It is well-known that hyperbolic space is delta-hyperbolic, but what is delta? 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.
 A: 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.)
A: 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.
A: 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.
A: See http://www.math.umn.edu/~am/book/outercircles.pdf, p.14-15 for detailed treatment.
