Prove that $\Delta ABC$ is an equilateral triangle if and only if the complex coordinates $a$,$b$, and $c$ satisfy a relation.

Question

Let $\triangle ABC$ be a triangle in the complex plane and let $a$,$b$ and $c$, respectively, be the complex coordinates of its vertices. Suppose that the tiangle is inscribed in the circle $C(0,1)$. Prove that $\triangle ABC$ is equilateral if and only if $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=0$$ and $$(a+b)(b+c)(c+a)\neq0\,.$$

I tried to convert $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=0$ into a simpler equation by multipling it by $(a+b)(b+c)(c+a)$

In the end I obtained $$a^2+b^2+c^2+3ab+3bc+3ca=0$$ but I don't know how to continue from here. Can you help me, please?

Answer 1

One direction (the direction assuming that $ABC$ is an equilateral triangle) is obvious. Here's a geometric proof of the converse.

Let $k$ be the given unit circle. If $D$, $E$, and $F$ are the midpoints of $BC$, $CA$, and $AB$, then note that the complex coordinates of the images $D'$, $E'$, and $F'$ of $D$, $E$, and $F$ under the inversion about $k$ are $\dfrac{2}{\bar{b}+\bar{c}}$, $\dfrac{2}{\bar{c}+\bar{a}}$, and $\dfrac{2}{\bar{a}+\bar{b}}$, respectively. Note that the origin $O$ is the incenter of the triangle $D'E'F'$. Your equation says that the centroid of the triangle $D'E'F'$ coincides with its incenter. This only happens when $D'E'F'$ is an equilateral triangle, whence $ABC$ is also an equilateral triangle.

Answer 2

Given equilateral triangle, i.e.

$$a= e^{i\alpha}, \>\>\>\>\>b = e^{i(\alpha+\frac{2\pi} 3)}, \>\>\>\>\>c= e^{i(\alpha-\frac{2\pi} 3)} $$

it is straightforward to verify $a^2+b^2+c^2+3ab+3bc+3ca=0$.

Conversely, given

$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=0$$

rewrite the equation in term of $b =a e^{i x}$, $c =a e^{ i y} $

$$\frac{1}{1+ e^{i x}}+\frac{1}{1+ e^{i y}}+\frac{1}{e^{i x} + e^{i y}}=0\tag1$$

Suppose $(x,y)$ is a solution, then $(y,x)$ is also a solution due to party, as well as $(-x,-y)$ via conjugate. which implies $x = -y$. Substitute into (1) to obtain $\cos x=-\frac12$, or $x= -y= \pm \frac {2\pi}3$. Thus

$$b = a e^{\pm i\frac{2\pi}3}, \>\>\>\>\>c=a e^{\mp i \frac{2\pi} 3} $$

hence, equilateral triangles.

Answer 3

During all the proof I will call $a=z_0$, $b=z_1$ and $c=z_2$.

$\Longrightarrow)$

If $z_0$, $z_1$ and $z_2$form a equilateral triangle inscribed in $C(0,1)$, then we can write $z_j$ as $$z_j=m\cdot e^{i\frac{\pi j}{3}};\quad \text{with } |m|=1.$$ Then, since $z_j$ are rotations of the cubic roots of $1$, then $$\sum\limits_{j=0}^2 m\cdot e^{i\frac{\pi j}{3}}=m\cdot \underbrace{\sum\limits_{j=0}^2 e^{i\frac{\pi j}{3}}}_{=0}=0.$$

The other condition is easy to proof since none of the $z_j$ are diametrically opposite (ie, an equilateral triangle inscribed in a circumference can't have two vertices diametrically opposite).

$\Longleftarrow)$

The proof of this statement can be found (with this same notation) here.