IMO 2015 · Problem 1

We say that a finite set \(S\) of points in the plane is balanced if, for any two different points \(A\) and \(B\) in \(S\), there is a point \(C\) in \(S\) such that \(AC = BC\). We say that \(S\) is centre-free if for any three different points \(A\), \(B\) and \(C\) in \(S\), there is no point \(P\) in \(S\) such that \(PA = PB = PC\).

(a) Show that for all integers \(n \geq 3\), there exists a balanced set consisting of \(n\) points.

(b) Determine all integers \(n \geq 3\) for which there exists a balanced centre-free set consisting of \(n\) points.

Official text of the problem. Official problem paper (PDF)

Answer

(a) Such a set exists for every integer \(n\geq 3\).

(b) Exactly the odd integers \(n\geq 3\).

Solution

Construction for odd \(n\), satisfying both conditions.

Let \(n\geq 3\) be odd, and let \(S\) be the vertices of a regular \(n\)-gon inscribed in a circle \(\omega\) with centre \(O\).

Take any two different vertices \(A,B\). The two arcs of \(\omega\) joining them contain, between them, \(n\) polygon sides. Since \(n\) is odd, exactly one arc contains an even number of sides. Its midpoint is therefore a vertex \(C\), different from \(A\) and \(B\). The arcs from \(C\) to \(A\) and \(B\) along this arc have equal lengths, so their chords satisfy \(AC=BC\). Thus \(S\) is balanced.

To show that \(S\) is centre-free, suppose that a vertex \(P\) were equidistant from three different vertices \(A,B,C\). Their common distance from \(P\) would be positive. The circle with centre \(P\) and this radius would contain \(A,B,C\), as does \(\omega\). These circles are distinct because their centres are \(P\) and \(O\), with \(P\neq O\). This contradicts the fact that two distinct circles have at most two common points.

Construction for even \(n\), proving the remaining cases of (a).

Write \(n=2m\), where \(m\geq 2\). Let \(\omega\) be a circle of radius \(1\) with centre \(O\). Choose three consecutive vertices \(A,B,C\) of a regular hexagon inscribed in \(\omega\). Then \[ AB=BC=1. \]

Choose \(m-2\) additional pairs of points \(X_i,Y_i\) on \(\omega\), with \[ X_iY_i=1, \] so that all chosen points are distinct. Such choices are always possible: rotate a unit chord around \(O\); only finitely many rotations make either endpoint coincide with a previously chosen point. When \(m=2\), no additional pairs are needed.

Set \[ S=\{O,A,B,C\}\cup\bigcup_{i=1}^{m-2}\{X_i,Y_i\}. \] This set has \(4+2(m-2)=n\) points.

Every point \(Q\in S\setminus\{O\}\) has another point \(R\in S\setminus\{O\}\) with \(QR=1\): use \(B\) for \(A\) or \(C\), use \(A\) for \(B\), and use the paired point for each \(X_i\) or \(Y_i\).

Now consider any two different points of \(S\). If both lie on \(\omega\), then \(O\) is equidistant from them. Otherwise, they are \(O\) and some \(Q\in\omega\). Choose its companion \(R\) as above; then \[ RO=1=RQ. \] Thus \(S\) is balanced, completing part (a).

Impossibility of even \(n\) in (b).

Suppose that \(S\) is balanced and centre-free, with even cardinality \(n\).

Fix \(P\in S\), and group the other \(n-1\) points according to their distances from \(P\). Each group has at most two points, since a group of three would violate the centre-free condition. Hence \(P\) can witness the equality \(PA=PB\) for at most \[ \left\lfloor\frac{n-1}{2}\right\rfloor=\frac{n-2}{2} \] unordered pairs \(\{A,B\}\). Here a witnessing point cannot equal either endpoint, because \(A\neq B\).

Count incidences consisting of an unordered pair and a point witnessing equal distances to its endpoints. Summing over all \(P\in S\), there are at most \[ \frac{n(n-2)}2 \] incidences. But balancedness requires at least one for every unordered pair, giving at least \[ \binom n2=\frac{n(n-1)}2 \] incidences. These bounds contradict each other.

Therefore \(n\) must be odd. The regular polygons constructed above establish existence for every odd \(n\geq 3\), proving (b).

The solution was checked against the official IMO shortlist solutions.