# Related to girth of planar graphs

From the paper https://web.math.princeton.edu/~pds/papers/girth6/paper.pdf, a cubic (every vertex is degree 3), planar graph has girth (smallest length face) strictly less than 6. It is also true that a cubic, bipartite (all faces even length) graph contains a square (or a 2-gon).

I am interested in planar graphs with exactly one degree 2 vertex, the rest degree 3. Is it true that such a graph must contain a face of length 1,2, 3 or 5? (In other words, a face of length less than or equal to 5 that is not a square.)

You can prove half of that. One way to conclude something about the girth of a planar graph is via an inequality like this one: if the graph has $$p$$ vertices, $$q$$ edges, and girth $$g$$, then $$q \le \frac{g(p−2)}{g−2}$$.
For a cubic planar graph, we have $$q = \frac32 p$$, and we get an upper bound on $$g$$ that way. The upper bound is always less than $$6$$ (but approaches $$6$$ as $$p \to \infty$$).
If you have a single vertex of degree $$2$$, then $$q = \frac32p - \frac12$$, and the same argument still works. Solving for $$g$$, we get $$g \le \frac{6p-2}{p+3} < \frac{6p}{p} = 6$$. So there is always a face of length $$1$$, $$2$$, $$3$$, $$4$$, or $$5$$.
However, it's possible for all such faces to be squares. For example, take the skeleton graph of the truncated octahedron (which is $$3$$-regular) and then, to create a vertex of degree $$2$$, subdivide an edge that separates two hexagonal faces. You'll get a graph with $$24$$ degree-$$3$$ vertices and one degree-$$2$$ vertex, in which $$6$$ faces are squares, $$6$$ are hexagons, and $$2$$ are heptagons.