56th International Mathematical Olympiad Shortlisted Problems

Part (a). Assume that $n$ is odd. Consider a regular $n$-gon. Label the vertices of the $n$-gon as $A_1,A_2,\dots ,A_n$ in counter-clockwise order, and set $\mathcal{V}=\{A_1,\dots ,A_n\}$. We check that $\mathcal{V}$ is balanced. For any two distinct vertices $A_i$ and $A_j$, let $k\in \{1,2,\dots ,n\}$ be the solution of $2k\equiv i+j(\mathrm{mod}n)$. Then, since $k-i\equiv j-k(\mathrm{mod}n)$, 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 $A_1,\dots ,A_{3n-6}$ in counter-clockwise order, and choose $\mathcal{V}=\{O,A_1,A_2,\dots ,A_{n-1}\}$. We check that $\mathcal{V}$ is balanced. For any two distinct vertices $A_i$ and $A_j$, we always have $O{A}_i=O{A}_j$. We now consider the vertices $O$ and $A_i$. First note that the triangle $O{A}_i{A}_{n/2-1+i}$ is equilateral for all $i⩽\frac{n}{2}$. Hence, if $i⩽\frac{n}{2}$, then we have $O{A}_{n/2-1+i}=A_i{A}_{n/2-1+i}$; otherwise, if $i>\frac{n}{2}$, then we have $O{A}_{i-n/2+1}=A_i{A}_{i-n/2+1}$. This completes the proof.

Part (b). We now show that there exists a balanced, center-free set containing $n$ points for all odd $n⩾3$, and that one does not exist for any even $n⩾3$. 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 $\lceil \frac{n(n-1)}{2}/n\rceil =\frac{n}{2}$ 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.