The mean distance to a simple closed curve on the sphere

Abstract: Kimberling's Problem 10 asks for a simple closed curve of prescribed length $L$ (in particular, $L=4π$) on the unit sphere minimizing the mean geodesic distance $\mathcal{J}$ from a point of the sphere to the curve. For a positive integer $n$, put $\vartheta_{n}=π/(2n)$ and $L_{n}=2π/\sin\vartheta_{n}$. We show that the minimum of $\mathcal{J}$ over rectifiable simple closed curves of length at most $L_{n}$ equals $\vartheta_{n}-\tan(\vartheta_{n}/2)$, that it is attained only by curves of length exactly $L_{n}$, and that the sphere-filling ropes $β<sup>{n,k}$ of Gerlach and von der Mosel attain it. Kimberling's case is $n=3$: at $L=4π$ the minimum is $π/6+\sqrt{3}-2=0.255649\ldots$, attained by an explicit six-arc curve and by its mirror image. For $L\le2π$ we determine $J(L)$, the infimum of $\mathcal{J}$ over curves of length $L$, exactly: it equals $π/2-L/(2π)$, attained precisely by the circles of length $L$. At the lengths $L_{n}$ we do not classify all minimizers, but show that every one of them bisects the sphere into two disks of area $2π$ and inradius $\vartheta_{n}$ whose inward collars have the largest possible area at every depth. The great circle is the only minimizer for $n=1$, and the $β<sup>{n,k}$ are, up to congruence, the only ones of thickness at least $\sin\vartheta_{n}$. For arbitrary $L$ the function $J$ is nonincreasing, and together with the above this brackets it between two explicit values.