# Is Lebesgue's “universal covering” problem still open?

The following problem has been attributed to Lebesgue. Let "set" denote any subset of the Euclidean plane. What is the greatest lower bound of the diameter of any set which contains a subset congruent to every set of diameter 1? There are a number of interesting geometric problems of this type. Is it possible that some of them may be difficult to solve because the solution is a real irrational number which (when expressed in decimal form) is not even recursive-and so cannot be approximated in the usual way?

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The question is still open. There are at least two versions. The most popular asks for the minimal-area convex subset of the plane such that every set with diameter 1 can be rotated and translated to fit inside it. Here is the best lower bound I know:

Their lower bound is 0.832, obtained through a rigorous computer-aided search for the convex set with the smallest area that contains a circle, equilateral triangle and pentagon of diameter 1.

Here is the best upper bound I know:

In 1920, Pál noted that a regular hexagon of area circumscribed around the unit circle does the job. This has area

$$\sqrt{3}/2 \approx 0.86602540$$

But in the same paper, he showed you could safely cut off two corners of this hexagon, defined by fitting a dodecagon circumscribed around the unit circle inside the hexagon. This brought the upper bound down to

$$2 - 2/\sqrt{3} \approx 0.84529946$$

He guessed this solution was optimal.

In 1936, Sprague sliced tiny pieces of Pal's proposed solution and bring the upper bound down to

$$\sim 0.84413770$$

(Image from Hansen's paper, added by J.O'Rourke.)

The big hexagon above is Pál's original solution. He then inscribed a regular dodecagon inside this, and showed that you can remove two of the resulting corners, say $B_1B_2B$ and $F_1F_2F,$ and get a smaller universal covering. But Sprague noticed that near $D$ you can also remove the part outside the circle with radius 1 centered at $B_1$, as well as the part outside the circle with radius 1 centered at $F_2.$

In 1975, Hansen showed you could slice off very tiny corners off Sprague's solution, each of which reduces the area by $6 \cdot 10^{-18}$.

In Hansen's 1992 paper, he did much better. He again sliced two corners off Sprague's solution, but now one reduces the area by a whopping $4 \cdot 10^{-11}$, while the other, the same shape as before, reduces the area by $6 \cdot 10^{-18}$.

One author, in a parody of the usual optimistic prophecies of accelerating progress, commented that

...progress on this question, which has been painfully slow in the past, may be even more painfully slow in the future.

In 1980, Duff considered nonconvex subsets of the plane with least area such that every set with diameter one can be rotated and translated to fit inside it. He found one with area

$$\sim 0.84413570$$

which is smaller than the best known convex solution:

• G. F. D. Duff, A smaller universal cover for sets of unit diameter, C. R. Math. Acad. Sci. 2 (1980), 37--42.

I've written a slightly more detailed account with some pictures here:

If anyone knows of further progress on this puzzle, please let me know!

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But the number in the 3rd display is bigger than the number in the 2nd display. –  Gerry Myerson Dec 8 at 0:32
Fixed, I think - thanks! Somehow I'd put down the wrong figure for $2−2/\sqrt{3}$, and this cures the problem you mention. As for Sprague's improvement, I'm getting the figure second-hand. –  John Baez Dec 8 at 7:37
You changed the second display, and then changed it back. $0.84413770$ is not an improvement on an upper bound of $0.8422038$, so what is going on here? Also, I edited in the ending page number for Brass-Sharifi, and you edited it out. –  Gerry Myerson Dec 8 at 23:10
I don't know how I screwed up that stuff. Sorry; it's my unfamiliarity with editing here. I fixed it up. –  John Baez Dec 9 at 1:29

It's Lebesgue Minimal Problem. It's still open, though there are some bounds to the area of such set, for example there are lower bound for area $S\ge \frac{\pi}{8}+\frac{\sqrt{3}}{4}$

it's not hard to show that such set must have diameter larger or equal to $\frac{\sqrt{3}}{3}+\frac{1}{2}=1.077350...$

Our set must have a subsets congruent to the equlateral triangle with side 1(ABC) and to the circle with radius 0.5(with center O). If I - the incenter of triangle ABC, and $O \in BIC$, then consider point D which lies on the radius perpendicular to BC. Then $AD \ge AI+OD=1.077...$.

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You say $O \in BIC$ but your diagrams seems to have $O \notin BIC$. Still, it obeys $AD \ge AI + OD$. –  John Baez Dec 6 at 8:13
The problem has been studied for various groups $G$ of isometries of $\mathbb R^n$. A set $K\subset \mathbb R^n$ is called $G$-universal cover iff every set of diameter 1 is contained in $gK$ for some $g\in G$.
V. Makeev proved that the mean width of a $T_n$-universal cover is greater or equal to $\sqrt{2n/(2n+1)}$, where $T_n$ is the group of translations of $\mathbb R^n$. For $n=2$ the estimate is sharp; the perimeter of a $T_2$-universal cover $\geq 2\pi/\sqrt{3}$ (link).
M. Kovalev obtained a rather explicit description of all minimal $D_2$-universal covers, where $D_2$ is the group of all isometries of $\mathbb R^2$.
Theorem. Every minimal universal $D_2$-cover $K$ is star-shaped. There is a polar coordinate system (with the centre in $K$) such that $$\partial K=\{(\phi,\rho(\phi)):\ 0 < \phi\leq 2\pi\},$$ where $\rho=\rho(\phi)$ is Lipschitz and for any $\phi\in[0,2\pi]$ $$c^2\leq \rho(\phi) \leq 1 - c^2,\qquad c=1-1/\sqrt{3}.$$
It's worth noting, for nonexperts, that "minimality" here means that the set $K$ doesn't contain a subset with the desired property. This is implied by, but does not imply, area minimization. Indeed, "minimal" solutions to the Lebesgue universal covering problem are known, while area-minimizing ones are not. –  John Baez Dec 8 at 7:36