Cylinder coverings below the half-area bound

Cylinders can cover a regular tetrahedron for less than half its smallest shadow.

4 papers September 27, 2026 Explained by AI

A cylinder cover wraps a solid in straight tubes, and its cost is the total area of the tubes' cross-sections. A classic two-tube cover of a regular tetrahedron costs exactly half the area of its smallest shadow. People asked whether any cover of a convex solid could ever cost less than that. This paper proves that one can, by splitting the two tubes into many slightly tilted ones (Lean proof listed). It also breaks a stronger 2016 conjecture that compares each tube with the shadow in its own direction.

First step

1951

Thøger Bang, the two-cylinder tetrahedron example (as later attributed)

Significance

Solid

It answers a covering question traced to 1951 and a 2016 conjecture, in a specialized area of geometry.

Trust

Formal proof listed

Field

Convex and metric geometry

4 papers

The explanation

Take a regular tetrahedron: a solid with four equal triangle faces. Make every edge 2 units long. Shine a distant light on it and look at its shadow. The shadow's size depends on the direction of the light. The smallest shadow has area √2, about 1.414. It appears when the light shines straight along one edge.

Now try to cover the solid with cylinders. Here a cylinder means a flat shape, its base, slid along a straight line forever in both directions. The base is measured at right angles to that line. The cost of a cover is the total area of its bases. There is a classic cheap cover with just two cylinders. Cut the tetrahedron halfway up. Cover the bottom half with a cylinder running along the bottom edge. Its base is a triangle of area √2/4, about 0.354. Cover the top half the same way, along the top edge. The two bases total √2/2, about 0.707. That is exactly half of the smallest shadow.

The question behind this comes from planks. A plank is the slab between two parallel planes. In 1951, Thøger Bang proved a rule about covering a convex body with planks. A convex body is a solid with no dents or holes. The planks' widths must add up to at least the body's smallest width. For cylinders, the tetrahedron example shows that the same rule with areas fails. Two cylinders cost only half the smallest shadow. Later authors trace this example, and a question, to Bang's 1951 paper. Must every finite cylinder cover cost at least half the smallest shadow? A 2026 survey by William Verreault still lists this as open.

There is also a stronger guess. Divide each base by the body's shadow in that cylinder's own direction, then add the results. In 2016, Károly Bezdek and Muhammad Khan recorded the guess that this sum is always at least one half. In 2009, Bezdek and Alexander Litvak proved it is at least one third. If the stronger guess were true, the half-area rule would follow.

This paper proves that both guesses fail for the regular tetrahedronSource 1. It starts from the classic two-cylinder cover. It slices each triangle base into many thin wedges. Each wedge gets its own slightly tilted lineSource 2. Tilting shrinks the base, because the base is measured across the new line. The one obstacle is gapsSource 3. Tilted neighbors can pull apart, and the two halves can separate near the middle.

Two choices close the gaps. Neighboring wedges are tilted so that their shared side lies in one common planeSource 4. Each wedge is also stretched a tiny amount past the halfway cutSource 5. The stretch costs area, but less than the tilt savesSource 6. Measure cost as a share of the smallest shadow, so the classic cover scores 1/2. Write ε for the size of the tilt. The tilt saves about ε²/240, and the stretch costs ε²/500. The net saving is 13ε²/6000, give or take at most 2ε⁴Source 7.

Take the largest tilt the theorem allows, ε = 1/2000. The cover then uses 16,000,000 cylinders, each with a triangle base. Adding up all their bases gives about 0.49999999946 of the smallest shadow. The saving is about 5.4 parts in ten billion. It is tiny, but the rule claimed no cover could save anything at all. The cover is always finite, but it needs more cylinders as the tilt shrinksSource 8. The result shows that one half is not a floor. It does not say where the true floor lies.

The stronger guess fails for every tetrahedron, not only regular ones. Stretching or squashing space changes each base and its matching shadow by the same factor. So each ratio in the sum stays the sameSource 9. The plain half-area claim is stated only for regular tetrahedraSource 10. Two companion papers build covers below one half by other routesSource 11. One uses parallelogram bases instead of trianglesSource 12. A third gives a general tool for turning smoothly tilting families of segments into finitely many cylindersSource 13.

