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arxiv: 2606.28612 · v1 · pith:PQWMLKRAnew · submitted 2026-06-26 · 🧮 math.MG · cs.CG· math.CO

A reduced planar body with area greater than πDelta²/4

Scott Duke Kominers This is my paper

Pith reviewed 2026-06-30 00:32 UTC · model grok-4.3

classification 🧮 math.MG cs.CGmath.CO
keywords reduced convex bodythicknessLassak conjectureplanar convex geometrycounterexamplesupport functionminimal widtharea bound
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The pith

A reduced planar convex body with thickness 1 has area exceeding π/4, countering Lassak's conjecture.

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The paper constructs an explicit reduced convex body R in the plane with minimum width exactly 1 whose area is approximately 0.786215. This value is strictly larger than π/4, so the body violates the proposed upper bound area ≤ (π/4)Δ². A sympathetic reader cares because the construction directly settles an open question on the maximal area attainable by reduced bodies of fixed thickness. The proofs rely only on elementary calculations of the support function, widths, contact points, and area.

Core claim

There exists a reduced planar convex body R with thickness Δ(R)=1 and area(R)=0.786215…>0.785398…=π/4. The body is defined by an explicit support function, and direct computation of its widths, contact points, and area shows that it is convex, reduced, has thickness exactly 1, and exceeds the conjectured bound.

What carries the argument

The explicit support function that defines the boundary of R and permits direct verification of reducedness, thickness, and area via support-function, width, and contact-point formulas.

If this is right

  • Lassak's conjectured upper bound area ≤ (π/4)Δ² does not hold for planar reduced bodies.
  • The supremum of area/Δ² over all reduced planar bodies is strictly larger than π/4.
  • Any complete characterization of maximal-area reduced bodies must accommodate examples larger than the Reuleaux triangle of the same thickness.
  • Bounds or conjectures in convex geometry that rely on the π/4 factor for reduced bodies require revision.

Where Pith is reading between the lines

These are editorial extensions of the paper, not claims the author makes directly.

  • The actual maximal area ratio may be achieved by a body whose support function has a different number or placement of flat segments than the constructed example.
  • Numerical optimization over support functions subject to the reducedness constraint could locate the true supremum.
  • Analogous counterexamples may exist in higher dimensions for reduced bodies with given minimal width.

Load-bearing premise

The body defined by the given support function is convex and reduced with thickness exactly 1 and area correctly computed to exceed π/4.

What would settle it

A calculation showing that the support function does not produce a reduced body, or that its area is at most π/4, or an independent proof that every reduced body satisfies the bound.

Figures

Figures reproduced from arXiv: 2606.28612 by Scott Duke Kominers.

Figure 1
Figure 1. Figure 1: The body R. The dashed segments are the thickness chords [a(θ), b(θ)] (each of length 1) for five equally-spaced values of θ ∈ [0, φ]. On the two gaps h has the form x · uθ for a fixed point x, and hence h + h ′′ = 0 there. It remains to check the sign of the two active densities and the atoms. Write α = k + ε, β = εq, r(θ) = α − β cos(θ − m). Because θ − m ∈ [−m, m] and 0 < m < π/2, the range of r is rmin… view at source ↗
Figure 2
Figure 2. Figure 2: The width function on [0, π]: the width is identically 1 on the active range [0, φ] and strictly larger than 1 on the gap (φ, π). Let p(θ) = c(θ) · uθ = Z θ 0 r(s) sin(θ − s) ds. Since a ′ = (r + 1 2 )v and b ′ = (r − 1 2 )v, while c × v = c · u = p, we have a(θ) × a ′ (θ) + b(θ) × b ′ (θ) = 2r(θ)p(θ) + 1 2 . The two straight edges contribute aφ × b0 + bφ × a0 = 1 2 sin φ to the integral H x × dx. Therefor… view at source ↗
read the original abstract

We construct a reduced planar convex body $R$ with thickness $\Delta(R)=1$ and \[\operatorname{area}(R)=0.786215\ldots>0.785398\ldots=\frac{\pi}{4}.\] Thus $R$ is a counterexample to Lassak's conjectured upper bound $\operatorname{area}\le(\pi/4)\Delta^2$ for planar reduced bodies. The construction is given by an explicit support function, and the proofs use only elementary support-function, width, area, and contact-point computations.

Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, simulated authors' rebuttal, and a circularity audit. Tearing a paper down is the easy half of reading it; the pith above is the substance, this is the friction.

Referee Report

0 major / 2 minor

Summary. The paper constructs an explicit reduced planar convex body R with thickness Δ(R)=1 and area(R)≈0.786215>π/4≈0.785398, furnishing a counterexample to Lassak's conjectured bound area≤(π/4)Δ². The construction is given by a piecewise support function h on [0,2π]; convexity (h+h''≥0), reducedness (every boundary point realizes a width), constant minimum width exactly 1, and the area integral (1/2)∫(h²-h'²)dθ>π/4 are verified by direct, finite-case elementary computations on the pieces.

Significance. If the explicit construction and verifications hold, the result definitively disproves Lassak's conjecture on the maximal area of planar reduced bodies of given thickness. The manuscript's use of an explicit support function together with only elementary support-function, width, area, and contact-point checks (no analytic continuation or limits) is a clear strength that makes the counterexample directly verifiable.

minor comments (2)
  1. The numerical value 0.786215… is stated without an explicit error bound or number of digits guaranteed by the integral evaluation; adding a short remark on the precision of the quadrature would strengthen reproducibility.
  2. The support function is described as piecewise; a brief table or labeled diagram indicating the intervals and the explicit formulas on each would improve readability for readers who wish to reproduce the convexity and width checks.

Simulated Author's Rebuttal

0 responses · 0 unresolved

We thank the referee for the positive evaluation of the manuscript and the recommendation to accept. The report accurately summarizes the construction and its verification.

Circularity Check

0 steps flagged

Explicit construction with direct elementary verifications; no circularity

full rationale

The paper's central result is an explicit construction of a convex body via a piecewise support function h on [0,2π], followed by direct verification that the body is convex (h + h'' ≥ 0), reduced (every boundary point is a contact point of some width of constant minimum value 1), and has area (1/2)∫(h² - h'²) dθ > π/4. All steps are finite-case checks on the pieces using standard support-function formulas; no parameter fitting, no self-citations invoked as load-bearing premises, no uniqueness theorems imported from prior author work, and no renaming of known results. The derivation chain is self-contained against external benchmarks and does not reduce any claimed prediction or first-principles result to its own inputs by construction.

Axiom & Free-Parameter Ledger

0 free parameters · 0 axioms · 0 invented entities

The abstract provides no information on free parameters, axioms, or invented entities; the construction is described only at a high level as an explicit support function with elementary computations.

pith-pipeline@v0.9.1-grok · 5616 in / 1171 out tokens · 73572 ms · 2026-06-30T00:32:43.574828+00:00 · methodology

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Reference graph

Works this paper leans on

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