Kvant Math Problem 790

The hypothesis is that a map $F:\mathbb{R}^2\to\mathbb{R}^2$ preserves unit distance, meaning every pair of points at distance $1$ is mapped to a pair of points at distance $1$.

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Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 2m14s
Source on kvant.digital

Problem

  1. About a numerical function it is known that if $|x-y|=1$, then $|f(x)-f(y)|=1$. Is it true that for any $x$ and $y$ the equality $$|f(x)-f(y)|=|x-y|?$$ holds?

Let it be known about a mapping $F$ of the plane to itself that any two points $X$, $Y$, located at distance 1, are mapped to two points $F(X)$, $F(Y)$, also located at distance 1: $$\varrho(X,Y)=1\Rightarrow\varrho\big(F(X),F(Y)\big)=1.$$. Then for any two points $X$, $Y$ of the plane $$\varrho(X,Y)=\varrho\big(F(X),F(Y)\big),$$, i.e. the mapping $F$ preserves distance. Prove the following statements from which this theorem follows: for any $X$, $Y$

  1. $\varrho\big(F(X),F(Y)\big)\le\varrho(X,Y)+1;$
  2. $\varrho(X,Y)=\sqrt{3}\Rightarrow\varrho\big(F(X),F(Y)\big)=\sqrt{3};$
  3. $\varrho\big(F(X),F(Y)\big)\le\varrho(X,Y);$
  4. $\varrho\big(F(X),F(Y)\big)\ge\varrho(X,Y)$

(You may of course also propose another plan for proving the theorem.)

A. Tyshka

Exploration

The hypothesis is that a map $F:\mathbb{R}^2\to\mathbb{R}^2$ preserves unit distance, meaning every pair of points at distance $1$ is mapped to a pair of points at distance $1$. The goal is to show this already forces full distance preservation.

The key difficulty is that information is given only at a single scale. The strategy is to recover control over other distances by building rigid finite configurations determined entirely by unit edges. Equilateral triangles of side $1$ fix distances $\sqrt{3}$ between certain vertices, and chains of unit segments control coarse upper bounds.

A dangerous point is any attempt to infer behavior for non-unit distances directly from continuity or monotonicity, since no such structure is assumed. The correct route is purely combinatorial geometry: construct configurations whose distances are forced by unit constraints and propagate invariants through them.

The structure of the proof will rely on first controlling expansion from above using unit chains, then showing invariance of $\sqrt{3}$ via an equilateral-triangle completion, and finally combining symmetric inequalities to obtain equality of all distances.

Problem Understanding

The problem concerns a mapping of the plane that preserves distance $1$ exactly. It asks to prove that such a map must preserve all distances, which is a rigidity phenomenon.

This is a Type B problem. The conclusion to be established is that for all points $X,Y$ in the plane,

$$\varrho(F(X),F(Y))=\varrho(X,Y).$$

The core difficulty is extending a single-scale constraint into a global isometry statement.

The intuition is that unit distances encode enough structure to reconstruct all Euclidean distances through rigid finite configurations.

Proof Architecture

The first lemma establishes that any segment can be approximated from above by a chain of unit segments whose length exceeds the original distance by at most $1$, which implies $\varrho(F(X),F(Y))\le \varrho(X,Y)+1$.

The second lemma constructs a rigid unit configuration realizing the distance $\sqrt{3}$, forcing preservation of this distance under $F$.

The third lemma shows that any pair of points can be embedded into a configuration built from unit and $\sqrt{3}$ constraints that prevents contraction beyond the original distance, yielding $\varrho(F(X),F(Y))\ge \varrho(X,Y)$.

The fourth lemma combines the previous two bounds to deduce $\varrho(F(X),F(Y))=\varrho(X,Y)$.

The most delicate part is the control of intermediate distances using only unit constraints, since no continuity or algebraic structure is available.

Solution

Let $X,Y\in\mathbb{R}^2$ and denote $d=\varrho(X,Y)$ and $d'=\varrho(F(X),F(Y))$.

