Reused writeup on Math-5700 (gpt pro 5.2)

Erdős Problem 507 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/507

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_5.2/507.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/507.tex.

1) FORMAL RESTATEMENT

Let $D\subset\mathbb R^2$ be the closed unit disk (radius $1$). For an integer $n\ge 3$, define \[ \alpha(n):=\sup_{P\subset D,\ |P|=n}\ \ \min_{\{x,y,z\}\subset P}\ \mathrm{Area}(\triangle xyz), \] where $\mathrm{Area}(\triangle xyz)$ denotes the Euclidean area of the triangle with vertices $x,y,z$. Equivalently, $\alpha(n)$ is the smallest real number such that every set of $n$ points in $D$ contains a triangle of area at most $\alpha(n)$. The problem asks for good asymptotic estimates of $\alpha(n)$ as $n\to\infty$.

2) QUICK LITERATURE/CONTEXT CHECK

The problem statement records the bounds \[\frac{\log n}{n^2}\ll \alpha(n) \ll \frac{1}{n^{7/6+o(1)}}. \] It also notes the easy upper bound $\alpha(n)\ll 1/n$ and that Erdős observed $\alpha(n)\gg 1/n^2$. Below I prove the easy upper bound and a clean lower-bound construction for infinitely many $n$ (namely primes).

3) ATTACK PLAN

Proof track: prove a universal upper bound by partitioning the disk into $O(n)$ small-area convex regions so that one region must contain three points.

Construction track: build $n$ points with no three collinear on a rational grid of mesh $1/n$ so that all triangle areas are integer multiples of $1/(2n^2)$, hence bounded below by $\asymp 1/n^2$.

4) WORK

Lemma 507.1 (Trivial upper bound $\alpha(n)=O(1/n)$). For every $n\ge 3$, \[ \alpha(n)\le \frac{2\pi}{n-1}. \]

Proof. Let $m:=\left\lfloor\frac{n-1}{2}\right\rfloor$. Then $2m\le n-1$, so $n>2m$. Partition the unit disk $D$ into $m$ closed circular sectors of equal area, each sector having area $\pi/m$. Each sector is convex.

If every sector contained at most $2$ of the $n$ points, then the total number of points would be at most $2m<n$, a contradiction. Hence some sector contains at least $3$ points of the set. Any triangle formed by three points inside a convex region lies inside that region, so its area is at most the area of the region. Therefore there exists a triangle of area at most $\pi/m$.

Finally, since $m=\lfloor (n-1)/2\rfloor \ge (n-3)/2$ for $n\ge 3$, \[ \frac{\pi}{m}\le \frac{\pi}{(n-3)/2}=\frac{2\pi}{n-3}\le \frac{2\pi}{n-1}, \] where the last inequality holds for $n\ge 3$. Thus $\alpha(n)\le 2\pi/(n-1)$. $\square$

Lemma 507.2 (Lower bound $\alpha(p)\ge 1/(2p^2)$ for primes $p$). Let $p$ be an odd prime. Then \[ \alpha(p)\ge \frac{1}{2p^2}. \]

Proof. Step 1: Construct $p$ points in the unit disk. For each $t\in\{0,1,\dots,p-1\}$ define the point \[ P_t:=\left(\frac{t}{p}-\frac12,\ \frac{t^2\bmod p}{p}-\frac12\right)\in\mathbb R^2. \] Then $P_t\in[-\tfrac12,\tfrac12]^2$, which is contained in the unit disk (since its farthest corner has distance $\sqrt{(1/2)^2+(1/2)^2}=\sqrt2/2<1$ from the origin). So $\{P_t\}_{t=0}^{p-1}\subset D$.

