Equilateral Triangle Property

Question

If the vertices of a triangle have integral coordinates how to prove that the triangle cannot be equilateral ?

Answer 1

Let's assume that the coordinate system is two-dimensional and orthonormal, and call the triangle $ABC$ and assume we have such coordinates. This will lead to a contradiction (of $\sqrt3$ being a rational number):

Extend the one of the sides $AC$ to the double length $AD$ forming the right triangle $ABD$, this will mean that $D$ has integral coordinates too. By pythagoras theorem the side $BD$ will be of $\sqrt3$ times as long as $AB$.

Now let's consider the vectors $u=(x_1, y_1)$ corersponding to $AB$ and $v=(x_2, y_2)$ corresponding to $BD$. Since we assumed integer cordinates then $x_1$, $x_2$, $y_1$, $y_2$ are all integers. Rotate $v$ by $90^\circ$ and get $w=(y_2, -x_2)$ (here I use the assumption on orthonormal coordinate system) and get a vector being parallell to $u$ and also $\sqrt3$ times as long, that is $w = \pm\sqrt3v$ so $y_2 = \pm\sqrt3x_1$ and $x_2 = \pm\sqrt3y_1$. Since $x_1$ or $y_1$ is non-zero it would mean that a non-zero integer times $\sqrt3$ is again an integer that is $\sqrt3$ is rational.

The assumption of orthonormal two-dimensional coordinate system is important as there are example if you drop one of them: having a coordinate system where the angle between the axes is $60^\circ$ it would be trivial in a otherwise normal coordinate system or a orthogonal coordinate system with different scales (eg differing a factor $\sqrt3$. The example with a three-dimensional orthonormal system would be $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ forming an equilateral triangle.

Answer 2

From the coordinates of the vertices we can compute the slopes of the sides. It follows that $\tan 60^\circ=\sqrt{3}$, being a rational expression in these slopes, would be rational.