## Wednesday, August 19, 2009

### Wednesday Math, Vol. 84: Pick's Theorem

Consider the Cartesian coordinate system, where we give every point on the plane a pair of numbers based on moving left or right from a point we arbitrarily designate the origin, the x coordinate, and then turning 90 degrees and moving up or down, the y coordinate. The points where both x and y are integer values are called lattice points. If you have graph paper, the lattice points would be where the grid lines intersect. In the picture on the left, the lines are left out and only the lattice points remain.

If I want to find the area of a polygon whose corners are all lattice points, one way to do it would be the slice the shape into triangles and find the each triangle area by means of determinants. A quicker way is as follows.

1. Count the number of lattice points on the interior, call that number I.
2. Count the number of lattice points on the perimeter, call that number P, and divide P by 2.
3. The area is I + P/2 - 1.

That's easy, yes?

In this example, there are 39 interior lattice points and 14 perimeter points, so the area is 39 + 14/2 - 1 = 45 square units. Any shape like this will either have a whole number for the area or a whole number plus one half.

I didn't find out about Pick's Theorem until I was a graduate student, but it's not a difficult idea or even that difficult to prove. It's just one of those things that doesn't fit perfectly into any class in the curriculum. It could find its way into a geometry course, but it hasn't yet at most schools.

Yay, Pick's Theorem!
~

#### 1 comment:

Dr. Zaius said...

What if the polygon doesn't like triangles or determinants? Maybe it has pointy edges, or a bad attitude.