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Timeline for Five points in spheres

Current License: CC BY-SA 2.5

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Mar 9, 2011 at 22:51 comment added Fedor Petrov Everybody interested in, see the link above in comments
Mar 4, 2011 at 19:28 comment added Fedor Petrov Sorry, I've been mistaken in calculations. The construction from my deleted comment:) is impossible by some clear reasons. @Gjergji: why none of them may be contained in the convex hull of others?
Mar 4, 2011 at 17:38 comment added Pace Nielsen @Fedor:Maybe I'm not understanding which are your five points, but I've plugged things into Mathematica, and I cannot get it to work with two points on the line through O perpendicular to a plane with 3 points in a regular triangle. To make things concrete, I center one of the spheres at (0,0,0). Then I can put one of the points (perhaps P in your notation) at (0,0,1). In other words, if I take my five points to be $(a,0,\sqrt{1-a^2}), ((\sqrt{3}/2)a,-(1/2)a,\sqrt{1-a^2}), (-(\sqrt{3}/2)a,-(1/2)a,\sqrt{1-a^2}), (0,0,1),(0,0,b)$ I cannot choose $a$ and $b$ to make all 5 spheres have radius 1
Mar 4, 2011 at 8:10 comment added Gjergji Zaimi @Fedor: How about insisting that none of the points is contained in the convex hull of the other four? It seems this should have a negative answer, but I haven't thought about it yet :P
Mar 3, 2011 at 17:14 comment added Fedor Petrov @Gjergji: it looks that the answer is positive (see comment to initial question), that makes question less interesting:)
Mar 2, 2011 at 18:54 comment added aaron @Gjergji et al: I think so too.
Mar 2, 2011 at 18:11 history edited Kevin O'Bryant CC BY-SA 2.5
added link to Helly's theorem
Mar 2, 2011 at 17:53 comment added Gjergji Zaimi I think the intended question is with "on a sphere", though it's hard to tell from the formulation. It would be an interesting question since it fails in 2d.
Mar 2, 2011 at 16:44 history answered Fedor Petrov CC BY-SA 2.5