Show that the Petersen graph, shown here, does not have a Hamilton circuit, but that the subgraph obtained by deleting a vertex $v$, and all edges incident with $v$, does have a Hamilton circuit.
My attempt:
I note that Ore's theorem${}^1$ is not satisfied. But this doesn't necessarily mean that no hamilton circuit exists. $\color{red}{\text{Is there any necessary condition which make it done}?}$ For the second part I think any inner or outer vertex removing done the Job because the original graph is symmetric. But is there any general way of thinking to removing a vertex from an arbitrary graph to come up with a Hamilton circuit.
Here is my main question: $\color{red}{\text{If the main graph isn't symmetric then how to know which vertex removing done the Job}?}$
$\color{red}{\text{Or any clues}?}$
${}^1$Ore states that: If $d(u) + d(v) \geq n$ for every pair of distinct nonadjacent vertices $u$ and $v$ of $G$, then $G$ is Hamiltonian.