IMO 2015 #1: "Balanced" and "Centre-Free" sets of points in the plane International Mathematical Olympiad 2015, Problem 1:

We say that a finite set $S$ of points in the plane is $\color{\red}{\text{balanced}}$ if, for any two different points $A$ and $B$ in $S$, there is a points $C$ in $S$ such that $AC = BC$. We say that $S$ is $\color{\red}{\text{centre-free}}$ if for any three different points $A$, $B$, and $C$ in $S$, there is no point $P$ in $S$ such that $PA = PB = PC$.
  
  
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*Show that for all integers $n\geq 3$, there exists a balanced set consisting of $n$ points
  
*Determine all integers $n\geq 3$ for which there exists a balanced $\color{\red}{\text{centre-free}}$ set consisting of $n$ points
  

 A: First, for people interested to know where this question come from : 
it was the first question of the International Mathematical Olympiad 2015 

Now a solution for the first part :
First notice that if $n$ is odd, the regular n-gon is a balanced centre free set.
Now for $2n$ points, take :


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*the point $O$ of affix $0$

*$0 = \theta_1 < \cdots < \theta_{n-1} = \frac{\pi}{2}$ 

*the points $z_k = e^{i \theta_k} $ and $z_k' = e^{i (\theta_k + \frac{\pi}{3} )} $

*the point $z_0 = e^{-i \frac{\pi}{3} }$ 


For each $0< k < n$, the points $O$, $z_k$ and $z_k'$ create an equilateral triangle, and $O$, $z_0$, $z_1$ create an equilateral triangle too : the set is balanced, and has $2n$ points
For second part :
Let's show that the only $n$ possible are odd. 
First, notice that if a set is balanced centre free, each circle centred on a point and passing by another point must contain exactly one third point.
This means, that, taking a point in the set, you can group all the others points by group of two : the total number of points is odd
