Note that Gk has a Hamilton path (with length β = 6(k+1)2 - 1 ),
but has no cycle of length exceeding α = 12k+6 and that α2 < 24β.
α < sqrt(24β) < 5sqrt(β).
The reason that I constructed this example is that the usual example (Voss and Zuluaga) which is used
to demonstrate that a 2-connected graph with a path of some length
β might have no cycle of length exceeding the ceiling of 2sqrt(β) has
a pair of vertices (in fact, many pairs) whose deletion leaves three components.
The graph Gk is 2-connected, but the deletion of any two vertices
leaves at most two components. (Which is why I call the graph (2,2)-connected.)
An astute reader might see an interesting question here, but I'd like to work on the next step for a while.....