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I have been struggling with this problem for quite some time now and I cannot think of a way to proceed with either part:

Suppose that G is a graph with $n > r + 1$ vertices and $t_r(n) + 1$ edges:

(a) Prove that for every $p$ with $r + 1 < p <= n$ there is a subgraph $H$ of $G$ with $|H| = p$ and $e(H) > t_r(p) + 1$.

(b) Prove that $G$ contains two copies of $K_{r+1}$ with exactly $r$ common vertices.

For part a), I have so far tried to use induction on $n$. For the base case with $n = r+2$ we have only $H = G$ and $n = p$ to check, which is clearly true. For the $n$th case I think we want to find a vertex $v$ in $G$ of degree $\delta(T_r(n))$, so that $H = G-v$ satisfies the induction hypothesis and gives all of the required subgraphs.

This is where I am stuck, I cannot seem to find anything close to this required vertex. As $e(G) = t_r(n) + 1$ we know $K_{r+1} \leq G$, so there is a vertex $v$ with $d(v) \geq r$, which is not enough as $(q-1)r \leq \delta(T_r(v)) \leq qr-1$.

Perhaps useful I thought would also be the fact that if $|G| = n, e(G) = t_r(n), K_{r+1}

leq G$ then $\delta(G) \leq \delta(T_r(v)) \leq \Delta(T_r(v)) \leq \Delta(G)$, this would give us a large enough vertex to delete, but we need $G$ to not contain $K_{r+1}$.

Likewise I did not get very far in the second part of the question, apart from the fact that a $K_{r+1}$ would exist in G in that case, but I cannot see how to get a second copy.

I noticed that someone has asked this identical problem a few years ago, but unfortunately, that one does not have any responses - which is why I have posted again as I do not have enough reputation to start a bounty without losing some rights on this site.