OPEN
What is the size of the largest $A\subseteq \mathbb{R}^n$ 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.
When $n=2$ the answer is $6$ (due to Kelly
[ErKe47] - an alternative proof is given by Kovács
[Ko24c]). When $n=3$ the answer is $8$ (due to Croft
[Cr62]). The best upper bound known in general is due to Blokhuis
[Bl84] who showed that
\[\lvert A\rvert \leq \binom{n+2}{2}.\]
Alweiss has observed a lower bound of $\binom{n+1}{2}$ follows from considering the subset of $\mathbb{R}^{n+1}$ 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 $\mathbb{R}^n$, and is easily checked to have the required property.
The fact that the truth for $n=3$ is $8$ suggests that neither of these bounds is the truth.
