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Let $G$ be a connected, planar graph for which every vertex has degree $3$ . Moreover, suppose there is a face of length at least $12$ .

Is it possible to construct an example of such $G $ whose only odd faces are six 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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  • $\begingroup$ Triangular prism? $\endgroup$ – Zachary Hunter Dec 30 '18 at 6:02
  • $\begingroup$ Thanks again. I've added that only 6 triangles are allowed. $\endgroup$ – Finallysignedup Dec 30 '18 at 6:08
  • $\begingroup$ Why have you repeated your question? The solution given in the first can be easily adapted to this. $\endgroup$ – Daniel Mathias Dec 30 '18 at 6:10
  • $\begingroup$ I've added some further restrictions, thanks. $\endgroup$ – Finallysignedup Dec 30 '18 at 6:20
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Simplest answer: triangular prism.

Niftier solution: the chamfer of any cubic polyhedral having only triangular odd faces. (Chamfering is a process where you replace each edge with a hexagon.)

Edit, if precisely 6 triangles are allowed, you want a chamfered cuboctohedron, or any further chamfering.

Edit 2: Start with a 12-gon, C. For 6 disjoint edges in C, create a triangle with a new vertex. Then, pair up triangles and create edges between their newest vertices.

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  • $\begingroup$ Sorry to edit, but there must (for my purposes) be even faces length at least 12. $\endgroup$ – Finallysignedup Dec 30 '18 at 6:16
  • $\begingroup$ Might I ask what this is for? $\endgroup$ – Zachary Hunter Dec 30 '18 at 6:20
  • $\begingroup$ Solved the new version. $\endgroup$ – Zachary Hunter Dec 30 '18 at 6:42
  • $\begingroup$ I have a planar tree with vertices of degree 3 except for one with degree 2. There is an odd, outer face of length at least 5, containing the degree 2 vertex. The inside faces consist of 3 mutually disjoint triangles,disjoint from the outer face, and the rest even. By reflecting this picture at the degree 2 vertex I wrote the above hypothesis. If you think this setup is actually impossible, I will repost a new question of you to answer. $\endgroup$ – Finallysignedup Dec 30 '18 at 13:25

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