SOLVED
Let $\alpha_n$ be the infimum of all $0\leq \alpha\leq \pi$ such that in every set $A\subset \mathbb{R}^2$ of size $n$ there exist three distinct points $x,y,z\in A$ such that the angle determined by $xyz$ is at least $\alpha$. Determine $\alpha_n$.
Blumenthal's problem. Szekeres
[Sz41] showed that
\[\alpha_{2^n+1}> \pi \left(1-\frac{1}{n}+\frac{1}{n(2^n+1)^2}\right)\]
and
\[\alpha_{2^n}\leq \pi\left(1-\frac{1}{n}\right).\]
Erdős and Szekeres
[ErSz60] showed that
\[\alpha_{2^n}=\alpha_{2^n-1}= \pi\left(1-\frac{1}{n}\right),\]
and suggested that perhaps $\alpha_{N}=\pi(1-1/n)$ for $2^{n-1}<N\leq 2^n$. This was disproved by Sendov
[Se92].
Sendov [Se93] provided the definitive answer, proving that $\alpha_N=\pi(1-1/n)$ for $2^{n-1}+2^{n-3}<N\leq 2^n$ and $\alpha_N=\pi(1-\frac{1}{2n-1})$ for $2^{n-1}<N\leq 2^{n-1}+2^{n-3}$.
