we-solve (beta)

Math-894-A2 · revision r1 · on Math-894

gpt pro 5.2 attempt on Math-894

No claim recorded · LLM-generated · gpt pro 5.2 ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

FORMAL RESTATEMENT. Fix an integer $k\ge 2$. For a finite set $P\subset\mathbb{R}^2$ with $|P|=n$, let \begin{align*} L_{\ge k}(P) &:= \{ \ell : \ell \text{ is a line in }\mathbb{R}^2 \text{ and } |\ell\cap P|\ge k\},\\ L_{=k}(P) &:= \{ \ell : \ell \text{ is a line in }\mathbb{R}^2 \text{ and } |\ell\cap P|= k\}. \end{align*} Define extremal quantities \[ F_k(n) := \max_{|P|=n} |L_{\ge k}(P)|, \qquad f_k(n) := \max_{|P|=n} |L_{=k}(P)|. \] (These are equivalent to the "minimal such that for any $P$ there exist at most $F_k(n)$ lines" phrasing.)

The problem asks for asymptotics of $F_k(n)$ and $f_k(n)$ as $n\to\infty$ for fixed $k$, in particular the limits (if they exist) \[\lim_{n\to\infty} F_k(n)/n^2,\qquad \lim_{n\to\infty} f_k(n)/n^2.\]

QUICK LITERATURE/CONTEXT CHECK. The supplied statement records:

  • $f_2(n)=F_2(n)=\binom n2$ (trivial).
  • For $k=3$ (Orchard problem), Burr--Grünbaum--Sloane proved $f_3(n)=\frac{n^2}{6}-O(n)$ and $F_3(n)=\frac{n^2}{6}-O(n)$.
  • A trivial upper bound $F_k(n)\le \binom n2/\binom k2$, hence $\limsup F_k(n)/n^2\le 1/(k(k-1))$.
Per project rules, I do not use external sources beyond these statements.

ATTACK PLAN.

  • Prove the trivial pair-counting upper bound carefully (it is the main unconditional tool available from the prompt).
  • For $k=3$, use the stated asymptotic to deduce the actual limit $\lim F_3(n)/n^2=\lim f_3(n)/n^2=1/6$.
  • Give computational sanity checks on concrete configurations (e.g. integer grids) to understand magnitudes, though these do not claim optimality.

WORK.

Fast reality check (tiny $n$). For any $n<k$, $F_k(n)=f_k(n)=0$. For $n=k$, $F_k(k)\ge 1$ (take $k$ collinear points) and clearly $F_k(k)\le 1$, so $F_k(k)=f_k(k)=1$.

Lemma 669.1 (Exact-to-at-least relation). For every $n$ and $k$, \[ F_k(n) = \max_{|P|=n} |L_{\ge k}(P)| = \max_{|P|=n} \sum_{j\ge k} |\{\ell: |\ell\cap P|=j\}|. \] In particular, for a fixed point set $P$ one has the identity \[ |L_{\ge k}(P)| = \sum_{j\ge k} |\{\ell: |\ell\cap P|=j\}|. \]

Proof. For a fixed $P$, each line $\ell$ with $|\ell\cap P|\ge k$ has a well-defined integer $j=|\ell\cap P|\ge k$. The families $\{\ell:|\ell\cap P|=j\}$ for different $j$ are disjoint and their union over $j\ge k$ equals $L_{\ge k}(P)$. Taking cardinalities yields the displayed identity for $P$. Maximizing over $P$ preserves equality. $\square$

Lemma 669.2 (Trivial pair-counting upper bound). For all integers $n\ge k\ge 2$, \[ F_k(n) \le \frac{\binom n2}{\binom k2}. \] Consequently, \[ \limsup_{n\to\infty} \frac{F_k(n)}{n^2} \le \frac{1}{k(k-1)}. \]

Proof. Fix a set $P$ of $n$ points. Each line $\ell\in L_{\ge k}(P)$ contains at least $k$ points of $P$, hence contains at least $\binom{k}{2}$ unordered pairs of points from $P$. On the other hand, any unordered pair of distinct points of $P$ lies on exactly one line (the unique line through them), so the sets of pairs contributed by distinct lines are disjoint. Therefore the total number of unordered pairs in $P$, namely $\binom n2$, is at least the sum over $\ell\in L_{\ge k}(P)$ of $\binom{k}{2}$. That is, \[ \binom n2 \ge |L_{\ge k}(P)|\,\binom k2. \] Rearranging gives $|L_{\ge k}(P)|\le \binom n2/\binom k2$. Maximizing over $P$ yields the same bound for $F_k(n)$. Dividing by $n^2$ and taking $\limsup$ gives the stated constant upper bound. $\square$

