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I did a lot of googling but I'm unable to find an example of a convex polyhedron in 3-dimensional space, such that its faces are all congruent irregular heptagons.

Is there a reason such a shape can't exist?

Also in parallel what is the word for a polyhedron, such that all of its faces are congruent but not necessarily face-transitive.

Sidharth Ghoshal
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    Euler relation for planar graphs, in this case spherical instead. – Will Jagy Nov 15 '17 at 20:23
  • Have a gander at the third picture here https://en.wikipedia.org/wiki/Uniform_tilings_in_hyperbolic_plane – Donald Splutterwit Nov 15 '17 at 20:25
  • https://en.wikipedia.org/wiki/Euler_characteristic#Plane_graphs – Will Jagy Nov 15 '17 at 20:26
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    If you start gluing heptagons together, you will get a hyperbolic surface. – Doug M Nov 15 '17 at 20:28
  • @WillJagy I don't know how to relate the # of vertices to the number of faces (it seems like there are some inequalities but no hard relationships) so that reduces to $ V - \frac{7}{2}F + F = 2$ and that equation still has a plethora of solutions in the natural numbers. I guess all we know is that $ V < \frac{7}{2} F$ – Sidharth Ghoshal Nov 15 '17 at 20:34
  • @DonaldSplutterwit thats a pretty image! Doug M interesting, its a different angle than what i had in mind but something to consider – Sidharth Ghoshal Nov 15 '17 at 20:36
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    Seems as though nobody addressed the vocabulary question. In my experience the usual word for a polyhedron with congruent faces that is not necessarily face-transitive is "monohedral." – Glen Whitney Mar 13 '18 at 08:01

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$$ V - E + F = 2 $$ let's see, $$ E = 7 F / 2 $$ Each vertex meets at least three faces, $$ V \leq 7F / 3. $$ $$ V - E + F \leq \frac{7F}{3} - \frac{5F}{2} = \frac{-F}{6} $$ $$ V - E + F \leq \frac{-F}{6} $$ $$ 2 \leq \frac{-F}{6} $$ which is bad

Will Jagy
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The Euler characteristic of a polyhedron $F + V - E = 2$

If we glue $n$ heptagons together we have

$F = n$

Since two faces meet at each edge

$E = \frac {7n}{2}$

And we must have at least 3 faces meeting at a vertex (unless you want to include degenerate heptagons with straight angles, and are really something with fewer sides)

$V \le \frac {7n}{3}$

and for any $n$

$F+V - E < 0$

You might be able to make some sort of torus, though.

Alternatively,

At each vertex the sum of the angles must be less than $360^\circ$ if the shape is convex.

And if we look at the differential between the sum of the angles and $360$ and sum it across all of the vertices, the sum equals $720^\circ$ if the surface is closed (and simply connected).

The average angle in a heptagon is $\frac {5}{7} 180^\circ$

The average vertex is concave. i.e. $ (1+ \frac {1}{7}) 360^\circ$

If you tile with heptagons you will either get "swiss cheese" i.e. a multi-holed torus, or you will get a model of a hyperbolic plane.

Doug M
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