Finite cylinder coverings below the tetrahedral half-area bound

Source and verification

This article explains family 100 of the October 6, 2026 OpenAI collection. Research conclusions are attributed to its manuscripts. The review covers statements, introductions, and selected proof mechanisms; it is not an independent verification of the complete proofs.

🏷️ Cylinder cost and two normalizations

A cylinder in ${\mathbb{R}}^3$ is $B+\mathbb{R}u$, where $u$ is a unit vector and the measurable base $B$ lies in $u^{\perp }$. Its cost is the perpendicular area $|B|$. Costs in a finite covering are added with multiplicity.

For a convex body $K$, define

$$A_{\min }(K)=\underset{|u|=1}{\min }|{\mathrm{proj}}_{u^{\perp }}K|.$$

The half-area question asks whether every finite cylinder cover has total cost at least $A_{\min }(K)/2$. A stronger, directionwise formulation asks whether

$$\mathcal{R}(\mathcal{C};K)=\sum _i\frac{|B_i|}{|{\mathrm{proj}}_{u_i^{\perp }}K|}\ge \frac{1}{2}.$$

Each denominator is at least $A_{\min }(K)$, so $\mathcal{R}\le {\sum }_i|B_i|/A_{\min }(K)$. A cover below the half-minimum bound therefore violates the directionwise bound too.

All four manuscripts in this family concern finite covers. A continuous family of tilted line segments is an intermediate construction, not yet the required finite collection. (OpenAI, 2026) (OpenAI, 2026b) (OpenAI, 2026c) (OpenAI, 2026c)

🏷️ The tetrahedral counterexample

Set $H=\sqrt{2}$ and

$$K=\left\{(x,y,Ht):\begin{array}{c}0\le t\le 1,\\ |x|\le 1-t,\\ |y|\le t\end{array}\right\}.$$

Its vertices are $(\pm 1,0,0)$ and $(0,\pm 1,H)$, a regular tetrahedron of edge length two. Its minimum projection area is $\sqrt{2}$.

Explicit statement in the angular-cover manuscript

For $0<\epsilon \le 1/2000$, this tetrahedron has a cover by $m=2\lceil 2/{\epsilon }^2\rceil$ cylinders with compact triangular perpendicular bases and

$$\begin{array}{c}\frac{1}{\sqrt{2}}\sum _i|B_i|\\ =\frac{1}{2}-\frac{13}{6000}{\epsilon }^2+R_{\epsilon },\\ |R_{\epsilon }|\le 2{\epsilon }^4.\end{array}$$

Hence its total cost is strictly below $A_{\min }(K)/2$. (OpenAI, 2026a)

The strict sign follows directly: throughout the given range, $2{\epsilon }^2<13/6000$, so the error cannot cancel the negative quadratic term. Every chosen $\epsilon$ gives a finite cover, though the number of cylinders grows as $\epsilon$ decreases.

🏷️ Tilting the equality arrangement

The familiar two-cylinder arrangement uses the $x$ direction below $t=1/2$ and the $y$ direction above it. Each perpendicular base is triangular, and their combined cost reaches the half-area threshold.

The angular manuscript subdivides these triangles into thin sectors and assigns each a slightly different axis. Tilt decreases perpendicular projection area, but can open gaps. Adjacent axes are chosen so that planes containing their common sector boundary agree.

At a point of the tetrahedron these planes give a finite sequence of intercept coordinates beginning at one side and ending at the other. Even if the sequence is not monotone, a first crossing brackets the point in some adjacent pair. This selects a sector without assuming a global ordering of all boundary intercepts.

The remaining possible gap lies near the interface $t=1/2$. The two radial heights have opposite first-order shifts, proportional to $-\epsilon xy$ and $+\epsilon xy$. Their sum changes only at second order. A quadratic radial enlargement closes the gap while costing less area than the tilt saves. The angular mesh is at most ${\epsilon }^2$, keeping the total discretization error at order ${\epsilon }^4$. (OpenAI, 2026a)

🏷️ A finite approximation theorem for ruled sets

The second manuscript treats a general field of segments

$$\begin{array}{c}E(D,V,L)\\ =\left\{(s,p+sV(p)):\begin{array}{c}p\in D,\\ |s|\le L\end{array}\right\},\end{array}$$

where $D\subset {\mathbb{R}}^2$ is compact and $V$ is $C^1$ nearby. Suppose

$$\begin{array}{c}\mathrm{tr}DV(p)=0,\\ -L^{-2}<\det DV(p)\le 0.\end{array}$$

It claims finite covers with square intercept tiles whose weighted area approaches ${\int }_D(1+|V(p){|}^2{)}^{-1/2}dp$ from above. After perpendicular projection the bases are parallelograms. No assumption that $\partial D$ has area zero is required. (OpenAI, 2026b)

For a constant velocity $g$, projection of an intercept tile contributes the factor $(1+|g{|}^2{)}^{-1/2}$. This is the cosine between the intercept-plane normal and the cylinder axis.

