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ERDőS #507 · PARTIAL

Erdős problem #507 — wave w041

Date: 2026-07-31 (UTC)

Claim labels

primary source are identified.

promoted to a theorem.

Step 0: live-page gate (mandatory)

I fetched https://www.erdosproblems.com/507 through the Bright Data browser path on 2026-07-31; this was a live browser extraction, not the stale tracker YAML.

Verbatim live statement:

Let α(n) be such that every set of n points in the unit disk contains three points which determine a triangle of area at most α(n). Estimate α(n).

(d), live-page observations. The page says OPEN and was last edited 30 December 2025. It displays:

results.

Thus neither stopping condition (claimed solution/falsification or current worker) was present, so the attempt was allowed to proceed.

(b), results stated on the live page. It identifies the problem as Heilbronn's triangle problem, states the trivial \(\alpha(n)\ll n^{-1}\), Erdős's \(\alpha(n)\gg n^{-2}\), and the current bounds

\[ \frac{\log n}{n^2}\ll \alpha(n) \ll \frac1{n^{7/6+o(1)}}. \]

The page attributes the lower bound to Komlós--Pintz--Szemerédi (1982), and the upper bound to Cohen--Pohoata--Zakharov (2024), following their 2023 improvement of the Komlós--Pintz--Szemerédi \(8/7\) exponent.

The original references exposed by the live page are:

Közl. (1961), 221–254, at p. 246, MR 177846.

Ann. Mat. Pura Appl. (4) (1975), 99–108, at p. 107, MR 411984.

Precise convention

Let

\[ D=\{(x,y)\in\mathbb R^2:x^2+y^2\leq 1\},\qquad \Delta(S)=\min_{\{p,q,r\}\in {S\choose3}}[pqr], \]

where \([pqr]\) is ordinary (possibly zero) triangle area, and put

\[ \alpha(n)=\sup_{\substack{S\subset D\\|S|=n}}\Delta(S). \]

(a). This is the standard extremal interpretation of “such that every set ... contains”. The linked formal statement independently confirms that the live page means the closed Euclidean ball of radius \(1\), not a disk of area \(1\): Metric.closedBall (0 : ℝ²) 1.

Literature audit

The following named papers/identifiers were checked against primary publisher or arXiv pages; none is an inferred or invented citation.

  1. (b) J. Komlós, J. Pintz, and E. Szemerédi,

On Heilbronn's Triangle Problem, J. London Math. Soc. (2) 24 (1981), 385–396. The paper defines the same extremal function (in a unit-area convex region) and is the cited \(8/7\)-exponent source.

  1. (b) J. Komlós, J. Pintz, and E. Szemerédi,

A Lower Bound for Heilbronn's Problem, J. London Math. Soc. (2) 25 (1982), 13–24. Its theorem gives a configuration with no triangle below \(C\log N/N^2\).

  1. (b) A. Cohen, C. Pohoata, and D. Zakharov,

A new upper bound for the Heilbronn triangle problem, arXiv:2305.18253 (2023). Its abstract states \(n^{-8/7-1/2000}\).

  1. (b) A. Cohen, C. Pohoata, and D. Zakharov,

Lower bounds for incidences, Invent. Math. 240 (2025), 1045–1118; preprint arXiv:2409.07658. Its abstract states the consequence \(n^{-7/6+o(1)}\) for Heilbronn's problem.

  1. (b) D. Svrtan, D. Veljan, and V. Volenec,

Geometry of pentagons: from Gauss to Robbins, arXiv:math/0403503. This is provenance for the Gauss pentagon identity used below. The identity is nevertheless rederived from coordinates here, so the finite result does not depend on trusting the citation.

Search miss reported honestly. Targeted searches for exact finite radius-one-disk results found a secondary MathWorld/Friedman table whose \(n=3,4,5\) entries agree with the values below after dividing by \(\pi\) to convert a radius-one disk to a unit-area disk. I did not locate a peer-reviewed disk-specific proof of its \(n=5\) entry in the limited search, so I make no novelty claim and do not use that table as evidence. The proof below is from scratch. The 2026 unit-square and unit-triangle certification papers found in the same search concern different domains.

