Problem: We say that a finite set $\mathcal{S}$ of points in the plane
is \emph{balanced} if,
for any two different points $A$ and $B$ in $\mathcal{S}$,
there is a point $C$ in $\mathcal{S}$ such that $AC=BC$.
We say that $\mathcal{S}$ is \emph{center-free} if for
any three different points $A$, $B$ and $C$ in $\mathcal{S}$,
there are no points $P$ in $\mathcal{S}$ such that $PA=PB=PC$.

\begin{enumerate}
\item[(a)] Show that for all integers $n\ge 3$,
  there exists a balanced set consisting of $n$ points.
\item[(b)] Determine all integers $n\ge 3$ for which
  there exists a balanced center-free set consisting of $n$ points.
\end{enumerate}

Solution: For part (a), take a circle centered at a point $O$,
and add $n-1$ additional points by adding pairs of points
separated by an arc of $60^{\circ}$ or similar triples.
An example for $n = 6$ is shown below.
\begin{center}
\begin{asy}
  size(4cm);
  draw(unitcircle);
  dotfactor *= 1.5;
  pair O = (0,0);
  dot("$O$", O, dir(-90), blue);
  pair A = dir(37);
  pair B = A*dir(-60);
  pair C = B*dir(-60);
  pair X = dir(117);
  pair Y = X*dir(60);
  draw(O--X--Y--cycle, deepgreen+dashed);
  dot(X, deepgreen);
  dot(Y, deepgreen);
  draw(B--O--A--B--C--O, red+dashed);
  dot(A, red);
  dot(B, red);
  dot(C, red);
\end{asy}
\end{center}

For part (b), the answer is odd $n$, achieved by taking a regular $n$-gon.
To show even $n$ fail, note that some point is on the perpendicular bisector of
\[ \left\lceil \frac 1n \binom n2 \right\rceil = \frac{n}{2} \]
pairs of points, which is enough.
(This is a standard double-counting argument.)

\begin{remark*}
  As an aside, there is a funny joke about this problem.
  There are two types of people in the world:
  \begin{itemize}
    \ii Those who solve (b) quickly and then take forever to solve (a),
    \ii those who solve (a) quickly and then can't solve (b) at all.
  \end{itemize}
  (Empirically true when the Taiwan IMO 2014 team was working on it.)
\end{remark*}