I've read that if the complex numbers $a_1$, $a_2$ and $a_3$ are the vertices of a triangle in the complex plane such that
$$
a_1^2+a_2^2+a_3^2=a_1a_2+a_2a_3+a_1a_3
$$
then the vertices are actually those of an equilateral triangle.
I tried to see why this by first shifting the vertex $a_1$ to the origin, and considering the triangle with vertices $0$, $a_2-a_1$ and $a_3-a_1$. The given equation holds if and only if
$$
(a_2-a_1)^2+(a_3-a_1)^2=(a_2-a_1)(a_3-a_1)
$$
which is equivalent to saying
$$
\frac{a_2-a_1}{a_3-a_1}+\frac{a_3-a_1}{a_2-a_1}=1.
$$
Letting $\frac{a_2-a_1}{a_3-a_1}=x$, this gives the quadratic $x+\frac{1}{x}=1$, and so I solve for $x$ to be
$$
\frac{a_2-a_1}{a_3-a_1}=x=\frac{1\pm\sqrt{-3}}{2}=e^{\pm i\pi/3}.$$
I'm not sure where I'm going now. How can I conclude that such a relation implies the vertices form an equilateral triangle? Thank you.
Best Answer
The sides $(a_1,a_2)$ and $(a_1,a_3)$ have the same length (because $|x|=1$) and their nonoriented angle is $\pi/3$ (because $\arg(x)=\pm\pi/3$). Since the triangle $a_1a_2a_3$ is isosceles at vertex $a_1$, its angles at vertices $a_2$ and $a_3$ are equal. The angle at $a_1$ is $\pi/3$ and the sum of the three angles is $\pi$. Hence the angle at each vertex is $\pi/3$. QED.