# What is the distance from the origin to a right angled regular hyperbolic octagon?

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?

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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.
Thanks Christian. We have never talked about hyperbolic triangles yet, so I assume we can't use it. We did a lot of stuff regarding cross ratios and how a hyperbolic distance is related to the euclidean distance $r$ by $d=log \frac{1+r}{1-r}$. Is there any way I can solve for this without using a triangle? Thanks in advance! – PandaMan Feb 10 '14 at 6:09