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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 $\epsilon=1$.

As a note, an earlier question here on MO gave a positive answer on the existence of non-convex simply connected manholes.

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.

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...indoor roof? –  meij Jun 21 '13 at 16:01
    
i.e. a ceiling. –  Yoav Kallus Jun 21 '13 at 16:06
    
Right, I knew there was a word for it, just didn't come to my mind at the moment :) –  Per Alexandersson Jun 21 '13 at 17:01
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I may be missing something, but the fact that ceiling tiles can be removed through their holes is an essential part of their design, allowing maintenance of the wiring/ductwork/whatever behind them, so you do not want them manhole-shaped. –  Emil Jeřábek Apr 30 at 13:14
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It's interesting that the equilateral triangle and the regular hexagon have the same roundness (as defined by @Per above) while the otherwise intermediate square has roundness much worse. –  Włodzimierz Holsztyński Sep 10 at 5:30

2 Answers 2

(EDIT (remarks after the actual answer below)   Technically word cover allows for arbitrary overlaps. However, despite the so familiar definition of notion cover, I was still under impression that the Question meant holes which had pairwise disjoint interiors. Thus my (modest) theorem makes this assumption. Actually, @Per himself (in his Question) had this clarification: there is no open ball in the interior of two different manholes.


I'll answer Question 1 for more general shapes, and in $\mathbb R^n\ (n>1)$:

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

By sloppy pronoun such in the question above we should understand arbitrary bounded convex bodies (closed sets) of constant width, where each body can have its own width.

Instead I'll prove more, by answering the question for arbitrary $n$-dimensional strictly convex proper subsets $B\subseteq \mathbb R^n$ (i.e. different from $\mathbb R^n$ itself)--they do not have to be bounded either:

THEOREM   Let $\ F\ $ be an arbitrary family of proper strictly convex closed sets $\ B\subseteq \mathbb R^n\ (n>1)\ $ such that the intersection of the interiors of every different two of these sets is empty. Then there exist arbitrarily small closed Euclidean balls $\ K\ $ such that $\ K\ $ is not contained in $\ \bigcup F$.

PROOF   If family $\ F\ $ is empty then the theorem holds. Thus let $\ B_0\in F\ $ and let $\ a\ $ belong to the boundary of $\ B_0.\ $ Then there exists an arbitrarily small real $\ r > 0\ $ (i.e. for every $\ \epsilon > 0\ $ there exists positive real $\ r < \epsilon\ $) such that

$$\forall_{B\ C\in F,\ B\ne C}\quad S\cap B\cap C\ =\ \emptyset$$

where $\ S\ $ is the boundary sphere of $\ K.\ $ Indeed, (1) F is countable due to the empty interior intersections; (2) the set of all intersections of the pairs of sets from $\ F\ $ is countable because each pair has at the most one intersection point, and (3) there are continuum of positive $\ r<\epsilon\ $ as needed. Now let's recall a classic Sierpiński's theorem (hm, all Sierpiński theorems are classic though):

If a connected (metric) compact space $\ X\ $ is covered by a countable family $\ F\ $ of pairwise disjoint closed subsets of $\ X\ $ then $\ F\ $ has at the most one member.

Since $\ S\ $ is compact and connected, and $\ S\ $ is not contained in $\ B_0,\ $ it follows from the Sierpiński's theorem that

$$\ \bigcup_{B\in F}\ (S\cap B)\ \ \ne\ \ S$$

Consequently, $\ F\ $ does not cover $\ K$.     END of PROOF


REMARK 1   Wacław Sierpiński formulated and proved his theorem for the metric compact spaces (also in general, Siepiński was not shy to use metric methods). But the theorem holds for arbitrary compact spaces. I am sure that Sierpiński proved his theorem before general compact spaces were introduced by Alexandrov and Urysohn (as bicompact spaces--perhaps even today general compact spaces are called bicompacts most of the time?).

REMARK 2   The assumption from the (full text of the) original question:

    all must have an area greater than some fixed uniform ϵ

was not necessary (as seen from my proof).

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You assume that all the shapes are (1) convex, (2) have non-empty intersection. By covering, we use the normal convention, a covering is an arrangement such that each point in the plane belong to at least one tile (generalized manhole), and no open ball is a subset of two different manholes. The manholes are considered to be closed sets. –  Per Alexandersson Sep 10 at 9:28
    
I meant--of course, that each intersection has at the most ONE point. I was prone to nasty typos. –  Włodzimierz Holsztyński Sep 10 at 9:34
    
Except for my sloppy typos my proof is clean. It is really very simple, elementary. –  Włodzimierz Holsztyński Sep 10 at 9:36
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I think his "such" does not insist on convexity. –  Hao Chen Sep 10 at 10:00
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You say the case of $n=1$ is "obvious", but why isn't $\mathbb{R}=\bigcup_{n\in\mathbb{Z}} [n,n+1]$ a counterexample? –  Joel David Hamkins Sep 10 at 10:40

I believe that @Per meant to ask (for $n=2$) about the following (for arbitrary $\ n>1$):

CONJECTURE A   There exists real $\delta_n > 0\ $ such that for every family $\ F\ $ of bounded constant width convex bodies $\ B\in F\ $ such that each such $\ B\ $ contains a ball of radius $\ 1\ $ there exists an open ball $\ K\ $ of radius $\ \delta_n\ $ such that $\ K\subseteq \mathbb R^n\setminus \bigcup F$.

Each convex body $\ B\in F\ $can have its own diameter (but it has to be $\ \ge 2\ $ due to the assumption about a ball). The constant should be just one constant $\ \delta_n\ $, the same for all said families F.

This seems to be an easy conjecture. The true challenge seems to compute the maximal possible $\ \delta_n.\ $ In a sophisticated case (not too realistic?) the $\ \sup \delta_n\ $ may be NOT attained.


COMMENTS   The original Question considered convex and non-convex manholes, etc. Already squares give a perfect tiling. Thus I thought that it is still sufficiently interesting to concentrate on a restriction to the manholes which are convex bodies of constant width.

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This does not appear to be an answer to the question. –  S. Carnahan Sep 13 at 23:21

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