1971 IMO Problems/Problem 5

Problem

Prove that for every natural number $m$ , there exists a finite set $S$ of points in a plane with the following property: For every point $A$ in $S$ , there are exactly $m$ points in $S$ which are at unit distance from $A$ .

Solution 1

I shall prove a more general statement about the unit distance graph( $V=\mathbb{R}^2$ , adjacency iff the Euclidean distance between the points is $1$ ) and this will follow as a consequence. if $G,H$ occur as unit distance graphs, then so does $G\times H$ ( here $G\times H$ is described as $V(G\times H)=V(G)\times V(H), (v_1,w_1)\leftrightarrow (v_2,w_2) \Leftrightarrow v_1=v_2,w_1\leftrightarrow w_2$ or $v_1\leftrightarrow v_2,w_1=w_2$ ). this is seen by describing the vertices by complex numbers. suppose there is an embedding of $G$ by the complex numbers $v_1,v_2\ldots,v_n$ and for $H$ by the numbers $w_1,w_2,\ldots, w_m$ . we claim that for some choice of $0\leq\theta<2\pi$ , $v_i+e^{i\theta}w_j$ will do the job(a suitable rotation). what we need is that $(v_i+e^{i\theta}w_j)\leftrightarrow (v_k+e^{i\theta}w_l)$ iff either $v_i=v_k,|w_j-w_l|=1$ or $w_j=w_l,|v_i-v_k|=1$ . clearly if the condition holds then the adjacency is satisfied. suppose $i\neq k,j\neq l$ and that the corresponding complex numbers are at a distance $1$ from one another. Then this gives a quadratic in $e^{i\theta}$ and hence $\theta$ can take only finitely many values here.ruling this out for each such set of $i,j,k,l$ hence rules out finitely many values of $\theta$ and therefore a suitable choice exists. for the given problem we need a unit distance graph which is regular of degree $m$ for any given $m$ . since we can form the graph $K_2$ , we can form $Q_n=Q_{n-1}\times K_2$ (the unit cube) and that solves the problem.

This solution was posted and copyrighted by seshadri. The original thread for this problem can be found here: [1]

Solution 2

Suppose $S$ has the form $S_m=\left\{\sum_{\mathbf v \in U} \mathbf v \;\middle\vert\; U \subseteq V_m\right\}=\left\{\sum_{i=1}^m b_i \mathbf v_i \;\middle\vert\; b_i \in \{0,1\}\right\}\!,$ where $V_m=\{\mathbf v_1,\mathbf v_2,\dots,\mathbf v_m\}$ is unknown set of distinct unit vectors in $\mathbb R^2$ . We can build $V_m$ inductively, starting from the empty set and adding vectors to it, one by one. We just need to make sure that two thing are respected: 1. All $2^m$ vectors in $S_m$ are distinct; 2. Two vector sums are at unit distance from one another if and only if they differ in presence of exactly one summand (i.e. one and only one coefficient $b_i$ in the sum changes from $0$ to $1$ ). If these two conditions are kept, then each of $2^m$ points at $S_m$ will be at unit distance from exactly $m$ points corresponding to sums at which one and only one of $m$ coefficients differs from coefficients of this point. However, respecting these conditions is not hard because $\left\|\sum_{\mathbf v \in U_1} \mathbf v - \sum_{\mathbf v \in U_2} \mathbf v\right\|=\left\|\sum_{\mathbf v \in U_1 \setminus U_2} \mathbf v \,-\! \sum_{\mathbf v \in U_2 \setminus U_1} \mathbf v\right\|$ and for each new vector being added to $V_m$ there is at most some finite set of forbidden endpoints given by sums/differences of already determined vectors but the rest of the (infinite) unit circle is permissible.

This solution was posted and copyrighted by Bandera. The original thread for this problem can be found here: [2]

Remarks (added by pf02, January 2025)

1. On the original thread at https://artofproblemsolving.com/community/c6h60830p there is a third solution, by Pgm03B. It is a very nice solution, simpler than the two above. It has a few flaws in the way it is presented. As a public service, I will add an edited version of it below.

2. Solutions 1 and 2 above are based on good ideas, but they are presented very poorly. If this page was a reviewed publication, and I were a reviewer, I would reject both of them saying "rewrite and resubmit".

2.1. Solution 1 suffers from undefined notation and terminology, from minor errors, and from unacceptable amount of hand waving replacing explanations of details.

2.2. Solution 2 suffers from poor explanation of details and from what seems to me to be an error (starting an inductive proof for a property of $m$ vectors from an empty set).

As a public service I will rewrite these proofs below, in what I hope is a much more presentable style.

3. Finally, I will give some examples. Philosophically speaking, these are nice, but "bad". They are bad for two reasons: on the one hand, I don't think they can be generalized. On the other hand, it appears from the three solutions we have, that the configurations of points in which every point $A$ has exactly $m$ neighbors at distance $1$ are good because they are not "nice". By contrast, the examples I give are "nice".

Definitions and terminology

It will be helpful (though not necessary) to imagine the set of points $S$ as a graph in the plane. Specifically, the vertices are the points in $S$ , and the edges are the line segments of length $1$ joining the points. All points at distance $1$ are joined, and only the points at distance $1$ are joined.

We will call such a graph $m$ -regular iff every $A \in S$ has exactly $m$ lines of the graph with one end at $A$ .

For the purposes of solution 1, it will be useful to think of the points $A \in S$ as points in the complex plane. For the purposes of solution 2, it will be useful to think of the plane as having on origin $O$ , the points $A \in S$ be all $\neq O$ , and $A$ being the end point of a vector in the plane.

Solution 1, rewritten

We will give a proof by induction on $m$ . First some notation:

Given two finite sets of points $S, T$ in the complex plane, define $U = S \oplus_\theta T = \{a + e^{i \theta} b, \text{\ for all\ } a \in S, b \in T\}$ .

$\mathbf{Proposition:}$ If $S$ is $m$ -regular and $T$ is $n$ -regular then we can choose $\theta$ so that $U = S \oplus_\theta T$ is $(m + n)$ -regular.

TO BE CONTINUED. SAVING MID WAY SO I DON'T LOSE TEXT WRITTEN SO FAR.

1971 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
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