OPEN - $500

Does every set of $n$ distinct points in $\mathbb{R}^2$ determine $\gg n/\sqrt{\log n}$ many distinct distances?

A $\sqrt{n}\times\sqrt{n}$ integer grid shows that this would be the best possible. Nearly solved by Guth and Katz [GuKa15] who proved that there are always $\gg n/\log n$ many distinct distances.

A stronger form (see [604]) may be true: is there a single point which determines $\gg n/\sqrt{\log n}$ distinct distances, or even $\gg n$ many such points, or even that this is true averaged over all points.

See also [661].

OPEN - $500

Does every set of $n$ distinct points in $\mathbb{R}^2$ contain at most $n^{1+O(1/\log\log n)}$ many pairs which are distance 1 apart?

The unit distance problem. This would be the best possible, as is shown by a set of lattice points. It is easy to show that there are $O(n^{3/2})$ many such pairs. The best known upper bound is $O(n^{4/3})$, due to Spencer, Szemerédi, and Trotter [SST84]. In [Er83c] and [Er85] Erdős offers \$250 for an upper bound of the form $n^{1+o(1)}$.

Part of the difficulty of this problem is explained by a result of Valtr (see [Sz16]), who constructed a metric on $\mathbb{R}^2$ and a set of $n$ points with $\gg n^{4/3}$ unit distance pairs (with respect to this metric). The methods of the upper bound proof of Spencer, Szemerédi, and Trotter [SST84] generalise to include this metric. Therefore to prove an upper bound better than $n^{4/3}$ some special feature of the Euclidean metric must be exploited.

See a survey by Szemerédi [Sz16] for further background and related results.

OPEN

Suppose $A\subset \mathbb{R}^2$ has $\lvert A\rvert=n$ and minimises the number of distinct distances between points in $A$. Prove that for large $n$ there are at least two (and probably many) such $A$ which are non-similar.

For $n=5$ the regular pentagon is the unique such set (Erdős mysteriously remarks this was proved by 'a colleague').

SOLVED - $500

Let $x_1,\ldots,x_n\in\mathbb{R}^2$ determine the set of distances $\{u_1,\ldots,u_t\}$. Suppose $u_i$ appears as the distance between $f(u_i)$ many pairs of points. Then for all $\epsilon>0$
\[\sum_i f(u_i)^2 \ll_\epsilon n^{3+\epsilon}.\]

OPEN

If $n$ points in $\mathbb{R}^2$ form a convex polygon then there are $O(n)$ many pairs which are distance $1$ apart.

Conjectured by Erdős and Moser. Füredi [Fu90] proved an upper bound of $O(n\log n)$. A short proof of this bound was given by Brass and Pach [BrPa01]. The best known upper bound is
\[\leq n\log_2n+4n,\]
due to Aggarwal [Ag15].

Edelsbrunner and Hajnal [EdHa91] have constructed $n$ such points with $2n-7$ pairs distance $1$ apart. (This disproved an early stronger conjecture of Erdős and Moser, that the true answer was $\frac{5}{3}n+O(1)$.)

OPEN - $100

Does every convex polygon have a vertex with no other $4$ vertices equidistant from it?

Erdős originally conjectured this with no $3$ vertices equidistant, but Danzer found a convex polygon on 9 points such that every vertex has three vertices equidistant from it (but this distance depends on the vertex), and Fishburn and Reeds [FiRe92] have found a convex polygon on 20 points such that every vertex has three vertices equidistant from it (and this distance is the same for all vertices).

If this fails for $4$, perhaps there is some constant for which it holds?

Erdős suggested this as an approach to solve [96]. Indeed, if this problem holds for $k+1$ vertices then, by induction, this implies an upper bound of $kn$ for [96].

The answer is no if we omit the requirement that the polygon is convex (I thank Boris Alexeev and Dustin Mixon for pointing this out), since for any $d$ there are graphs with minimum degree $d$ which can be embedded in the plane such that each edge has length one (for example one can take the $d$-dimensional hypercube graph on $2^d$ vertices). One can then connect the vertices in a cyclic order so that there are no self-intersections and no three consecutive vertices on a line, thus forming a (non-convex) polygon.

