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 $C$.

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.

My hunch is that geometers should have encountered this question before. Does anyone have a reference, an argument, or (gasp) a counterexample?


Hello Dave,

Three disks of equal radius in Euclidean plane with centers on a circle of sufficiently large radius seems to be an easy counter-example.

  • $\begingroup$ Right you are! Apparently, it was a brain fart on my part to believe that this is true. $\endgroup$ – Dave Futer Sep 26 '10 at 20:17

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