Problem Statement
What is the size of the largest $A\subseteq \mathbb{R}^d$ such that every three points from $A$ determine an isosceles triangle? That is, for any three points $x,y,z$ from $A$, at least two of the distances $\lvert x-y\rvert,\lvert y-z\rvert,\lvert x-z\rvert$ are equal.
Categories:
Geometry Distances
Progress
When $d=2$ the answer is $6$ (due to Kelly [ErKe47] - an alternative proof is given by Kovács [Ko24c]). When $d=3$ the answer is $8$ (due to Croft [Cr62]). The best upper bound known in general is due to Blokhuis [Bl84] who showed that\[\lvert A\rvert \leq \binom{d+2}{2}.\]Alweiss has observed a lower bound of $\binom{d+1}{2}$ follows from considering the subset of $\mathbb{R}^{d+1}$ formed of all vectors $e_i+e_j$ where $e_i,e_j$ are distinct coordinate vectors. This set can be viewed as a subset of some $\mathbb{R}^d$, and is easily checked to have the required property.Weisenberg observed in the comments that an additional point can be added to Alweiss' construction, giving a lower bound of $\binom{d+1}{2}+1$.
The fact that the truth for $d=3$ is $8$ suggests that neither of these bounds is the truth.
See also [1088] for a generalisation.
Source: erdosproblems.com/503 | Last verified: January 15, 2026