Main finite result

Theorem

(a). For the closed disk of radius \(1\),

\[ \boxed{\alpha(3)=\frac{3\sqrt3}{4}},\qquad \boxed{\alpha(4)=1}, \]

and

\[ \boxed{\alpha(5) =2\sin^2\!\frac{\pi}{5}\sin\!\frac{2\pi}{5} =\sqrt{\frac{5(5-\sqrt5)}{32}} =0.657163890148917\ldots}. \]

In all three cases a regular \(n\)-gon on the boundary is optimal.

The nontrivial item is \(n=5\). The proof is split into reusable lemmas.

Lemma 1: maximum small-polygon area in the disk

(a). For \(m=3,4,5\), every convex \(m\)-gon in \(D\) has area at most

\[ M_m=\frac m2\sin\frac{2\pi}{m}, \]

with equality for the regular inscribed \(m\)-gon.

Proof. A maximum exists. A maximizing polygon has exactly \(m\) strict vertices: if an edge is redundant, a point of the circular arc outside that edge can be inserted and increases area. If a strict vertex \(p_i\) were in the interior of \(D\), then, with its two neighbours fixed, signed polygon area has the nonzero linear variation

\[ \frac12\det(\delta,p_{i+1}-p_{i-1}). \]

A sufficiently small interior perturbation in the sign that makes this positive preserves strict convexity and increases area. Hence every vertex of a maximizer lies on the circle.

Let the successive central gaps be \(\theta_1,\ldots,\theta_m>0\), with sum \(2\pi\). Twice the polygon area is \(\sum_i\sin\theta_i\). If every gap is at most \(\pi\), concavity of sine on \([0,\pi]\) and Jensen give

\[ \sum_i\sin\theta_i\leq m\sin(2\pi/m). \]

If one gap exceeds \(\pi\), its sine is negative and each of the other \(m-1\) sines is at most \(1\). For \(m=3,4,5\),

\[ m-1<m\sin(2\pi/m), \]

so such a polygon is not maximal. Equality in Jensen forces equal gaps.

\(\square\)

In particular,

\[ M_3=\frac{3\sqrt3}{4},\qquad M_4=2,\qquad M_5=\frac52\sin\frac{2\pi}{5}. \]

Lemma 2: a sharp ear in every convex pentagon

For a counterclockwise convex pentagon \(P=p_0p_1p_2p_3p_4\), let \(Q=[P]\), and let

\[ e_i=[p_{i-1}p_ip_{i+1}] \]

be its five ear areas (indices modulo \(5\)).

(a). Some ear satisfies

\[ e_i\leq cQ,\qquad c=\frac{5-\sqrt5}{10}. \]

Coordinate derivation of Gauss's identity. Areas scale uniformly under similarities, so take

\[ p_0=(0,0),\ p_1=(1,0),\ p_2=(x,y),\ p_3=(u,v),\ p_4=(w,t). \]

Let \(q=2Q\) and \(d_i=2e_i\). Direct shoelace determinants give

\[ \begin{aligned} q&=tu-uy-vw+vx+y,\\ d_0&=t,\quad d_1=y,\quad d_2=-uy+vx-v+y,\\ d_3&=tu-tx-uy-vw+vx+wy,\quad d_4=tu-vw. \end{aligned} \]

Substitution and expansion (independently expanded to the zero polynomial by the checker) gives

\[ q^2-q\sum_i d_i+\sum_i d_id_{i+1}=0. \]

Dividing by \(4\) yields the Gauss identity

\[ \tag{G} Q^2-Q\sum_i e_i+\sum_i e_ie_{i+1}=0. \]

Sharp inequality. Scale to \(Q=1\). From

\[ \sum_i e_i^2-\sum_i e_ie_{i+1} =\frac12\sum_i(e_i-e_{i+1})^2\geq0 \]

and (G),

\[ \tag{1} 1=\sum_i e_i-\sum_i e_ie_{i+1} \geq\sum_i e_i(1-e_i). \]

