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ERDőS #352 · PARTIAL

Erdős problem 352 — wave w041

Date: 2026-07-31 (UTC)

Claim labels

here.

are identified.

theorem.

checker.

the linked source; this is not a mathematical-proof label.

Step 0: mandatory live-page check

I fetched the live problem page, its complete discussion thread, and its proof-claim page through the Bright Data browser on 2026-07-31. Direct extraction was allowed to finish before any mathematical work. [source-checked]

The live statement, verbatim apart from normalizing its visual line breaks, is:

Is there some 𝑐 >0 such that every measurable 𝐴 ⊆ℝ² of measure ≥𝑐 contains the vertices of a triangle of area 1?

The same page displayed all of the following:

claims have been submitted yet”;

J_Koizumi_144**.

Thus neither mandatory stop condition applied. In particular, “interested in collaborating” is recorded but is not the page's “currently working” marker. [source-checked]

Known results displayed on the live page

The page lists the following facts.

  1. Erdős proved (unpublished) that an infinite-measure planar set, or an

unbounded positive-measure set, contains a triangle of every prescribed positive area. The cited Erdős sources say this follows easily from the Lebesgue density theorem. [b, source-checked]

  1. Erdős suggested the sharp constant

\[ c_0=\frac{4\pi}{\sqrt{27}}\approx2.418. \] An open disk of radius \(<2\cdot3^{-3/4}\) is the obstruction: its largest inscribed triangle is equilateral and has area \(<1\). [a for the disk calculation; source-checked for Erdős's conjecture]

  1. Freiling and Mauldin proved that outer measure \(>c_0\) forces a triangle

of area \(>1\). The page notes that this gives the original threshold for a compact convex set. [b: Mauldin 2002, source-checked]

  1. It suffices to settle the question for sets that are unions of the

interiors of finitely many compact convex sets; Freiling and Mauldin proved the sharp \(c_0\) result for unions of at most three such sets. [b: Mauldin 2013, source-checked]

What the 17 live comments add

The site warns that comments are user-supplied and not verified. I read all 17 and used them only after the checks described below. [source-checked]

references, now incorporated into the main page.

\(S\subset\mathbb Z^2\) is called \(N\)-avoiding when \[ \bigl|\operatorname{area}(\triangle pqr)-N\bigr| >\operatorname{diam}(\triangle pqr) \tag{1} \] for every triple \(p,q,r\in S\), with repetitions allowed. If \(f(N)\) is the maximum size of such a set, the continuous problem is equivalent to \(f(N)=O(N)\). A linked note contains the proof.

KentaKitamura's public repository rechecks those and supplies additional witnesses through \(N=40\).

and exact values only for \(N=1,\ldots,7\): \[ 1,4,4,7,9,10,12. \] The comment says the readable exact-chromatic implementation became too slow beyond \(7\).

asymptotic \((4\pi/\sqrt{27})N+O(\sqrt N)\). Other comments discuss dyadic discretization, horizontal sections, toy grid computations, disk-like extremizers, and widely separated “local extremizers.” These are useful structural context, not claimed proofs of the full problem.

2026 preprint of Bulj and Kovač discussed below.

All names, dates, links, tables, and the previous \(N\le7\) exact range in this inventory were checked in the rendered thread. [source-checked]

Primary-source literature audit

  1. Paul Erdős, *Set-theoretic, measure-theoretic, combinatorial, and

number-theoretic problems concerning point sets in Euclidean space*, Real Analysis Exchange 4 (1978/79), 113–138, author archive PDF. Pages 121–123 state the infinite-measure observation, the finite constant question, the disk obstruction, and the proposed value \(4\pi/\sqrt{27}\). [b, source-checked]

  1. R. Daniel Mauldin, Some Problems in Set Theory, Analysis and Geometry,

in Paul Erdős and His Mathematics I, Bolyai Society Mathematical Studies 11 (2002), 493–506, author-uploaded PDF. Section 1 states the Freiling–Mauldin outer-measure theorem and the convex/covering reductions used on the live page. [b, source-checked]

  1. R. Daniel Mauldin, *Some Problems and Ideas of Erdős in Analysis and

Geometry, in Erdős Centennial* (2013), 365–376, author-hosted PDF, DOI. Problems 5.1 and 5.2 give the continuous problem and finite-union reduction; the following paragraphs prove the sharp constant for one, two, and three compact convex bodies. [b, source-checked]

