Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

Given a hyperbolic triangle $T$ and two points $p$ and $q$ in Poincare disk. Note that $p$ and $q$ are outside the triangle. If $p$ has shorter distances to the three vertices of $T$ than $q,$ can we claim that $p$ has shorter distances to all points inside $T$ than $q$?

If the answer is true, can we extend ths claim from a hyperbolic triangle to a hyperbolic convex hull?

share|cite|improve this question

Yes, this is true.

For a point $p$ not in the triangle, there is a unique point $x$ in the triangle $T$ such that $d(p,x)=d(p,T)$ (where $d$ denotes the hyperbolic distance function). If $x$ is a vertex of $T$, then by your hypothesis, $q$ is closer to $x$ than $p$, and therefore is closer to $T$ than $p$.

If $x$ is not a vertex of $T$, then it must lie in the interior of an edge $e\subset T$, lets say with endpoints $a, b$. Let $R=d(p,e)=d(p,T)$. The intersection of the two disks of radius $d(a,p), d(b,p)$ about $a$ and $b$ respectively lies inside the tube of radius $R$ about $e$. This in turn follows because if you have a circle, and you take a radius of it, and a chord perpendicular to the radius which cuts off a lune, then every point in the lune will be closer to the radius than the two extreme points of the lune (it might be helpful to consider this in euclidean space before thinking about it in hyperbolic space, since the argument is essentially the same).

share|cite|improve this answer

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.