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I try to understand: is there a smallest in area convex set that every smooth curve with length 1 can be placed inside it by translation and rotation?

I only have a upper bound $S \leq \frac14+\frac{\pi}{16}$ because of convex hull of two circles radius $\frac14$ and simple lower bound $S\geq\frac1{4\pi}$.

Does this set exist and what is its length?

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Is your question the same as this one on MathOverflow? Smallest area shape that covers all unit length curve –  Rahul Dec 24 '11 at 3:19
    
Oh, I've searched here, not on the MathOverflow because this problem looks so pretty simple. Thx. –  sas Dec 24 '11 at 4:33
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up vote 3 down vote accepted

This is entry D18, "The worm problem," in the book Unsolved Problems in Geometry (Croft, Falconer, and Guy, 1991):

Leo Moser asked what are the minimal comfortable living quarters for a "unit worm"?

They credit the smallest cover yet discovered to Gerriets & Poole (in 1973 or 1974), and describe it as a "certain truncated rhombus of area less than $0.286$...". Maybe someone will have a link or reference to the particular shape? For your bounds, I would only point out that the convex hull of a semicircle of arclength $1$, with radius $1/\pi$ and area $1/(2\pi)$, gives a better lower bound (of $0.159...$).

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