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Math-6140-A1 · revision r1 · on Math-6140

Reused writeup on Math-6140 (gpt pro 5.2)

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

Contents
  1. Erdős Problem 1086 --- collection record
  2. PROVENANCE AND STATUS
  3. REUSED SOURCE 1
  4. Erdős Problem 1086 (many equal-area triangles)
  5. FORMAL RESTATEMENT
  6. QUICK LITERATURE/CONTEXT CHECK
  7. ATTACK PLAN
  8. WORK
  9. Fast reality check.
  10. VERIFICATION
  11. FINAL
  12. COMPLETION ESTIMATE

Erdős Problem 1086 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/1086

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/1086.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/1086.tex.

Erdős Problem 1086 (many equal-area triangles)

FORMAL RESTATEMENT

For a finite set $P\subset\mathbb{R}^2$ with $|P|=n$, and a real number $A>0$, let $T_A(P)$ be the set of unordered triples $\{p,q,r\}\subset P$ with (nonzero) triangle area $\mathrm{Area}(pqr)=A$. Define \[ g(n):=\max\{\,|T_A(P)|: P\subset\mathbb{R}^2,\ |P|=n,\ A>0\,\}. \] Equivalently (by scaling), $g(n)$ is the maximum possible number of unit-area triangles determined by $n$ points in the plane. The problem asks for the asymptotic growth of $g(n)$.

QUICK LITERATURE/CONTEXT CHECK

The problem statement records that Erdős--Purdy proved $n^2\log\log n \ll g(n) \ll n^{5/2}$, and that the best known upper bound recorded there is $g(n)\ll n^{20/9}$. I do not use further literature.

ATTACK PLAN

  • Proof track (partial). Prove two basic structural lemmas: $1$ for any fixed base segment $pq$, unit-area third vertices must lie on two parallel lines; $2$ if the point set has bounded collinearity, this yields an $O(n^2)$ (or $O(Ln^2)$) upper bound. Then give an explicit construction with $\Omega(n^2)$ unit-area triangles.
  • Disproof track. Not applicable.

WORK

Fast reality check.
For the explicit two-line construction below, brute-force counts of unit-area triangles (unordered triples) give: $n=6\mapsto 9$, $n=8\mapsto 20$, $n=10\mapsto 35$. (Computed exactly.)
Lemma (Third vertex lies on two parallel lines): Fix distinct points $p,q\in\mathbb{R}^2$ and an area $A>0$. The set of points $r\in\mathbb{R}^2$ such that $\mathrm{Area}(pqr)=A$ is the union of two lines parallel to the line $pq$, each at perpendicular distance $\frac{2A}{\|p-q\|}$ from $pq$ (one on each side).
Proof: Let $\ell$ be the line through $p$ and $q$. For any point $r$, $\mathrm{Area}(pqr)=\frac12\|p-q\|\cdot \mathrm{dist}(r,\ell)$, where $\mathrm{dist}(r,\ell)$ is the perpendicular distance from $r$ to $\ell$. Thus $\mathrm{Area}(pqr)=A$ iff $\mathrm{dist}(r,\ell)=\frac{2A}{\|p-q\|}$. The locus of points at a fixed distance $h$ from a line is the union of the two lines parallel to it at distances $h$ on either side. Taking $h=\frac{2A}{\|p-q\|}$ gives the claim. ∎
Lemma (Bound under limited collinearity): Let $P\subset\mathbb{R}^2$ be a set of $n$ points and suppose no line contains more than $L$ points of $P$. Then for any $A>0$, the number of triangles of area exactly $A$ determined by $P$ satisfies \[ |T_A(P)|\le 2L\binom{n}{2}. \] In particular, if $P$ is in general position (no three collinear, so $L=2$), then $|T_A(P)|\le 4\binom{n}{2}=2n(n-1)$.
Proof: Count triangles by choosing their base $\{p,q\}\subset P$. For fixed $p\ne q$, by the previous lemma, any point $r\in P$ with $\mathrm{Area}(pqr)=A$ must lie on one of two lines parallel to $pq$. Each such line contains at most $L$ points of $P$ by hypothesis, so for the base $\{p,q\}$ there are at most $2L$ valid choices of $r$. Summing over all $\binom{n}{2}$ choices of base gives $|T_A(P)|\le 2L\binom{n}{2}$.

For general position, set $L=2$. ∎

Lemma (Two-line construction with $\Omega(n^2)$ unit-area triangles): For each integer $m\ge 3$, let $P$ consist of $m$ points on the line $y=0$ at $(0,0),(1,0),\dots,(m-1,0)$, and $m$ points on the line $y=1$ at $(0,1),(1,1),\dots,(m-1,1)$. Then $|P|=n=2m$, and $P$ determines at least $2m(m-2)$ triangles of area exactly $1$. Consequently, $g(n)\ge \frac12 n^2 - O(n)$.
Proof: Consider triangles with base on $y=0$ and third vertex on $y=1$. If $p=(i,0)$ and $q=(i+2,0)$, then $\|p-q\|=2$ and the height from any $r=(j,1)$ to the base line is $1$, so $\mathrm{Area}(pqr)=\frac12\cdot 2\cdot 1=1$, provided $r$ is not collinear with $p,q$ (which cannot happen since $r$ lies on a different horizontal line). There are $m-2$ choices of the base pair $(i,0),(i+2,0)$ with $0\le i\le m-3$, and $m$ choices of $r$ on $y=1$, giving $m(m-2)$ unit-area triangles. By symmetry, using bases on $y=1$ and third vertices on $y=0$ gives another $m(m-2)$ unit-area triangles, for a total of at least $2m(m-2)$. In terms of $n=2m$, this is $\frac12 n^2-2n$. ∎

VERIFICATION

  • Lemma 1 is exact and uses only the elementary base$\times$height formula.
  • Lemma 2 is conditional on the maximum number $L$ of collinear points; the conclusion specializes cleanly to general position.
  • For the two-line construction, brute-force enumeration confirms counts exceed the proven lower bound: for $n=6,8,10$ the exact numbers of unit-area triangles are $9,20,35$ respectively (vs. the lower bounds $6,16,30$).

FINAL

UNRESOLVED.
  1. (i) Strongest proved partial result here: unconditional construction yields $g(n)\ge \frac12 n^2-O(n)$; additionally, under the hypothesis "no line contains more than $L$ points" one has $g(n)\le 2L\binom{n}{2}$ (in particular $O(n^2)$ in general position).
  2. (ii) First gap: obtain a nontrivial unconditional upper bound better than $O(n^3)$ without importing deep incidence machinery beyond what is stated in the problem file.
  3. (iii) Top 3 next moves: $1$ analyze contributions from "heavy" lines (with many points) and show triangles of fixed area force many incidences between points and parallel lines; $2$ attempt a double-counting scheme that reduces the problem to bounding incidences between points and a family of lines determined by pairs; $3$ computationally search for configurations that beat the two-line construction substantially to understand extremizers.
  4. (iv) Minimal counterexample structure (to the heuristic that $g(n)$ is near $n^2$ up to polylogs): point sets with many points on many parallel lines, arranged so that for many bases $pq$, both parallel offset lines at the required height each contain $\Theta(n)$ points.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.