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On a euclidean plane, what is the minimal area shape S, such that for every unit length curve, a translation and a rotation of S can cover the curve.

What are the bounds of the shape's area if this is a open problem?

When I asked this problem few years ago, someone told me it's open. I don't know if this is still open and I can't find any reference on it.

I don't even know what branch of mathematics it falls under. so I can't even tag this question.

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Stupid question: Does a semi-circle of diameter 1 work? If so, can we provably do better than $\pi /8$? –  Tony Huynh Jul 19 '10 at 12:44
as to the tag, I would suggest: Geometric Measure Theory, Calculus of Variations, and maybe Geometric Probability (if you had the solution, I would be even more precise). –  Pietro Majer Jul 19 '10 at 12:59
arxiv.org/abs/math/0701391 –  Piero D'Ancona Jul 19 '10 at 13:36
See also a related MO question concerning the Lebesgue minimal problem: mathoverflow.net/questions/31315/… –  Andrey Rekalo Jul 19 '10 at 13:50
Note that the quoted paper assumes convexity of the cover... If convexity is not assumed it does not even seems obvious that the infimum of the area be positive. –  Pietro Majer Jul 19 '10 at 14:59

6 Answers 6

up vote 14 down vote accepted

Whereas I don't know of any recent progress in this problem, let me mention one result for closed curves.

Theorem. A closed plane curve of length $L$ and curvature bounded by $K$ can be contained inside a circle of radius $L/4 - (\pi - 2)/2K$.

This was proved in 1974 by H.H. Johnson (link 1) who used calculus of variations methods. A geometric proof was given a bit later by Chakerian, Johnson and Vogt (link 2).

Edit. Apparently the problem is still open. Here's an article (arXiv link), which contains a survey of some known results as of 2009. From the Introduction:

In 1966, Leo Moser asked for the region of smallest area which can accommodate every planar arc of length one. The problem is known as “Moser’s worm problem” and is a variation of universal cover problems. In Moser’s problem, a cover is a set which contains a copy of any rectifiable planar arc of unit length, and is usually assumed to be convex. Such a minimal cover is known to have area between 0.2194 and 0.2738. However, the original problem remains unsolved.

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The lower bound (initially provided by Khandawit and Sriwasdi) was improved in 2009 by Dimitrios Pagonakis. The bound was improved from 0.227498 to 0.231999.

Tirasan Khandhawit, Dimitrios Pagonakis, Sira Sriswasdi. Lower Bound for Convex Hull Area and Universal Cover Problems. Int.J.Comput.Geom.Appl. 23 (2013) 197-212. arXiv:1101.5638. DOI: 10.1142/S0218195913500076

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The result of mine alluded to in A B's answer, giving an improved lower bound, is now on arxiv: http://arxiv.org/abs/1101.5638

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Reportedly R. Norwood, G. Poole, M. Laidacker: The Worm Problem of Leo Moser, Discrete & Computational Geometry 7 (1992), 153-162. has an example of area $\sqrt{3}/12+\pi/24$ (a 60 degree sector of a circle with two triangular "wings"), and this was the best known in 1999.

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P.A.P. Moran proved in 1946, in "On a Problem of S. Ulam" [J. London Math. Soc. 1946 s1-21: 175-179] this theorem:

If $C$ is a curve of unit length in the plane, and $|K$| is the area of its smallest convex cover $K$, then $|K| \le 1/(2\pi)$, and this is the best possible result, since this limit is attained for a semicircle of unit length.

This may not answer your questions entirely, but perhaps it can seed your search.

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There's a chapter in Ian Stewart's 'Game, Set and Math' that covers this problem in a very accessible way (in the guise of a blanket for a worm, IIRC). I'm pretty sure that Joseph's semicircle is the right answer, but it's been ages since I've read that book.

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