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.
Proof by Contradiction
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.
CONTRADICTION
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
kmatematikatolyesi.com 