OPEN

Let $h(n)$ be such that any $n$ points in $\mathbb{R}^2$, with no three on a line and no four on a circle, determine at least $h(n)$ distinct distances. Does $h(n)/n\to \infty$?

Erdős could not even prove $h(n)\geq n$. Pach has shown $h(n)<n^{\log_23}$. Erdős, Füredi, and Pach have improved this to
\[h(n) < n\exp(c\sqrt{\log n})\]
for some constant $c>0$.

OPEN - $100

Let $A\subseteq\mathbb{R}^2$ be a set of $n$ points with minimum distance equal to 1, chosen to minimise the diameter of $A$. If $n$ is sufficiently large then must there be three points in $A$ which form an equilateral triangle of size 1?

The minimal such diameter is achieved (asymptotically) by the points in a triangular lattice intersected with a circle. In general Erdős believed such a set must have very large intersection with the triangular lattice (perhaps as many as $(1-o(1))n$).

The stated problem is false for $n=4$, for example taking the points to be vertices of a square. The behaviour of such sets for small $n$ is explored by Bezdek and Fodor [BeFo99].

OPEN

Let $A\subseteq\mathbb{R}^2$ be a set of $n$ points with distance set $D=\{\lvert x-y\rvert : x\neq y \in A\}$. Suppose that if $d_1\neq d_2\in D$ then $\lvert d_1-d_2\rvert \geq 1$. Is there some constant $c>0$ such that the diameter of $A$ must be at least $cn$?

Perhaps the diameter is $\geq n-1$ for all large $n$ (Piepmeyer has an example of $9$ such points with diameter $<5$).

OPEN

Let $A$ be a set of $n$ points in $\mathbb{R}^2$ such that all pairwise distances are at least $1$ and if two distinct distances differ then they differ by at least $1$. Is the diameter of $A$ $\gg n$?

Perhaps the diameter is even $\geq n-1$ for sufficiently large $n$. Kanold proved the diameter is $\geq n^{3/4}$. The bounds on the distinct distance problem [89] proved by Guth and Katz [GuKa15] imply a lower bound of $\gg n/\log n$.

OPEN - $100

Let $A\subset \mathbb{R}^2$ be a set of $n$ points. Must there be two distances which occur at least once but between at most $n$ pairs of points? Must the number of such distances $\to \infty$ as $n\to \infty$?

Asked by Erdős and Pach. Pannowitz and Hopf proved that the largest distance between points of $A$ can occur at most $n$ times, but it is unknown whether a second such distance must occur.

It may be true that there are at least $n^{1-o(1)}$ many such distances. In [Er97e] Erdős offers \$100 for 'any nontrivial result'.

OPEN - $250

Let $A\subset \mathbb{R}^2$ be a set of $n$ points such that any subset of size $4$ determines at least $5$ distinct distances. Must $A$ determine $\gg n^2$ many distances?

Erdős also makes the even stronger conjecture that $A$ must contain $\gg n$ many points such that all pairwise distances are distinct.

OPEN

Is there a dense subset of $\mathbb{R}^2$ such that all pairwise distances are rational?

Conjectured by Ulam. Erdős believed there cannot be such a set. This problem is discussed in a blogpost by Terence Tao, in which he shows that there cannot be such a set, assuming the Bombieri-Lang conjecture. The same conclusion was independently obtained by Shaffaf [Sh18].

Indeed, Shaffaf and Tao actually proved that such a rational distance set must be contained in a finite union of real algebraic curves. Solymosi and de Zeeuw [SdZ10] then proved (unconditionally) that a rational distance set contained in a real algebraic curve must be finite, unless the curve contains a line or a circle.

Ascher, Braune, and Turchet [ABT20] observed that, combined, these facts imply that a rational distance set in general position must be finite (conditional on the Bombieri-Lang conjecture).