The derivative condition says the two eigenvalues are opposite reals of magnitude less than $1/L$. In suitable orthonormal coordinates, the affine derivative has zero diagonal. A rounded fixed-point argument shows that translated square tiles still cover under the associated affine motion, even when one shear coefficient is large.

The nonlinear proof uses a coarse affine approximation and a much finer tile grid. Taylor errors and the inverse affine map confine the necessary tiles to a thin collar. Outer-neighborhood measure convergence handles irregular compact label sets. Simply freezing the velocity on a small patch would not, by itself, control gaps or area excess.

🏷️ Slope fields and prescribed triangular approximations

The third manuscript constructs rational, singular-power, and limiting angular slope fields that lower the tetrahedral cost. It then invokes finite approximation to obtain compact parallelogram bases, or bounded open bases. The cutoff near a singular tip is part of the cost budget; an undefined apex cannot be ignored. (OpenAI, 2026d)

The fourth manuscript gives a more specialized theorem with triangular bases and prescribed sampling directions. Let $a,b$ be independent vectors in a plane, $e$ its unit normal, and consider

$$\begin{array}{c}{p}_0+\rho (a+tb)+s(e+w(t)),\\ 0\le \rho \le R(t),|s|\le L.\end{array}$$

Here $R\ge 0$ is continuous and $w$ is $C^1$. If

$$w^{\prime }(t)=\lambda (t)(a+tb),$$

with continuous $\lambda$, each interval of any tagged partition can be covered by one cylinder in the direction sampled at its tag. As the mesh tends to zero, total triangular-base area tends uniformly to

$$|\det (a,b)|\int \frac{R(t{)}^2}{2\sqrt{1+|w(t){|}^2}}dt.$$

The condition makes first-order directional changes move along an existing radial ray. Absorbing that displacement into $\rho$ leaves only a quadratic transverse error. Extended triangles can therefore cover the frozen-direction intercepts at vanishing total area cost. The theorem includes the apex and allows $R$ or $\lambda$ to vanish. (OpenAI, 2026c)

🏷️ Affine transfer and its limit

For an invertible linear map $A$, set $v=Au/|Au|$. The induced map between perpendicular planes is $L={\mathrm{proj}}_{v^{\perp }}A{|}_{u^{\perp }}$. It multiplies both a base area and the corresponding shadow of $K$ by the same determinant factor. Each directionwise ratio is therefore invariant.

Every nondegenerate tetrahedron is an affine image of the regular one, so the directionwise counterexample extends to every tetrahedron. A single common minimum projection area does not transform by that same direction-independent factor. The half-minimum conclusion is asserted here for regular tetrahedra; the all-tetrahedron conclusion uses the directionwise normalization.

🏷️ Key steps and proof outlines

Coverage is proved before the area is optimized

The finite angular construction perturbs two families of cylinder directions. Given a point in the tetrahedron, project it along a candidate axis to the corresponding intercept plane. Its two intercept coordinates depend continuously on an angular parameter. At the extreme parameter values, an auxiliary scalar function lies on opposite sides of the required coordinate. A first-crossing choice therefore supplies an admissible sector.

The proof then checks that the intercept lies inside the enlarged base assigned to that sector, including face, edge, and apex cases. The angular discretization creates an error, so the base has an explicit radial margin. Its order is chosen smaller than the area improvement. Only after this coverage calculation does the source sum projected base areas. Taylor expansion with a bounded second derivative gives a strict decrease whose leading term dominates the accumulated margins. A smaller formal integral would not suffice unless every point had already been assigned to a cylinder. (OpenAI, 2026a)

