Finite Incidence Geometry Questions Definition of finite incidence geometry given: Consists of a finite set $P$ of points and a set of nonempty subsets of $P$ called lines, that satisfy the axioms:
(F1) Two points determine exactly one line joining them;
(F2) There are at least two lines and every line contains at least two points;
(F3) There is a constant $n$ such that every line contains $n$ points.
Question 1: Prove that every finite incidence geometry is a balanced incomplete block design.
Question 2: Prove that in a finite incidence geometry, there is a number $t$ such that for any point $p$ and any line $l$ not through $p$ there are exactly $t$ lines containing $p$ that have no common point with $l$.
Update on my attempt:
We want to establish parameters $(v,b,r,k,\lambda)$. (F3) gives us $k=n$ and I think (F1) gives us $\lambda = 1$. Now, if we select any line $l$ and suppose that it contains points $a_1,\dots,a_n$ and then select a point $p$ that is not on $l$, it follows that $p$ lies on exactly $n$ lines that meet $l$, namely $pa_1,\dots,pa_n$. No lines are parallel to $l$ be the axioms (I'm not sure if this is true!) Therefore, these are all of the lines that $p$ lies on $\implies$ $r=n$. So then using the BIBD formulas, we can solve for $v$ and $b$ to arrive at a $(n^2-n+1, n^2-n+1, n, n, 1)$-design.
For problem 2, I had two ideas:  I see that t=0 for a triangle, but it also equals zero for the Fano plane (which is an example of an incidence geometry). So I suspect that t=0 for any incidence geometry (I'm not sure about that!). I'm struggling to show this. My approach was to start with n^2-n+1 lines (blocks) and subtract off r (the lines that contain p). Also, we subtract off L. Leaving us with n^2-2n lines that do not contain p and L is not one of them. Somehow, I have subtract off everything, to give me t=0.
My second idea was to say that if t>0, then this implies that there exists a line parallel to L, which implies that r = n+1 (for one parallel line), which contradicts our findings from 4ii). I can't say this for certain, because I'm not sure if incidence geometries can have parallel lines!
As always, I appreciate any thoughts, ideas, and assistance with regards to either or both problems!
 A: Question 1: Let $v$ be the number of points and $b$ be the number of lines in the incidence geometry. We are given that each line has $k=n$ points, and that any two points determine exactly $\lambda = 1$ line. The only thing we need to show is that there exists a constant $r$ such that every point is incident with exactly $r$ lines (notice that we do not need to obtain a formula for $v$ or $b$ in terms of $n$, they will not be uniquely determined).
If we fix a point $p$, we have that there are $(v-1)$ ordered pairs $(p, p^{\prime})$ of distinct points. Each of these pairs are incident with exactly one line. Since each line on $p$ contains $n-1$ such pairs, we have that $p$ must be incident with $r=(v-1)/(n-1)$ lines, independent of the choice of $p$. Thus we have that the points and lines of an incidence geometry form a $(v,b,(v-1)/(n-1),n,1)$ BIBD. (Double counting can also be used to show that $b = \frac{v(v-1)}{n(n-1)}$, but this is not necessary to solve the problem.)
Question 2: Take a point $p$ and a line $\ell$ not on $p$. Then there are $n$ points $v_{1}$, $\ldots$, $v_{n}$ on $\ell$, and a single line containing each of $(p, v_{1})$, $(p,v_{2})$, $\ldots$, $(p,v_{n})$. These $n$ lines are all distinct, because $\ell$ is the only line containing $(v_{i}, v_{j})$ for $i \neq j$. Then, since there are $(v-1)/(n-1)$ total lines on $p$, there are $(v-1)/(n-1) - n$ lines on $p$ which do not meet $\ell$.
Note that you cannot assume the existence or nonexistence of "parallel" lines (and you need to be clear exactly how these are defined). You may or may not have parallel lines depending on the type of incidence geometry. You get the design regardless.
A: From the axioms for an incidence geometry, show that two lines l and l ′, l ≠ l ′, either share a point in common or no points in common.
