In every tree, does there exists a vertex $v$ such that every path with maximum length contains $v$?

Solution 1:

In addition to the method proposed by Manuel Lafond in its comment, which gives an elementary proof, here is a more elaborated proof as it requires to know the Helly property of subtrees of a tree.

Definition: a family of sets $\mathcal{S} \subseteq 2^X$ has the Helly property if for any subfamily $\mathcal{S}' \subseteq \mathcal{S}$ of pairwise intersecting sets (that is for any $S,T \in \mathcal{S}'$, $S \cap T \neq \emptyset$), $\bigcap_{S \in \mathcal{S}'} S \neq \emptyset$.

It is known that the vertex sets of subtrees of any tree has the Helly property.

Back to your question. First prove that any pair of paths of maximum lengths intersects. Then by the Helly property of subtrees of a tree, all the paths of maximum lengths have a common intersection.

Actually, it is enough to know that the vertex sets of subpaths of a tree have the Helly property, which is not hard to prove.