most recent 30 from http://mathoverflow.net 2013-06-19T20:51:47Z http://mathoverflow.net/feeds/question/57115 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57115/five-points-in-spheres José Araujo 2011-03-02T14:31:24Z 2011-10-04T06:29:15Z

<p>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?</p> <p>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?</p>

http://mathoverflow.net/questions/57115/five-points-in-spheres/57131#57131 Fedor Petrov 2011-03-02T16:44:17Z 2011-03-02T18:11:35Z

<p>If you mean "any four lie in some ball of radius 1", then the same holds for all five points due to <a href="http://en.wikipedia.org/wiki/Helly%27s_theorem" rel="nofollow">Helly's theorem</a> (the unit balls centered in these points must have a common point). </p>

http://mathoverflow.net/questions/57115/five-points-in-spheres/57139#57139 Myke 2011-03-02T17:24:01Z 2011-03-02T17:24:01Z

<p>Imagine a space filled at random with points everywhere, then take the unit sphere as a test vehicle and enclose all the four point sets that you can. Is it possible to cover the whole space without missing any one point? </p>

http://mathoverflow.net/questions/57115/five-points-in-spheres/57160#57160 Mark Bennet 2011-03-02T19:41:20Z 2011-03-05T08:02:51Z

<p>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.</p> <p>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.</p> <p>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.</p> <p><em>[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]</em></p> <p>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]</p>