Proving Schweikart's constant is $\log(1+\sqrt{2})$ http://imgur.com/Uzzq9zB This link contains the problem and it's referenced lemma and figures.
Forgive me for not wanting to type the entire problem out along with the figures.
As provided in the hint, I got the length $OP=\sqrt{2}-1$. I assumed the radius was 1 and made a square. The hypotenuse of the square was $\sqrt{2}$ via Pythagorean Theorem. Because the radius is 1 for the square, it was 1 for the circle. By definition of a circle, the radius is 1 all around. Using the figure 7.52 given in the problem, I subtracted 1 from $\sqrt{2}$ to get length OP to be $\sqrt{2}$-1. 
Now I am stuck. How to I get length OP (which is equal to d) to equal $\log(1+(\sqrt{2})$?  Using Lemma 7.4, I got $$d= \frac{e^{\sqrt{2}-1}+1}{e^{\sqrt{2}-1}-1}$$ but this did not get me very far even with log rules. Did I miss something? I provided the problem and everything the problem references. The class is called History of Geometry, Euclidean and non-Euclidean using the Greensberg text. 
 A: First note that $|ab|$ will denote the Euclidean length of the line $ab$. Secondly note that I didn't use lemma 7.4 in your image, sorry.
Your image mentions the "distance function defined for the Poincaré disk model". I assume this is defined as follows:
Let $p, q$ be points in the disk. Then the (unique) hyperbolic line passing through $p, q$ intersects the boundary at two points $a, b$. These are the two points at infinity attained by extending the line joining $p$ and $q$. (Make sure $a$, $b$ are labelled so that $|aq|≥|ap|$.) Then the hyperbolic distance from $p$ to $q$ is defined as:
\begin{align*}
d(p, q) = \ln\bigg(\frac{|aq||pb|}{|ap||qb|}\bigg)
\end{align*}
In your specific case we have chosen $p=P$ and $q=O$. Moreover, the hyperbolic line passing through $O, P$ is just a "straight" line as it goes through the origin. This means $a$ is the point on the boundary of the disk collinear to $O,P$, and nearer to $P$, while $b$ is the point on the boundary collinear to $O,P$ but nearer to $O$. Now we can find the individual distances:
\begin{align*}
|aO| &= |1-0| = 1\\
|aP| &= |1-(\sqrt{2}-1)| = 2-\sqrt{2}\\
|Pb| &= |(\sqrt{2}-1) +1| = \sqrt{2}\\
|Ob| &= |0 + 1| = 1
\end{align*}
Thus the hyperbolic distance from $O$ to $P$, which is Schweikart's Constant, is equal to:
\begin{align*}
d(O,P) = \ln\bigg(\frac{|aO||Pb|}{|aP||Ob|}\bigg) = \ln\bigg(\frac{\sqrt{2}}{2-\sqrt{2}}\bigg) = \ln\bigg(\frac{\sqrt{2}(2+\sqrt{2})}{(2-\sqrt{2})(2+\sqrt{2})}\bigg) = \ln(1+\sqrt{2})
\end{align*}