OPEN

Let $n\geq 4$. Are there $n$ points in $\mathbb{R}^2$, no three on a line and no four on a circle, such that all pairwise distances are integers?

Harborth constructed such a set when $n=5$. The best construction to date, due to Kreisel and Kurz [KK08], has $n=7$.

Ascher, Braune, and Turchet [ABT20] have shown that there is a uniform upper bound on the size of such a set, conditional on the Bombieri-Lang conjecture. Greenfeld, Iliopoulou, and Peluse [GIP24] have shown (unconditionally) that any such set must be very sparse, in that if $S\subseteq [-N,N]^2$ has no three on a line and no four on a circle, and all pairwise distances integers, then \[\lvert S\rvert \ll (\log N)^{O(1)}.\]

See also [130].

SOLVED

Let $S\subset \mathbb{R}^2$ be such that no two points in $S$ are distance $1$ apart. Must the complement of $S$ contain four points which form a unit square?

The answer is yes, proved by Juhász [Ju79], who proved more generally that the complement of $S$ must contain a congruent copy of any set of four points. This is not true for arbitrarily large sets of points, but perhaps is still true for any set of five points.

OPEN

For which $n$ are there $n$ points in $\mathbb{R}^2$, no three on a line and no four on a circle, which determine $n-1$ distinct distances and so that (in some ordering of the distances) the $i$th distance occurs $i$ times?

An example with $n=4$ is an isosceles triangle with the point in the centre. Erdős originally believed this was impossible for $n\geq 5$, but Pomerance constructed a set with $n=5$ (see [Er83c] for a description), and Palásti has proved such sets exist for all $n\leq 8$. Erdős believes this is impossible for all sufficiently large $n$.

OPEN

Let $d\geq 2$ and $n\geq 2$. Let $f_d(n)$ be maximal such that, for any $A\subseteq \mathbb{R}^d$ of size $n$, with diameter $1$, the distance 1 occurs between $f_d(n)$ many pairs of points in $A$. Estimate $f_d(n)$.

SOLVED

For $A\subset \mathbb{R}^2$ we define the upper density as
\[\overline{\delta}(A)=\limsup_{R\to \infty}\frac{\lambda(A \cap B_R)}{\lambda(B_R)},\]
where $\lambda$ is the Lebesgue measure and $B_R$ is the ball of radius $R$.

Estimate \[m_1=\sup \overline{\delta}(A),\] where $A$ ranges over all measurable subsets of $\mathbb{R}^2$ without two points distance $1$ apart. In particular, is $m_1\leq 1/4$?

A question of Moser [Mo66]. A lower bound of $m_1\geq \pi/8\sqrt{3}\approx 0.2267$ is given by taking the union of open circular discus of radius $1/2$ at a regular hexagonal lattice suitably spaced aprt. Croft [Cr67] gives a small improvement of $m_1\geq 0.22936$.

The trivial upper bound is $m_1\leq 1/2$, since for any unit vector $u$ the sets $A$ and $A+u$ must be disjoint. Erdős' question was solved by Ambrus, Csiszárik, Matolcsi, Varga, and Zsámboki [ACMVZ23] who proved that $m_1\leq 0.247$.

SOLVED

What is the size of the largest $A\subseteq \mathbb{R}^n$ such that there are only two distinct distances between elements of $A$? That is,
\[\# \{ \lvert x-y\rvert : x\neq y\in A\} = 2.\]

Asked to Erdős by Coxeter. Erdős thought he could show that $\lvert A\rvert \leq n^{O(1)}$, but later discovered a mistake in his proof, and his proof only gave $\leq \exp(n^{1-o(1)})$.

Bannai, Bannai, and Stanton [BBS83] have proved that \[\lvert A\rvert \leq \binom{n+2}{2}.\] A simple proof of this upper bound was given by Petrov and Pohoata [PePo21].

