3
$\begingroup$

A theorem of Jung states that, given n pairwise distinct points in the Euclidean plane E, there is a unique circle of smallest radius in E that contains all the points and its radius is less than or equal to d/(3^(0.5)) where d is the maximum distance between any pair of the points. Is there an analogous theorem that applies to the 2-dimensional surface of a sphere in 3-dimensional Euclidean space?

$\endgroup$
2
  • $\begingroup$ It seems that the problem is open. See Ю.Д.Бураго. ."Задача о круге Юнга для сферы" Математическое просвещение (II). № 6, math.ru/lib/book/djvu/mp2/mp2-6.djvu $\endgroup$ Jul 21, 2011 at 9:45
  • $\begingroup$ Yet one more ref: Dekster, B. V. The Jung theorem for spherical and hyperbolic spaces. Acta Math. Hungar. 67 (1995), no. 4, 315--331. $\endgroup$ Jul 21, 2011 at 10:29

1 Answer 1

2
$\begingroup$

Triangle with side $d$ is the worst example for $d< \arccos(-\tfrac13)$. In this case you also have uniqueness.

For $d=\arccos(-\tfrac13)$, the four point set with pairwise distance $\arccos(-\tfrac13)$ is the worse; no uniqueness and $r=d$.

It is not clear for me what happens for $d>\arccos(-\tfrac13)$. I guess $r=d$, but I do not see the worse configuration except for particular values of $d$. For example, concentric circles with radii $$2\alpha, 4\alpha, \dots, 2k\alpha$$ give the worse configuration for $\alpha=\pi/(2k+1)$, $k$ is a positive integer and $d=\pi-\alpha$.

$\endgroup$
2
  • $\begingroup$ For the regular tetrahedral configuration ($d=\cos^{-1}(-\frac{1}{3})$), surely it is impossible to cover all points by a single circle! $\endgroup$ Jul 19, 2011 at 21:09
  • $\begingroup$ I think that if there is a theorem about points on the surface of a sphere-call it the S theorem-which is analogous to Jung's theorem about points in a plane, then the role of circles in Jung's theorem would be played by Spherical Caps in the S theorem. Spherical caps are those subsets of the whole spherical surface that lie on one side of any plane which intersects the sphere. We assume the standard metric for the spherical surface-the same metric used to measure the sides of triangles in spherical trigonometry. $\endgroup$ Jul 20, 2011 at 22:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.