Lines through four points when no five are collinear

Take $n$ points in the plane with no five on a line. How many lines can pass through exactly four of them? Counting pairs gives an upper bound of order $n^2$ , and constructions come remarkably close to it. Erdős conjectured that the number is nevertheless $o(n^2)$ , and offered $100 for a proof.

Problem

Let $A\subset\mathbb{R}^2$ be a set of $n$ points, no five of which lie on a line. Is it true that the number of lines containing four points of $A$ is $o(n^2)$ as $n\to\infty$ ?

Since each such line contains $\binom42=6$ pairs of points and different lines share no pair, there are at most $\binom n2/6$ such lines; the question is whether this trivial bound can be beaten by more than a constant factor.

Background

The analogous question for three points has a different answer: there are configurations of $n$ points with no four on a line and $\sim n^2/6$ collinear triples, given by Burr, Grünbaum and Sloane (1974) and by Füredi and Palásti (1984). This is the classical orchard problem. Erdős asked whether four points per line behave differently, and speculated, after Grünbaum found configurations with $\gg n^{3/2}$ four-point lines, that $n^{3/2}$ might be the right order. The problem is Problem 71 on Ben Green's list of open problems, and a related question asks the same for $k$ points per line with no $k+1$ collinear.

What is known

Lower bound. Grünbaum (1976) constructed configurations with $\gg n^{3/2}$ four-point lines. Solymosi and Stojaković (2013) disproved Erdős's guess by constructing sets with no five on a line and at least $n^{2-O(1/\sqrt{\log n})}$ lines containing exactly four points, so the exponent $2$ cannot be lowered; their construction works for $k$ points per line for every fixed $k\ge4$ .
Upper bound. The trivial bound $\binom n2/6$ is the only general upper bound recorded on the source page; the question is whether it is attained up to a constant factor. General incidence bounds do not help here: the Szemerédi–Trotter theorem bounds the number of lines through at least $k$ of $n$ points by $O(n^2/k^3+n/k)$ , which for $k=4$ is again of order $n^2$ .
Small cases. The maximum number of lines through exactly four of $n$ points, the orchard-planting numbers, is OEIS sequence A006065 for $n\le20$ .

What would count as progress

A proof that the number of four-point lines is $o(n^2)$ , or even at most $(1/6-\varepsilon)\binom n2$ for large $n$ .
A construction with $cn^2$ four-point lines and no five collinear points, which would disprove the conjecture; for explicit coordinates this can be checked by exact arithmetic.
Exact or computer-assisted values of the maximum for small $n$ .
A formal proof against the Lean statement in formal-conjectures.

References

S. A. Burr, B. Grünbaum and N. J. A. Sloane, The orchard problem, Geom. Dedicata 2 (1974), 397–424. doi:10.1007/BF00147569
Z. Füredi and I. Palásti, Arrangements of lines with a large number of triangles, Proc. Amer. Math. Soc. 92 (1984), 561–566. doi:10.2307/2045427
J. Solymosi and M. Stojaković, Many collinear $k$ -tuples with no $k+1$ collinear points, Discrete Comput. Geom. 50 (2013), 811–820. doi:10.1007/s00454-013-9526-9
OEIS Foundation, Maximal number of 4-tree rows in $n$ -tree orchard problem, sequence A006065. oeis.org/A006065

Source

Imported from Erdős Problems 101, curated by Thomas Bloom. Open according to the source as of 2026-10-20. The statement above is written in our own words; see the source for the original wording, history and discussion.