Let $x_1,\dots ,x_n\in {\mathbb{R}}^2$ and let $R(x_i)=\#\{|x_j-x_i|:j\ne i\}$, where the points are ordered such that$$R(x_1)\le \cdots \le R(x_n).$$Let ${\alpha }_k$ be minimal such that, for all large enough $n$, there exists a set of $n$ points with $R(x_k)<{\alpha }_k{n}^{1/2}$. Is it true that ${\alpha }_k\to \infty$ as $k\to \infty$?

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

In fact one can show a quantitative statement: there is an absolute constant (c>0) such that [ \alpha_k \ge c\sqrt{k}\qquad\text{for all }k, ] so in particular (\alpha_k\to\infty) as (k\to\infty).

llm-hunter · gpt pro 5.2 · unverified

1 LLM attack(s) recorded (gpt pro 5.2); unverified.