A problem about set system, equivalent to minimum number of edges s.t. each pair of vertices have a common neighbor

Question

I find the following problem in Bela Bollobas' book "Combinatorics": Let $\mathcal{F}\subset X^{(2)}$, where $|X|=n$, be such that if $Y\subset X,|Y|=n-2$, then there exist distinct sets $F_1,F_2\in\mathcal{F}$ including the same subset of $Y:F_1\cap Y=F_2\cap Y.$ Then $|\mathcal{F}|\ge\lceil3(n-1)/2\rceil$. Show also that for each $n\ge 3$ there is a system with the equality holds.

In my opinion, if we view it as a graph with vertex set $X$ and $\mathcal{F}$ as its edge set. Then it says for any $Y$ with order $n-2$, there exist a vertex $v\in Y$ connecting to the other 2 vertices out of $Y$. Therefore it seems we need to show average degree of this graph is about 3. Suppose the conclusion does not hold, it can be shown that at least 3 vertices have degree 2. I don't know what to do next.

Answer 1

The problem is about the minimum possible number of edges in a graph of order $n\ge3$ with the property that each pair of distinct vertices has a common neighbor; it claims that the minimum is exactly $\left\lceil\frac{3(n-1)}2\right\rceil.$

First I will construct, for each $n\ge3,$ a graph with the stated property which has exactly $n$ vertices and $\left\lceil\frac{3(n-1)}2\right\rceil$ edges.

Suppose $n$ is odd, $n=2t+1\ (t\ge1).$ The friendship graph $F_t$ (consisting of $t$ triangles with a common vertex) has the stated property, and it has $n=2t+1$ vertices and $\left\lceil\frac{3(n-1)}2\right\rceil=3t$ edges.

Suppose $n$ is even, $n=2t+2\ (t\ge1).$ If we add a new vertex to $F_t,$ and two new edges joining the new vertex to the central vertex of $F_t$ and one other vertex, we get a graph wich has the stated property and has $n=2t+2$ vertices and $\left\lceil\frac{3(n-1)}2\right\rceil=3t+2$ edges.

Now I have to show that this construction is optimal. Given a graph $G$ of order $n\ge3$ in which each pair of distinct vertices has a common neighbor, I have to show that $e\ge\frac{3(n-1)}2$ where $e$ is the number of edges; in other words, I have to show that $\sum_{v\in V(G)}\deg(v)\ge3n-3.$ Let $V$ be the vertex set of $G.$

Clearly, every vertex has degree $\ge2.$ If every vertex has degree $\ge3$ then $\sum_{v\in V}\deg(v)\ge3n$ and we're done. Thus we may assume that there is a vertex $u$ of degree $2.$ Let $w_1,w_2$ be the two neighbors of $u.$ Since the vertex $u$ is connected to each vertex in $V\setminus\{u\}$ by a path of length $2,$ the vertices $w_1,w_2$ must be joined to each other, and every vertex in $V\setminus\{u,w_1,w_2\}$ must be joined to at least one of the vertices $w_1,w_2.$ It follows from this that $\deg(w_1)+\deg(w_2)\ge n+1.$ Since the $n-2$ vertices in $V\setminus\{w_1,w_2\}$ have degree $\ge2,$ it follows that $\sum_{v\in V}\deg(v)\ge(n+1)+2(n-2)=3n-3,$ Q.E.D.