Let $\alpha>0$ be a constant (can be sufficiently small if necessary) and $n$ be sufficiently large. What can we say about the cardinality of a family of subsets of $\{1,2,\ldots,n\}$, each of size $k$, such that any two of the subsets intersect in at most $\alpha k$ elements, when $k$ is of size linear in $n$?
I was able to find relevant results only when $k = O(\sqrt{n})$ (see Find a family $\mathcal{A} \subseteq {[n] \choose k}$ such that $|\mathcal{A}| \geq (n/2k)^h$ and $|A_i \cap A_j| \leq h-1$) and when $k$ is fixed ("On a Packing and Covering Problem", Rodl, 1985).
Any idea if results for other ranges of $k$ are available? I am more interested in constructions rather than upper bounds, since the context in Rodl's work anyway gives $\binom{n}{\alpha k}/\binom{k}{\alpha k}$ as a reasonable upper bound.
Any help appreciated!
Maybe the following simple construction can be useful for some choice of parameters, although I afraid it anyway provides too weak lower bound. Let $q$ be any natural power of any prime number, $\alpha=\frac 1{q}$, $\mathbb F_q$ be a field of order $q$, $s<r$ be any natural numbers, $n=|\mathbb F_q^r|=q^r$, $k=q^s$, and $\mathcal A$ be the family of all $s$-dimensional subspaces of $\mathbb F_q^r$. Then $$|\mathcal A|=\binom r s_q=\dfrac{(q^r-1)(q^r-q)\cdots(q^r-q^{s-1})}{(q^s-1)(q^s-q)\cdots(q^s-q^{s-1})},$$ see this answer, and the intersection of any distinct members of $\mathcal A$ has size at most $q^{s-1}=\alpha k$. In particular, when $s=r-1$ then $|\mathcal A|=\frac{q^r-1}{q-1}=\frac{n-1}{\alpha^{-1}-1}$.