Finite hyperbolic geometry with ideal points I was browsing "Thinking Geometricly: A Survey in Geometries" by Thomas Q. Sibley, 2015 and on page  388 it mentions a finite hyperbolic geometry of order 3 (3 points per line) consisting of 13 (ordinary) points.
I was wondering are there  finite hyperbolic geometries that also contain ideal points? (for exploring parallel and hyperparallel lines, parallel lines share an ideal point, hyperparallel lines are not sharing any point)  
So I guess this would mean: 


*

*There are two types of points 


*

*ordinary points

*ideal points 


*Each lines contains 


*

*3 ordinary points and 

*2 ideal points 


*Each pair of two points (ideal or ordinary) are on one and only one line.

*Given an ordinary point P not on line l there are at least two lines not containing an ordinary point of l.  (is this even needed?) 


Does such a finite geometry exist?
any references welcome
Added later:
following the remarks (thanks) 
i guess the last condition " given an ordinary point P not on line l there are at least two lines not containing an ordinary point of l." is not needed.
 A: As I understand the research, there are two approaches to making finite hyperbolic planes. 
The one in my textbook comes from balanced incomplete block designs, where each line (block) has the same number of points on it.  
These don't lend themselves naturally to ideal (omega) points.  
The other approach starts from the projective geometry idea.
In that approach one starts with a conic in a finite projective plane and consider the points interior to the conic as the hyperbolic points.
From this approach, every hyperbolic line will have two ideal points.  However, lines do not all have the same number of points on them.
I don't know of any finite example that satisfies both approaches.
A: Unfortunately, the requirement of each line having the same number of ideal and ordinary points is a bit too strong -- it forces too much "regularity" into the structure.
Let $L$ denote the number of lines, $R$ the total number of ordinary and $D$ the number of ideal points. Let $r$ and $d$ be the number of ordinary and ideal points on each line (so $r=3$ and $d=2$ in the original problem statement).
There are $\binom{R+D}{2}$ pairs of points in total. Each line contains $(r+d)$ points and thus $\binom{r+d}{2}$ pairs. Since there are $L$ lines and each pair of points lies on exactly one line, we have $$\binom{r+d}{2}L=\binom{R+D}{2}$$
The same kind of reasoning applies if we restrict our attention to just the ordinary points: There are $\binom{R}{2}$ pairs of them and they are divided into $L$ lines evenly, $\binom{r}{2}$ per line. Finally, ideal points are not different either, so we have:
$$\begin{eqnarray}\binom{r}{2}L = \binom{R}{2} & \ & \mathrm {and} & \ &\binom{d}{2}L = \binom{D}{2}\end{eqnarray}$$
The rest is matter of simple algebraic manipulation: Subtracting these two equalities from the first one yields $$rdL = RD$$ which we can divide the same two equalities with to obtain:
$$\begin{eqnarray}\frac{r-1}{d} = \frac{R-1}{D} & \ & \mathrm {and} & \ &\frac{d-1}{r} = \frac{D-1}{R}\end{eqnarray}$$
Now, multiply the first one by $d\times D$ and the second one by $r\times R$ and add the resulting expressions to obtain $$(r+d)=(R+D)$$
Plugging this back to the very first equality yields $L=1$, which means the resulting system is trivial: it can only consist of a single line which can contain any number of ordinary and ideal points. Obviously, no such structure can satisfy the last condition which deals with multiple lines.
