11 solved out of 38 shown

$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 possibe. Nearly solved by Guth and Katz [GuKa15] who proved that there are always $\gg n/\log n$ many distinct distances. It may be true that there is a single point which determines $\gg n/\sqrt{\log n}$ distinct distances, or even that there are $\gg n$ many such points, or even that this is true averaged over all points.

$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?

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 [SpSzTr84]. In [Er83c] Erdős offers \$250 for an upper bound of the form $n^{1+o(1)}$.

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').

If $n$ distinct points in $\mathbb{R}^2$ form a convex polygon then they determine at least $\lfloor n/2\rfloor$ distinct distances.

Solved by Altman [Al63]. The stronger variant that says there is one point which determines at least $\lfloor n/2\rfloor$ distinct distances is still open.

Suppose $n$ points in $\mathbb{R}^2$ determine a convex polygon and the set of distances between them is $\{u_1,\ldots,u_t\}$. Suppose $u_i$ appears as the distance between $f(u_i)$ many pairs of points. Then
\[\sum_i f(u_i)^2 \ll n^3.\]

Solved by Fishburn [Al63]. Note it is trivial that $\sum f(u_i)=\binom{n}{2}$. The stronger conjecture that $\sum f(u_i)^2$ is maximal for the regular $n$-gon (for large enough $n$) is still open.

$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 $\epsilon>0$
\[\sum_i f(u_i)^2 \ll_\epsilon n^{3+\epsilon}.\]

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 has proved an upper bound of $O(n\log n)$. Edelsbrunner and Hajnal have constructed $n$ such points with $2n-7$ pairs distance $1$ apart.

Open even for convex polygons.

Danzer has 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 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?

$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?

In fact Erdős believes such a set must have very large intersection with the triangular lattice. This is false for $n=4$, for example a square. The behaviour of such sets for small $n$ is explored by Bezdek and Fodor [BeFo99].

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 $<5$).

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 $\gg n/\log n$.

$50

Let $A,B\subset \mathbb{R}^2$ be disjoint sets of size $n$ and $n-3$ respectively, with not all of $A$ contained on a single line. Is there a line which contains at least two points from $A$ and no points from $B$?

$500

Let $f(n)$ be minimal such that any $f(n)$ points in $\mathbb{R}^2$, no three on a line, contain $n$ points which form the vertices of a convex $n$-gon. Prove that $f(n)=2^{n-2}+1$.

The Erdős-Klein-Szekeres 'Happy Ending' problem. The problem originated in 1931 when Klein observed that $f(4)=5$. Turán and Makai showed $f(5)=9$. Erdős and Szekeres proved the bounds
\[2^{n-2}+1\leq f(n)\leq \binom{2n-4}{n-2}+1.\]
([ErSz60] and [ErSz35] respectively). There were several improvements of the upper bound, but all of the form $4^{(1+o(1))n}$, until Suk [Su17] proved
\[f(n) \leq 2^{(1+o(1))n}.\]
The current best bound is due to Holmsen, Mojarrad, Pach, and Tardos [HMPT20], who prove
\[f(n) \leq 2^{n+O(\sqrt{n\log n})}.\]

$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.

In any $2$-colouring of $\mathbb{R}^2$, for all but at most one triangle $T$, there is a monochromatic congruent copy of $T$.

For some colourings a single equilateral triangle has to be excluded, considering the colouring by alternating strips. Shader [Sh76] has proved this is true if we just consider a single right-angled triangle.

A finite set $A\subset \mathbb{R}^n$ is called Ramsey if, for any $k\geq 1$, there exists some $d=d(A,k)$ such that in any $k$-colouring of $\mathbb{R}^d$ there exists a monochromatic copy of $A$. Characterise the Ramsey sets in $\mathbb{R}^n$.

Erdős, Graham, Montgomery, Rothschild, Spencer, and Straus [EGMRSS73] proved that every Ramsey set is 'spherical': it lies on the surface of some sphere. Graham has conjectured that every spherical set is Ramsey. Leader, Russell, and Walters [LRW12] have alternatively conjectured that a set is Ramsey if and only if it is 'subtransitive': it can be embedded in some higher-dimensional set on which rotations act transitively.

Sets known to be Ramsey include vertices of $k$-dimensional rectangles [EGMRSS73], non-degenerate simplices [FrRo90], trapezoids [Kr92], and regular polygons/polyhedra [Kr91].

What is the smallest $k$ such that $\mathbb{R}^2$ can be red/blue coloured with no pair of red points unit distance apart, and no $k$-term arithmetic progression of blue points?

Juhász [Ju79] has shown that $k\geq 5$. It is also known that $k\leq 10000000$ (as Erdős writes, 'more or less').

