Real Mathematics – Killer Numbers #5

Equilateral Triangle and Irrational Number

In the previous article I asked you to prove whether it is possible or not to draw and equilateral triangle on the system of lattice points. Here is one of the possible proofs for that.

Let’s assume that we have an equilateral triangle that has sides of length 2 units:

Here, we will make a critical assumption: Corners of this triangle sits on the system of lattice points. Because of that the triangle must have an area that is rational. Why?

Because Pick’s theorem says that whenever we are inside the system of lattice points the numbers of points and area of the polygon are directly related with each other. And since we can’t count irrational number of points (eg. we can’t have √3 points, can we?!), area of the polygon must be rational too.

We already know that we can find a triangle’s area: ½(height x base). Then let me draw the height of the base and find its length using Pythagoras’ theorem:

h2 + 12 = 22

h2 = 4 – 1

h2 = 3

h = √3.

We just found the height of our equilateral triangle an irrational number. From here we will find the area

1/2(2*√3) = √3 units.

This result is a contradiction. Despite what Pick’s theorem says (polygons inside the system of lattice points must have rational areas) this result shows an irrational number. Then we can conclude that this or any equilateral triangle can never be drawn on the system of lattice points.

One wonders…

Can you find another polygon which you can’t draw on the system of lattice points?

M. Serkan Kalaycıoğlu

Real Mathematics – Killer Numbers #4

Lattice Points

Imagine a page from a squared notebook. Take all the edges of little squares and leave the vertices. If you assume that horizontally and vertically distance between two consecutive points is exactly 1 unit. I name this plane as the system of lattice points.

In the system of lattice points every point is represented by a number duo:

Pick’s Theorem: An alternative method in order to find the area of a polygon.

In the system of lattice points one can construct as many polygons as wanted with joining points together. Pick’s theorem is handy for finding the areas of these polygons.

According to the theorem only two things are needed to find the area of a given polygon: Number of points polygon has on its edges (let’s call it e) and the number of points staying inside the polygon (let’s call it i). Pick’s theorem gives the following formula for the area of a polygon sitting on the system of lattice points:

Area = i + (e/2) – 1

Example 1: Triangle.

Assume that there is a triangle as shown above. Number of points on its edges e is equal to 4 as the number of points inside the triangle i is equal to 0. Hence Pick’s theorem says the area of this triangle is:

Area = 0 + (4/2) – 1

= 0 + 2 – 1

= 1 unit.

We know from basic geometry that the area of a triangle is the half of height times base: (2*1)/2 = 1 unit.

Example 2: Square.

In the picture we can see that e=12 and i=4. Thus the area is:

4 + (12/2) – 1 = 4 + 6 – 1 = 9 units.

We know that area of a square is the square of the lenght of its edge which makes 3*3=9 units.

Square and triangle are easy examples and perhaps you are thinking that Pick’s theorem is redundant. Then let’s continue drawing a more complex polygon and find its area.

Example 3: Polygon.

Now Pick’s theorem shows its strength. Without the theorem we’d have to divide this polygon into various polygons and then calculate areas one by one. But with Pick’s theorem it is just counting points:

e=12 and i=72. Hence the area of the polygon can be found with:

Area = 72 + (12/2) – 1 = 72 + 6 – 1 = 77 units.

Equilateral Triangle

Up to this point it looks like we are dealing with geometry but the headlines said that it is an article about numbers. Let me change the course of the article with a question.

Q: Is it possible to draw an equilateral triangle on our points system while the corners of the triangle sits on the lattice points?

For instance let me draw the base of an equilateral triangle that has 2 units of length:

As seen above third corner of the triangle won’t be on a lattice point.

One wonders…

Can you prove that this is the case for each and every equilateral triangle on the system of lattice points?

To be continued…

M. Serkan Kalaycıoğlu