  1. J. Koizumi, Triangles in a Planar Measurable Set,

linked note. Theorem 2 states and proves the equivalence with \(f(N)=O(N)\). This is an informal linked note rather than a journal publication, so I independently reconstruct the reduction below. [source-checked; mathematical use is [a]]

  1. Aleksandar Bulj and Vjekoslav Kovač, *On hyperbolic corners and unit-area

triangles in planar sets of large measure*, arXiv:2605.30033v1, submitted 28 May

  1. The identifier, title, authors, date, and full 25-page paper all

exist. Its Theorem 2 says that if \(M_\triangle(R)\) is the supremum of measures of unit-area-triangle-avoiding measurable subsets of \([0,R]^2\), then for \(R\ge10\) \[ M_\triangle(R)\lesssim R^2\left(\frac{\log\log R}{\log R}\right)^{1/2}. \tag{2} \] The paper explicitly says the desired \(O(1)\) bound remains open. [b, source-checked]

Targeted searches for the exact statement, “triangle of area one,” “unit-area triangles” with measurable sets, Freiling–Mauldin, and “Erdős Problem #352” found no later primary source claiming the full problem or the exact lattice values below. This is a search miss and not a proof of novelty. [c, source-checked]

New finite result

Let \(f(N)\) have the precise meaning in (1).

Theorem

The exact values for \(8\le N\le16\) are

\[ \begin{array}{c|rrrrrrrrr} N &8&9&10&11&12&13&14&15&16\\ \hline f(N) &14&16&18&21&23&24&26&28&32. \end{array} \tag{3} \]

[d: exhaustive integer certificates]

The lower-bound point sets were already present in the live thread/witness repository; this run independently verifies them. The new contribution of this run is the matching exhaustive upper bound for every \(N=8,\ldots,16\). Thus it extends the posted exact range \(N\le7\), but it does not give a new continuous counterexample or solve the uniform problem. [d, source-checked]

The complete standalone checker is runs/erdos352_wavew041_reverify.py. It uses only the Python standard library, embeds every lower witness, constructs every upper certificate from scratch, and then audits the certificates independently. Its SHA-256 is

3f79b4cfca1db2ada15e38ce9dac317336ae92ffb5b5a816464bd6ce41f11791

[d]

Proof of the finite upper bounds

1. An exact integer predicate

For lattice points \(p,q,r\), let

\[ A_2=\left|\det(q-p,r-p)\right|, \qquad D_2=\max\{|p-q|^2,|p-r|^2,|q-r|^2\}. \]

Thus the triangle area is \(A_2/2\) and its diameter is \(\sqrt{D_2}\). Both sides of (1) are nonnegative, so multiplying by two and squaring shows that (1) is equivalent to

\[ (A_2-2N)^2-4D_2>0. \tag{4} \]

Every comparison in the checker is therefore an exact integer comparison. [a]

2. Exhausting diameter pairs

Repetitions in (1) matter. Applying it to \((p,p,q)\) gives

\[ |p-q|<N. \tag{5} \]

Every finite avoiding set has a diameter pair. Translate one endpoint to \(A=(0,0)\), and apply a sign change, coordinate swap, and reflection to put the other endpoint at

\[ B=(b_x,b_y),\qquad b_x\ge b_y\ge0,\quad 0<b_x^2+b_y^2<N^2. \tag{6} \]

These are all lattice-preserving isometries and preserve absolute triangle area. The checker enumerates every vector in (6). An independent audit starts with every signed vector in the square \([-(N-1),N-1]^2\), normalizes its full dihedral orbit, and checks that the two lists coincide. [a for completeness; d for the list comparison]

Write \(D=|A-B|\). Since \(A,B\) are a diameter pair, every further point \(q\) lies in the finite lens

\[ |q-A|\le D,\qquad |q-B|\le D. \tag{7} \]

In particular every coordinate of \(q\) lies in \([-(N-1),N-1]\). The checker exhausts that entire integer box, imposes (7), and tests \((A,B,q)\) using (4). There is no truncation heuristic. [a,d]

3. The exhaustive low/high branch split

For a surviving \(q\), the diameter of \(\triangle ABq\) is \(D\), so (1) gives exactly one of

\[ \operatorname{area}(\triangle ABq)<N-D \quad\text{or}\quad \operatorname{area}(\triangle ABq)>N+D. \tag{8} \]

Call these candidates low and high.

