Equation of a Equilateral Triangle in $\mathbb{R}^{2}$

Question

I need to find a metric where a Sphere of radius $r$ is a equilateral triangle, but I think it is necessary to find a way to describe a equilateral triangle in $\mathbb{R}^{2}$, just like we do with the ellipse and hyperbole.

Answer 1

Let us assume that $ABC$ has direct orientation.

Let us make a detour by complex considerations.

Let $\omega=\tfrac12(-1+i \sqrt{3})$ (a cubic root of unity). [please note that $\overline{\omega}=\omega^2$].

A classical necessary and sufficient condition for a triangle with direct orientation with vertices $a,b,c \in \mathbb{C}$ to be equilateral is

$$a+b\omega+c \overline{\omega}=0\tag{1}$$

(see here).

Now convert (1), after having multiplied it by 2, under a real form:

$$2(x_A+iy_A)+(x_B+iy_B)(-1+i \sqrt{3})+(x_C+iy_C)(-1-i \sqrt{3})=0\tag{2}$$

We can stay there or expand (2) and equate separately its real and imaginary part to $0$, giving two simultaneous (necessary and sufficient) conditions for being an equilateral triangle:

$$\begin{cases}2x_A-x_B-\sqrt{3}y_B-x_C+\sqrt{3}y_C&=&0\\ 2y_A-y_B+\sqrt{3}x_B-y_C-\sqrt{3}x_C&=&0 \end{cases}\tag{3}$$