gpt pro 5.2 attempt on Math-5795

Problem #652

1) FORMAL RESTATEMENT

Let $n\ge 2$ and let $X=\{x_1,\dots,x_n\}\subset \mathbb{R}^2$ be a set of $n$ distinct points. For a point $x\in X$ define \[ R_X(x)\;:=\;\left\lvert \{\,\mathrm{dist}(x,y): y\in X\setminus\{x\}\,\} \right\rvert\in\{1,2,\dots,n-1\}, \] the number of distinct distances from $x$ to the other points. Order the points of $X$ so that \[ R_X(x_{(1)})\le R_X(x_{(2)})\le \cdots \le R_X(x_{(n)}), \] where $x_{(k)}$ denotes a point attaining the $k$th smallest value (break ties arbitrarily). Fix $k\in\mathbb{N}$.

Define \[ \alpha_k\;:=\;\inf\left\{\,\alpha>0:\exists N_0\ \forall n\ge N_0\ \exists X\subset\mathbb{R}^2,\ \left\lvert X \right\rvert=n,\ \text{with }R_X(x_{(k)})\le \alpha\sqrt n\right\}. \] (Using $\le$ instead of $<$ is immaterial asymptotically and avoids the issue that a "minimum" need not be attained.)

Question. Is it true that $\alpha_k\to\infty$ as $k\to\infty$?

2) QUICK LITERATURE/CONTEXT CHECK

• A construction of Elekes ("circle grid") shows that for every fixed $k$ one can realize $R_X(x_{(k)})=O_k(\sqrt n)$, i.e. $\alpha_k<\infty$ for every fixed $k$.
• The bipartite distinct-distances paper of Mathialagan (EJC 2021) contains a theorem (their Theorem 14) implying a lower bound of order $\sqrt{k}\,\sqrt n$ for $R_X(x_{(k)})$ once $n$ is sufficiently large compared to $k$. This yields $\alpha_k=\Omega(\sqrt{k})$, and in particular $\alpha_k\to\infty$.

3) ATTACK PLAN

1. Let $X$ be an arbitrary $n$-point set and let $P$ be the subset of the $k$ points with smallest $R_X(\cdot)$, i.e. $P=\{x_{(1)},\dots,x_{(k)}\}$. Let $Q=X\setminus P$.
2. Apply the following bipartite theorem: for $\left\lvert P \right\rvert=m$ and $\left\lvert Q \right\rvert=N$ with $m\le N^{1/3}$, there exists $p\in P$ that determines $\Omega(\sqrt{mN})$ distinct distances to $Q$.
3. Since $p\in P$, we have $R_X(p)\le R_X(x_{(k)})$; but also $R_X(p)\ge R_Q(p)$ (distinct distances from $p$ to $Q$). This will force $R_X(x_{(k)})\ge \Omega(\sqrt{k(n-k)})$.
4. Divide by $\sqrt n$ and choose $n$ large compared to $k^3$ to conclude $\alpha_k\ge c\sqrt{k}$ for an absolute constant $c>0$.

4) WORK

Step 1: A bipartite lower bound (quoted).

We quote the following theorem from Mathialagan [Mathialagan2021, Theorem 14].

Theorem (Mathialagan). Let $P,Q\subset\mathbb{R}^2$ be finite point sets with $\left\lvert P \right\rvert=m$ and $\left\lvert Q \right\rvert=n$ satisfying $2\le m\le n^{1/3}$. Then there exists a point $p\in P$ such that the number of distinct distances from $p$ to points of $Q$ is $\Omega(\sqrt{mn})$.

Here the implied constant in $\Omega(\cdot)$ is absolute. (For completeness: Mathialagan proves this via a crossing-number argument on a multigraph built from concentric circles centered at points of $P$ passing through points of $Q$; see [Mathialagan2021, pp. 6--7].)

Step 2: Apply the bipartite theorem to the $k$ smallest $R$-values.

