Neighboring Triples of Complex Numbers

From above condition, The quadruple $(z_1,z_2,z_3,x)$ and triple $(w,w,w)$ are neighbor. so $\frac{w-z_1}{z_2-z_1}, \frac{w-z_2}{z_3-z_2}, \frac{w-z_3}{z_1-z_3}$ ,

$i\frac{w-x}{z_2-z_1} \ldots\ldots ⓐ,$

$i\frac{w-x}{z_3-z_2} \ldots\ldots ⓑ,$

$i\frac{w-x}{z_1-z_3}$ $\in \Re$

$\frac{1}{ⓐ} / (\frac{1}{ⓐ} + \frac{1}{ⓑ})$ = $\frac{z_2-z_1}{z_3-z_1}$ $\in \Re$ so $\Im(\frac{z_2-z_1}{z_3-z_1})$ = 0

I posed this problem for the Proofathon Complex Numbers / Trigonometry contest. Unfortunately it didn't make it to the final 8.

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Consider any quadruple $(z_1, z_2, z_3, z_4)$ whose neighbor is $(\omega_1, \omega_2, \omega_3)$ . Let $A, B, C, D, P, Q, R$ denote the images of $z_1, z_2, z_3, z_4, \omega_1, \omega_2, \omega_3$ in the complex plane respectively. We shall show that $P, Q, R$ are the feet of perpendiculars from $D$ to $AB, BC, CA$ respectively.

First, note that if $z$ is any point whose image $Z$ lies on line $AB$ , the vector joining $Z$ and $B$ must be a multiple of the vector joining $Z$ and $A$ , so $\dfrac{\omega_1 - z_1}{z_2-z_1} \in \mathbb{R}.$ Now, if $z$ is any point whose image $Z$ lies on the perpendicular from $D$ to $AB$ , the image of $e^{i \pi / 2} z = iz$ must lie on $AB$ , and similarly we find out that $i \dfrac{\omega_1 - z_4}{z_2-z_1} \in \mathbb{R}$ . Since $\omega_1$ satisfies all these conditions, we conclude that $P$ is the foot of the perpendicular from $D$ to $AB$ . Similarly, we find out that $Q, R$ are the feet of perpendiculars from $D$ to $BC$ and $D$ to $CA$ respectively.

In case of a popular triple, we have that $\omega_1 = \omega_2 = \omega_3$ , so the feet of perpendiculars from $D$ to $AB, BC, CA$ coincide. In case of an attractive triple, this happens regardless of the position of $D$ , which is possible if and only if $A, B, C$ are collinear. This implies $\dfrac{z_2-z_1}{z_3-z_2} \in \mathbb{R},$ so $\Im \left( \dfrac{z_2-z_1}{z_3-z_2} \right) = \boxed{0}.$

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My problem had a second part: when is $\dfrac{\omega_3-\omega_2}{\omega_3-\omega_1} \in \mathbb{R}?$ This is left to the reader. Hint-- Simson lines.

Assume: $z_i=z_i.a+iz_i.b => I(\frac{w-z_1}{z_2-z_1})=0<=>(w.b-z_1.b)(z_2.a-z_1.a)=(w.a-z_1.a)(z_2.b-z_1.b)$ $I(\frac{w-z_2}{z_3-z_2})=0<=>(w.b-z_2.b)(z_3.a-z_2.a)=(w.a-z_2.a)(z_3.b-z_2.b)$ $I(\frac{w-z_3}{z_1-z_3})=0<=>(w.b-z_3.b)(z_1.a-z_3.a)=(w.a-z_3.a)(z_1.b-z_3.b)$ $I(i\frac{w-z_4}{z_2-z_1})=0<=>(w.a-z_4.a)(z_2.a-z_1.a)=-(w.b-z_4.b)(z_2.b-z_1.b)$ $I(i\frac{w-z_4}{z_3-z_2})=0<=>(w.a-z_4.a)(z_3.a-z_2.a)=-(w.b-z_4.b)(z_3.b-z_2.b)$ $I(i\frac{w-z_4}{z_1-z_3})=0<=>(w.a-z_4.a)(z_1.a-z_3.a)=-(w.b-z_4.b)(z_1.b-z_3.b)$ $=>\frac{z_2.a-z_1.a}{z_2.b-z_1.b}=\frac{z_3.a-z_2.a}{z_3.b-z_2.b}=\frac{z_1.a-z_3.a}{z_1.b-z_3.b}$ $=>I(\frac{z_2-z_1}{z_3-z_1})=\frac{(z_2.b-z_1.b)(z_3.a-z_1.a)-(z_2.a-z_1.a)(z_3.b-z_1.b)}{(z_3.a-z_1.a)^2+(z_3.b-z_1.b)^2}=\boxed{0}.$