We have a tree T on n vertices. Prove that the centre of the tree T is isomorphic to a complete graph on 2 vertices if and only if diam(T)=2*rad(T)-1

Question

I know that the tree $T$ has $n-1$ edges, one component (it is connected), and does not contain a cycle (that is, from each vertex to each there is a single path, or between every two vertices of the tree there is a single path, otherwise there would be a cycle) and we want to show the equivalence between the statements: The center of the tree $T$ is isomorphic to $K_2$ (a complete graph on 2 vertices) and $diam(T) = 2\;rad(T)-1$ holds, so we have to prove implications in both directions: The first implication: If the center of the tree $T$ is isomorphic with $K_2$ (a complete graph on 2 vertices), then $diam(T) = 2\;rad(T)-1$ holds. Well, we know from the assumption that the Centre of the tree $T$ (that is, the induced subgraph of the tree T) is induced only on two vertices, let's denote them $x$, $y$, which are with eccentricity $ ecc(x)=ecc(y)=rad(T)$ and the other vertices necessarily have an eccentricity greater than $rad(G)$, we also know that for every connected graph: $rad(G)\leq diam(G)\leq 2\;rad(G)$, and if we want to show that $diam(T)= 2\;rad(T)-1$, so that $rad(T)=(diam(T)+1)/2$. But if we use the fact that in every tree there is a single path between any two vertices and its length is at most as much as the number of edges in the tree, i.e. $n-1$, which is basically also the maximum possible eccentricity that we can achieve in any tree, therefore, in addition to the fact that $diam(T)\leq 2\;rad(T)$, $diam(T)\leq n-1$ also holds, after adding the inequalities we have $2\;diam(G)\leq 2\;rad(T)+ n-1$ and after dividing by two we have $diam(T)\leq rad(T)+(n-1)/2$, so $rad(T)\geq diam(T)-(n-1)/2$, but how would I show the equality, and how would I prove the opposite implication? Thanks for any help.

Answer 1

I will turn our exchange into an answer.

Let $d$ be the diameter and $r$ be the radius. Let $p$ be a shortest path of length $d$.

Clearly, vertices on the center of the path have eccentricity $\lceil \frac{d}{2}\rceil$, because if there exists a vertex further, then redirecting $p$ in this direction would increase the diameter, which is a contradiction.

Let's show that any vertex of minimal eccentricity belongs to $p$. Let $p_1$ and $p_2$ be the paths linking this vertex to the two extremities of $p$. These two paths cover $p$, so at least one is strictly longer than $\frac{d}{2}\rceil$, hence it can't be a vertex of minimum eccentricity.

We conclude that the center of the graph is the same as the center of the path $p$, which is made of one vertex if $p$ has an even length and two vertices if $p$ has an odd length. In the latter case, we have $d = 2r-1$.