Every avoiding set for a fixed \(B\) belongs to one of these exhaustive branches:

candidates remain;

candidate as \(H\) and fixes \(A,B,H\).

In either branch, candidates incompatible with the fixed points or with the chosen diameter are first removed. [a,d]

4. Checkable colorings, not chromatic optimization

For a fixed set \(F\), where \(F=(A,B)\) or \(F=(A,B,H)\), form a graph on the remaining candidates. Two candidates are adjacent precisely when they can coexist with \(F\): their mutual distance does not exceed \(D\), and every triple in \(F\cup\{p,q\}\) passes (4).

Every actual avoiding extension of \(F\) is a clique in this compatibility graph. Consequently, any proper \(k\)-coloring—not necessarily an optimal one—proves that at most \(k\) candidates can be added:

\[ |S|\le |F|+k. \tag{9} \]

The program uses a deterministic degree-ordered greedy coloring. The slow exact-chromatic search in the posted reference implementation is therefore unnecessary for these upper bounds. [a,d]

The crucial certification direction is checked independently. For each color class and every same-color pair \(p,q\), the audit verifies directly that either \(|p-q|>D\), or a full combinations-with-repetition scan of \(F\cup\{p,q\}\) finds a failed inequality (4). Hence no compatible pair has the same color, which is exactly the property needed in (9). [a,d]

5. Taking all maxima

For every normalized \(B\), the checker takes the larger of the no-high bound and all fixed-\(H\) bounds. It then takes the maximum over all \(B\). Sections 2–4 show that every \(N\)-avoiding lattice set occurs in one of the audited cases, up to an allowed lattice isometry. The resulting upper bound matches the embedded witness size for each row of (3), proving equality. [a,d]

Independent lower-bound and certificate audit

For a witness of size \(m\), the checker evaluates all \(\binom{m+2}{3}\) triples with repetition. The following table records the smallest exact value of the left side of (4), so every positive entry is a strict certificate. It also gives the exhaustive upper-search counts. [d]

| \(N\) | \(f(N)\) | witness triples | min gap | diameter \(B\) cases | high-\(H\) branches | color certificates | same-color pairs audited | tight \(B\) | |---:|---:|---:|---:|---:|---:|---:|---:|:---| | 8 | 14 | 560 | 24 | 30 | 98 | 128 | 28 | \((3,3)\) | | 9 | 16 | 816 | 29 | 38 | 268 | 306 | 50 | \((4,1),(3,3)\) | | 10 | 18 | 1140 | 13 | 46 | 518 | 564 | 383 | \((4,3)\) | | 11 | 21 | 1771 | 28 | 55 | 926 | 981 | 2806 | \((4,2)\) | | 12 | 23 | 2300 | 9 | 64 | 1466 | 1530 | 10809 | \((4,4)\) | | 13 | 24 | 2600 | 5 | 75 | 2348 | 2423 | 50654 | \((5,1),(5,2),(6,1)\) | | 14 | 26 | 3276 | 16 | 87 | 3572 | 3659 | 148701 | \((5,2),(5,3)\) | | 15 | 28 | 4060 | 17 | 99 | 5094 | 5193 | 373340 | \((4,4),(5,3),(5,4)\) | | 16 | 32 | 5984 | 16 | 112 | 7094 | 7206 | 814708 | \((5,3)\) |

The script has deliberately separate producer and audit paths:

  1. the producer computes area with a based-vector determinant; the audit uses

the shoelace formula;

  1. the producer computes distances from coordinate differences; the audit

uses a dot-product expansion;

  1. the producer lists normalized \(B\)'s directly; the audit normalizes every

signed raw vector;

  1. candidate pools are regenerated with the audit predicate and compared

exactly;

  1. the audit does not trust the compatibility graph: it checks every

same-color pair by full triple enumeration and separately checks the fixed diameter.

