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Lebesgue's universal covering problem

Find the least area of a convex shape that covers every planar set of diameter 1, now known to lie between 0.832 and 0.8440936.

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  1. 开设2026年10月21日
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What is the smallest area of a convex shape that can cover every planar set of diameter 1, after moving the set rigidly? Lebesgue asked this in 1914. The answer is known only to lie between 0.832 and 0.8440935944, and both bounds come from explicit geometric constructions checked by computer.

Problem

A universal cover is a convex set Ω⊂R2\Omega \subset \mathbb{R}^2 that contains a congruent copy of every planar set of diameter 1. Let

a=inf⁡{area⁡(Ω):Ω is a universal cover}.a = \inf \{\operatorname{area}(\Omega) : \Omega \text{ is a universal cover}\}.

Determine aa, or narrow the known range 0.832≤a≤0.84409359440.832 \le a \le 0.8440935944.

A smallest universal cover exists (by the Blaschke selection theorem), and it suffices to cover the sets of constant width 1 (Vrećica, 1981).

Background

Lebesgue posed the question in 1914. Pál (1920) observed that the regular hexagon circumscribed about a disk of diameter 1 is a universal cover, and that it can be truncated further using an inscribed regular dodecagon. Sprague (1936) and Hansen (1992) removed further, much smaller pieces; Baez, Bagdasaryan and Gibbs (2015) showed that Hansen's last cut was even smaller than claimed and found a better cover with computer assistance. Lower bounds come from showing that any convex set covering a few fixed shapes of diameter 1, such as the disk and the equilateral triangle, must have a certain area.

What is known

  • Upper bounds: 3/2≈0.8660\sqrt{3}/2 \approx 0.8660, the hexagon, and 2−2/3≈0.845302 - 2/\sqrt{3} \approx 0.84530, the truncated hexagon (Pál, 1920); 0.8441377084360.844137708436 (Sprague, 1936); 0.8441377083980.844137708398 (Hansen, 1992, as corrected in 2015); 0.84411529710.8441152971 (Baez, Bagdasaryan and Gibbs, 2015, verified to high precision by Greg Egan); 0.84409359440.8440935944 (Gibbs, 2018).
  • Lower bounds: π/4≈0.7854\pi/4 \approx 0.7854 (the unit disk); 0.82570.8257 and 0.82710.8271 (Elekes, 1994, using the disk, the triangle and regular 3j3^j-gons); 0.8320.832 (Brass and Sharifi, 2005, a rigorous computer search using the disk, the equilateral triangle and the regular pentagon).

What would count as progress

  • A convex set of area below 0.84409359440.8440935944, with a proof that it contains a congruent copy of every set of constant width 1. The area is computed exactly; the covering proof is the substantive part, and the recent records rely on computer-assisted geometry checked to high precision.
  • A finite family of sets of diameter 1 whose simultaneous covering forces area above 0.8320.832, with a certified search over the placements of the shapes (in the style of Brass and Sharifi).

References

  1. G. Elekes, "Generalized breadths, circular Cantor sets, and the least area UCC", Discrete & Computational Geometry 12 (1994), 439–449. doi:10.1007/BF02574391
  2. P. Brass and M. Sharifi, "A lower bound for Lebesgue's universal cover problem", International Journal of Computational Geometry & Applications 15 (2005), 537–544. doi:10.1142/S0218195905001828
  3. J. C. Baez, K. Bagdasaryan and P. Gibbs, "The Lebesgue universal covering problem", Journal of Computational Geometry 6 (2015), 288–299. arXiv:1502.01251
  4. P. Gibbs, "An upper bound for Lebesgue's covering problem" (2018). arXiv:1810.10089

Source

Imported from Optimization Constants in Mathematics, constant 13b, curated by Terence Tao and contributors. Open according to the source as of 2026-10-20. The statement above is written in our own words; see the source for the original wording, history and discussion.

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