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Let G be a connected, planar graph for which every vertex has degree 3, except that one vertex has degree 2.

Is it possible to construct an example of such G whose only odd faces are triangles, and for which no two such triangles share a common vertex?

(For my purposes I may assume there are no faces of length 1 or 2, in which case there must be an odd face of length at least 3. By the handshake lemma, there are an even number of, hence at least 2, such odd faces.)

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  • $\begingroup$ For my own edification.. what is an odd face ? $\endgroup$ – T. Fo Dec 30 '18 at 5:04
  • $\begingroup$ Face with odd number of sides. $\endgroup$ – Zachary Hunter Dec 30 '18 at 5:09
  • $\begingroup$ I think "those bounded by an odd length cycle" is a sensible definition. $\endgroup$ – Finallysignedup Dec 30 '18 at 5:11
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Here is one such graph - or, more precisely, one such plane embedding of a graph, since the lengths of the faces are not properties of the graph itself. It has two faces of length 3 (including the external face), two faces of length 4, and two faces of length 6.

enter image description here

(Motivation: we start with a triangular pyramid, which has all the required properties except for the degree-2 vertex, and modify it a little to make it work.)

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  • $\begingroup$ Thanks. I now wish to only have degree 3 vertices. I've made a new (hopefully final) question. $\endgroup$ – Finallysignedup Dec 30 '18 at 6:03
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Take two cycles of the same order 2n+1, $C_a$ and $C_b$ and number each in clockwise fashion.

"Glue together" $v_{1a}$ and $v_{1b}$, and then do the same for $v_{2a}$ and $v_{2b}$, so that there is just one edge between $v_1$ and $v_2$.

Then, for each $i > 2$ create an edge between $v_{ia}$ and $v_{ib}$.

Finally, subdivide the edge between $v_1$ and $v_2$.

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  • $\begingroup$ Thanks. Sorry that I could not tick both answers. $\endgroup$ – Finallysignedup Dec 30 '18 at 6:02
  • $\begingroup$ Understandable, it is the fault of my laziness in creating diagrams. $\endgroup$ – Zachary Hunter Dec 30 '18 at 6:04

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