Perimeter restriction on a hyperbolic polygon with all angles $\geq \theta$ and all side lengths $\geq l$ Let $P$ be a regular hyperbolic polygon, with $n$ sides, with all interior angles equal to $\theta$, and all side-lengths equal to $l$.
So, its perimeter is $nl$.
Now let $P'$ be a hyperbolic polygon with all interior angles at least $\theta$ and all side-lengths at least $l$.
I would like to know that the perimeter of $P'$ is at least $nl$.
This seems 'obvious', at least in the Euclidean case. Can anyone provide a proof in the hyperbolic case? Much much gratitude to any such person!
 A: The polygon $P'$ must have more than $n$ sides, and hence its perimeter is indeed greater than $nl$, unless it is itself a regular $n$-sided polygon with angles $\theta$ and side lengths $l$. From what I can see, the proof is the same as it would be in the Euclidean plane.
For the proof, let us enumerate the vertices and edges of $P$ in counterclockwise order as
$$V_1,E_1,V_2,E_2,....,V_n,E_n
$$
On $P'$, let us similarly enumerate the vertices and edges in counterclockwise order as
$$V'_1,E'_1,V'_2,E'_2,...
$$
with the difference that we do not know when the enumeration ends.
Now position $P'$, moving it by a translation of the hyperbolic plane $\mathbb H^2$, so that $V_1$ and $V'_1$ coincide, and so that the initial direction of $E_1$ coincides with the initial direction of $E'_1$. Since $\text{Length}(E'_1) \ge l = \text{Length}(E_1)$, it follows that the point $V'_2$ occurs on or beyond the point $V_2$ on the ray determined by $E_1$ and $E'_1$.
The rough idea is that as you go around the edges of $P'$ starting from $V'_1$, you cannot close up the polygon. To make this more rigorous, at each vertex $V_k$ let's consider the "external angular region" of $P$ at $V_k$, which is the region $A_k \subset \mathbb H^2$ bounded by two rays that meet  angle $\pi-\theta$, namely: the ray starting at $V_k$ and extending in the direction beyond $E_{k-1}$; and the ray starting at $V_k$ and extending in the direction of $E_k$. 
Now let's trace around the boundary of $P'$, to see how it interacts with the sequence external angular regions $A_k$. Moving out from $V'_1=V_1$ in the direction of $E'_1$, one traces along the $E_1$ side of $A_1$ out to the point $V'_2$, which lies on that side on or beyond $V_2$. At the point $V'_2$, one makes an external angle of at most $\pi - \theta$, entering into the angular region $A_2$, and one then travels a distance at most $l$. The key observation is that the opposite endpoint $V'_3$ will not even reach the other side of the angular region $A_2$ where it meets $A_3$, unless $V'_2=V_2$ and $\text{Length}(E'_2) = l = \text{Length}(E_2)$. 
And now one continues by induction. The point $V'_k$ will not even reach the  side of $A_{k-1}$ where it meets $A_k$, unless all the angles up to that point have been exactly equal to $\theta$ and all the side lengths exactly equal to $l$. 
Thus, after $n$ steps, the polygon $P'$ will not have closed up unless $P'=P$.
