Let $T$ be an undirected tree. Can we always find a leaf vertex $s$ such that every longest path of $T$ has its other endpoint in $s$? It's easy to see that every longest path passes through the center of $T$. However, it is of course not true that there is a longest path between any two arbitary leaves. 

Here's an example, where every longest path ends in $6$.

![example](https://upload.wikimedia.org/wikipedia/commons/thumb/2/24/Tree_graph.svg/180px-Tree_graph.svg.png)

Maybe this could be shown by contradiction. So one could suppose there is no such "sink" leaf $s$. Then there is longest path starting at $v$, not ending in $s$. Can this be extended to a proof? Is there a simpler argument?