IMO 1984 Problem 3

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 5m56s

Problem

Given points $O$ and $A$ in the plane. Every point in the plane is colored with one of a finite number of colors. Given a point $X$ in the plane, the circle $C(X)$ has center $O$ and radius $OX+{\angle AOX\over OX}$, where $\angle AOX$ is measured in radians in the range $[0,2\pi)$. Prove that we can find a point $X$, not on $OA$, such that its color appears on the circumference of the circle $C(X)$.

Exploration

Fix $O$ and $A$. Write a point $X$ in polar coordinates centered at $O$ as $(r,\theta)$, where $\theta=\angle AOX \in [0,2\pi)$. The radius of the circle $C(X)$ is

$$R(r,\theta)=r+\frac{\theta}{r}.$$

Thus $C(X)$ is the circle centered at $O$ passing through the point whose distance from $O$ is $R(r,\theta)$.

The problem asks for a point $X$ such that the color of $X$ appears somewhere on the circle of radius $R(r,\theta)$. This couples the color of a point on one circle with colors appearing on a different circle whose radius depends on both $r$ and $\theta$.

A useful reduction is to restrict attention to a single ray from $O$ not containing $A$, so that $\theta$ becomes constant. Then the dependence simplifies to a one-variable function $R(r)=r+\frac{\alpha}{r}$, where $\alpha\in(0,2\pi)$ is fixed. This function is continuous and has a well-controlled monotonic structure: it decreases on $(0,\sqrt{\alpha})$ and increases on $(\sqrt{\alpha},\infty)$.

Along such a ray, points are totally ordered by distance from $O$, while their images under $r\mapsto R(r)$ lie on the same ray but farther out. This creates a natural discrete dynamical system along the ray.

The central obstacle is that the coloring is arbitrary, so no continuity can be used. Any successful argument must extract a combinatorial repetition along an infinite sequence of points whose radii are linked by the function $R$.

A promising strategy is to construct a sequence $X_1,X_2,\dots$ on a fixed ray such that $OX_{n+1}=R(OX_n,\alpha)$, then use finiteness of colors to force a structured repetition between consecutive terms.

Problem Understanding

The problem is a Type D existence problem. One must prove that there exists a point $X$ not lying on the line $OA$ such that the color of $X$ also appears on the circumference of the circle centered at $O$ with radius $OX+\frac{\angle AOX}{OX}$.

Equivalently, for some point $X$, if $R=OX+\frac{\angle AOX}{OX}$, then some point $Y$ with $OY=R$ has the same color as $X$.

The key difficulty is that the radius of the target circle depends on the angular position of $X$, while the coloring is completely arbitrary. No geometric regularity of the coloring is available, so the construction must force a repetition purely through combinatorial inevitability along a structured geometric sequence.

Proof Architecture

First, a lemma will establish that a recurrence of the form $r_{n+1}=r_n+\frac{\alpha}{r_n}$ produces a strictly increasing sequence tending to infinity.

Second, a lemma will show that if among a sequence of colored points indexed along a strictly increasing sequence of radii there exists a repeated color, then one may choose a repetition with minimal index gap, and this minimality forces a consecutive repetition.

Third, the main construction defines points $X_n$ on a fixed ray with $OX_{n+1}=R(OX_n,\alpha)$ and applies the previous lemma to obtain consecutive indices with identical colors.

The most delicate step is the minimal-gap argument, since it is the only place where the arbitrary coloring interacts with the rigid geometric recurrence.

Solution

Fix the point $A$ and choose a ray $OA'$ emanating from $O$ such that $OA'$ is not the same line as $OA$. Let $\alpha=\angle AO A'$ so that $\alpha\in(0,2\pi)$ is constant along this ray.

For any point $X$ on this ray with $OX=r$, the angle satisfies $\angle AOX=\alpha$, hence

$$R(r)=r+\frac{\alpha}{r}.$$

Define a sequence of points $X_1,X_2,\dots$ on the ray $OA'$ by choosing $OX_1>0$ arbitrarily and then defining inductively

$$OX_{n+1}=OX_n+\frac{\alpha}{OX_n}.$$

Lemma 1

The sequence $(OX_n)$ is strictly increasing and unbounded.

Since $\frac{\alpha}{OX_n}>0$, one has $OX_{n+1}>OX_n$ for all $n$. For unboundedness, assume $OX_n\le M$ for all $n$. Then

$$OX_{n+1}-OX_n=\frac{\alpha}{OX_n}\ge \frac{\alpha}{M},$$

so $OX_n\ge OX_1+(n-1)\frac{\alpha}{M}$, which diverges as $n\to\infty$. This contradiction implies unboundedness. ∎

This establishes that the construction generates infinitely many distinct points moving outward along the ray without accumulation.

Color each point $X_n$ by its given color in the plane. Since only finitely many colors exist, there exist indices $i<j$ such that $X_i$ and $X_j$ have the same color.

Among all such pairs $(i,j)$ with $i<j$ and $X_i,X_j$ having the same color, choose one with minimal $j-i$.

Lemma 2

For this minimal pair, one has $j=i+1$.

Assume $j\ge i+2$. Then consider the intermediate point $X_{i+1}$. If the color of $X_{i+1}$ equals the color of $X_i$, then $i$ and $i+1$ form a closer repeated pair, contradicting minimality. If the color of $X_{i+1}$ equals the color of $X_j$, then $i+1$ and $j$ form a repeated pair with smaller index difference $j-(i+1)<j-i$, again contradicting minimality. Both cases exhaust all possibilities for the color of $X_{i+1}$ among the finitely many colors. Hence no such $j\ge i+2$ exists, and $j=i+1$. ∎

This step enforces that a repetition of color must occur at consecutive indices in the constructed sequence.

Let $X=X_i$. Then $OX_i=r$ and $OX_{i+1}=r+\frac{\alpha}{r}$. By construction, $X_{i+1}$ lies on the circle centered at $O$ of radius

$$R=r+\frac{\alpha}{r}=OX_i+\frac{\angle AOX_i}{OX_i}.$$

Since $X_i$ and $X_{i+1}$ have the same color, the color of $X$ appears on the circumference of $C(X)$.

Thus there exists a point $X$ not on $OA$ such that its color appears on the circumference of $C(X)$.

Verification of Key Steps

The construction of the sequence uses only radial geometry along a fixed ray, ensuring the angle $\angle AOX$ remains constant and the radius recursion is well-defined.

The unboundedness argument relies on a uniform positive lower bound for the increments under a boundedness assumption, which forces divergence.

The minimal-gap argument depends critically on the fact that any index gap larger than one allows insertion of an intermediate point whose color must coincide with one of the endpoints, contradicting minimality of the chosen repetition.

Alternative Approaches

A different approach replaces the discrete recurrence by a continuous spiral curve defined by $r'(\theta)=r(\theta)+\frac{\theta}{r(\theta)}$ along varying angles. One then attempts to compare color partitions induced on two parametrized curves: a fixed circle and a spiral-like curve. The main difficulty in that approach is the absence of any regularity in the coloring, which makes intersection arguments less direct than the discrete recurrence method used above.