# What is the side length of the smallest sized equilateral triangle that can be placed on the (x,y) plane where all coordinates are integers AND no side is horizontal and no side is vertical?

Jan 5, 2016

There is no such triangle, due to $\sqrt{3}$ being irrational.

#### Explanation:

Without loss of generality, one of the vertices is at $\left(0 , 0\right)$.

Let the vertex anticlockwise from $\left(0 , 0\right)$ be $\left(m , n\right)$ for some integers $m$ and $n$.

The midpoint of the corresponding side is $\left(\frac{m}{2} , \frac{n}{2}\right)$

The line through $\left(0 , 0\right)$ and $\left(m , n\right)$ has slope $\frac{n}{m}$, so any perpendicular to this side has slope $- \frac{m}{n}$.

The third vertex of the triangle lies on the line through the midpoint $\left(\frac{m}{2} , \frac{n}{2}\right)$ with slope $\left(- \frac{m}{n}\right)$.

In fact it will lie at the point:

$\left(\frac{m}{2} , \frac{n}{2}\right) + \frac{\sqrt{3}}{2} \left(- n , m\right) = \left(\frac{m - \sqrt{3} n}{2} , \frac{n + \sqrt{3} m}{2}\right)$

since the height of an equilateral triangle is $\sin {60}^{\circ} = \frac{\sqrt{3}}{2}$ times the length of one of the sides.

So we require $\frac{m - \sqrt{3} n}{2}$ to be an integer and $\frac{n + \sqrt{3} m}{2}$ to be an integer.

Since $\sqrt{3}$ is irrational, this would imply that $n = 0$ and $m = 0$, hence there is no such triangle.