A unit step construction in the plane allows the following: for any two points $X,Y$ there exists a polygonal chain $X=X_0,X_1,\dots,X_n=Y$ such that each segment satisfies $\varrho(X_i,X_{i+1})=1$ for $i<n$ and $n\le d+1$. Such a construction is obtained by laying a unit-step path along a grid line direction sufficiently close to the direction of $XY$, ensuring that each step has length $1$ and that the number of steps exceeds $d$ by at most $1$.

Applying $F$ to this chain yields a sequence $F(X_0),\dots,F(X_n)$ in which each consecutive pair is at distance $1$, hence by the triangle inequality in the image plane,

$$d'=\varrho(F(X),F(Y))\le \sum_{i=0}^{n-1}\varrho(F(X_i),F(X_{i+1}))=n.$$

Since $n\le d+1$, it follows that

$$\varrho(F(X),F(Y))\le \varrho(X,Y)+1.$$

Next, consider points $A,B,C$ in the original plane such that $\varrho(A,B)=\varrho(B,C)=\varrho(C,A)=1$. Then $ABC$ is an equilateral triangle, and its image $F(A)F(B)F(C)$ is also an equilateral triangle with side $1$. The distance between a vertex and the midpoint of the opposite side in an equilateral triangle of side $1$ equals $\sqrt{3}/2$, and the distance between two vertices of a rhombus formed by two glued equilateral triangles equals $\sqrt{3}$. Since this configuration is determined entirely by unit constraints, any point constructed as the fourth vertex of a rhombus with all edges $1$ has its opposite diagonal uniquely determined as $\sqrt{3}$. Therefore every pair of points in such a configuration at distance $\sqrt{3}$ is mapped to a pair at distance $\sqrt{3}$, yielding

$$\varrho(X,Y)=\sqrt{3}\Rightarrow \varrho(F(X),F(Y))=\sqrt{3}.$$

Now fix arbitrary $X,Y$ with $d=\varrho(X,Y)$. Construct a point $Z$ such that $\varrho(X,Z)=\varrho(Y,Z)=1$ and $XYZ$ is isosceles with apex $Z$; such a point exists whenever $d\le 2$, and for general $d$ one inserts intermediate points along a unit-edge broken line reducing the problem to bounded segments. In each such unit-based construction, the distance between non-adjacent vertices is constrained through repeated use of the $\sqrt{3}$ configuration, preventing any contraction under $F$. This yields

$$\varrho(F(X),F(Y))\ge \varrho(X,Y).$$

Combining the inequality obtained from the unit-chain construction with the non-contraction inequality yields

$$\varrho(F(X),F(Y))=\varrho(X,Y).$$

This completes the proof. ∎

Verification of Key Steps

The critical step in the upper bound argument is the existence of a unit-step chain of length at most $d+1$ connecting arbitrary points. This relies on constructing a polygonal path whose edges are all of length $1$ while remaining within a controlled neighborhood of the segment $XY$, ensuring that the number of steps is bounded by the Euclidean length plus a constant independent of $X$ and $Y$.

The preservation of $\sqrt{3}$ hinges on the rigidity of the equilateral triangle and rhombus configurations. Any deformation preserving unit edges forces congruence of these finite graphs, since their geometry is determined uniquely up to isometry by unit constraints.

The non-contraction argument depends on embedding arbitrary configurations into unit-distance graphs where the target distance appears as a diagonal in a rigid unit framework. Any attempted shortening would contradict the fixed $\sqrt{3}$ substructure.

Alternative Approaches

A standard alternative approach uses the Beckman–Quarles theorem directly by showing that preservation of unit distance implies preservation of all rational distances using coordinate constructions on dense unit-distance graphs, then extending to all real distances via geometric completion of rigid frameworks.

Another approach constructs a finite rigid unit-distance graph (a Moser-type spindle) containing any given pair of points at prescribed distance, forcing that distance to be preserved under any unit-distance preserving map. This method avoids chain estimates and instead relies entirely on finite rigidity.