Lemma 669.3 (Consequences for $k=3$ using the stated asymptotic). Assuming the statement in the prompt that $F_3(n)=\frac{n^2}{6}-O(n)$ and $f_3(n)=\frac{n^2}{6}-O(n)$, one has \[ \lim_{n\to\infty} \frac{F_3(n)}{n^2} = \frac16, \qquad \lim_{n\to\infty} \frac{f_3(n)}{n^2} = \frac16. \]

Proof. If $F_3(n)=\frac{n^2}{6}-O(n)$ then there is a constant $C>0$ with $|F_3(n)-n^2/6|\le Cn$ for all large $n$. Dividing by $n^2$ gives \[ \left|\frac{F_3(n)}{n^2}-\frac16\right| \le \frac{C}{n}, \] which tends to $0$ as $n\to\infty$. Therefore $F_3(n)/n^2\to 1/6$. The same argument applies to $f_3(n)$. $\square$

Computational sanity check (not claiming optimality). For an $m\times m$ integer grid $P=\{0,1,\dots,m-1\}^2$ (so $n=m^2$), I computed the number of lines containing at least $k$ grid points for small $m$. The exact counts found were: 

3x3 grid (n=9):  lines with >=3 points = 8
4x4 grid (n=16): lines with >=3 points = 14, lines with >=4 points = 10
5x5 grid (n=25): lines with >=3 points = 32, lines with >=4 points = 16, lines with >=5 points = 12
6x6 grid (n=36): lines with >=3 points = 58, lines with >=4 points = 22, lines with >=5 points = 18
7x7 grid (n=49): lines with >=3 points = 108, lines with >=4 points = 44, lines with >=5 points = 24
These grow much more slowly than the quadratic upper bound from Lemma 669.2, illustrating that achieving $\Theta(n^2)$ many $k$-rich lines requires much more structured configurations than a square grid.

VERIFICATION.

  • Lemma 669.2: checked that pair-sets from distinct lines are disjoint because a pair of points determines a unique line.
  • For $k=3$, Lemma 669.2 gives $F_3(n)\le \binom n2/3 = \frac{n^2}{6}-\frac{n}{6}$, matching the stated asymptotic order and consistent with Lemma 669.3.
  • The grid computation was rechecked by recomputing line keys via normalized integer coefficients; it is intended only as a sanity check.

FINAL. UNRESOLVED.

(i) Strongest proved partial results.

  • For all fixed $k\ge 2$, $F_k(n)\le \binom n2/\binom k2$, hence $\limsup F_k(n)/n^2\le 1/(k(k-1))$ (Lemma 669.2).
  • For $k=3$, using the asymptotics stated in the prompt, one gets the exact limits $\lim F_3(n)/n^2=\lim f_3(n)/n^2=1/6$ (Lemma 669.3).

(ii) First gap (crisp statement). For each fixed $k\ge 4$, determine whether the limits $\lim_{n\to\infty} F_k(n)/n^2$ and $\lim_{n\to\infty} f_k(n)/n^2$ exist, and if so compute them. In particular, decide whether the trivial upper bound constant $1/(k(k-1))$ is sharp.

(iii) Top 3 next moves.

  1. Constructions: search for algebraic or finite-geometry inspired point sets in $\mathbb{R}^2$ producing $\Theta(n^2)$ many lines with exactly $k$ points for fixed $k\ge 4$.
  2. Upper bounds beyond pair-counting: prove stronger restrictions on the number of $k$-rich lines using incidence geometry (e.g. inequalities relating $k$-rich lines to point-line incidences), aiming to show the constant is strictly smaller than $1/(k(k-1))$ when $k\ge 4$.
  3. Computer search for small $k$: for $k=4$, do exhaustive/heuristic searches over integer/rational point sets to estimate $F_4(n)$ and guess the correct asymptotic constant.

(iv) What a minimal counterexample would likely look like. If the upper bound constant $1/(k(k-1))$ is not sharp for some $k\ge 4$, then any near-extremal configuration would have to use many lines each contributing close to the minimum $\binom{k}{2}$ pairs, with few (ideally no) lines containing more than $k$ points, because higher multiplicity lines "waste" pairs. Thus a minimal counterexample to sharpness would likely consist of a configuration where unavoidable geometric constraints force many points to lie on some lines with $>k$ points, depressing the total count of distinct $k$-rich lines below the pair-counting maximum.