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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? $\endgroup$ Commented 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. $\endgroup$ Commented 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. $\endgroup$ Commented 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. $\endgroup$ Commented Feb 23, 2017 at 1:10
  • $\begingroup$ Ahh, yes, I see - I said 'vertex' the first time, but then 'face' the second; mea culpa. $\endgroup$ Commented Feb 23, 2017 at 1:11

2 Answers 2

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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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To disprove the pentagonal 14-hedron:

Draw the order-4 vertex. It is surrounded by four faces which together take up 16 of the edges and twelve additional vertices. Eight of the additional vertices must have another edge emerging from it, producing an additional face betwen adjacent pairs of these edges. This accounts for a total of 24 edges and 12 faces. The remaider of the polyhedron, closing up the "bottom" side, must have two faces supplying 11 additional edges — but the faces are pentagonal and thus can supply a maximum of only ten more edges. $\rightarrow\leftarrow$

The contradiction disappears with both 12 and 16 pentagons. With 12 there is no order-4 vertex to start the chain of arguments. With 16 we have two order-4 vertices; when we construct the surrounding faces and edges from one of them as described above we find that the closure requiring four faces and 16 edges becomes satisfied by the second order-4 vertex. The number 14 is the only even number greater than 12 for which no pentagonal polyhedron has that number of faces.

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