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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?

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1 Answer 1

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).

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