Pick’s Theorem

Longer Introductory Piece to start off a kind of discussion within the Mathematical section of STEM. Any and all submissions welcome!

Pick’s Theorem is a very useful theorem in analytic geometry which can be used for finding the area of certain types of polygons. Before we delve into the mathematics though, let’s start off with some definitions. 

Definitions:

  • Lattice point = a point (x,y) where x and y are integers
  • Lattice polygon = a polygon whose vertices are all lattice points 
  • Boundary points = lattice points which lie on the edges of the polygon 
  • Interior points = lattice points which lie inside the polygon (not including the edges)

Statement of Pick’s Theorem:

If P is any lattice polygon, let B(P) be the boundary points of P and I(P) be the interior points of P. Then:

Proof:

  1. Show that it holds for rectangles whose sides are parallel to the axes

Let the side-lengths of the rectangle be a and b. A side-length of 1 implies that you have two lattice points connected, a side-length of 2 implies you have 3 lattice points connected together (and two gaps between them). Similarly, a side-length of a has a+1 lattice points, and a side-length of b has b+1 lattice points. The area of this rectangle as well all know is . Looking at Pick’s Theorem, we know B(P) already: . We add together the number of lattice points on each side, but subtract 4 because we have counted each one of the vertices twice (once as part of a vertical edge and once as part of a horizontal edge). Now, . We know that there are a-1 lattice points along one side, and b-1 along the other. We want to calculate the number of interior points, and we can do this by constructing a rectangle of lattice points and calculating the ‘area’ of that rectangle. We can let the a above be the width and b be the length as they are interchangeable. Then, the width of this new rectangle is (a+1) -2 as we have to subtract off the two boundary points, and similarly with b. Then, we multiply together as (number of interior points in one column) x (number of interior points in one row) = total number of interior points. Plugging into Pick’s Theorem to see if it works: 

Which is, as we know, correct!

  • Show it is true for right triangles 

Any right triangle can be formed from a rectangle by splitting it in half, as shown above. Let’s calculate B(P) and I(P) again now. , in the same way as above, letting ‘h’ be the number of points which are on the hypotenuse of the right-triangle, subtracting 2 because we don’t want to count points A and C above in the diagram again, and subtracting 1 because we don’t want to count B twice. . We proceeded in the same way as with the rectangle, but we subtracted off the points on the hypotenuse, taking away 2 from them as we don’t want to count points A and C in the diagram again. We then half as the area of a right-triangle is half that of the rectangle it is made from. 

Testing Pick’s Theorem:

Which we know is correct!

  • Show it is true for all triangles 

Given any irregular triangle, we can construct a rectangle around it such that all 3 of the vertices of the triangle lie on the rectangle’s sides. Then, as Pick’s Theorem applies to each of those 3 triangles and the rectangle, we can perform a subtraction to show that Pick’s Theorem works for the internal triangle within, i.e. for any triangle. As we know that that the area of the irregular triangle is equal to the area of the rectangle minus the areas of the three right triangles, we can perform this subtraction for each of the interior points, boundary points, and one’s at the end. 

Considering the Interior points first, we end up with

When we add up all of the interior points of the rectangle and subtract off the interior points of each of the right triangles, we are left with the interior points of P (the triangle we care about) and its boundary points, as these are not part of the interior points of the right-triangles, but they are part of the interior points of the rectangle, with the exception of the vertices which are boundary points of the rectangle, so we subtract 3. 

Considering the Boundary points now, we end up with: 

            When we subtract the boundary points of each of the triangles from the boundary points of the rectangle, we remove all of the boundary points of the rectangle, as the right-triangles’ vertical and horizontal sides are part of the rectangle’s sides, but we also remove all of the boundary points of the irregular triangle, as they consist of the points on the hypotenuse of the right triangles!

            Considering the 1s on the end: 

            When we add everything together, we get:

             which is Pick’s Theorem!

  • Show it holds for all lattice polygons 

Now, to finish off and show that Pick’s Theorem holds for all polygons, we observe that we can construct a rectangle around the whole polygon, and because it is a lattice polygon, form the area on the outside of the polygon but within the rectangle into right triangles and rectangles, and then by similar reasoning to the above, we can show that the polygon obeys Pick’s Theorem. 

                                                                                                                        Q.E.D.

Interesting Extensions:

The area of any lattice polygon is always an integer or half of an integer.  

Proof:

By Pick’s Theorem, 

As I(P) and B(P) are always integers, it follows that B(P)/2 is either an integer if B(P) is even or half an integer if B(P) is odd, hence the conclusion above. 

 No equilateral triangle can ever be drawn as a lattice polygon. 

Proof:

The area of all equilateral triangles is . This is never an integer due to the irrational number in the numerator, and by extension of the claim that the area of any lattice polygon is always an integer or half of an integer, no equilateral triangle can ever be drawn as a lattice polygon as if it could be, then it would violate the proof above. 

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