The main problem in this lesson is to determine the values of the coefficients and the constant term in Pick’s Theorem. In particular, what are the values of coefficients
a and
b, as well as the constant term
c, in the following equation:
Area = a (Number of Perimeter Pins) + b (Number of Interior Pins) + c
For simpler notation, A = aP + bI + c.
It may be worth a brief discussion to talk about the difference between the coefficients a, b, and c, and the variables P and I. Students need to understand that a, b, and c are merely placeholders that stand for specific, though as yet unknown, values; whereas, P and I are true variables, in that they change from case to case.
This is a problem equivalent to finding a parabola given three points. However, the setting seems a bit more interesting to most students.
Have students create three different figures on their geoboards. (Students can use geoboards with rubber bands, dot paper, or the geoboard applet available from the NCTM E‑Examples.) The figures should have differ in the number of perimeter pins, the number of interior pins, and the area. As noted in Rediscovering the Pattern of Pick’s Theorem, the most difficult part is correctly determining the area. Students may check each other’s area calculations to ensure that they are correct.
With three figures and the corresponding values for P, I, and A, students can create and solve a system of three equations.
For example, consider three figures with the following values:
- P = 4, I = 0, A = 1
- P = 8, I = 1, A = 4
- P = 8, I = 4, A = 7
These values yield the following system of equations:
1 = a(4) + b(0) + c
4 = a(8) + b(1) + c
7 = a(8) + b(4) + c
which reduces to:
1 = a(2) + b(0) + c(0)
1 = a(0) + b(1) + c(0)
1 = a(0) + b(0) + c(-1)
In other words, a = ½, b = 1, and c = ‑1, so Pick’s Theorem must be A = ½P + I ‑ 1.
This is one method, but other methods could involve substitution, linear combinations, or matrices. (With matrices especially, the use of a graphing calculator could be particularly helpful. The system of equations could be described by matrices as follows:
The solution, then, can be found by multiplying the inverse of the coefficient matrix by the matrix of constants, which yields the following result:
As above, this solution indicates that a = ½, b = 1, and c = ‑1.)
Then, have students check another figure to make sure their equation works. Errors may have occurred either in the manipulation of the system of equations or in determining the original area, both of which could result in an incorrect formula.
