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I'm trying to solve my own question. I almost got the answer (which I'll post in a few days), but there are two things I'm not able to prove but that I know are true (On Polyhedra with Specified Types of Face):

  • There does not exists a convex heptahedron with only quadrilateral faces
  • There does not exists a convex $14$-hedron with exactly $14$ pentagonal faces

The heptahedron if it exists would have $9$ vertices, $14$ edges and $7$ faces. I can prove that it has $8$ vertices of degree $3$ and one of degree $4$.

The $14$-hedron if it exists would have $23$ vertices, $35$ edges and $14$ faces. I can prove that it has $22$ vertices of degree $3$ and one of degree $4$.

Question: Why do such polyhedra not exist?

Remark: An argument of the non-existence of only one polyhedron is good enough for an answer. I could extrapolate it to the other case.

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    $\begingroup$ It seems as though you should be able to prove impossibility by combinatorially eliminating the vertex of degree four: unless I'm missing something, the faces adjacent to it have to (at least combinatorially) account for all of the other vertices, and then by eliminating that face you get a contradiction. Is there any way that any of the eight vertices around it can coincide? $\endgroup$
    – Steven Stadnicki
    Feb 23, 2017 at 0:40
  • $\begingroup$ @StevenStadnicki There isn't a way for the eight vertices around to coincide, otherwise their degree would not be $3$. But I don't understand what face you are eliminating. $\endgroup$
    – Simon Marynissen
    Feb 23, 2017 at 0:44
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    $\begingroup$ I'm not eliminating a face; I'm eliminating a vertex, but I think I can make the argument even more direct. Imagine the one vertex of degree four, call it $v$. Then it seems as though we must have four vertices $a, b, c, d$ adjacent to $v$ with four quadrilaterals $(v, a, x, b)$, $(v, b, y, c)$, $(v, c, z, d)$ and $(v, d, w, a)$. But now we've accounted for all of the vertices of the polyhedron; the final two edges must be either $(x, y)$ and $(z, w)$ or $(x, z)$ and $(y, w)$, and neither of these cases has correct face structures. $\endgroup$
    – Steven Stadnicki
    Feb 23, 2017 at 1:05
  • $\begingroup$ That must have been a typo in your first comment. I think I got the gist of it now. $\endgroup$
    – Simon Marynissen
    Feb 23, 2017 at 1:10
  • $\begingroup$ Ahh, yes, I see - I said 'vertex' the first time, but then 'face' the second; mea culpa. $\endgroup$
    – Steven Stadnicki
    Feb 23, 2017 at 1:11

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The convex heptahedron seems easy to eliminate. Consider the vertex of degree 4, call it $v$; then $v$ is adjacent to four vertices $a, b, c, d$ and four quadrilaterals $(v, a, x, b)$, $(v, b, y, c)$, $(v, c, z, d)$ and $(v, d, w, a)$. But this accounts for all nine vertices of the polyhedron, and 12 of the 14 edges; to meet the degree constraints we must then have the last two edges (up to isomorphism) as either $(x, y)$ and $(z, w)$ or $(x, z)$ and $(y, w)$. But neither of these choices leads to the correct face structure.

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