I'm trying to prove a theorem which is stated as follows:
If G is a tournament graph and x is a vertex with the maximum out-degree,
then for any other vertices in G, say y, there is a directed path from x
to y of length at most 2.
A tournament is a complete graph with directions - so I can see how a selected vertex with the highest number of outdegree can reach any other vertices within 2 steps. But I have no idea how to prove this.
$x$ must have won at least half the games-at least half the edges from $x$ must be outward. Assume there is a vertex $y$ that cannot be reached from $x$ in two steps. Clearly the edge between $x$ and $y$ is directed toward $x$. If there are $n$ participants the number of outgoing edges from $x$ is at least $\frac {n-1}2$. If we cannot reach $y$ in two steps all the edges from $y$ to these vertices are directed from $y$. Then the outgoing edges from $y$ number $\frac {n-1}2+1$, which is more than the number of outgoing edges from $x$, contradicting the statement that $x$ is the vertex with maximum outgoing degree.