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I have a perfect rigid sphere of diameter 1.

I have infinite supply of rope. The rope is infinitely flexible and can be cut or glued without losing or adding length. The rope can be glued at any place (extreme to extreme, or extreme to middle). The cross-section of the rope has area zero (i.e. the rope has only length but no other dimensions).

The objective is to use the rope to build a net from which a sphere of diameter 1 cannot escape.

What is the minimum amount of rope needed?

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    $\begingroup$ My first guess: make the net of the shape of the 1-skeleton of a regular tetrahedron (or rather its projection onto the surface of the sphere). $\endgroup$ – André Henriques Dec 30 '14 at 16:52
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    $\begingroup$ My earlier MO question, "Hanging a ball with string," is quite close. And in fact, in that question's Addendum, I quoted Croft: "Besicovitch [1] has shown: if a net of inextensible string encloses a sphere of unit radius in such a way that the sphere cannot slip out, then the length of the string is strictly greater than 3π, and this is false with any greater constant replacing 3π." $\endgroup$ – Joseph O'Rourke Dec 30 '14 at 17:40
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    $\begingroup$ @AndréHenriques, apparently this is not optimal. $\endgroup$ – Anton Petrunin Dec 30 '14 at 21:34
  • $\begingroup$ @AndréHenriques: the tetrahedron achieves total length of $12 \tan^{-1}{\sqrt{2}} \approx 3.649\pi$. You can use less string by shrinking one face of the tetrahedron: have a very small circle around the north pole, from which 3 meridians (at $120^{\circ}$ from each other) radiate and go all the way down to the south pole. The perimeter of each large hole is still less than $2\pi$ and the sphere cannot escape. $\endgroup$ – Yaakov Baruch Dec 30 '14 at 23:16
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    $\begingroup$ There is a picture of the arrangement that Yaakov is describing here, taken from that earlier question. $\endgroup$ – Joseph O'Rourke Dec 31 '14 at 0:19