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OpenGeometry·1914

Lebesgue's Universal Cover

Find the convex set of minimum area that contains a congruent copy of every planar set of diameter 1.

Formula

Universal cover
min{C:  C convex,  SR2,  diam(S)1     rigid motion f,  f(S)C}\min\bigl\{|C|:\;C\text{ convex},\;\forall S\subset\mathbb{R}^2,\;\operatorname{diam}(S)\le 1\;\Rightarrow\;\exists\text{ rigid motion }f,\;f(S)\subseteq C\bigr\}

Find the minimum-area convex set containing a congruent copy of every planar set of diameter ≤ 1.

Pál hexagon
APaˊl=320.8660A_{\mathrm{Pál}} = \frac{\sqrt{3}}{2} \approx 0.8660

A regular hexagon of width 1 covers all unit-diameter sets; Pál's corner trimming reduces this to ≈ 0.8453.

Known bounds
0.832A0.84411530.832 \le A^* \le 0.8441153

The optimal area lies in this interval after a century of work; the gap of ≈ 0.012 remains open.

Summary

In a 1914 letter to Gyula Pál, Henri Lebesgue asked for the minimum-area convex subset of the plane that can accommodate, by rigid motion, every planar set of diameter at most 1. Pál's 1920 answer showed a regular hexagon of width 1 (area √3/2 ≈ 0.8660) is sufficient; by trimming two opposite corners he reduced this to ≈ 0.8453. Sprague (1936) improved the upper bound further to ≈ 0.8441, and Baez, Bagdasaryan, and Gibbs (2015) reached 0.8441153 using computer-assisted corner trimming. Gibbs (2018) refined this to ≈ 0.8440936. The best lower bound, 0.832, was proved by Brass and Sharifi (2005) through a computational search over configurations of circles, triangles, and pentagons. Despite over a century of effort the gap of roughly 0.012 between the bounds remains fully open, making this one of the most stubborn unsolved problems in combinatorial geometry.

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