An Investigation of Pick's Theorem

By Kyle Schultz

What is Pick's Theorem?

Pick's Theorem is a useful method for determining 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 uses a square lattice to examine Pick's Theorem.

Examples of polygons whose area can be computed by Pick's Theorem are shown below

Although the area of each figure can be calculated using other methods (e.g., partitioning it into smaller pieces or using a surrounding rectangle), Pick's theorem provides a relatively simple alternative. To understand this formula, two definitions are needed:

Boundary Point (B): a lattice point on the polygon (including vertices)

Interior Point (I): a lattice point in the polygon’s interior region

Pick's Theorem uses these definitions to state the area of a polygon whose vertices are lattice points:

For example, the area of the yellow polygon above requires counting the number of border points (5) and interior points (5). These values are used in the formula:

As an exercise, use Pick’s Theorem to calculate the area of the green and blue figures. Solutions

I had used Pick's Theorem in a classroom setting regularly, but did not understand the mathematics behind it. Why does it work? Can I prove it? The work presented below is my attempt to understand Pick’s Theorem and answer these questions.

Building a Proof: Rectangles

To construct my understanding of the underlying mechanics of Pick's theorem, I began with very simple polygons—rectangles with vertical and horizontal edges. Examining these rectangles revealed several relationships:

For a j x k rectangle, B = 2j + 2k.

For a j x k rectangle, I = (j – 1)(k – 1).

Using these relationships, I could verify Pick's Theorem accurately calculates the area for rectangles with vertical and horizontal sides.

If desired, use the 5 x 3 rectangle above as an exercise. Use the formulas to verify it has 16 boundary points and 8 interior points as well as an area of 15 square units

This understanding of Pick’s Theorem with rectangles helped launch my investigation of triangles below.

Triangles with a Vertical and a Horizontal Leg

Given the shared properties with the rectangles described above, I began examining right triangles with one vertical and one horizontal leg. Consider the figure below:

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, where h represents the number of lattice points intersected by the hypotenuse of the triangle not including the hypotenuse endpoints. 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. gcf(4, 8) = 4, so h = 3. The figure verifies this calculation.

Recall the number of interior points for a j x k rectangle is I = (j - 1)(k - 1). When partitioning this rectangle into two triangles, interior points coinciding with the hypotenuse become boundary points. These points have been quantified 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 triangle, I halved this value:

Inputting the formulas for the boundary points and interior points of a triangle with a vertical and a horizontal leg into Pick’s Theorem, I verified correct calculation of the area:

Knowing the accuracy of Pick's Theorem in this case enables investigating other types of triangles created on a lattice.

Triangles with a Vertical Side or a Horizontal Side

This triangle differs from the triangles above because only one of its sides is vertically or horizontally aligned.

To verify Pick's Theorem’s accuracy with respect to this triangle, I constructed a j x k rectangle such that two of the triangle’s vertices coincided with vertices of the rectangle. I then labeled the sides according the scheme below.

The area of the rectangle is the sum of the areas of the three triangles. To find the area of the desired (brown) triangle, I calculated the number of boundary and interior points. Note h1 and h2 represent the number of lattice points coinciding with the interior of the marked segment, not the length of the segment.

Using these formulas and Pick's Theorem produces the correct area:

Triangles With No Vertical or Horizontal Sides

The final case of triangles constructed on a lattice consisted of those triangles with no horizontal or vertical sides (adding complexity due to the base and height of these triangles not being aligned with the lattice). Two subcases exist, each involving creating a surrounding rectangle. For the first case, the surrounding rectangle is composed of the original triangle and three additional right triangles. In the second case (suggested by Maxime Lapointe on 12/11/14), the surrounding rectangle is composed of the original triangle, three additional right triangles, and a rectangle. I present each subcase below with verification of Pick’s Theorem.

Subcase 1

Once again, my strategy involved creating a surrounding rectangle such that this rectangle and the triangle share one vertex and the other two triangle vertices coincide with the rectangle’s sides. 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 segment, not the length of the segment.

The calculations below show the use of Pick’s Theorem to determine the area. These calculations verify the area of the blue triangle equals the area of the rectangle less the areas of the three yellow triangles.

[This next step introduces some new terms whose combined sum is zero.

To determine these terms, I worked backwards from the desired conclusion.]

Subcase 2

The creation and labeling of the triangle and surrounding rectangle are similar to that in Subcase 1.

The boundary points are counted in the same manner as in Subcase 1.

To calculate the interior points of the purple triangle, I counted the interior points of the surrounding rectangle and subtracted points that are not part of the purple triangle. These points are (1) the points on the sides of the triangle, (2) the interior points of the three yellow surrounding triangles, (3) the interior points of the small green rectangle created by the surrounding triangles, and (4) the border points of the small rectangle that are in the interior of the surrounding rectangle (two sides and one vertex). These elements are listed in order in the statement of the number of interior points in our triangle below.

Substituting these expressions into Pick’s Theorem and simplifying:

[Similar to Subcase 1, this next step involves inserting additional terms (summing to zero).

These terms were determined by considering the result I wanted to achieve.]

[Some rearranging and combining reveals area formulas confirming the accuracy of Pick’s Theorem.]

The area of the purple triangle is the area of the surrounding rectangle with the areas of the three yellow triangles and the green rectangle subtracted away.

Because of these verified cases, I concluded Pick's Theorem accurately calculates the area for all triangles.

Partitioning a Polygon

So far, I have demonstrated 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 verify the second condition, I created a polygon A partitioned into two polygons, Q and R, by connecting two non-adjacent vertices of A with an arbitrary lattice point path through the interior of A. This path includes p interior points of A.

For the calculations below, B and I indicate boundary and interior points, respectively. Subscripts indicate the attributed polygon.

Because (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 Pick's Theorem will correctly calculate the area of any polygon constructed on a square lattice.

An Extension:   Polygons with Holes

A modification of Pick's Theorem incorporates the presence of holes in a polygon. For example, consider the figure below.

One way to calculate the shaded green area is to determine the area of the polygon (P) less the area of the hole (T).

Alternatively, using Pick's Theorem on the green polygon with an interior (triangular) and exterior (pentagonal) border and interior points inside the green shaded produces the following calculations.

Notice the discrepancy between these formulas. 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 analyze this case in a similar manner, first by calculating the area of the pentagon and subtracting the area of the holes.

Next, we calculate the area with Pick's Theorem. The figure in question has three boundaries, exterior pentagonal, interior triangular, and interior quadrilateral. 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. The empirical evidence uncovered here leads to a conjecture regarding how to incorporate the number of holes into Pick's Theorem. A modified formula is shown below, where H is the number of holes in the figure.

This essay does not verify the accuracy of this formula for all cases, but it all calculations above (and some not included) support the formula’s accuracy. A proof might involve chopping a hole-laden polygon into two separate hole-free polygons by creating a path involving of a chain of segments connecting two arbitrary exterior boundary points and one or more boundary points of each hole.

More Extensions: Different Lattice Systems

What would this investigation look like if it were conducted on a different isometric lattice system? What if the lattice points had a triangular arrangement on a plane? Could a similar investigation produce formulas be used to predict volume and surface area in a three-dimensional lattice system?

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