Some shapes, such as the disk or the Releaux triangle can be used as manholes, that is, it is a curve of constant width. (The width between two parallel tangents to the curve are independent of the orientation of the curve.)

**(1) Is it possible to tile the plane with such shapes?**

The shapes should be simply connected, and all must have an area greater than some fixed uniform $\epsilon.$ Otherwise, we'd just tile the plane with disks of various diameters where some are arbitrarily small, similar to the Apollonian circle. By scaling the tiling, we can take $\epsilon=1$.

*Some clarifications:* By tiling, we mean that all manholes used are closed sets,
and there is no open ball that is simultaneously in the interior of two different manholes. Thus, boundaries of manholes may intersect.
Note, we may use several different manholes in a tiling (otherwise, we are essentially asking for a solution for the open einstein problem).

If the answer is negative, a more general question is the following. Define the roundness of an object as the minimum width divided by the maximum width of the shape (width is the distance between two non-equal parallel tangents). The roundness of a circle or a Releaux triangle is 1, and the square has roundness $1/\sqrt{2}.$ Define the roundness of a tiling as the minimum of the roundness of all shapes in the tiling.

For non-convex shapes, roundness can be defined as follows: It is the factor I need to re-size the hole with, so that the original shape cannot fall through that hole. For example, a square with sides $1$ cannot fall through a square hole with sides $1/\sqrt{2}.$

**(2) What is the best possible roundness $R$ a tiling of the plane can have?**

Trivially, $\sqrt{3}/2 \leq R \leq 1$ since we can tile the plane with equilateral triangles, and a positive answer to question 1 gives $R=1.$

The applications are evident: This would be a nice way to tile a ceiling, instead of using regular square tiles, that sometimes falls down.

canbe removed through their holes is an essential part of their design, allowing maintenance of the wiring/ductwork/whatever behind them, so you donotwant them manhole-shaped. $\endgroup$ – Emil Jeřábek Apr 30 '14 at 13:148more comments