IMO 1986 · Problem 2

A triangle \(A_1A_2A_3\) and a point \(P_0\) are given in the plane. We define \(A_s=A_{s-3}\) for all \(s\geq 4\). We construct a set of points \(P_1,P_2,P_3,\ldots\), such that \(P_{k+1}\) is the image of \(P_k\) under a rotation with center \(A_{k+1}\) through angle \(120^\circ\) clockwise (for \(k=0,1,2,\ldots\)). Prove that if \(P_{1986}=P_0\), then the triangle \(A_1A_2A_3\) is equilateral.

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

Solution

Identify the plane with the complex plane, and denote the complex coordinates of \(A_j\) and \(P_k\) by \(a_j\) and \(p_k\), respectively. Put \[ \omega=-\frac12-\frac{\sqrt3}{2}i. \] Multiplication by \(\omega\) represents a clockwise rotation through \(120^\circ\), and \[ \omega^3=1,\qquad 1+\omega+\omega^2=0,\qquad |\omega|=1. \]

A rotation with center \(a\) therefore sends \(z\) to \[ a+\omega(z-a)=\omega z+(1-\omega)a. \] Consequently, the composition of the three successive rotations with centers \(a_1,a_2,a_3\) sends \(z\) to \[ \begin{aligned} \omega\bigl(\omega(\omega z+(1-\omega)a_1) +(1-\omega)a_2\bigr)+(1-\omega)a_3 &=\omega^3z+(1-\omega)(\omega^2a_1+\omega a_2+a_3)\\ &=z+t, \end{aligned} \] where \[ t=(1-\omega)(\omega^2a_1+\omega a_2+a_3). \]

The centers repeat every three steps, so \[ p_{3m+3}=p_{3m}+t \] for every integer \(m\geq0\). Induction gives \(p_{3m}=p_0+mt\). Since \(1986=3\cdot662\), the hypothesis implies \[ 0=p_{1986}-p_0=662t, \] hence \(t=0\). As \(1-\omega\neq0\), it follows that \[ \omega^2a_1+\omega a_2+a_3=0. \]

Using \(1+\omega+\omega^2=0\), we obtain \[ a_3-a_1=-\omega(a_2-a_1), \qquad a_3-a_2=\omega^2(a_2-a_1). \] Taking absolute values yields \[ |a_3-a_1|=|a_2-a_1|=|a_3-a_2|. \] Thus all three sides of \(A_1A_2A_3\) are equal, so the triangle is equilateral.

All rotation formulas remain valid when a point coincides with a rotation center; no distinctness assumption on the points \(P_k\) was needed. \(\square\)

The solution was checked against John Scholes' IMO solution archive (Kalva).