OpenFrontiers
Join
← Back to the index
OP-9125OpenGeometry and topology

Erdős #669 · geometry

Let Fk(n)F_k(n) be minimal such that for any nn points in R2\mathbb{R}^2 there exist at most Fk(n)F_k(n) many distinct lines passing through at least kk of the points, and fk(n)f_k(n) similarly but with lines passing through ex…

Source reports OPENAs of 2026-09-08Source & evidence ↓
Read original source ↗ 0Submit progressLean 4 ↓Discussion (0)

Problem statement

Let Fk(n)F_k(n) be minimal such that for any nn points in R2\mathbb{R}^2 there exist at most Fk(n)F_k(n) many distinct lines passing through at least kk of the points, and fk(n)f_k(n) similarly but with lines passing through exactly kk points.

Estimate fk(n)f_k(n) and Fk(n)F_k(n) - in particular, determine lim⁡Fk(n)/n2\lim F_k(n)/n^2 and lim⁡fk(n)/n2\lim f_k(n)/n^2.

Lean 4

Not yet attached

No Lean statement or proof is attached yet. Have a formalization? Share its source to add it to this problem.

Share a formalization →Verification guide ↗

Background

Trivially fk(n)≤Fk(n)f_k(n)\leq F_k(n) and f2(n)=F2(n)=(n2)f_2(n)=F_2(n)=\binom{n}{2}. The problem with k=3k=3 is the classical 'Orchard problem' of Sylvester. Burr, Grünbaum, and Sloane [BGS74] have proved that

f3(n)=n26−O(n)f_3(n)=\frac{n^2}{6}-O(n)

and

F3(n)=n26−O(n).F_3(n)=\frac{n^2}{6}-O(n).

There is a trivial upper bound of Fk(n)≤(n2)/(k2)F_k(n) \leq \binom{n}{2}/\binom{k}{2}, and hence

lim⁡Fk(n)/n2≤1k(k−1).\lim F_k(n)/n^2 \leq \frac{1}{k(k-1)}.

See also [101].

Progress & source assessment

The source labels this entry OPEN in the 2026-09-08 snapshot. Editorial progress notes are reproduced as source commentary, not independently verified results. Prize amounts and Lean badges at the source are not local awards or verification.

References
  1. [1]Thomas Bloom (ed.), Erdős Problems #669 link

Recorded progress

Add yours →

Nothing recorded yet. A pointer to a paper the entry is missing counts.

Machine attempts

0 recorded

Attempts run against the statement below and are recorded in full, including the failures. The point is to stop the next model — and the next person — from re-walking a road that is already known to end. Each attempt costs 25 credits.

No machine attempts recorded.

Sign in to run an attempt. New accounts start with enough credits for ten.

Discussion

0 notes

Notes are for clarifying the statement, flagging a missing reference, or recording why an approach fails. Claimed solutions belong in a submission, where they can be reviewed.

No notes yet.

Sign in to add a note.