# Center of a triangle free connected graph atmost two vertex

• Any graph $G$, $d(u,v)$ denote the length of the shortest path from $u$ to $v$.
• $\epsilon (u) =$ max$_{v \in V(G)}d(u,v)$ is called eccentricity of $u$ and $r$ = min$\epsilon (u)$ denote the radius of a graph $G.$

• Center of a graph $G$ is the set of all vertex $u$ such that $\epsilon (u) = r$.

Let $G$ be a triangle free graph. Then center of $G$ has atmost two vertices.

I want to show by contradiction. Suppose that the center of $G$ has atleast three vertex says $u,v,w$. Then there exist $x,y,z \in V(G)$ such that $\epsilon (u) = \epsilon (v) = \epsilon (w) = d(u,x) = d(v,y) = d(w,z)$. If $u$ is not adjacent with $v$, then $d(u,v) > d(u,x )$, which is a contradiction. Thus $u,v,w$ form a triangle which is a contradiction.

• Not sure how you determined $d(u, v)>d(u, x)$. – Dark Logician Jul 26 '17 at 4:56
• If $G$ is a regular $n$-gon, isn't every vertex in the center by symmetry? – stewbasic Jul 26 '17 at 6:11