What is the radius of the inscribed circle of an ideal triangle I wanted to calculate the radius of the inscribed circle of an ideal triangle.
and when i dat calculate it i came to $\ln( \sqrt {3}) \approx  0.54 $ (being arcos(sec (30^o)) but then
at https://en.wikipedia.org/w/index.php?title=Ideal_triangle&oldid=668440011 
it says that the equilateral triangle that is made by connecting the intersections of the ideal triangle and its [[inscribed circle]]  has a side length of $ 4\ln \left( \frac{1+\sqrt 5}{2}\right) \approx 1.925 $
but this would mean that the radius is less than half the side which is impossible 
so didI make a mistake, or is wikipedia wrong? 
PS the wikipedia page is corrected now and gives the right values)
 A: I think the Beltrami-Klein model is particularly useful here since a hyperbolic line is just a straight chord there. For simplicity use a regular triangle as the inscribed ideal triangle. Its inscribed circle has an Euclidean radius which is half that of the fundamental circle. So the chord along the radius gets divided $1:1$ by the midpoint but $3:1$ by the point on the circle, therefore the radius is
$$\frac12\ln\left(\frac11\cdot\frac31\right)\approx0.54$$
just as you computed.
Now to the edge length of the inscribed triangle. A bit of messing around with the Pythagorean theorem will tell you that the ratio between the Euclidean edge length and the length of its supporting chord is $1:\sqrt5$. Which means one endpoint divides the chord in a ratio of $\sqrt5-1:\sqrt5+1$ and the other in the reciprocal ratio $\sqrt5+1:\sqrt5-1$ so the length between them is
$$\frac12\ln\left(\frac{\sqrt5+1}{\sqrt5-1}\cdot\frac{\sqrt5+1}{\sqrt5-1}\right)
=\ln\frac{\sqrt5+1}{\sqrt5-1}
\approx0.96$$
This looks like a factor of two mistake. Some extra computation shows that this is in fact the case, since
$$\frac{\sqrt5+1}{\sqrt5-1} = \frac{\sqrt5 + 3}2 =
\left(\frac{\sqrt5 + 1}2\right)^2 = \varphi^2$$
I will edit Wikipedia.
A: In the upper half plane model, we take the ideal triangle as $x=-1,$ $x=1,$ and $x^2 + y^2 = 1.$ For this problem, you want the inscribed circle, which is $$ x^2 + (y-2)^2 = 1,  $$
which meets the ideal triangle at $(-1,2),$ $(1,2),$ $(0,1).$ 
The diameter along the $y$ axis goes from $(0,1)$ to $(0,3).$ The geodesic center of the circle is at the geometric mean of $1$ and $3,$ therefore $(0, \sqrt 3).$ Recall from my 2012 answer that this geodesic is $(0,e^t).$ Therefore the radius is $\log {\sqrt 3} - \log 1$ = $(1/2) \log 3.$
One edge of the inscribed triangle passes through $(1,2),$ $(0,1).$ This segment is unit speed parametrized by
$$ 2 + \sqrt 5 \tanh t, \sqrt 5 \operatorname{sech} t  $$
and both ending $t$ values will be negative, because $0 < 2$ and $1 < 2.$
Let us just use the reciprocal of $\cosh t$ and be careful a bout signs. 
One endpoint has
$$ \frac{\sqrt 5}{ \cosh t} = 1  $$ and the other has
$$ \frac{\sqrt 5}{ \cosh t} = 2,  $$ so
$$ \cosh t = \frac{\sqrt 5}{2} $$ and
$$ \cosh t = \sqrt 5 $$
Looked it up, $$ \operatorname{arcosh} x = \log \left(x + \sqrt {x^2 - 1}\right) $$ so the absolute value of the $t$ difference is
$$ \log \left( \sqrt 5 + 2 \right) -  \log \left( \frac{\sqrt 5 + 1}{2} \right) = \log (\varphi^2 + \varphi) - \log \varphi = \log (\varphi + 1) = \log (\varphi^2) = 2 \log \varphi \approx 0.96242365 $$
