IMO 1999 · Problem 1

Determine all finite sets \(S\) of at least three points in the plane which satisfy the following condition:

for any two distinct points \(A\) and \(B\) in \(S\), the perpendicular bisector of the line segment \(AB\) is an axis of symmetry for \(S\).

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

Answer

Precisely the vertex sets of regular \(n\)-gons, for every integer \(n\ge 3\).

Solution

Let \(S=\{P_1,\ldots,P_n\}\), where \(n\ge 3\), and let \(O\) be its centroid: \[ O=\frac1n\sum_{k=1}^{n}P_k. \] Every reflection preserving \(S\) fixes \(O\). Indeed, a reflection \(\sigma\) is an affine map, and it permutes the points of \(S\), so \[ \sigma(O)=\frac1n\sum_{k=1}^{n}\sigma(P_k) =\frac1n\sum_{k=1}^{n}P_k=O. \]

Thus, for any distinct \(A,B\in S\), the point \(O\) lies on the perpendicular bisector of \(AB\), giving \(OA=OB\). Consequently, all points of \(S\) lie on a circle \(\Gamma\) with centre \(O\). Its radius is positive because the points are distinct.

Label the points \(P_1,\ldots,P_n\) in counterclockwise order around \(\Gamma\), taking indices modulo \(n\). Fix an index \(i\), and consider the arc from \(P_i\) to \(P_{i+2}\) that passes through \(P_{i+1}\) in counterclockwise order. The only point of \(S\) in the interior of this arc is \(P_{i+1}\).

Let \(\sigma\) be reflection in the perpendicular bisector of \(P_iP_{i+2}\). By hypothesis, \(\sigma\) preserves \(S\). It also preserves \(\Gamma\), because its axis passes through \(O\). Moreover, it preserves each of the two arcs joining \(P_i\) and \(P_{i+2}\): its axis is perpendicular to the line \(P_iP_{i+2}\), so reflection preserves each half-plane bounded by that line.

Therefore \(\sigma\) maps the interior of our chosen arc to itself. Since \(P_{i+1}\) is its only point belonging to \(S\), we have \[ \sigma(P_{i+1})=P_{i+1}. \] As \(\sigma\) exchanges \(P_i\) and \(P_{i+2}\), it exchanges the two subarcs from \(P_i\) to \(P_{i+1}\) and from \(P_{i+1}\) to \(P_{i+2}\). These subarcs consequently have equal lengths.

This holds for every \(i\), so all \(n\) consecutive arcs between points of \(S\) have equal lengths. Their central angles are therefore all \(2\pi/n\), proving that \(S\) is the vertex set of a regular \(n\)-gon.

Conversely, suppose \(S\) is the vertex set of a regular \(n\)-gon. Write its vertices as \[ P_k=O+R\left( \cos\left(\alpha+\frac{2\pi k}{n}\right), \sin\left(\alpha+\frac{2\pi k}{n}\right) \right), \qquad k=0,\ldots,n-1. \] For distinct \(i,j\), the perpendicular bisector of \(P_iP_j\) is the line through \(O\) with direction angle \[ \alpha+\frac{\pi(i+j)}{n}\pmod{\pi}. \] Reflection in this line sends the polar angle of \(P_k\) to \[ 2\left(\alpha+\frac{\pi(i+j)}n\right) -\left(\alpha+\frac{2\pi k}n\right) =\alpha+\frac{2\pi(i+j-k)}n. \] It therefore sends \(P_k\) to \(P_{i+j-k}\), with indices taken modulo \(n\), and preserves \(S\). This also covers antipodal pairs when \(n\) is even. Hence every stated set satisfies the required condition.

The solution was checked against Evan Chen's published IMO solution notes.