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Do there exist five points in the euclidean space ${\mathbb R}^3$ such that every four of these points are in a spherical ball of radius 1, but that the five points are not in a ball of radius 1?

Do there exist five points in the euclidean space ${\mathbb R}^3$ such that every four of these points are on a sphere of radius 1, but that the five points are not on a sphere radius 1?

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When you say "in a sphere" do you mean "on a sphere"? – aaron Mar 2 '11 at 14:52
The OP is looking for a set of five points so that, for every four of them, the four points are contained in a ball (or maybe on a sphere) of radius 1, and yet all five are not. I think we have a foreign-language issue here, in that the English is not precise enough to tell whether he is asking for "on the surface of a sphere" or "in the interior of a ball". – Peter Shor Mar 2 '11 at 15:43
Some background/motivation? – Qfwfq Mar 2 '11 at 15:49
Note that vertices of triangle plus orthocenter give a solution of the analog problem in 2D. – ε-δ Mar 2 '11 at 21:51
As for the second question, the answer is also positive, proved by Hiroshi Maeharaa and Norihide Tokushigea, in European Journal of Combinatorics (Volume 30, Issue 5, July 2009, Pages 1337-1351). – Fedor Petrov Mar 9 '11 at 22:51

If you mean "any four lie in some ball of radius 1", then the same holds for all five points due to Helly's theorem (the unit balls centered in these points must have a common point).

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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. – Gjergji Zaimi Mar 2 '11 at 17:53
@Gjergji et al: I think so too. – aaron Mar 2 '11 at 18:54
@Gjergji: it looks that the answer is positive (see comment to initial question), that makes question less interesting:) – Fedor Petrov Mar 3 '11 at 17:14
@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 – Gjergji Zaimi Mar 4 '11 at 8:10
@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 – Pace Nielsen Mar 4 '11 at 17:38

Clearly we can't have three collinear points on a sphere. Look at the "on a sphere" case and assume there is a configuration where they do not all lie on a single sphere.

Any three points define a plane. The locus of points equidistant from these three is a line perpendicular to the plane (through the circumcentre). There are at most two points on such a line which are unit distance from the original three, $P$ and $Q$ say. These are the centres of two unit circles, and one of the remaining two points must lie on each sphere.

Note that $P$ and $Q$ are related by a reflection in the original plane. There are five sets of four points in the original configuration. Each set defines a unit sphere, and if two spheres are the same, then all are. So there are five spheres and the centres are related by reflections in the planes defined by triangles.

[From here is a bathtime intuition which, per comments doesn't work. However note that each pair of centres is related by a reflection, which may not map other centres to centres, and therefore creates an infinite group. I thought I could see the group acting on the centres, but I can't make it work - thanks for comments to put me right]

There is just one group of order 5 - cyclic - and this would imply that the centres of the five spheres formed a regular pentagon. Since this does not provide a suitable configuration, none exists. [Would need three points in each of five planes meeting in a single line, no three collinear]

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Why do the reflections form a finite group? They don't in the two-dimensional case. – Peter Shor Mar 2 '11 at 20:39
To add a point to Peter Shor's objection: you constructed five reflexions each echanging two centers and fixing three of the given points, but these reflexions do not map the other three centers to centers, so the group they generate does not act on the set of five centers. – Benoît Kloeckner Mar 3 '11 at 12:51

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