The ears \(e_j\) and \(e_{j+2}\) have disjoint interiors, so \(e_j+e_{j+2}\leq1\). Suppose for contradiction every \(e_i>c\). If some \(e_j\geq1-c\), then \(e_{j+2}\leq1-e_j\leq c\), a contradiction. Consequently every \(e_i\in(c,1-c)\). But

\[ c(1-c)=\frac15 \]

and \(x(1-x)>1/5\) on \((c,1-c)\), so the right side of (1) is greater than \(1\), again a contradiction. Thus some \(e_i\leq cQ\). \(\square\)

Proof of the theorem for \(n=5\)

(a), upper bound. Let \(S\) be any five-point set in \(D\), and let \(h\) be the number of vertices of its convex hull. A collinear triple already has area \(0\), so assume nondegeneracy where needed.

partitions the hull triangle into three triangles. One has area at most \(M_3/3=\sqrt3/4\).

partitions the hull quadrilateral into four triangles. One has area at most \(M_4/4=1/2\).

\[ cQ\leq cM_5 =\frac{5-\sqrt5}{10}\cdot\frac52\sin\frac{2\pi}{5} =2\sin^2\frac{\pi}{5}\sin\frac{2\pi}{5}. \]

Both \(\sqrt3/4\) and \(1/2\) are strictly smaller than the last number. This proves the universal upper bound.

(a), matching construction. Put the five points at the fifth roots of unity. Each of the five consecutive triples has area

\[ A_5=2\sin^2(\pi/5)\sin(2\pi/5). \]

The other five triangles have area

\[ 2\sin(\pi/5)\sin^2(2\pi/5) =2\cos(\pi/5)\,A_5>A_5. \]

Thus the construction's minimum is exactly \(A_5\), proving equality.

\(\square\)

The cases \(n=3,4\)

(a). Lemma 1 immediately gives \(\alpha(3)\leq M_3=3\sqrt3/4\), and an equilateral inscribed triangle attains it.

For four points, if the hull has at most three vertices, the interior-point partition above gives a triangle of area at most \(\sqrt3/4<1\). If the hull is a quadrilateral, either diagonal divides it into two triangles. Moreover, if its diagonal vectors are \(d_1,d_2\), then

\[ [P]=\frac12|\det(d_1,d_2)|\leq\frac12(2)(2)=2, \]

so one of the two triangles has area at most \(1\). A regular inscribed square has all four of its three-vertex triangles of area \(1\). Hence \(\alpha(4)=1\).

Exact all-boundary subproblem

Define

\[ \beta(n)= \sup_{\substack{S\subset\partial D\\|S|=n}}\Delta(S). \]

Boundary theorem

(a). For every \(n\geq3\),

\[ \boxed{\beta(n)= 2\sin^2\frac{\pi}{n}\sin\frac{2\pi}{n}}. \]

A regular \(n\)-gon attains the value.

Upper bound. Order arbitrary boundary points cyclically and let their positive angular gaps be \(a_i\), with \(\sum_i a_i=2\pi\). Since

\[ \frac1n\sum_i(a_i+a_{i+1})=\frac{4\pi}{n}, \]

some \(s=a_i+a_{i+1}\leq4\pi/n\). The corresponding consecutive triple has area

\[ T=2\sin\frac{a_i}{2}\sin\frac{a_{i+1}}2\sin\frac{s}{2}. \]

For fixed \(s\),

\[ \sin^2\frac{s}{4} -\sin\frac{a_i}{2}\sin\frac{s-a_i}{2} =\sin^2\frac{2a_i-s}{4}\geq0. \]

Therefore \(T\leq g(s)\), where

\[ g(s)=2\sin^2(s/4)\sin(s/2),\qquad g'(s)=\sin^2(s/4)\bigl(3-4\sin^2(s/4)\bigr). \]

Because \(0\leq s\leq4\pi/n\leq4\pi/3\), \(g'(s)\geq0\), and hence

\[ T\leq g(4\pi/n) =2\sin^2(\pi/n)\sin(2\pi/n). \]

Matching regular construction. A triple of vertices of a regular \(n\)-gon has positive integral cyclic gaps \(r,s,t\) summing to \(n\). Its area, using circumradius \(1\), is

\[ 2\sin\frac{r\pi}{n}\sin\frac{s\pi}{n}\sin\frac{t\pi}{n}. \]

