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Recall that a space is $\delta$-hyperbolic if there is some number $\delta$ with the property that every point on an edge of a geodesic triangle lies within $\delta$ of another edge. For example a tree is $0$-hyperbolic. One of the basic facts about standard hyperbolic space is that it is $\delta$-hyperbolic for some $\delta$, and I am looking for the smallest delta which makes this true.

Full disclosure: I stole this question from Dima Burago, who brought it up as an example of of a useless problem about which he is nevertheless a little curious. I haven't exactly burned the midnight oil, but I can't solve it.

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    $\begingroup$ This is a well-known elementary exercise mentioned in Bridson-Heafliger and other places. Incidentally, the optimal delta is not totally useless: it could be helpful when doing explicit computations in $H^n$. For example, a cool fact is that in $H^n$ any $r$-quasi-convex set lies in bounded Hausdorff distance from its convex hull, and I would like to know this distance as a function of $r$; I suspect the optimal delta will be part of the formula. $\endgroup$
    – Igor Belegradek
    Commented Apr 30, 2010 at 2:53
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    $\begingroup$ @Will -- the distance between the midpoints is not the desired delta. Paul is asking for the slightly smaller distance from one midpoint to the other edges. $\endgroup$
    – Sam Nead
    Commented Apr 30, 2010 at 3:34

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We can use the isometry group of $H^n$ to reduce to the case of an ideal triangle in the upper half plane, with vertices at -1, 1, and infinity. We want to find the distance between $i$ and the vertical geodesic with real part 1. To find the shortest geodesic, we reflect $i$ in the vertical line, and take half the distance between $i$ and $i+2$. The distance formula yields $\tanh^{-1}\left(\frac{|(i+2)-i|}{|(i+2) + i|} \right) = \tanh^{-1}(1/\sqrt2)$. This is about 0.8813735.

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  • $\begingroup$ There is a Mobius transformation turning your proof into my proof. Of course, Mobius transformations have inverses... :) $\endgroup$
    – Sam Nead
    Commented Apr 30, 2010 at 4:06
  • $\begingroup$ since a geodesic triangle is always contained in a totally geodesic plane why can't we deduce immediately that the $\delta$-hyperbolicity constant of $\mathbb{H}^2$ is optimal, i.e. $\operatorname{log}(2)$? $\endgroup$
    – Dinisaur
    Commented Apr 28, 2022 at 19:07
  • $\begingroup$ ok i just realised that the difinition of hyperbolicity used here is not the $4$-points definition for which the constant is $\operatorname{log}(2)$, but can the same reasoning still apply? are there precise relations between the constants of the different definitions or are these equivalent only under upper and lower linear bounds? $\endgroup$
    – Dinisaur
    Commented Apr 28, 2022 at 19:16
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Consider the ideal triangle with vertices at infinity, zero, and one. Let $C$ be the semicircle perpendicular to the vertical line $[0, \infty]$ and meeting $1/2 + i/2$ (ie the midpoint of the semicircle $[0,1]$). So $C$ meets $[0, \infty]$ at the point $i \cdot \sqrt{2}/2$. Scale down by a factor of $\sqrt{2}/2$ to get the point $1/\sqrt{2} + i/\sqrt{2}$. Use a Mobius transformation to rotate this by $\pi/2$ about $i$ to get $i(1+\sqrt{2})$. Now integrate $dy/y$ to get $\log(1 + \sqrt{2})$. This is approximately 0.88137358702.

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  • $\begingroup$ Thanks, Sam. I started drawing pictures after Matthew seemed to get something different but didn't get very far. Just as well I did not finish the calculation. $\endgroup$
    – Will Jagy
    Commented Apr 30, 2010 at 4:12
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Let $T$ be a triangle in $\mathbb{H}^2$. Its area is $\pi - \alpha - \beta - \gamma$, where $\alpha$, $\beta$, and $\gamma$ are the interior angles. You can find how slim this triangle is by considering an inscribed circle in $T$. The radius of this triangle, thus $\delta$, are bounded above by the area, so to find the $\delta$ that works for all triangles, you take the limit and consider an ideal triangle $T_\infty$. You can explicitly compute that the inscribed circle minimizing distance between the sides has length $4 \log \phi$, where $\phi$ is the golden ratio. (See here and here.)

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  • $\begingroup$ If you make the ideal triangle in the disk model, the $4 \log \phi$ comes from connecting the midpoints. It's the minimum distance between the sides. $\endgroup$
    – Matthew Stover
    Commented Apr 30, 2010 at 2:50
  • $\begingroup$ Ahh, I see the confusion. I said radius instead of hyperbolic length. Edit forthcoming. $\endgroup$
    – Matthew Stover
    Commented Apr 30, 2010 at 3:05
  • $\begingroup$ That second link explains the correct way forward - you want to inscribe a semicircle inside of the triangle, not a circle. Doing the latter will give the in-radius instead of the slimness constant. I don't know what the problem with Wikipedia is - it looks like they are computing the distance between midpoints, but is too large by a factor of two. $\endgroup$
    – Sam Nead
    Commented Apr 30, 2010 at 4:01
  • $\begingroup$ Well, that's what I get for not checking it... $\endgroup$
    – Matthew Stover
    Commented Apr 30, 2010 at 4:21
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See http://www.math.umn.edu/~am/book/outercircles.pdf, p.14-15 for detailed treatment.

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