If $\mathbb{R}^2$ is finitely coloured then must there exist some colour class which contains the vertices of a rectangle of every area?

Graham [Gr80] has shown that this is true if we replace rectangle by right-angled triangle. The same question can be asked for parallelograms. It is not true for rhombuses.

This is false; Kovač [Ko23] provides an explicit (and elegantly simple) colouring using 25 colours such that no colour class contains the vertices of a rectangle of area $1$. The question for parallelograms remains open.

Let $S\subseteq \mathbb{Z}^3$ be a finite set and let $A=\{a_1,a_2,\ldots,\}\subset \mathbb{Z}^3$ be an infinite $S$-walk, so that $a_{i+1}-a_i\in S$ for all $i$. Must $A$ contain three collinear points?

Originally conjectured by Gerver and Ramsey [GeRa79], who showed that the answer is yes for $\mathbb{Z}^3$, and for $\mathbb{Z}^2$ that the number of collinear points can be bounded.

Let $A$ be a finite collection of $d\geq 4$ non-parallel lines in $\mathbb{R}^2$ such that there are no points where at least four lines from $A$ meet. Must there exist a 'Gallai triangle': three lines from $A$ which intersect in three points, and each of these intersection points only intersects two lines from $A$?

The Sylvester-Gallai theorem implies that there must exist a point where only two lines from $A$ meet. This problem asks whether there must exist three such points which form a triangle (with sides induced by lines from $A$). Füredi and Palásti [FuPa84] showed this is false when $d\geq 4$ is not divisible by $9$. Escudero [Es16] showed this is false for all $d\geq 4$.

Let $f(n)$ be minimal such that the following holds. For any $n$ points in $\mathbb{R}^2$, not all on a line, there must be at least $f(n)$ many lines which intersect exactly 2 points. Does $f(n)\to \infty$? How fast?

Conjectured by Erdős and de Bruijn. The Sylvester-Gallai theorem states that $f(n)\geq 1$. That $f(n)\to \infty$ was proved by Motzkin [Mo51]. Kelly and Moser [KeMo58] proved that $f(n)\geq\tfrac{3}{7}n$ for all $n$. This is best possible for $n=7$. Motzkin conjectured that for $n\geq 13$ there are at least $n/2$ such lines. Csima and Sawyer [CsSa93] proved a lower bound of $f(n)\geq \tfrac{6}{13}n$ when $n\geq 8$. Green and Tao [GrTa13] proved that $f(n)\geq n/2$ for sufficiently large $n$.

$100

Let $1\leq k<n$. Given $n$ points in $\mathbb{R}^2$, at most $n-k$ on any line, the number of distinct lines formed joining two points is $\gg kn$.

In particular, given any $2n$ points with at most $n$ on a line there are $\gg n^2$ many lines formed by the points. Solved by Beck [Be83].

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.

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.

Does there exist $S\subseteq \mathbb{R}^2$ such that every set congruent to $S$ (that is, $S$ after some translation and rotation) contains exactly one point from $\mathbb{Z}^2$?

An old question of Steinhaus. Erdős was 'almost certain that such a set does not exist'.

In fact, such a set does exist, as proved by Jackson and Mauldin [JaMa02]. Their construction depends on the axiom of choice.

Let $g(k)$ be the smallest integer (if any such exists) such that any $g(k)$ points in $\mathbb{R}^2$ contains a convex $k$-gon with no point in the interior. Does $g(k)$ exist? If so, estimate $g(k)$.

A variant of the 'happy ending' problem, which asks for the same without the 'no point in the interior' restriction. Erdős observed $g(4)=5$ (as with the happy ending problem) but Harborth [Ha78] showed $g(5)=10$. Nicolás [Ni07] and Gerken [Ge08] independently showed that $g(6)$ exists. Horton [Ho83] showed that $g(n)$ does not exist for $n\geq 7$.

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 'a Hungarian high school student' gave an unspecified construction for $n=6$. Is this possible for $n=7$?

If $A\subseteq \mathbb{R}^d$ is any set of $2^d+1$ points then some three points in $A$ determine an obtuse angle.

For $d=2$ this is trivial. For $d=3$ there is an unpublished proof by Kuiper and Boerdijk. The general case was proved by Danzer and Grünbaum [DaGr62].

Is there some $c>0$ such that every measurable $A\subseteq \mathbb{R}^2$ of measure $\geq c$ contains the vertices of a triangle of area 1?

Erdős (unpublished) proved that this is true if $A$ has infinite measure, or if $A$ is an unbounded set of positive measure.

Let $A\subseteq \mathbb{R}^2$ be a measurable set with infinite measure. Must $A$ contain the vertices of an isosceles trapezoid of area $1$?

Erdős and Mauldin (unpublished) claim that this is true for trapezoids in general, but fails for parallelograms.