most recent 30 from http://mathoverflow.net 2013-05-22T16:25:54Z http://mathoverflow.net/feeds/question/40046 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40046/equidistant-points-in-negatively-curved-metric-spaces Dave Futer 2010-09-26T19:55:28Z 2010-09-26T20:10:23Z

<p>Suppose that $X$ is a simply connected metric space, with a non-positively curved metric (for example, Euclidean or hyperbolic space). Let $A,B,C$ be disjoint, convex sets in $X$, and suppose that the shortest path from $A$ to $B$ passes through $C$. Under these hypotheses, it should follow that there does not exist a point in $X$ that is equidistant to $A$, $B$, and $C$.</p> <p>In the special case where $A,B,C$ are points, this statement amounts to checking inequalities between the sides of a triangle. That is, for any $D \in X$, one of the triangles $ACD$ or $BCD$ -- say, $ACD$ -- will have an obtuse angle at $C$. Then the side $AD$ is longer than $CD$, hence $D$ is not the equidistant point. But I'm stumped about how to show this for more general convex sets.</p> <p>My hunch is that geometers should have encountered this question before. Does anyone have a reference, an argument, or (gasp) a counterexample?</p>

http://mathoverflow.net/questions/40046/equidistant-points-in-negatively-curved-metric-spaces/40047#40047 Oleg Eroshkin 2010-09-26T20:10:23Z 2010-09-26T20:10:23Z

<p>Hello Dave,</p> <p>Three disks of equal radius in Euclidean plane with centers on a circle of sufficiently large radius seems to be an easy counter-example.</p>