Can a planar graph with a 3-cycle always be drawn with a 3-sided face?

Question

In general a planar graph can be drawn in many different ways. Can a planar graph that has a cycle of length three nevertheless not have any planar representation which includes a face of degree three?

Answer 1

I started by guessing that any graph with a 3-cycle can be drawn in a planar embedding with a 3-face. I started to consider how to "flip" sections of the graph from inside the cycle to outside, and realized that if there is both an exterior and interior graph connected to all three points of the 3-cycle, this cannot be achieved.

Therefore I believe this graph, with its (only) 3-cycle highlighted in green, cannot be drawn with a 3-face:

(If the degree-2 nodes are considered undesirable, you can add another 4-cycle, connected pointwise, inside each of the interior and exterior 4-cycles (turn the 4-cycles into cube nets).)

As proof I suppose that we can, indeed, draw this graph with the 3-cycle $abc$ as a face. Then we can also add a point $d$ in the middle of the face and connect to the points of the 3-cycle and that too will be planar. However if we reduce the 4-cycles that are currently wholly exterior and wholly interior to the 3-cycle to points $e$ and $f$ then we can see there is a $K_{3,3}$ minor with $(a,b,c)$ on one part and $(d,e,f)$ in the other, contradicting that the graph with the extra point is planar.