MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer
Right you are! Apparently, it was a brain fart on my part to believe that this is true. – Dave Futer Sep 26 '10 at 20:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.