# Hyperbolic Lambert quadrilateral

A hyperbolic Lambert quadrilateral is a 4-sided polygon in the hyperbolic plane with at least three right angles. A hyperbolic Lambert quadrilateral

Consider any hyperbolic Lambert quadrilateral. As in the picture on the right, label its sides $$a$$, $$b$$, $$c$$, and $$d$$ in the clockwise direction, with the not necessarily right angle $$\theta$$ between $$c$$ and $$d$$.

It can be shown that any two of these five quantities determine the other three. Thus, any three of them satisfy a relation. There are six of these relations up to symmetry, as follows:

1. $$\sinh a \cdot \sinh b = \cos \theta$$
2. $$\cosh a = \cosh c \cdot \sin \theta$$
3. $$\tanh a \cdot \cosh b = \tanh c$$
4. $$\sinh a \cdot \cosh d = \sinh c$$
5. $$\tanh a \cdot \sinh d = \cot \theta$$
6. $$\tanh c \cdot \tanh d = \cos \theta$$

Consider equation (1). Since the hyperbolic sine $$\sinh x$$ is positive for positive values of $$x$$, the left side of the equation is always positive. This means that $$\theta$$ must always be acute (less than $$\pi / 2$$), and in particular non-right. Thus, a hyperbolic Lambert quadrilateral always has exactly three right angles. As $$a$$ and $$b$$ approach $$0$$, the hyperbolic Lambert quadrilateral approaches a Euclidean rectangle.

Also from equation (1), since $$\sinh x$$ approaches $$0$$ as $$x$$ approaches $$0$$, you can see that as $$a$$ and $$b$$ approach $$0$$, the angle $$\theta$$ approaches a right angle of $$\pi / 2$$. (This also holds if just one of the two approaches $$0$$ while the other stays bounded.) Moreover, you can see from equation (2) that as $$\theta$$ approaches $$\pi / 2$$, the side lengths $$a$$ and $$c$$ approach each other (and similarly for $$b$$ and $$d$$). Thus, as $$a$$ and $$b$$ approach $$0$$, the hyperbolic Lambert quadrilateral approaches a Euclidean rectangle. This is illustrated on the right.