MHB Triangle with Complex Number Vertices SATISFIES Equilateral Relation

[anemone]

Here is this week's POTW:

-----

If the vertices of a triangle are represented by the complex numbers $a,\,b,\,c$ and these numbers satisfy $\dfrac{a-b}{c-b}+\dfrac{c-a}{b-a}=2\left(\dfrac{b-c}{a-c}\right)$, prove that the triangle is equilateral.

-----

 

[anemone]

Congratulations to Opalg for his correct solution(Cool), which you can find below:

Spoiler

Let the sides of the triangle be $x = b-c$, $y = c-a$ and $z = a-b$. Then $x+y+z = 0$, and the given equation becomes $\dfrac zx + \dfrac yz = 2\dfrac xy$. Multiply by $xyz$ to get $$z^2y + y^2x = 2x^2z.$$ Now substitute $z = -(x+y)$: $$ (x+y)^2y + y^2x + 2x^2(x+y) = 0, $$ $$ y^3 + 3y^2x + 3yx^2 + 2x^3 = 0.$$ Divide by $x^3$ and let $\lambda = \dfrac yx$: $$\lambda^3 + 3\lambda^2 + 3\lambda + 2 = 0,$$ $$(\lambda+1)^3 = -1.$$ Let $\omega = e^{2\pi i/3}$. The complex cube roots of $1$ are $\omega^k \ (k = 0,1,2)$. Take the cube roots of the previous equation to see that the three values of $\lambda + 1$ are $-1$, $-\omega$ and $-\omega^2$. Therefore $\lambda = -2$, $\lambda = -1 - \omega = \omega^2$ or $\lambda = -1-\omega^2 = \omega$. If $\lambda=-2$ then $y=-2x$ and $z=x$ (because $x+y+z=0$). So $x$, $y$ and $z$ lie on a straight line and cannot be the sides of a triangle. If $\lambda=\omega$ then $y=\omega x$ and $z=\omega^2x$ (because $1 + \omega + \omega^2=0$). Similarly, if $\lambda=\omega^2$ then $y=\omega^2 x$ and $z=\omega x$. In both cases, $x$, $y$ and $z$ all have the same absolute value, and their arguments differ by $120^\circ$. So they form the sides of an equilateral triangle (with vertices at $a$, $b$ and $c$).