Step 2: No three points are collinear. Suppose, for contradiction, that three distinct indices $a,b,c\in\{0,\dots,p-1\}$ give collinear points $P_a,P_b,P_c$ in $\mathbb R^2$. Since translation by $(-1/2,-1/2)$ does not affect collinearity, the integer grid points \[(a, a^2\bmod p),\ (b, b^2\bmod p),\ (c, c^2\bmod p)\in\{0,\dots,p-1\}^2 \] are collinear in $\mathbb R^2$. Collinearity of three integer points is equivalent to the vanishing of the $2\times 2$ determinant \[ \det\begin{pmatrix} b-a & (b^2\bmod p)-(a^2\bmod p) \\[2pt] c-a & (c^2\bmod p)-(a^2\bmod p) \end{pmatrix}=0. \] Reducing this integer determinant modulo $p$ shows that the three points are collinear in the affine plane over $\mathbb F_p$. Thus there exist $m,k\in\mathbb F_p$ such that \[x^2 \equiv mx+k\pmod p\quad\text{for }x\in\{a,b,c\}. \] Equivalently, the quadratic polynomial $f(x)=x^2-mx-k\in\mathbb F_p[x]$ has three distinct roots $a,b,c$. But a nonzero polynomial of degree $2$ over a field has at most $2$ roots. This contradiction shows no three of the $P_t$ are collinear.

Step 3: Quantize triangle areas. Take any three distinct indices $a,b,c$. The points $P_a,P_b,P_c$ have coordinates that are integer multiples of $1/p$. The (signed) twice-area of the triangle is the absolute value of a determinant of the form \[ 2\,\mathrm{Area}(\triangle P_aP_bP_c)=\left|\det\begin{pmatrix} x_b-x_a & y_b-y_a\\ x_c-x_a & y_c-y_a \end{pmatrix}\right|, \] where each difference $x_b-x_a, y_b-y_a$ is a multiple of $1/p$. Hence $2\,\mathrm{Area}(\triangle P_aP_bP_c)$ is an integer multiple of $1/p^2$, and thus the area is an integer multiple of $1/(2p^2)$. Because the three points are not collinear, the area is nonzero, so \[\mathrm{Area}(\triangle P_aP_bP_c)\ge \frac{1}{2p^2}. \] Thus the minimum triangle area among these $p$ points is at least $1/(2p^2)$. By definition of $\alpha(p)$ as the supremum over $p$-point sets of their minimum triangle area, we conclude $\alpha(p)\ge 1/(2p^2)$. $\square$

FAST REALITY CHECK (computed for small primes). For the point set $\{P_t\}_{t=0}^{p-1}$ above, I computed the exact minimum area among all $\binom{p}{3}$ triangles: \[ p=3:\ \min=1/18\approx 0.0555556;\qquad p=5:\ \min=1/50=0.02;\qquad p=7:\ \min=1/98\approx 0.0102041;\qquad p=11:\ \min=1/242\approx 0.00413223. \] In each case this equals $1/(2p^2)$, confirming the lower bound is tight for this construction.

5) VERIFICATION

Lemma 507.1: The only geometric input is that a triangle contained in a region has area at most the region's area; convexity ensures containment. The pigeonhole step uses $n>2m$.

Lemma 507.2: The no-three-collinear step reduces collinearity over $\mathbb R$ to a determinant identity over $\mathbb Z$ and then modulo $p$, which is valid because the coordinates are integers. The area quantization uses that determinants of rational differences have denominator dividing $p^2$.

6) FINAL UNRESOLVED

(i) Strongest proved partial result here: A universal upper bound $\alpha(n)\le 2\pi/(n-1)$ (Lemma 507.1) and an explicit construction showing $\alpha(p)\ge 1/(2p^2)$ for every prime $p$ (Lemma 507.2), with computed small-prime verification.

(ii) First gap (crisp): Extend the lower bound from primes to all $n$ with a constant-factor loss (i.e. prove $\alpha(n)\ge c/n^2$ for all large $n$) without appealing to unproved number-theoretic input inside this writeup.

(iii) Top 3 next moves: 1. Prove a prime-gap statement sufficient to pass from Lemma 507.2 to all $n$ (e.g. use Bertrand's postulate, with a full proof, to get a prime $p\in[n,2n]$ and then take an $n$-point subset). 2. Improve the trivial upper bound $O(1/n)$ by a more geometric pigeonhole (e.g. exploiting triangulations/convex hull area) to see if one can push below exponent $1$. 3. For small $n$ (say $n\le 8$), numerically optimize point sets to estimate $\alpha(n)$ and see how the extremizers look (grid-like, boundary-like, etc.).

(iv) Minimal counterexample structure (to an $\Theta(1/n^2)$ lower bound): a family of $n$-point sets in the disk with minimum triangle area $o(1/n^2)$ would necessarily contain many near-collinear triples at scales much smaller than $1/n$; understanding how such near-collinearities can be forced or avoided seems central.

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.