Problem 1957

Part (a).
Assume that $n$ is odd. Consider a regular $n$-gon. Label the vertices of the $n$-gon as $A_{1}, A_{2}, \ldots, A_{n}$ in counter-clockwise order, and set $\mathcal{V} = \{A_{1}, \ldots, A_{n}\}$. We check that $\mathcal{V}$ is balanced. For any two distinct vertices $A_{i}$ and $A_{j}$, let $k \in \{1,2, \ldots, n\}$ be the solution of $2k \equiv i + j \pmod{n}$. Then, since $k - i \equiv j - k \pmod{n}$, we have $A_{i}A_{k} = A_{j}A_{k}$, as required.

Now assume that $n$ is even. Consider a regular $(3n - 6)$-gon, and let $O$ be its circumcenter. Again, label its vertices as $A_{1}, \ldots, A_{3n-6}$ in counter-clockwise order, and choose $\mathcal{V} = \{O, A_{1}, A_{2}, \ldots, A_{n-1}\}$. We check that $\mathcal{V}$ is balanced. For any two distinct vertices $A_{i}$ and $A_{j}$, we always have $OA_{i} = OA_{j}$. We now consider the vertices $O$ and $A_{i}$. First note that the triangle $OA_{i}A_{n/2-1+i}$ is equilateral for all $i \leqslant \frac{n}{2}$. Hence, if $i \leqslant \frac{n}{2}$, then we have $OA_{n/2-1+i} = A_{i}A_{n/2-1+i}$; otherwise, if $i > \frac{n}{2}$, then we have $OA_{i-n/2+1} = A_{i}A_{i-n/2+1}$. This completes the proof.

Part (b).
We now show that there exists a balanced, center-free set containing $n$ points for all odd $n \geqslant 3$, and that one does not exist for any even $n \geqslant 3$.
If $n$ is odd, then let $\mathcal{V}$ be the set of vertices of a regular $n$-gon. We have shown in part (a) that $\mathcal{V}$ is balanced. We claim that $\mathcal{V}$ is also center-free. Indeed, if $P$ is a point such that $PA = PB = PC$ for some three distinct vertices $A, B$ and $C$, then $P$ is the circumcenter of the $n$-gon, which is not contained in $\mathcal{V}$.

Now suppose that $\mathcal{V}$ is a balanced, center-free set of even cardinality $n$. We will derive a contradiction. For a pair of distinct points $A, B \in \mathcal{V}$, we say that a point $C \in \mathcal{V}$ is associated with the pair $\{A, B\}$ if $AC = BC$. Since there are $\frac{n(n-1)}{2}$ pairs of points, there exists a point $P \in \mathcal{V}$ which is associated with at least $\left\lceil \frac{n(n-1)}{2} / n \right\rceil = \frac{n}{2}$ pairs. Note that none of these $\frac{n}{2}$ pairs can contain $P$, so that the union of these $\frac{n}{2}$ pairs consists of at most $n-1$ points. Hence there exist two such pairs that share a point. Let these two pairs be $\{A, B\}$ and $\{A, C\}$. Then $PA = PB = PC$, which is a contradiction.