Affine motion of squares is governed by a discrete fixed point

For the general ruled-set approximation, the affine model has a derivative $A$ with trace zero and $-L^{-2}<\det A\le 0$. In suitable orthonormal coordinates it has zero diagonal, so the transformed grid is controlled by two shear coefficients $\alpha ,\beta$. To cover a target point $u$, the source seeks integers satisfying

$$i=R(u_1-\alpha j),j=R(u_2-\beta i),$$

where $R(t)=\lfloor t+1/2\rfloor$. Substituting the second into the first gives an integer-valued fixed-point problem. At sweep parameter $s\in [0,L]$, the product is $\alpha \beta =-s^2\det A\in [0,1)$, which supplies monotonicity of the substituted map. This monotonicity and the strict determinant margin prevent the iteration from jumping past all fixed points or escaping to infinity. This handles large individual shear coefficients; small operator norm of the derivative is not assumed. (OpenAI, 2026b)

The nonlinear construction freezes the derivative on coarse patches and uses a much finer square grid. Taylor’s theorem bounds the difference from affine motion. The strict determinant margin controls the inverse affine map, so only tiles in a thin collar of the original parameter set need be added. Compactness makes the estimates uniform over finitely many patches.

Projection and collar estimates control the total area

For an intercept tile in the horizontal plane and axis $(V,1)$, orthogonal projection onto the axis-perpendicular plane multiplies area by $(1+|V{|}^2{)}^{-1/2}$. Thus a Riemann-sum-like total tends to the weighted area integral in the theorem. The parameter set can have an irregular boundary: outer neighborhoods of a compact set decrease to it, and finite measure continuity controls their excess area. No boundary-measure-zero assumption is inserted.

The slope-field companion constructs rational and singular-power fields for which the weighted area is strictly below the reference cost. It proves that the two swept families leave no gap by explicitly inverting the tilted coordinate maps. A singular tip is cut off and covered separately; its cost is included before the cutoff is removed. The coordinate dictionary tracks physical scaling, so a parameter-plane integral is not mistaken for perpendicular base area. (OpenAI, 2026d)

Radial compatibility makes triangular freezing accurate enough

In the radial sweep theorem, freeze the direction at a tag $t_0$. The displacement caused by replacing $w(t)$ by $w(t_0)$ has first-order term

$$(t-t_0)w^{\prime }(t_0)=(t-t_0)\lambda (t_0)(a+t_{0}b).$$

This lies along the existing radial ray and can be absorbed into the radial coordinate $\rho$. The remaining transverse error is of smaller order, controlled by continuity of $w^{\prime }$ and the variation of the ray. A slightly extended triangle therefore contains every frozen intercept from the parameter interval. Summing the extensions gives vanishing excess as the mesh tends to zero, uniformly over the choice of tags. (OpenAI, 2026c)

The proof allows $R$ or $\lambda$ to vanish and treats the common apex explicitly. Continuity of $R$ controls radial endpoint changes. The determinant $|\det (a,b)|$ supplies the parameter-to-physical area factor, and perpendicular projection supplies the cosine factor. Their product gives the stated integral with $R(t{)}^2/2$.

Finally, affine transfer acts separately in each cylinder direction. The induced map between perpendicular planes multiplies both that base area and that directional shadow of the tetrahedron by the same determinant. Their ratio is invariant. A minimum taken over all directions does not have one common determinant factor, explaining the distinction between the regular-tetrahedron half-area statement and the directionwise statement for every tetrahedron.

See Also

Covering densities also requires exact control of admissible covering arrangements. Projection bodies packages directional shadow areas into a convex body.

🏷️ Manuscripts and verification scope

• Finite angular cylinder covers below the half-area bound — pinned manuscript and companion source.
• Finite cylinder approximation of ruled sets — pinned manuscript and companion source.
• Slope-field perturbations of the two-cylinder covering — pinned manuscript and companion source.
• Finite triangular approximation of radial sweeps — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. The scope distinguishes the finite angular counterexample, the ruled-set approximation hypotheses, the slope-field cover, and the prescribed-tag triangular approximation. In particular, the slope-field scope does not include its affine extension to all tetrahedra. No Lean build, axiom audit, or manuscript-to-formal-statement equivalence check was performed for these notes.

OpenAI collection provides prerequisites and the reading sequence; coverage register records completed and pending families.

🐻  OpenAI 2026a. Finite angular cylinder covers below the half-area bound.

🐻  OpenAI 2026b. Finite cylinder approximation of ruled sets.

🐻  OpenAI 2026c. Finite triangular approximation of radial sweeps.

🐻  OpenAI 2026d. Slope-field perturbations of the two-cylinder covering.