Short proof for the non-Hamiltonicity of the Petersen Graph It is well known that the Petersen Graph is not Hamiltonian. I can show it by  case distinction, which is not too long - but it is not very elegant either.
Is there a simple (short) argument that the Petersen Graph does not contain a Hamiltonian cycle? 

Motivated from the wikipedia page I will add an answer to the question by myself. It is still a little case distinction, but it is small.
We know that the Petersen graph is 3-regular and has girth 5. Suppose it has a Hamiltonian cycle $H$ and we draw the graph such that the $H$ is drawn as cycle. The edges that are not in $H$ are chords of $H$. If there would be two chords that do not intersect, then these two chords are part of two disjoint 5 cycles. But in this case the two chords and the two edges of $H$ not in the 5-cycles form a 4-cycle. Hence all chords cross pairwise. The only possibility for this is shown in the second picture. Clearly, we have a 4-cycle. Hence, we have a contradiction.