Thus a search error can at worst produce a certificate that fails; no optimizer's assertion is accepted as a proof. [d]

As a cross-implementation check, I also ran the public upper-bound repository at commit 7a4a19468771766cb4f20e10dfa5cf770cc2487b with exact chromatic minimization disabled (--max-color-vertices 0). Its separate implementation returned the same upper totals and diameter-case counts for \(N=8,\ldots,16\). The embedded lower points were compared with the public witness repository at commit 09e5c5393943c881ab60118aa32fe5660539c8d5. The standalone checker imports neither repository. [d, source-checked]

For reproducibility, the SHA-256 digests of the canonical audited certificate streams are:

N= 8  82a246077f367d140d7e977167a203c15fa2fa0c3e158de41727bac94317ad41
N= 9  8c4251f00f2c14475832a7568dde89080796a126fae4a80b96a0ad549388e225
N=10  0a0c50ab0e12bca0fa487ba6f77beaaed3a64d5e78d8541f595c2d66133cbf3a
N=11  8687754c9fcf1433a3bfdc1435c3bffb1a097ae89374a44b842b209980c883c3
N=12  395fb6971b90dd2b0463323e02382890b75b4b36f91ebb621b8774e067cf2ae0
N=13  903a23b95da5c4ac678570123f410ed88138ab5fb9777bfd5244524672a52e3a
N=14  d9aa00efb67719feb1b8254925b3a321e338d52e56db068a0bfe1ae193168d9b
N=15  31650a5d06b64a1c161e117318bdab9d3103d277ab293790c27f82db7d60453f
N=16  839008d566ff41e4801c2f63fe61af5beb3081ad475a8b94864db84b7c5085d3

[d]

Reproduction

From the repository root:

python3 runs/erdos352_wavew041_reverify.py

The completed run printed:

N= 8 f(N)=14 witness_triples= 560 min_gap= 24 B_cases= 30 high_branches=   98 certs=  128 same_color_pairs=    28 tight_B=3,3 sha256=82a246077f367d140d7e977167a203c15fa2fa0c3e158de41727bac94317ad41 seconds=0.053
N= 9 f(N)=16 witness_triples= 816 min_gap= 29 B_cases= 38 high_branches=  268 certs=  306 same_color_pairs=    50 tight_B=4,1;3,3 sha256=8c4251f00f2c14475832a7568dde89080796a126fae4a80b96a0ad549388e225 seconds=0.163
N=10 f(N)=18 witness_triples=1140 min_gap= 13 B_cases= 46 high_branches=  518 certs=  564 same_color_pairs=   383 tight_B=4,3 sha256=0a0c50ab0e12bca0fa487ba6f77beaaed3a64d5e78d8541f595c2d66133cbf3a seconds=0.409
N=11 f(N)=21 witness_triples=1771 min_gap= 28 B_cases= 55 high_branches=  926 certs=  981 same_color_pairs=  2806 tight_B=4,2 sha256=8687754c9fcf1433a3bfdc1435c3bffb1a097ae89374a44b842b209980c883c3 seconds=1.045
N=12 f(N)=23 witness_triples=2300 min_gap=  9 B_cases= 64 high_branches= 1466 certs= 1530 same_color_pairs= 10809 tight_B=4,4 sha256=395fb6971b90dd2b0463323e02382890b75b4b36f91ebb621b8774e067cf2ae0 seconds=2.363
N=13 f(N)=24 witness_triples=2600 min_gap=  5 B_cases= 75 high_branches= 2348 certs= 2423 same_color_pairs= 50654 tight_B=5,1;5,2;6,1 sha256=903a23b95da5c4ac678570123f410ed88138ab5fb9777bfd5244524672a52e3a seconds=6.082
N=14 f(N)=26 witness_triples=3276 min_gap= 16 B_cases= 87 high_branches= 3572 certs= 3659 same_color_pairs=148701 tight_B=5,2;5,3 sha256=d9aa00efb67719feb1b8254925b3a321e338d52e56db068a0bfe1ae193168d9b seconds=13.828
N=15 f(N)=28 witness_triples=4060 min_gap= 17 B_cases= 99 high_branches= 5094 certs= 5193 same_color_pairs=373340 tight_B=4,4;5,3;5,4 sha256=31650a5d06b64a1c161e117318bdab9d3103d277ab293790c27f82db7d60453f seconds=28.996
N=16 f(N)=32 witness_triples=5984 min_gap= 16 B_cases=112 high_branches= 7094 certs= 7206 same_color_pairs=814708 tight_B=5,3 sha256=839008d566ff41e4801c2f63fe61af5beb3081ad475a8b94864db84b7c5085d3 seconds=69.480
ALL CHECKS PASSED for N=8..16; exact table=[14, 16, 18, 21, 23, 24, 26, 28, 32]; total_seconds=122.418

/usr/bin/time independently reported 122.48 seconds elapsed, 91% CPU, and 21,796 KiB maximum resident memory. python3 -m py_compile also passed. [d]

Why this is relevant to the continuous problem

For completeness, here is a reconstruction of Koizumi's equivalence.

