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. 
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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?
Best Answer
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.