Here's the question:
Let G be a connected graph. (Remember that this means that every two vertices of G can be joined by a path starting at one and ending at the other.)
Suppose also that G has the following property: there is no path of length more than 100 in G. (Remember also that the length of path is the number of edges it traverses, which is one less than the number of vertices in the path.)
Prove that if P1 and P2 are two paths in G with length exactly 100, then there must be some vertex that belongs to both P1 and P2.
Isn't every vertex connected in a connected graph? (ie, the longest path will touch every vertex) In which case the maximum length of the path would be the number of vertices minus 1. Wouldn't that mean that P1 and P2 share a vertex because they share every vertex?
If I'm wrong I guess my question would be "What is example of a graph with a maximum path length in which two paths that are equal to the maximum path length only share one vertex?"
Take a path $P_1$ of length $100$; it has $101$ vertices. Let $v$ be the middle vertex, and attach to it two more disjoint paths of length $50$ each. You now have a big X in which each of the four arms is a path of length $50$. This graph has $\binom42=6$ paths of length $100$ and no longer path, and as long as you pick one of the three pairs of maximal paths that share only the vertex $v$, you have two paths of length $100$ with a single vertex in common.
Added: For the actual result, suppose that $P_1$ and $P_2$ have no vertex in common. Let $Q$ be a shortest path between some vertex of $P_1$ to some vertex of $P_2$; note that the endpoints of $Q$ are the only vertices that it shares with $P_1\cup P_2$. Show that if $u$ is the endpoint of $Q$ on $P_1$, and $v$ is the endpoint of $Q$ on $P_2$, then there are an endpoint $x$ of $P_1$ and an endpoint $y$ of $P_2$ such that the path from $x$ to $u$ along $P_1$, from $u$ to $v$ along $Q$, and from $v$ to $y$ along $P_2$ has length greater than $100$.