Let $n$ be a positive integer and consider $2009$ (not necessarily distinct) subsets of $\{1,2,\ldots,n\}$ such that each of them has size $45$ and any two of them have exactly $1$ common element. Prove that there is an integer which belongs to all $2009$ subsets.
Progress: 1) By Fisher's inequality the number of distinct subsets is at most $n$ and so by Pigeonhole there are some $\lceil 2009/n \rceil$ of the $2009$ subsets which coincide.
- Suppose the desired statement is false, let $k_i$ be the number of subsets containing $i$ and let $k = \max_{i} k_i$. We claim that $k\leq 44$. Suppose $k = k_j$ and pick sets $A$ and $B$ which contain $j$ and $X$ which does not contain $j$. Then $A\cap B = \{j\}$ and $A \cap X \neq B \cap X$. Now by fixing $A$ and $X$ and letting $B$ run through all subsets which contain $j$, the claim follows.
Any help appreciated! (Hopefully with an entirely combinatorial proof.)