Erdős Problem 659

References:

• erdosproblems.com/659
• [MoOs06] Moree, Pieter and Osburn, Robert, Two-dimensional lattices with few distances. Enseign. Math. (2) (2006), 361--380
• [ErFi96] Erdős, Paul and Fishburn, Peter, Maximum planar sets that determine {$k$} distances. Discrete Math. (1996), 115--125.
• Gr26: Benjamin Grayzel, Solution to a Problem of Erdős Concerning Distances and Points

theorem Erdos659.erdos_659 : True ↔ ∃ (A : ℕ → Finset (EuclideanSpace ℝ (Fin 2))), (∀ (n : ℕ), (A n).card = n ∧ ∀ S ⊆ A n, S.card = 4 → 3 ≤ EuclideanGeometry.distinctDistances S) ∧ (fun (n : ℕ) => ↑(EuclideanGeometry.distinctDistances (A n))) =O[Filter.atTop] fun (n : ℕ) => ↑n / √(Real.log ↑n)

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 $\ll \frac{n}{\sqrt{\log n}}$?

There does exist such a set: a suitable truncation of the lattice $\{(a,b\sqrt{2}): a,b\in\mathbb{Z}\}$ suffices. This construction appears to have been first considered by Moree and Osburn \cite{MoOs06}, who proved that it has $\ll \frac{n}{\sqrt{\log n}}$ many distinct distances. This construction was independently found by Lund and Sheffer, who further noted that this configuration contains no squares or equilateral triangles.

There are only six possible configurations of $4$ points which determine only $2$ distances (first noted by Erdős and Fishburn [ErFi96]), and five of them contain either a square or an equilateral triangle. The remaining configuration contains four points from a regular pentagon, and Grayzel [Gr26] (using Gemini) has noted in the comments that this configuration can also be ruled out, thus giving a complete solution to this problem. Boris Alexeev using Aristotle provides a formalisation of the proof.