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Let $S$ be a sphere in $\mathbb{R}^3$. Let $C$ be a closed curve in $\mathbb{R}^3$ disjoint from and exterior to $S$ which has the property that every point $x$ on $S$ is visible to some point $y$ of $C$, in the sense that the segment $xy$ intersects $S$ at precisely the one point $x$. I am interested in the shortest $C$ with this property. In computational geometry, such paths are called watchman tours, and there are many results concerning polygons in the plane finding such tours.

This question arose at a conference I'm attending, and I was pointed to a paper by V. A. Zalgaller:

"Shortest Inspection Curves for the Sphere" (Journal of Mathematical Sciences, Volume 131, Number 1, 5307-5320; Translated from Zapiski Nauchnykh Seminarov POMI, Vol. 299, 2003, pp. 87–108.)

I cannot access the paper from the conference, but from the abstract it appears he focused on open rather than closed curves.

Has anyone heard of this natural question? Can you point me to relevant literature? Thanks!

Addendum. Here is the $4\pi$ saddle / baseball-stitches curve suggested by Gjergji Zaimi:

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My guess is $4\pi R$, which I can achieve gluing 4 semicircles of radius $R$ (saddle shape). My attempts at finding a clever way to prove the lower bound using a Crofton formula have failed though... – Gjergji Zaimi Jun 29 '11 at 9:51
It is also the single boundary curve remaining on this figure: take the favorite cylinder of Archimedes, having height and diameter $2R.$ Take a saw and cut it in half on a plane that contains the axis of rotation. Now you have two shapes, each a cylinder over a semicircle, and each having a flat square as one face. Rotate one piece 90 degrees with respect to the other, so that the squares match up, and glue back together. – Will Jagy Jun 29 '11 at 17:57
(of course if we allow unions of closed curves, the infimum of length is 0, e.g. C = 6 small circles, each passing through a vertex of the cube $[−1,+1]^3$, or even C = the vertices themselves, as a degenerate case). – Pietro Majer Jun 30 '11 at 8:35
It is not difficult to see that if the curve is at constant distance from the sphere, then its length is at least $4\pi$. Indeed suppose that it is a constant distance $r$ from the center. The area of the set of points newly seen from a short segment of length $ds$ is $a(r)ds$, with $a(r)=2\sqrt{r^2-1}/r^2$. This is maximal for $r=\sqrt{2}$, then $a=1$. Since each point of the sphere is "newly seen" from at least one point of the curve, the result follows. The curve proposed by Gjergji Zaimi is at constant distance from the center, so it is optimal at least in this restricted sense. – Jean-Marc Schlenker Jul 2 '11 at 6:24
This problem appears in the last lines of a math popularization article by Jean-Baptiste Hiriart-Urruty, Du calcul différentiel au calcul variationnel, in Quadrature 70(2008):8-18, see There the problem is presented as open and attributed to Alain Grigis. – Jean-Marc Schlenker Jul 5 '11 at 6:15

1 Answer 1

I am quite sure that the problem is open, but we can play math-sport --- who makes a better constant.

Let me make a long remark, mostly based on the Zalgaller's paper. I will describe a family of examples, which includes the $4{\cdot}\pi$-example of Gjergji Zaimi, but I can not see if $4{\cdot}\pi$ is the best constant in this family. (It seems that you are a friend of computer and it would be easy for you to check.)

The curve can be viewed a $S^1$-family of circles on the sphere. In the example of Gjergji Zaimi, each circle in the family has exactly one point on one half-equator and one on the opposite meridian. Instead of half-equator and meridian one can choose two curves and consider corresponding $S^1$-family of circles; such a pair of curves is described by few parameters. (The centers of the circles in this family also can be described as envelop-line for circles of half-radius in the Zalgaller's family)

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Does math-sport include free generalizations? I would then ask the best (=minimum) constant $C$ such that, for any curve $\gamma:\mathbb{S}^1\to\mathbb{R}^3$ with $\|\gamma(t)\|\ge 1$, the area of the portion of the unit sphere which is visible from $\gamma$ is at most $2\pi+ C \mathrm{length}(\gamma)$. A nice guess is $C=1/2$, that would produce $\mathrm{length}(\gamma)\ge4\pi$ for any inspecting curve. – Pietro Majer Jun 30 '11 at 15:08

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