For \(n\geq4\), all three sine factors are at least \(\sin(\pi/n)\), and at least one gap lies in \(\{2,\ldots,n-2\}\), whose sine factor is at least \(\sin(2\pi/n)\). Thus every such triangle has area at least the displayed bound, with equality for three consecutive vertices. The \(n=3\) case is immediate. \(\square\)

(a), structural consequence.

\[ \beta(n)=\frac{4\pi^3}{n^3}+O(n^{-5}). \]

Thus an all-on-one-circle construction is asymptotically much worse than the KPS lower bound \(\gg(\log n)/n^2\). Any asymptotically competitive construction must genuinely exploit interior/radial structure. This boundary theorem does not improve the known asymptotic lower bound for \(\alpha(n)\).

Computational reconnaissance and independent verification

Before the proof was found, a differential-evolution search over five arbitrary disk points returned the regular pentagon with minimum \(0.657163890088\). (d) This was hypothesis generation only and is not used in any proof.

The standalone checker is runs/erdos507_wavew041_verify.py (SHA-256 ba8476a0565a55e7982b0ef29ce86480fd82f77dd9da6874a85265b2658a33ea). It uses SymPy only for exact symbolic expansion and standard Python for the independent numerical layer. Its core exact recomputation is:

# q = twice the shoelace area; d[i] = twice the i-th ear.
q = expand(sum(det(P[i], P[(i + 1) % 5]) for i in range(5)))
d = [
    expand(det(P[i] - P[(i - 1) % 5], P[(i + 1) % 5] - P[i]))
    for i in range(5)
]
assert expand(
    q**2 - q * sum(d) + sum(d[i] * d[(i + 1) % 5] for i in range(5))
) == 0

A5 = 2*sin(pi/5)**2*sin(2*pi/5)
assert simplify(A5**2 - 5*(5-sqrt(5))/32) == 0
assert simplify(
    (5-sqrt(5))/10 * Rational(5, 2)*sin(2*pi/5) - A5
) == 0

The full source additionally:

\(3\leq n\leq40\);

fixed-sum sine identity, and the derivative formula used above.

Reproduction:

$ python runs/erdos507_wavew041_verify.py
random disk sanity: largest sampled five-point minimum = 0.328521708977 < A5 = 0.657163890149
symbolic Gauss identity: PASS
symbolic constants/inequalities: PASS
regular n-gons, 3 <= n <= 40: PASS
random boundary configurations: PASS (3800 tests)
random five-point disk configurations: PASS (5000 tests)
ALL CHECKS PASSED

Runtime on this VM was 1.38 seconds. No heavy computation was run.

What remains / exact wall

(b). The asymptotic problem remains open at the live-page gap

\[ \frac{\log n}{n^2}\ll\alpha(n)\ll n^{-7/6+o(1)}. \]

The finite proof above does not contain a hidden uniformity step: its sharp ingredient is the five-term quadratic Gauss identity, specific to pentagons. The boundary theorem also shows why merely extending the regular-polygon construction cannot help—it is only \(\Theta(n^{-3})\).

(b), named methodological barrier. Cohen--Pohoata--Zakharov explain that their high--low method reaches a Szemerédi--Trotter-type barrier at exponent \(7/6\). Their 2023 paper further notes that continuing the relevant incidences down to scale about \(n^{-1}\), itself currently out of reach, would yield only about \(n^{-3/2+o(1)}\), still short of the conjectural \(n^{-2+o(1)}\). Consequently the exact missing asymptotic ingredient is not a larger finite search: it is an incidence/projection estimate exploiting the special dependence between the point set and its pair-generated lines beyond the general high--low/Szemerédi--Trotter obstruction.

PARTIAL: Proved from scratch the exact radius-one-disk values α(3)=3√3/4, α(4)=1, and α(5)=√(5(5−√5)/32), plus the sharp all-boundary formula β(n)=2 sin²(π/n) sin(2π/n); the asymptotic Erdős problem remains open at the stated KPS–CPZ gap.

This is the AI working report, labelled by outcome — not an independently verified claim unless marked PROVED. ← ledger