Common sizes of intersections I'm trying to come up with the largest family of sets that obeys the following properties:
Consider $X = \{1,\dots,n\}$ and take $\mathcal{F} \subset 2^X$ such that for any three subsets $A,B,C \in \mathcal{F}$ we have that (at least) two of the numbers $|A \cap B|, |B \cap C|, |A \cap C|$ are the same size.
There is a trivial example here, $\{1\}, \{1,2\}, \{1,2,3\},\dots$ which gives a family $\mathcal{F}$ of size $n$, but I don't see how to do better (I've tried using modular arithmetic, which doesn't seem to help all that much.)
In this problem it doesn't particularly matter (to me at least) what $n$ is, eg, for the modular arithmetic way I was trying I set $n = L^2$ for some integer.
So, my question is as follows, is there a better choice for $\mathcal{F}$, eg, one that is larger than the trivial?
 A: Consider $\ \binom A2\ $ for an arbitrary finite set $\ A.\ $ Then you get a proper family with $\ \binom n2\ $ members for $\ n:=|A|.\ $ You get an improvement for every $\ n>3$.
EDIT:   Let me coin the name "Three Is a Crowd" (or TIsC for short) for the families introduced in the Question by James Kilbane. Now thanks to the comments let me record certain improvements:


*

*The Masked Avenger has proposed to add the whole set $\ A,\ $ and the empty set too--let's be greedy. Thus the enlarged Three Is a Crowd family looks like this:


$$ \binom A2\cup\{A\,\ \emptyset\} $$
for a new record $\ \binom n2+2\ $ for $\ n=|A|$.


*

*Tony Huynh has successfully added the singletons (and the empty set). This gives the following increased Three Is a Crowd family:


$$\binom{A}{2} \cup \binom A 1 \cup \binom A 0$$
for a new record $\ \binom {n+1}2$.


*

*The semi-dual example is just as good:


$$\binom A{n-2}\cup\binom A{n-1}\cup\{\emptyset\}$$
for $\ n:=|A|.\ $ We get $\ \binom {n+1}2\ $ different members again.

It'd be nice to find the maximum.

