In every tree, does there exists a vertex $v$ such that every path with maximum length contains $v$? I'm trying to prove that every tree contains a vertex which is included on every path of maximum length. Intuitively I can see that every path of max length ends in a leaf. Where I'm struggling though is how to prove it generally. For example, in this graph, you can see that the central node is on every path of max length. There are trees without such a node though, and I'm unsure how to generalize my argument to include all trees. .
I can also intuit that a proof by contradiction will be best, but I'm struggling to think of what contradiction I'll prove. Perhaps that if there does not exist such a vertex, the graph cannot be a tree?
Am I on the right path here? Any suggestions on approaches here?
 A: 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.
A: Any two longest paths in a tree must share a common vertex, otherwise, given two such paths, because a tree is connected, the two paths must be connected, and this contradicts the premise that both the paths are longest, as traversing the connecting path produces a longer path.
Continuing the argument with a new path that doesn't share the common vertex proves the result.
