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Math-675-A2 · revision r1 · on Math-675

gpt pro 5.2 attempt on Math-675

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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

Contents
  1. Erdős Problem #101
  2. FORMAL RESTATEMENT
  3. QUICK LITERATURE/CONTEXT CHECK
  4. ATTACK PLAN
  5. WORK
  6. VERIFICATION
  7. FINAL

Erdős Problem #101

FORMAL RESTATEMENT

Verbatim problem statement (from file). "Given $n$ points in $\R^2$, no five of which are on a line, the number of lines containing four points is $o(n^2)$."

Definitions and conventions. Let $P\subset\R^2$ with $|P|=n$. Assume: \[\text{(No-5-collinear)}\qquad \text{every line in }\R^2\text{ contains at most 4 points of }P. \] Let $L_4(P)$ be the number of distinct (geometric) lines $\ell$ such that $|\ell\cap P|=4$. (The "no five collinear" hypothesis implies "containing four points" is the same as "containing at least four points".)

Claim to prove (asymptotic). \[\frac{L_4(P)}{n^2}\to 0\quad\text{as }n\to\infty,\] uniformly over all such $P$. Equivalently: for every $\varepsilon>0$ there exists $N(\varepsilon)$ such that if $n\ge N(\varepsilon)$ then $L_4(P)\le \varepsilon n^2$.

QUICK LITERATURE/CONTEXT CHECK

No external browsing was used. The file states that there are constructions with $L_4(P)\ge n^{2-O(1/\sqrt{\log n})}$, so any $o(n^2)$ upper bound would have to be very close to sharp. I do not attempt to reproduce those constructions.

ATTACK PLAN

  • Upper bound track: Use double counting (pairs, incidences) and possibly the Szemerédi--Trotter theorem to bound the number of 4-rich lines.
  • Reduction track: Relate 4-point lines to counts of collinear triples or to rich-line incidence graphs.
  • Disproof track: Try to construct a configuration with $\Theta(n^2)$ many 4-point lines while keeping no five collinear; this would disprove the conjectured $o(n^2)$ bound.

WORK

FAST REALITY CHECK (very small $n$).
  • $n\le 3$: $L_4(P)=0$.
  • $n=4$: if all 4 points are collinear then $L_4(P)=1$; if not, $L_4(P)=0$.
  • $n=5$: one can have $L_4(P)=1$ (four collinear, fifth off the line).
  • $n=7$: one can realize $L_4(P)=2$ by taking two 4-point lines meeting in one point (uses $1+3+3=7$ points).
Lemma (Pair-counting upper bound): If $P\subset\R^2$ has $n$ points with no five collinear, then \[L_4(P)\le \frac{\binom{n}{2}}{\binom{4}{2}}=\frac{n(n-1)}{12}. \]
Proof: Each line $\ell$ with $|\ell\cap P|=4$ contains exactly $\binom{4}{2}=6$ unordered pairs of points from $P$. Distinct lines contribute disjoint sets of pairs because a pair of distinct points determines a unique line. Therefore \[6\,L_4(P)\le \binom{n}{2}, \] which rearranges to the stated bound. ∎
Proposition (4-point lines through a fixed point): Let $P\subset\R^2$ be a set of $n$ points with no five collinear. Fix a point $p\in P$ and let $r(p)$ denote the number of lines $\ell$ such that $p\in\ell$ and $|\ell\cap P|=4$. Then \[r(p)\le \frac{n-1}{3}.\] Consequently, \[L_4(P)=\frac14\sum_{p\in P} r(p)\le \frac14\cdot n\cdot \frac{n-1}{3}=\frac{n(n-1)}{12},\] recovering the pair-counting bound from a different viewpoint.
Proof: Each such 4-point line through $p$ contains exactly three further points of $P$ besides $p$. Moreover, two distinct lines through $p$ cannot share any other point of $P$ (if they shared $q\ne p$ then both lines would be the unique line determined by $p$ and $q$). Hence the $r(p)$ distinct 4-point lines through $p$ collectively contain at least $3r(p)$ distinct points of $P\setminus\{p\}$. Since $|P\setminus\{p\}|=n-1$, we have $3r(p)\le n-1$, i.e. $r(p)\le (n-1)/3$.

For the second claim, observe that $\sum_{p\in P} r(p)$ counts incidences between points and 4-point lines, and each 4-point line has exactly $4$ incidences. Thus $\sum_{p\in P} r(p)=4L_4(P)$. Substituting the bound $r(p)\le (n-1)/3$ gives the displayed inequality for $L_4(P)$. ∎

VERIFICATION

  • The pair-counting lemma is a direct double count of unordered point pairs on 4-point lines; it uses only that distinct point pairs determine unique lines.
  • The proposition bounds the number of 4-point lines through a fixed point by noting that distinct lines through that point use disjoint sets of the remaining points; no external theorems are invoked.
  • Both bounds are $\Theta(n^2)$, so they do not approach the conjectured $o(n^2)$; they are consistent with the near-quadratic constructions mentioned in the file.

FINAL

UNRESOLVED.
  1. Strongest proved partial result here: For any such $P$, $L_4(P)\le n(n-1)/12$ by pair counting.
  2. First gap (crisp): Prove that for every $\varepsilon>0$ and all sufficiently large $n$, every $n$-point set in $\R^2$ with no five collinear has $L_4(P)\le \varepsilon n^2$.
  3. Top 3 next moves (concrete):
    1. Try to strengthen Szemerédi--Trotter-type bounds using the additional constraint "no five collinear" to get a saving over the trivial $\Theta(n^2)$.
    2. Convert 4-point lines into a 4-uniform hypergraph on $P$ (edges = 4-tuples on a line) and attempt to exploit geometric restrictions to bound the number of edges.
    3. Computational experiment: for moderate $n$ (say $n\le 50$), search over known high-incidence constructions (grids, projected grids, algebraic curves) and measure $L_4(P)/n^2$ to see decay rates.
  4. What a minimal counterexample would likely look like: A family $P_n$ with no five collinear and with $L_4(P_n)\ge c n^2$ for some fixed $c>0$ (i.e. a positive density of all $\binom{n}{2}$ pairs lie in 4-point lines). Such a configuration would resemble a near-Steiner system of 4-point lines geometrically realizable in $\R^2$.