of interior points for the two triangles
is (j - 1)(k - 1) - h. To get the number of interior points for one quantity,
I halved this value:
The number of interior points for any triangle with a vertical and a horizontal leg can also be calculated. Recall that the number of interior points for a j x k rectangle is I = (j - 1)(k - 1).When the rectangle is partitioned into two triangles, some of its interior points may become boundary points if they coincide with the hypotenuse. This quantity has already been defined as h. Thus, the number of interior points for the two triangles is (j - 1)(k - 1) - h. To get the number of interior points for one quantity, I halved this value:
Using the formulas for the boundary points and interior points of a triangle with a vertical and a horizontal leg, I implemented Pick's Theorem and verified that it correctly calculates the area of this type of triangle:
As I have shown Pick's Theorem to be accurate in this case, I will now proceed to the two other types of triangles that can be created on a lattice.
Triangles with a Vertical Side or a Horizontal Side
The triangle below is different from the triangles described above. Only one of its sides is vertically or horizontally aligned.
To verify that Pick's Theorem will accurately calculate the area of such a triangle. I constructed a j x k rectangle around the triangle such that two of the triangle's vertices were also vertices of the rectangle. I then labeled the sides according the scheme below.
The rectangle can be viewed as the sum of three triangles. To find the area of the desired triangle (brown), I calculated the number of boundary and interior points. Note that h1 and h2 represent the number of lattice points coinciding with the interior of the marked segement, not the length of the segment.
Next I calculated the correct area of the triangle using these formulas and Pick's Theorem.
Triangles With No Vertical or Horizontal Sides
The final case of triangles that can be constructed on a lattice has no horizontal or vertical sides. An example of such a triangle is shown below. I found this case to be the most difficult to verify of the triangle cases due to the fact that base and height of the triangle are not aligned with the lattice.
Once again, my strategy is to create a rectangle surrounding the triangle. In this case, the optimal rectangle and triangle share one vertex, while the other two vertices of the triangle conicide with the sides of the rectangle. The formulas for the boundary and interior points are below. Note that h1, h2, and h3 represent the number of lattice points coinciding with the interior of the marked segement, not the length of the segment.
The calculations for the area are below. As you can see, the area of the triangle (in blue) is the area of the rectangle less the area of the three yellow triangles.
This formula troubled me at first, as I could not see how it related to the diagram. With some analysis, I discovered that the terms that included the product jk were important.
The area of the blue triangle is the area of the rectangle with the area of each yellow triangle subtracted away.
As all three cases have been verified, it can be concluded that Pick's Theorem is accurate for all triangles.
Partitioning a Polygon
Up to this point, I have demonstrated that Pick's Theorem correctly calculates the area of any triangle. This can be generalized to say that Pick's theorem correctly calculates the area of any polygon whose vertices are points on a lattice IF two conditions are met:
1. Any polygon on the lattice can be partitioned into triangles. Click Here for a nice proof of this conjecture.
2. The area of any polygon is equal to the sum of the areas of its partitions. To demonstrate this, I created a polygon A that has been partitioned into two polygons, Q and R by connecting two non-adjacent vertices of A with an arbitrary lattice path through the interior of A. The path drawn will include p interior points of A.
For the proof below, boundary and interior points will by indicated with B and I respectively. Subscripts will indicate the polygon being described.
Since (1) Pick's Theorem shows the sum of the areas of the partitions of a polygon equals the area of the entire polygon, (2) any polygon can be partitioned into triangles, and (3) Pick's Theorem is accurate for any triangle, then it can be concluded that Pick's Formula will correctly calculate the area of any polygon constructed on a lattice.
An Extension: Polygons with Holes
Pick's Formula can be modified to incorporate the presence of holes in a polygon. For example, consider the figure below:
One method to calculate the area that is shaded green is to calculate the area of the polygon (P) less the area of the hole (T).
Another is to use Pick's Theorem, where the boundary points include both the boundary points of the pentagon and triangle, while the interior points are only those inside the green shaded area (the pentagon interior points less the boundary and interior points of the triangle.
Notice that these formulas do not give the same value for the area. The Pick's Theorem calculation is one less than the actual value. To further investigate this discrepancy, consider a figure with two holes.
We can calculate the area shaded in green in a similar manner. First, the area of the pentagon is calculated, less the area of the holes.
Now, the area is calculated with Pick's Theorem. The boundary points include the boundary points of the pentagon along with the boundary points of the holes. The interior points include the interior points of the pentagon less the boundary and interior points of the holes.
With two holes, there is a discrepancy of two between the calculations. This leads to the conjecture that the number of holes must be incorporated into Pick's Theorem. A new formula that incorporates the presence of holes is shown below, where B is the number of boundary points, I is the number of interior points, and H is the number of holes in the figure.
It is important to note that I have not proven this conjecture, though it has correctly calculated the area of several polygons with holes that I tested.
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