Dice Question Streamline Icon: https://streamlinehq.com

Subquadratic upper bound for isosceles‑triangle‑free subsets of the n×n grid

Prove an upper bound f(n) ≤ n^{1.99} for the function f(n), where f(n) denotes the maximum size of a subset S ⊆ [n]^2 such that no three distinct points a, b, c ∈ S form the vertices of an isosceles triangle (including degenerate/collinear cases, i.e., d(a,b) ≠ d(b,c) for all distinct a,b,c ∈ S with d denoting Euclidean distance).

Information Square Streamline Icon: https://streamlinehq.com

Background

The problem asks for the maximal number of points one can choose from the n×n grid [n]2 while avoiding any triple of points forming an isosceles triangle. Writing f(n) for this maximum, a trivial upper bound is f(n) ≤ n2. Using the observation that horizontal 3-term arithmetic progressions yield isosceles triangles and recent bounds on progression-free sets, this can be improved to f(n) ≤ e{-c (\log n){1/9}} * n2 for some absolute constant c.

Despite these improvements, any polynomially subquadratic bound remains out of reach. In particular, establishing an upper bound of the form f(n) ≤ n{1.99} would demonstrate genuinely subquadratic growth, and the authors explicitly note that this remains open.

References

Remarkably, even proving an upper bound of the form f(n)\leq n{1.99} is an open problem!

PatternBoost: Constructions in Mathematics with a Little Help from AI (2411.00566 - Charton et al., 1 Nov 2024) in Subsection “No isosceles triangles” (Section “Other problems”)