I have three pairs of points in 3D space. These may or may not be coplanar. I want to find a point such that it is equidistant from each pair of points. I know that may or may not be possible depending on the positions of the points. What I want is the best average point, which I can take safely as the centre and draw a sphere from there whose radius is the maximum distance of this point from any of the six points, then I want all the points to remain inside the sphere.
[Math] the average center of six points in space
mg.metric-geometry
Related Solutions
I took a pane of clear glass
and touched two balls at once
I put my light, perhaps, by chance,
above the pane. Alas,
the shining pile on the same side
in its arrangement lay,
and no matter what I tried
(I tried a whole day)
Some darkness (though not too much)
remained around points of touch...
I believe the same algorithm works. Note the remarkable fact, first understood by Kevin Brown, that the Voronoi diagram of points $P$ on a sphere is combinatorially identical to the (3D) convex hull of $P$: each face of the hull corresponds to a Voronoi vertex (and to an empty circle). This leads to an $O(n \log n)$ algorithm, as explained in this paper:
Hyeon-Suk Na, Chung-Nim Lee, Otfried Cheong, "Voronoi diagrams on the sphere." Computational Geometry: Theory and Applications, 23 (2002) 183–194. (ACM link)
Here is a recent implementation article:
Zheng, Xiaoyu, et al. "A Plane Sweep Algorithm for the Voronoi Tessellation of the Sphere." Electronic-Liquid Crystal Communications [online] (2011). (PDF download.)
I can't resist re-sharing this image of a Vornoi diagram on a sphere from an earlier MO question:
Mathematica image by Maxim Rytin.
Best Answer
Distance from a pair of points instead of from each one of a pair of points does not seem to be well defined. However, you could simply take the centroid of all six points (add the coordinates and divide by 6 in each axis), then compute the distance to each of the six points, and use the maximum for the radius of the sphere. It is not clear how you want us to use the fact that the points come in pairs instead of six single points.