$\Gamma = (V,E)$ is a simple undirected graph with $|V| = n$ odd vertices. $\Gamma \cong \bar{\Gamma}$ so $\Gamma$ is isomorph with its complement graph. How can i prove that there is a $i \in V$ vertex where $\deg(i) = \frac{n-1}{2}$
I know that the graph with odd number of vertices should have at least one vertex with even degree. For example the pentagon isomorph with the pentagram, and it has got more than one vertices with $\deg = \frac{5-1}{2} = 2$
My problem is, i can't prove, that the vertex has exactly $\frac{n-1}{2}$ edges connected.
Sketch:
Suppose that $G$ is isomorphic to its complement $G'$ via $\phi$. Then, for every $i\in V$, there is some $j\in V$ so that $\phi(i)=j$. Then since $\deg_G(i)=\deg_{G'}(j)$ and $\deg_G(j)+\deg_{G'}(j)=n-1$, it follows that $\deg_G(i)+\deg_G(j)=n-1$.
Consider the set $\{i_1,\dots,i_k\}$ of all vertices of $G$ with $\deg_G(i_\ell)<\frac{n-1}{2}$. Each $i_\ell$ is paired with $j_\ell$ via $\phi$, i.e., $\phi(i_\ell)=j_\ell$.
If there are no vertices of degree $\frac{n-1}{2}$, then one can show that this is a complete list of the vertices of $G$ (and the two sets are disjoint). This is a contradiction, since $G$ has a odd number of vertices.