How sure is this? A computer proof checker called Lean has checked formal versions of the regular tetrahedron results. Every regular tetrahedron has a finite cover costing strictly less than half its smallest shadow. The stronger ratio sum is also below one half. The companion approximation results are in the same Lean files. The extension to every tetrahedron by stretching is outside the checked statements. No specialists have published a review yet.

What it doesn't show. This does not find the cheapest possible cover, and the saving it gives is tiny.

Where it came from

This result comes from an unreleased OpenAI model, published in the openai/math release on 6 October 2026.

Hype check

Solid within its area. The saving is about five parts in ten billion. Lean checks the regular tetrahedron statements. The extension to all tetrahedra is outside that check. No specialist review is published.

Words used

Regular tetrahedron: A solid with four faces that are identical equilateral triangles.
Convex body: A bounded solid with no dents or holes, including its surface.
Cylinder: A flat base shape slid along a straight line forever in both directions.
Perpendicular base: The cross-section of a cylinder cut at right angles to its line; its area is the cylinder's cost.
Projection: The shadow a solid casts on a plane when light comes from one fixed direction.
Plank: The slab of space between two parallel planes; its width is their distance apart.
Directionwise relative area: The sum, over all cylinders, of each base area divided by the shadow in that cylinder's direction.
Lean: A computer program that checks every step of a formal mathematical proof.

The original papers (4)

Finite angular cylinder covers below the half-area bound (PDF, opens in a new tab)
September 27, 2026
Finite cylinder approximation of ruled sets (PDF, opens in a new tab)
September 27, 2026
Slope-field perturbations of the two-cylinder covering (PDF, opens in a new tab)
September 27, 2026
Finite triangular approximation of radial sweeps (PDF, opens in a new tab)
September 27, 2026

Lean notes

All quoted lines (13)

The remainder has absolute value at most $2\eps^4$ , and the total base area is strictly less than $A_{\min}(K)/2$ .
Finite angular cylinder covers below the half-area bound Back to text
We subdivide these triangles into narrow sectors and give each sector its own tilted axis. Tilting reduces the perpendicular area, but it can also open gaps between neighboring sectors and between the two families.
Finite angular cylinder covers below the half-area bound Back to text
Tilting reduces the perpendicular area, but it can also open gaps between neighboring sectors and between the two families.
Finite angular cylinder covers below the half-area bound Back to text
First, the axes adjacent to a shared sector side make the planes containing that side coincide.
Finite angular cylinder covers below the half-area bound Back to text
Second, we increase each sector's radial extent by a quantity of order $\eps^2$ .
Finite angular cylinder covers below the half-area bound Back to text
The increases can therefore be chosen small enough that their area cost is less than the quadratic gain from tilting.
Finite angular cylinder covers below the half-area bound Back to text
Finally, $\int_{-1}^1d(q)\,dq=\frac1{15},\qquad \int_{-1}^1A(q)\,dq=\frac{17}{30},$ so this integral equals $2\eta-1/240$ . The combined error is at most $(59/64+1)\eps^4<2\eps^4$ , as claimed.
Finite angular cylinder covers below the half-area bound Back to text
Every covering in the theorem is finite, although its number of cylinders increases as the tilt decreases.
Finite angular cylinder covers below the half-area bound Back to text
The directionwise ratios are invariant under invertible affine maps.
Finite angular cylinder covers below the half-area bound Back to text
The conclusion of Theorem …, which uses one minimum projection area for all directions, is asserted here for regular tetrahedra.
Finite angular cylinder covers below the half-area bound Back to text
Our construction lowers the tetrahedron's two-cylinder cost strictly by varying the directions, while keeping coverage after finite approximation.
Finite triangular approximation of radial sweeps Back to text
Every regular tetrahedron $K\subset\R^3$ of positive edge length admits a finite cylinder cover with compact parallelogram perpendicular bases such that $\sum_i|B_i|<\tfrac12 A_{\min}(K), \qquad \sum_i\frac{|B_i|}{|\pi_{u_i^\perp}K|}<\tfrac12.$
Slope-field perturbations of the two-cylinder covering Back to text
A useful way to search for inexpensive covers is to perturb the lines in a simple configuration, compare the added label area with the saving in perpendicular projection, and only then make the family finite. This paper supplies that last step with its exact hypotheses and area factor.
Finite cylinder approximation of ruled sets Back to text