3-polytope with 9 vertices and 8 facets

Question

My guess is that this is a 3-polytope, since the existence of the two facets {357} and {048} rules out the possibility of dimensionality 2 and 4.

How can I go about sketching it?

Answer 1

There are many ways of sketching it, but they all are quite similar. I started with the triangle $\{3, 5, 7\}$:

Then I looked at all facets (faces, really) which shared a side with this one. They are the ones in the top row. I filled them in to get this:

(Note that the top triangle isn't a face; it's "open".) Then I took the remaining three quadrilateral faces, put them on:

And thus the sketch was finished.

Answer 2

Identify shared edges, and use that to draw a graph of the polytope connections.

For instance, 3-5-7 and 1-2-3-7 have the shared edge 3-7. So, putting 3-5-7 in the center of your graph, you have 3 and 7 next to 1 and 2 in some order. Similarly 3 and 5 are next to 2 and 6, and 5 and 7 ate next to 1 and 6. This combibation is consistent with the graph having 3 directly connected to 2, 5 connected to 6, and 7 connected to 1. You also connect 1, 2 and 6 to each other to close the quadrilateral faces.

Now you have the open edges 1-2, 1-6, and 2-6. Find the faces that share those edges and draw them in using similar logic to that above. Your complete graph now shows the two triangular faces (3-5-7 in the center and 0-4-8 in the exterior region) sandwiched around two layers of quadrilaterals. Can you see now what sort of polyhedron has layers of quadrilaterals capped by triangles at opposite ends?

Answer 3

The sketch of @arthur already shows the shape.
If you'd set up the 6 tetragons as trapezia, you'd recognize this figure to be a bitruncated triagular bipyramid.

--- rk