Continuous statement implies \(f(N)=O(N)\)

Assume some constant \(c\) forces a unit-area triangle in every measurable set of measure \(>c\). Given an \(N\)-avoiding \(S\), put a disk of radius \(1/(3\sqrt N)\) around each \(q/\sqrt N\), \(q\in S\). The disks are disjoint by (5).

If their union contained a unit-area triangle \(p_1p_2p_3\), set \(q_i'=\sqrt N\,p_i\). Then \(\operatorname{area}(\triangle q_1'q_2'q_3')=N\) and \(|q_i-q_i'|\le1/3\). The elementary determinant perturbation bound

\[ |\operatorname{area}(\triangle q_1q_2q_3) -\operatorname{area}(\triangle q_1'q_2'q_3')| \le 2\varepsilon\operatorname{diam}(\triangle q_1q_2q_3) +2\varepsilon^2 \]

with \(\varepsilon=1/3\) contradicts (1); the lattice diameter is at least one unless all \(q_i\) coincide, and the latter is impossible because one such small disk cannot contain a unit-area triangle. Hence the disk union avoids unit area and

\[ |S|\frac{\pi}{9N}\le c, \qquad\text{so}\qquad f(N)\le\frac{9c}{\pi}N. \tag{10} \]

[a]

\(f(N)=O(N)\) implies the continuous statement

Conversely, assume \(f(N)\le CN\). By inner regularity, it is enough to treat a compact \(A\) of measure \(>4\pi C\). For each integer \(N\), let \(S_N\) contain those \(q\in\mathbb Z^2\) for which the disk of radius \(2/\sqrt N\) about \(q/\sqrt N\) meets \(A\). These disks cover \(A\), and the sum of their areas gives

\[ |S_N|\frac{4\pi}{N}>4\pi C, \quad\text{hence}\quad |S_N|>CN\ge f(N). \tag{11} \]

Thus \(S_N\) has a triple failing (1). After scaling by \(1/\sqrt N\), its triangle area differs from \(1\) by at most its diameter divided by \(\sqrt N\). Moving the vertex opposite a diameter side within its radius \(2/\sqrt N\) disk changes the area continuously through an interval large enough to hit \(1\).

This gives a unit-area triangle in the disk union for every \(N\). Each union lies within distance \(4/\sqrt N\) of compact \(A\); compactness and area continuity give a convergent subsequence whose limiting three vertices lie in \(A\) and span area \(1\). Therefore the continuous problem is equivalent to \(f(N)=O(N)\). The live page uses a threshold \(\ge c\), while the note uses \(>c\); existence of some threshold is unchanged by increasing the constant. [a]

Exact remaining obstruction

The finite table (3) supplies no uniformity in \(N\). It therefore cannot, even in principle, close the continuous problem. [a]

Within this diameter-pair method, a sufficient missing lemma is:

There is an absolute \(C\) such that, for every \(N\), every normalized diameter vector \(B\), and every no-high or fixed-high branch \(F\), the associated compatibility graph has a proper coloring with at most \(CN-|F|\) colors.

Such a uniform coloring lemma would imply \(f(N)=O(N)\) by (9) and hence solve the continuous problem by (10)–(11). The live discussion's all-low argument does not control all fixed-high branches; those branches are the precise remaining obstacle in this framework. [a for the implication; c for the prospect that such a coloring exists]

The computational growth already shows why extending a table is not a substitute. At \(N=16\) the audit processed 7,094 high branches and 814,708 same-color pairs, taking 69.48 seconds. Extrapolating the last three timing ratios suggests an \(N=17\) audit would cost roughly 2–3 one-core minutes, but that is only an engineering estimate, not a complexity theorem; it was not run under the present CPU cap. No finite extension, regardless of cost, would prove the required bound for all \(N\). [d for measured costs; c for the estimate]

PARTIAL: Exact exhaustive certificates prove f(N)=14,16,18,21,23,24,26,28,32 for N=8,...,16 in Koizumi's equivalent lattice model, extending the posted exact range N<=7; Erdős problem 352 remains open because no uniform f(N)=O(N) bound is proved.

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