Hamilton Cycles

Let us prove that it has no 10-cycle, so the circumference is 9. We think of the Petersen graph as an outside 5-cycle and an inside 5-cycle, connected by 5 links. A 10-cycle would have to contain an even number of such links, and not 0 since then we would not get a connected subgraph. Up to isomorphism, this leaves the two cases below: two links or four links (depicted with thick black edges); note that in the case of two links, they must hit adjacent vertices on one of the two 5-cycles, so after possibly swapping the 5-cycles, this is the only case with two links. In each case, we mark edges that cannot be in the cycle with red, and edges that must be in the cycle with green. Whenever a vertex has a red edge, its other two edges must be green or black. And if two edges of a vertex are green or black, then the third edge must be red. In this way we get a contradiction in both cases, either because of a vertex of degree 3 or because of a 5-cycle.