Diameteral paths Prove that in a tree, all diametral paths go through all vertices in the center of the tree.
My thoughts were to use the fact that all diametral paths in a tree begin and end at leaves and do some kind of induction, separating the cases where the tree has one or two centers. But I'm not really sure how to putt this all together. Can anyone help?
 A: The following lemma seems to solve it:
Let $T$ be a tree with radius $r$, then the diameter $D$ is $2r$ or $2r-1$.
It is clear that the diameter cannot be more than $2r$.
Assume the diameter is less than $2r-1$ and take a diameteral path $v_1,\dots,v_{D+1}$. Take a point in the middle of the path, and notice that there must be a point $x$ at distance $r$ from it. Notice the distance from $v_1$ to $x$ or the distance from $v_{D+1}$ to $x$ must be greater than $D$ (because one of those paths goes through the central vertex).

We now assume we have a graph with radius $r$ where $v_1,\dots,v_{D+1}$ is a diametral path and $c$ is a vertex not on the center. Then the distance between $v$ to one of the endpoints of the diametral path must be at least $r+1$,  which means $c$ is not in the center.
A: A diametral path is the shortest path through a graph with a length equal to the diameter of the graph.
The center of a tree is always the middle one or two vertices in every longest path in a tree--the diameter.
Because a diametral path inherently has a length equal to the diameter, and the center of a tree is always on the diameter, any diametral path will go through the center of a tree.
