Problem source

The distinct points $L,M,P$ and $Q$ of the Argand diagram lie on a circle $S$ centred on the origin and the corresponding complex numbers are $l,m,p$ and $q$. By considering the perpendicular bisectors of the chords, or otherwise, prove that the chord $LM$ is perpendicular to the chord $PQ$ if and only if $lm+pq=0.$ 
	Let $A_{1},A_{2}$ and $A_{3}$ be three distinct points on $S$.	For any given point $A_{1}'$ on $S$, the points $A_{2}',A_{3}'$ and $A_{1}''$ are chosen on $S$ such that $A_{1}'A_{2}',A_{2}'A_{3}'$ and $A_{3}'A_{1}''$ are perpendicular to $A_{1}A_{2},A_{2}A_{3}$ and $A_{3}A_{1},$ respectively. Show that for exactly two positions of $A_{1}',$ the points $A_{1}'$ and $A_{1}''$ coincide. 
	If, instead, $A_{1},A_{2},A_{3}$ and $A_{4}$ are four given distinct points on $S$ and, for any given point $A_{1}',$ the points $A_{2}',A_{3}',A_{4}'$ and $A_{1}''$ are chosen on $S$ such that $A_{1}'A_{2}',A_{2}'A_{3}',A_{3}'A_{4}'$ and $A_{4}'A_{1}''$ are respectively perpendicular to $A_{1}A_{2},A_{2}A_{3},A_{3}A_{4}$ and $A_{4}A_{1},$ show that $A_{1}'$ coincides with $A_{1}''.$ 
	Give the corresponding result for $n$ distinct points on $S$.

Solution source

The perpendicular bisector of the chords runs through the origin, therefore $LM$ is perpendicular to $PQ$ if and only if $\frac{l+m}{2}$ is perpendicular to $\frac{p+q}{2}$, ie 

\begin{align*}
&& (l+m) &= it (p+q) \\
\Leftrightarrow && \frac{l+m}{p+q} & \in i \mathbb{R} \\
\Leftrightarrow && 0 &= \frac{l+m}{p+q} + \frac{l^*+m^*}{p^*+q^*} \\
&&&= \frac{l+m}{p+q} + \frac{\frac{r^2}{l}+\frac{r^2}{m}}{\frac{r^2}{p}+\frac{r^2}{q}} \\
&&&=\frac{l+m}{p+q} + \frac{l+m}{p+q} \frac{pq}{lm} \\
&&&= \frac{l+m}{p+q} \left ( \frac{lm+pq}{lm} \right)
\end{align*}

Therefore as long as $l+m, p+q \neq 0$ $lm+pq = 0$ is equivalent to the chords being perpendicular. In the case where (say) $l,m$ is a diameter, then the condition for the chords to be perpendicular is that $p,q$ is also a diameter and at right angles, but clearly this is also equivalent to our condition.


Suppose $A_1, A_2, A_3$ are distinct points on $S$, and $A_1'$ is given and suppose $a_i, a_i'$ are the corresponding complex numbers, then the conditions are:

\begin{align*}
A_1'A_2' \perp A_1A_2: && 0 &= a_1'a_2' + a_1a_2  \\
A_2'A_3' \perp A_2A_3: && 0 &= a_2'a_3' + a_2a_3  \\
A_3'A_1'' \perp A_3A_1: && 0 &= a_3'a_1'' + a_3a_1 \\
\\
\Rightarrow && a_2' &= -\frac{a_1a_2}{a_1'} \\
&& a_3' &= -\frac{a_2a_3}{a_2'} \\
&&&= \frac{a_1'a_2a_3}{a_1a_2} \\
&&&= \frac{a_1'a_3}{a_1} \\
&& a_1'' &= - \frac{a_3a_1}{a_3'} \\
&&&= \frac{a_3a_1a_1}{a_1'a_3} \\
&&&= \frac{a_1^2}{a_1'} \\
\Rightarrow && a_1'a_1'' &= a_1^2
\end{align*}

Therefore $a_1' = a_1''$ if $a_1' = \pm a_1$

Suppose we have $4$ points, then

\begin{align*}
A_1'A_2' \perp A_1A_2: && 0 &= a_1'a_2' + a_1a_2 \\
A_2'A_3' \perp A_2A_3: && 0 &= a_2'a_3' + a_2a_3 \\
A_3'A_4' \perp A_3A_4: && 0 &= a_3'a_4' + a_3a_4 \\
A_4'A_1'' \perp A_4A_1: && 0 &= a_4'a_1'' + a_4a_1 \\
\\
\Rightarrow && a_4' &= -\frac{a_3a_4}{a_3'} \\
&&&= -\frac{a_1a_3a_4}{a_1'a_3} \\
&&&= -\frac{a_1a_4}{a_1'} \\
\Rightarrow && a_1'' &= -\frac{a_4a_1}{a_4'} \\
&&&= \frac{a_4a_1a_1'}{a_1a_4} \\
&&&= a_1'
\end{align*}

So they coincide.

For $n$ points if there are an even number of points they coincide, an odd number and there are two points when they coincide.