Prove that coordinates of all vertices of an equilateral triangle cant be rational.

Question

We have to find whether the all vertices can have rational coordinates or not
i have proved this question via contradiction but i want a solution not using contradiction ,Please help me find that, i dont find an idea for that.proof via contradiction(image as i dont know latex ,sorry)

Answer 1

If we can assume that if $k$ and $c$ are rational, then $k\sqrt{3}+c$ is irrational, then:

Without loss of generality, suppose one vertex is at the origin, and another vertex has coordinates $(a,b)$ where $a,b\in\mathbb{Q}$.

Then the third vertex of the equilateral triangle (going anti-clockwise, without loss of generality) has position vector $$\left(\begin{matrix}p\\q\end{matrix}\right)=\left(\begin{matrix}\cos60&-\sin60\\\sin60&\cos60\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)$$ So $$p=\frac12a-b\frac{\sqrt{3}}{2}$$ and $$q=a\frac{\sqrt{3}}{2}+\frac12b$$ neither of which are rational.

Answer 2

Without loss of generality, we may assume that the coordinates of two vertices are $(0,0)$ and $(1,0)$. The third vertex then will be $(1/2, \sqrt 3 /2)$ Consider The equilateral triangle with vertices $$ (0,0), (1,0),(1/2, \sqrt 3 /2)$$

All three sides have length $1$ and two vertices have integer coordinates while the third side does not have integer or even rational coordinates.