Shengtong Zhang has observed that a simple lower bound of $\binom{n}{2}$ is given by considering all points with exactly two coordinates equal to $1$ and all others equal to $0$.

OPEN - $500

Given $n$ distinct points $A\subset\mathbb{R}^2$ must there be a point $x\in A$ such that
\[\#\{ d(x,y) : y \in A\} \gg n^{1-o(1)}?\]
Or even $\gg n/\sqrt{\log n}$?

The pinned distance problem, a stronger form of [89]. The example of an integer grid show that $n/\sqrt{\log n}$ would be best possible.

It may be true that there are $\gg n$ many such points, or that this is true on average. In [Er97e] Erdős offers \$500 for a solution to this problem, but it is unclear whether he intended this for proving the eixstence of a single such point or for $\gg n$ many such points.

In [Er97e] Erdős wrote that he initially 'overconjectured' and thought that the answer to this problem is the same as for the number of distinct distances between all pairs (see [89]), but this was disproved by Harborth. It could be true that the answers are the same up to an additive factor of $n^{o(1)}$.

The best known bound is \[\gg n^{c-o(1)},\] due to Katz and Tardos [KaTa04], where \[c=\frac{48-14e}{55-16e}=0.864137\cdots.\]

SOLVED

Is there some function $f(n)\to \infty$ as $n\to\infty$ such that there exist $n$ distinct points on the surface of a two-dimensional sphere with at least $f(n)n$ many pairs of points whose distances are the same?

See also [90]. This was solved by Erdős, Hickerson, and Pach [EHP89]. For $D>1$ and $n\geq 2$ let $u_D(n)$ be such that there is a set of $n$ points on the sphere in $\mathbb{R}^3$ with radius $D$ such that there are $u_D(n)$ many pairs which are distance $1$ apart (so that this problem asked for $u_D(n)\geq f(n)n$ for some $D$).

Erdős, Hickerson, and Pach [EHP89] proved that $u_{\sqrt{2}}(n)\asymp n^{4/3}$ and $u_D(n)\gg n\log^*n$ for all $D>1$ and $n\geq 2$ (where $\log^*$ is the iterated logarithm function).

This lower bound was improved by Swanepoel and Valtr [SwVa04] to $u_D(n) \gg n\sqrt{\log n}$. The best upper bound for general $D$ is $u_D(n)\ll n^{4/3}$.

OPEN

Let $x_1,\ldots,x_n\in \mathbb{R}^2$ and let $R(x_i)=\#\{ \lvert x_j-x_i\rvert : j\neq i\}$, where the points are ordered such that
\[R(x_1)\leq \cdots \leq R(x_n).\]
Let $\alpha_k$ be minimal such that, for all large enough $n$, there exists a set of $n$ points with $R(x_k)<\alpha_kn^{1/2}$. Is it true that $\alpha_k\to \infty$ as $k\to \infty$?

It is trivial that $R(x_1)=1$ is possible, and that $R(x_2) \ll n^{1/2}$ is also possible, but we always have
\[R(x_1)R(x_2)\gg n.\]
Erdős originally conjectured that $R(x_3)/n^{1/2}\to \infty$ as $n\to \infty$, but Elekes proved that for every $k$ and $n$ sufficiently large there exists some set of $n$ points with $R(x_k)\ll_k n^{1/2}$.

OPEN

Let $x_1,\ldots,x_n\in \mathbb{R}^2$ and let $R(x_i)=\#\{ \lvert x_j-x_i\rvert : j\neq i\}$, where the points are ordered such that
\[R(x_1)\leq \cdots \leq R(x_n).\]
Let $g(n)$ be the maximum number of distinct values the $R(x_i)$ can take. Is it true that $g(n) \geq (1-o(1))n$?

Erdős and Fishburn proved $g(n)>\frac{3}{9}n$ and Csizmadia proved $g(n)>\frac{7}{10}n$. Both groups proved $g(n) < n-cn^{2/3}$ for some constant $c>0$.

