IMO 1986 Problem 2

Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 4m15s

Problem

Given a point $P_0$ in the plane of the triangle $A_1A_2A_3$. Define $A_s=A_{s-3}$ for all $s\ge4$. 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 center $A_{k+1}$ through an angle $120^o$ clockwise for $k=0,1,2,\ldots$. Prove that if $P_{1986}=P_0$, then the triangle $A_1A_2A_3$ is equilateral.

Exploration

The construction applies successive $120^\circ$ clockwise rotations about the vertices $A_1,A_2,A_3$ in a periodic pattern. After every three steps, the center sequence repeats, so the behavior is governed by the third iterate of a fixed affine transformation.

A natural approach is to encode the plane as the complex plane, where rotations become affine maps of the form $z \mapsto \omega(z-a)+a$ with $\omega=e^{-2\pi i/3}$. Composing three such maps yields an affine transformation of the form $z \mapsto z + v$, hence a translation. The long iteration reduces to repeated application of a single translation.

If the $1986$-th point returns to the starting point, then this translation must be trivial, so its translation vector vanishes. That condition reduces to a linear relation among $A_1,A_2,A_3$ with coefficients involving $\omega$. The structure of those coefficients forces one side vector to be a $60^\circ$ rotation of another, which determines an equilateral configuration.

The key difficulty lies in passing from the vanishing of a complex linear combination to a geometric characterization of the triangle.

Problem Understanding

This is a Type B problem, requiring a proof that a given periodicity condition forces a geometric conclusion.

A point moves under repeated $120^\circ$ rotations about $A_1,A_2,A_3$ in cyclic order. After $1986$ steps it returns to its initial position. The goal is to deduce that the triangle $A_1A_2A_3$ must be equilateral.

The mechanism of motion suggests grouping steps into blocks of three, producing a single affine transformation applied repeatedly. If a nontrivial translation arises from one block, repeated application would prevent periodic return. Therefore the block transformation must be the identity, forcing a rigid algebraic constraint on the vertices, which in turn enforces equality of side lengths and $60^\circ$ angles.

Proof Architecture

The argument proceeds through complex coordinates, identifying the plane with $\mathbb{C}$ and writing $\omega=e^{-2\pi i/3}$.

A first lemma describes the complex form of a $120^\circ$ rotation about a point.

A second lemma computes the composition of three successive rotations about $A_1,A_2,A_3$ and shows it equals a translation $z \mapsto z+v$ for an explicit vector $v$ depending linearly on $A_1,A_2,A_3$.

A third lemma shows that if a translation applied repeatedly has a fixed point after a positive number of iterations, then the translation vector must vanish.

A fourth lemma analyzes the condition $v=0$ and shows it is equivalent to a relation of the form $A_2-A_3=-\omega(A_1-A_3)$, which forces the triangle to be equilateral by interpreting multiplication by $-\omega$ as a rotation by $60^\circ$.

The most delicate step is the equivalence between $v=0$ and a geometric rotation relation between side vectors.

Solution

The plane is identified with the complex plane. Let $\omega=e^{-2\pi i/3}$, so that $\omega^3=1$ and $\omega\neq 1$. A $120^\circ$ clockwise rotation corresponds to multiplication by $\omega$.

For any point $a\in\mathbb{C}$, the rotation of a point $z$ about $a$ by $120^\circ$ clockwise is given by $z \mapsto \omega(z-a)+a$, which expands to $\omega z+(1-\omega)a$.

Let $R_k$ denote the rotation about $A_k$ by $120^\circ$ clockwise. The construction gives $P_{k+1}=R_{k+1}(P_k)$ with indices taken modulo $3$.

Lemma 1

The composition $T=R_3\circ R_2\circ R_1$ is a translation of the form $T(z)=z+v$ for a constant vector

$$v=(1-\omega)\bigl(\omega^2 A_1+\omega A_2+A_3\bigr).$$

The computation follows by applying the affine form of each rotation successively. The coefficient of $z$ becomes $\omega^3=1$, and the remaining terms combine linearly in $A_1,A_2,A_3$ with weights $\omega^2,\omega,1$.

This step establishes that the dynamics over three steps is independent of the current point and depends only on the triangle.

Lemma 2

For any integer $n\ge 1$, the map $T^n$ is the translation $z\mapsto z+n v$.

Since $T$ is a translation, iteration adds the same vector each time, yielding $n v$ after $n$ applications.

This step reduces long iterates to a linear growth of a single vector.

Lemma 3

If a translation $z\mapsto z+v$ satisfies $T^n(z)=z$ for some $z$ and some positive integer $n$, then $v=0$.

Indeed, $T^n(z)=z+n v$, so the equality $z+n v=z$ forces $n v=0$, hence $v=0$.

This step shows that periodicity is possible only when the translation is trivial.

Lemma 4

The condition $v=0$ is equivalent to

$$\omega^2 A_1+\omega A_2+A_3=0.$$

Since $1-\omega\neq 0$, vanishing of $v$ is equivalent to vanishing of the parenthetical expression.

This step converts dynamical periodicity into an algebraic constraint on the vertices.

Lemma 5

If $\omega^2 A_1+\omega A_2+A_3=0$, then the triangle $A_1A_2A_3$ is equilateral.

Translate coordinates so that $A_3=0$. The relation becomes $\omega^2 A_1+\omega A_2=0$, hence

$$A_2=-\omega A_1.$$

The complex number $-\omega$ has modulus $1$ and argument $-60^\circ$, since $\omega=e^{-2\pi i/3}$ implies $-\omega=e^{-i\pi/3}$. Multiplication by $-\omega$ is therefore a rotation by $60^\circ$ clockwise about the origin, preserving magnitude.

Thus $|A_2|=|A_1|$ and the angle between $A_1$ and $A_2$ equals $60^\circ$. These conditions imply that $A_1A_3A_2$ forms an equilateral triangle with $A_3$ as the vertex.

This step converts a complex linear relation into a rigid geometric configuration.

Completion of the proof

Since $1986=3\cdot 662$, the construction applies the translation $T$ exactly $662$ times, giving $P_{1986}=T^{662}(P_0)=P_0+662v$. The assumption $P_{1986}=P_0$ yields $662v=0$, hence $v=0$.

Applying Lemma 5 forces the triangle $A_1A_2A_3$ to be equilateral.

This completes the proof. ∎

Verification of Key Steps

The first delicate point is the computation of the translation vector in Lemma 1. Recomputing directly from the affine form of each rotation confirms that the coefficient of $z$ is $\omega^3=1$, and all dependence on the $A_k$ remains linear, so no nonlinear terms appear.

The second delicate point is the deduction $A_2=-\omega A_1$ after translation. Any oversight in choosing the origin would distort the relation, so fixing $A_3=0$ is essential for isolating two independent vectors.

The third delicate point is the geometric interpretation of multiplication by $-\omega$. Writing $-\omega=e^{-i\pi/3}$ confirms that it is a pure rotation by $60^\circ$, ensuring equal magnitudes and fixed angle.

Alternative Approaches

A synthetic approach avoids complex numbers by interpreting each rotation as a composition of directed angle changes and using the fact that three $120^\circ$ rotations compose to a translation. One can then argue using vector addition of directed sides and properties of equiangular spirals.

Another approach uses barycentric coordinates, expressing the rotation maps as affine transformations in $\mathbb{R}^2$ and computing the eigenstructure of the induced linear part. This leads to the same translation condition but requires heavier coordinate manipulation compared to the complex formulation.