Question

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?

Answer 1

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 $AD \ge AI+OD=1.077...$.

Answer 2

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 $\phi\in[0,2\pi]$ $$c^2\leq \rho(\phi) \leq 1 - c^2,\qquad c=1-1/\sqrt{3}.$$