OPEN

Let $x_1,\ldots,x_n\in \mathbb{R}^2$ be such that no circle whose centre is one of the $x_i$ contains three other points. Are there at least
\[(1+c)\frac{n}{2}\]
distinct distances determined between the $x_i$, for some constant $c>0$ and all $n$ sufficiently large?

A problem of Erdős and Pach. It is easy to see that this assumption implies that there are at least $\frac{n-1}{2}$ distinct distances determined by every point.

OPEN

Is it true that if $A\subset \mathbb{R}^2$ is a set of $n$ points such that every subset of $4$ points determines at least $5$ distances then $A$ must determine $\gg n^2$ distances?

A problem of Erdős and Gyárfás. Erdős could not even prove that the number of distances is at least $f(n)n$ where $f(n)\to \infty$.

More generally, one can ask how many distances $A$ must determine if every set of $p$ points determines at least $q$ points.

See also [657].

OPEN

Is it true that if $A\subset \mathbb{R}^2$ is a set of $n$ points such that every subset of $3$ points determines $3$ distinct distances (i.e. $A$ has no isosceles triangles) then $A$ must determine at least $f(n)n$ distinct distances, for some $f(n)\to \infty$?

Investigated by Erdős, Füredi, Ruzsa, and Pach.

See also [656].

OPEN

Is there a set of $n$ points in $\mathbb{R}^2$ such that every subset of $4$ points determines at least $3$ distances, yet the total number of distinct distances is
\[\ll \frac{n}{\sqrt{\log n}}?\]

Erdős believed this should be possible, and should imply effective upper bounds for [658] (presumably the version with no alignment restrictions on the squares).

OPEN

Let $x_1,\ldots,x_n\in \mathbb{R}^3$ be the vertices of a convex polyhedron. Is there some constant $c>0$ such that, for all $n$ sufficiently large, the number of distinct distances determined by the $x_i$ is at most
\[(1-c)\frac{n}{2}?\]

OPEN

Are there, for all large $n$, some points $x_1,\ldots,x_n,y_1,\ldots,y_n\in \mathbb{R}^2$ such that the number of distinct distances $d(x_i,y_j)$ is
\[o\left(\frac{n}{\sqrt{\log n}}\right)?\]

One can also ask this for points in $\mathbb{R}^3$. In $\mathbb{R}^4$ Lenz observed that there are $x_1,\ldots,x_n,y_1,\ldots,y_n\in \mathbb{R}^4$ such that $d(x_i,y_j)=1$ for all $i,j$, taking the points on two orthogonal circles.

See also [89].

OPEN

Consider the triangular lattice with minimal distance between two points $1$. Denote by $f(t)$ the number of distances from any points $\leq t$. For example $f(1)=6$, $f(\sqrt{3})=12$, and $f(3)=18$.

Let $x_1,\ldots,x_n\in \mathbb{R}^2$ be such that $d(x_i,x_j)\geq 1$ for all $i\neq j$. Is it true that, provided $n$ is sufficiently large depending on $t$, the number of distances $d(x_i,x_j)\leq t$ is less than or equal to $f(t)$ with equality perhaps only for the triangular lattice?

In particular, is it true that the number of distances $\leq \sqrt{3}-\epsilon$ is less than $1$?

A problem of Erdős, Lovász, and Vesztergombi.

This is essentially verbatim the problem description in [Er97e], but this does not make sense as written; there must be at least one typo. Suggestions about what this problem intends are welcome.

Erdős also goes on to write 'Perhaps the following stronger conjecture holds: Let $t_1<t_2<\cdots$ be the set of distances occurring in the triangular lattice. $t_1=1$ $t_2=\sqrt{3}$ $t_3=3$ $t_4=5$ etc. Is it true that there is an $\epsilon_n$ so that for every set $y_1,\ldots,$ with $d(y_i,y_j)\geq 1$ the number of distances $d(y_i,y_j)<t_n$ is less than $f(t_n)$?'

Again, this is nonsense interpreted literally; I am not sure what Erdős intended.