Source: IMO-2015-P1
Category: Existence and Construction

Problem:
We say that a finite set $\mathcal{S}$ of points in the plane is *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 *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:
**Part (a)**

We show that a balanced set exists for all integers $n \ge 3$. Let $O$ be a point and $C$ be a circle centered at $O$. We construct the set $\mathcal{S}$ by choosing $O$ and $n-1$ additional points on $C$. We can group these $n-1$ points into disjoint pairs and triples as follows:
- A *pair* consists of two points $A, B$ on $C$ separated by an arc of $60^\circ$.
- A *triple* consists of three points $A, B, C$ on $C$ such that $A$ and $B$ are separated by $60^\circ$, and $B$ and $C$ are separated by $60^\circ$ (so $A, B, C$ are consecutive vertices of a regular hexagon).

Since $n \ge 3$, we have $n-1 \ge 2$. Any integer $k \ge 2$ can be expressed as $2x + 3y$ for some non-negative integers $x, y$. Thus, we can place $x$ pairs and $y$ triples on $C$ such that all $n-1$ points are distinct.

We now verify that $\mathcal{S}$ is balanced. For any two distinct points $X, Y \in \mathcal{S} \setminus \{O\}$, they lie on $C$, so $OX = OY$. Thus, $O$ serves as the required third point. For $O$ and a point $X \in \mathcal{S} \setminus \{O\}$, we need a point $Y \in \mathcal{S}$ such that $YX = YO$. Since $Y$ must lie on $C$, $YO$ is the radius of $C$, which means $\triangle OXY$ must be equilateral, so $X$ and $Y$ must be separated by $60^\circ$. 
- If $X$ belongs to a pair $\{A, B\}$, then $A$ and $B$ are separated by $60^\circ$, so $A$ works for $O$ and $B$, and $B$ works for $O$ and $A$.
- If $X$ belongs to a triple $\{A, B, C\}$, then $B$ works for $O$ and $A$, $B$ works for $O$ and $C$, and both $A$ and $C$ work for $O$ and $B$.

Thus, the set $\mathcal{S}$ is balanced.

**Part (b)**

The answer is all odd integers $n \ge 3$.

First, we show that for any odd $n \ge 3$, the set of vertices of a regular $n$-gon is balanced and center-free.
- **Balanced:** For any two distinct vertices $A$ and $B$, their perpendicular bisector passes through the center of the $n$-gon and exactly one vertex $C$ of the $n$-gon (since $n$ is odd). Thus, $CA = CB$, so the set is balanced.
- **Center-free:** Any three distinct vertices $A, B, C$ uniquely determine a circumcircle, which is the circumcircle of the regular $n$-gon. The only point equidistant from $A, B,$ and $C$ is the center of this circumcircle. However, the center is not a vertex of the $n$-gon, so it is not in the set. Thus, the set is center-free.

Next, we use a standard double-counting argument to show that no balanced center-free set exists for even $n$. Assume for the sake of contradiction that $\mathcal{S}$ is a balanced center-free set of $n$ points, where $n$ is even.

By the balanced property, for each of the $\binom{n}{2}$ pairs of distinct points $\{A, B\}$ in $\mathcal{S}$, there exists at least one point $C \in \mathcal{S}$ such that $CA = CB$. We assign one such point $C$ to each pair. By the Pigeonhole Principle, since there are $n$ points available to be assigned, some point $P \in \mathcal{S}$ must be assigned to at least
$$ \left\lceil \frac{1}{n}\binom{n}{2} \right\rceil = \left\lceil \frac{n-1}{2} \right\rceil $$
pairs of points. Because $n$ is even, $\left\lceil \frac{n-1}{2} \right\rceil = \frac{n}{2}$.

Thus, $P$ is on the perpendicular bisector of at least $\frac{n}{2}$ pairs of points in $\mathcal{S} \setminus \{P\}$. This means $P$ is equidistant from the two points in each of these $\frac{n}{2}$ pairs. 

However, because $\mathcal{S}$ is center-free, $P$ cannot be equidistant from any three points in $\mathcal{S}$. This implies that the points in $\mathcal{S} \setminus \{P\}$ can be partitioned into equivalence classes based on their distance to $P$, and each class can contain at most $2$ points. The pairs assigned to $P$ correspond to these classes of size exactly $2$.

Since there are exactly $n-1$ points in $\mathcal{S} \setminus \{P\}$, the maximum number of disjoint pairs of points equidistant from $P$ is
$$ \left\lfloor \frac{n-1}{2} \right\rfloor = \frac{n-2}{2}. $$
This contradicts the fact that $P$ is assigned to $\frac{n}{2}$ pairs, because $\frac{n}{2} > \frac{n-2}{2}$. Therefore, no balanced center-free set exists for even $n$.

Answer: Part (a): Exists for all integers $n \ge 3$. Part (b): All odd integers $n \ge 3$.

Grading guidelines:
(Partial) 1. For (a), gave a valid construction idea on a circle with a distinguished point O (e.g. O plus points on a circle) and correctly verified the balanced condition for pairs of points lying on the circle (using O as the equidistant point). 2. For (a), correctly reduced the remaining balanced requirement to ensuring that for each point X on the circle there is another chosen point Y with ∠XOY=60° (equivalently, an equilateral triangle with OX=OY=XY). 3. For (b), proved that for odd n the vertices of a regular n-gon form a balanced set (e.g. argued that for any two vertices A,B there is a vertex C on the perpendicular bisector of AB, or an equivalent symmetry argument). 4. For (b), established a key obstruction for even n, e.g. started the assignment/double-counting argument by assigning to each unordered pair {A,B} a point C with CA=CB and deduced that some P is assigned to at least n/2 pairs when n is even. (Almost) 1. Solved (a) completely and proved existence for all odd n in (b), and for even n carried out the assignment/pigeonhole step, but did not finish the final contradiction from center-free (e.g. did not justify why P cannot be assigned to more than ⌊(n-1)/2⌋ pairs). 2. Solved (b) completely (odd n works and even n impossible), but in (a) gave the circle construction without fully justifying that any n−1≥2 can be decomposed into disjoint 2- and 3-blocks (or an equivalent bookkeeping step ensuring all points are distinct). 3. Provided the full intended strategy for both parts, but left one localized gap such as: uniqueness/existence of the vertex C on a perpendicular bisector in the odd n regular n-gon, or the claim that center-free implies no distance-to-P class has size ≥3.