Given a right angled regular hyperbolic octagon centered at origin, what is the distance from the origin to any vertex?
I know that the distance between the origin and the point $p=(a,0)$, $a>0$, can be solved using the metric $g = \frac{4}{(1-x^2-y^2)^2}(dx^2+dy^2)$. But how will I figure out where $g$ is?
On
"The Geometry and Spectra of Compact Riemann Surfaces" by Peter Buser has a great section on Hyperbolic trigonometry. Chapter Two has the relevant material, although the whole book is great.
Connecting the origin with two adjacent vertices of the octagon you obtain an isosceles triangle, and halving that a right triangle. The remaining angles of the latter are $45^\circ$ and $22.5^\circ$. Using the formulas from here you should be able to compute the hypotenuse. Verify the normalization used by your $g$.
The problem can also be solved using "euclidean analytic geometry" in the following way: Consider two rays through the origin with slopes $\pm22.5^\circ$. These represent two subsequent spikes of your octagon. Now determine the circle with center $(a,0)$ and radius $r>0$ which intersects the two lines at an angle of $45^\circ$ and the unit circle orthogonally. The points of intersection with the two lines are vertices of your octagon.