Let $X\subset\mathbb{R}^2$ be any $n$-point set and write $x_{(1)},\dots,x_{(n)}$ for the ordering by $R_X(\cdot)$. Fix $k\ge 2$ and set \[ P:=\{x_{(1)},\dots,x_{(k)}\},\qquad Q:=X\setminus P. \] Then $\left\lvert P \right\rvert=k$ and $\left\lvert Q \right\rvert=n-k$. Assume that \begin{equation}\label{wsref-0-65-71-3a-6d-6b-5f-63-6f-6e-64-69-74-69-6f-6e} 2\le k\le (n-k)^{1/3}. \end{equation} Under (eq:mk_condition), Mathialagan's theorem yields a point $p\in P$ such that \begin{equation}\label{wsref-0-65-71-3a-6c-6f-77-65-72-5f-62-69-70} \left\lvert \{\mathrm{dist}(p,q): q\in Q\} \right\rvert\;\ge\; c\sqrt{k(n-k)} \end{equation} for some absolute constant $c>0$.

But \[ R_X(p)\;\ge\;\left\lvert \{\mathrm{dist}(p,q): q\in Q\} \right\rvert, \] because the set of distances from $p$ to $Q$ is a subset of the set of distances from $p$ to $X\setminus\{p\}$. Therefore, from (eq:lower_bip) we get \begin{equation}\label{wsref-0-65-71-3a-6c-6f-77-65-72-5f-52-70} R_X(p)\ge c\sqrt{k(n-k)}. \end{equation} Since $p\in P$ and $P$ consists of the $k$ points with the smallest $R_X$-values, \[ R_X(p)\le R_X(x_{(k)}). \] Combining this with (eq:lower_Rp) gives the universal bound \begin{equation}\label{wsref-0-65-71-3a-6c-6f-77-65-72-5f-52-6b} R_X(x_{(k)})\ge c\sqrt{k(n-k)}\qquad\text{whenever }2\le k\le (n-k)^{1/3}. \end{equation}

Step 3: Convert to a lower bound on $\alpha_k$.

Fix $k\ge 2$ and choose $n$ sufficiently large so that $n-k\ge k^3$; equivalently $n\ge k^3+k$. Then $k\le (n-k)^{1/3}$, so (eq:lower_Rk) applies and yields \[ R_X(x_{(k)})\ge c\sqrt{k(n-k)}\ge c\sqrt{k}\,\sqrt{n-k}. \] For example, if $n\ge 2k^3$ then $n-k\ge n/2$ and we get \begin{equation}\label{wsref-0-65-71-3a-6c-6f-77-65-72-5f-72-61-74-69-6f} \frac{R_X(x_{(k)})}{\sqrt n}\ge \frac{c}{\sqrt 2}\,\sqrt{k}\,. \end{equation} Since (eq:lower_ratio) holds for every $n$-point set $X$ once $n\ge 2k^3$, any constant $\alpha$ for which one can have $R_X(x_{(k)})\le \alpha\sqrt n$ for large $n$ must satisfy $\alpha\ge (c/\sqrt2)\sqrt{k}$. Therefore \[ \alpha_k\ge c'\sqrt{k}\qquad\text{for some absolute }c'>0. \] In particular $\alpha_k\to\infty$ as $k\to\infty$.

5) VERIFICATION / FAST REALITY CHECK

• Sanity check for small $k$. For $k=2$ and large $n$, (eq:lower_Rk) gives $R_X(x_{(2)})\ge c\sqrt{2(n-2)}=\Omega(\sqrt n)$, consistent with the known phenomenon that one can force at least constant$\times\sqrt n$ distinct distances for the second-smallest $R$.
• Scaling. If $n\asymp k^3$, then (eq:lower_Rk) gives $R_X(x_{(k)})=\Omega(k^2)$ while $\sqrt n\asymp k^{3/2}$, so $R_X(x_{(k)})/\sqrt n=\Omega(\sqrt k)$, matching the claimed growth.
• Compatibility with Elekes. Elekes' construction yields $R_X(x_{(k)})\le C_k\sqrt n$ with $C_k$ depending on $k$; our lower bound shows $C_k$ must grow at least on the order of $\sqrt k$.

6) FINAL

FULL SOLUTION
PROVED

Using Mathialagan's bipartite theorem (Theorem 14 in [Mathialagan2021]), one obtains the universal bound $R_X(x_{(k)})\ge c\sqrt{k(n-k)}$ for all $n$-point sets $X$ whenever $k\le (n-k)^{1/3}$. Taking $n\ge 2k^3$ gives $R_X(x_{(k)})/\sqrt n\ge c'\sqrt{k}$. Consequently $\alpha_k\ge c'\sqrt{k}$ for an absolute $c'>0$, and hence $\alpha_k\to\infty$ as $k\to\infty$.

7) COMPLETION ESTIMATE

100%.