Erdős Problem #1090 — Monochromatic Collinear Sets in Two-Colorings

About 51 years old · traced to

For every natural number k≥3k\geq 3, does there exist a finite set A⊂R2A\subset\mathbb R^2 such that, for every coloring of AA with two colors, there is a subset S⊆AS\subseteq A satisfying: SS is collinear, ∣S∣≥k|S|\geq k, every point of AA lying in the affine span of SS already belongs to SS, and all points of SS have the same color?

References

Progress summary

Refreshed
Open

The two-color problem remains open, and the known three-color obstruction does not answer it.

The problem asks whether, for every k≥3k \ge 3, some finite planar point set forces every two-coloring to contain a monochromatic line with at least kk points. No proposer or date is identified in the available sources.

Known results

  • Gruslys showed that the analogous statement fails for three colors, and therefore for every number of colors at least 33.
  • His construction gives, for every n≥2n \ge 2, an nn-point set with no four collinear points that admits a coloring with no monochromatic line.
  • The case k=2k=2 follows from the Motzkin–Rabin theorem; the problem concerns k≥3k \ge 3.
  • The three-color construction does not settle the two-color case.

Current status (as of September 2026): The two-color problem remains open; no proof or counterexample for the stated formulation has been found.

Sources

Solutions 0

No solutions have been posted yet.