Is there more than one pseudo-Catalan solid?

Question

This question was asked on MSE a year ago. Motivation for this question can be found in other MSE questions here, here or here.

Convex solids can have all sorts of symmetries:

• the platonic solids are vertex and face-transitive, meaning there is a subgroup of the rotations of 3-dimensional space which can bring any vertex onto another one (and the same for faces). The list there is limited to the 5 platonic solids.

• face transitive (or isohedral) solids include the Catalan solids, the (infinite family) of dipyramids and the (infinite family) of trapezohedra. Note that without further restricitions these solids can come in infinite families: the rhombic dodecahedron has an infinite number of deltoidal cousins (see deltoidal dodecahedron); it also fits in a one-parameter family of dodecahedra called pyritohedra; the dodecahedron and the triakis tetrahedron fit in the one-parameter family called tetartoid; dipyramids and trapezohedron also admit all sorts of deformations beside the number of faces.

• there is a much weaker symmetry one can ask for. Let's call it pseudo-Catalan (for lack of a better name). Fix a "centre" $C$. The convex solid is pseudo-Catalan, if each face can be sent to another face by a rotation with centre $C$ or a reflection (whose plane goes through $C$). Note that there is no requirement that this rotation (+ reflection) preserve the whole solid. An example of such a solid which is not a Catalan solid is the gyrate deltoidal icositetrahedron.

Question: is there a list of solids which are pseudo-Catalan but not Catalan? [More desperately: is there any such solid beside the gyrate deltoidal icositetrahedron?]

• note that there would be a last category, where the solid is convex and all the faces are congruent (a convex monohedral solid). The difference with the previous category is that translations are now allowed. In particular, to check that a solid belongs to the previous category, the choice of $C$ (and the fact that all rotations and reflections are constrained by this point) is important. Examples of such solids are the the triaugmented triangular prism and the gyroelongated square dipyramid.

Answer 1

This is just a detailed version of the comments.

As M. Winter pointed out there is a family of polyhedra with $4k$-faces which fit the bill ($k=5$ is the icosahedra). Here is an image for the case $k=4$ and $k=6$.

Start with an antiprism over a $k$-gon (say the lower $k$-gon has vertices with coordinate $(e^{i \pi (2j+1)k},0)$ and the upper vertices $(e^{i \pi 2j k},h)$ where $0 \leq j <k$ and $h$ is a real number; I'm using complex numbers for the $x$ and $y$ coordinates). Glue a pyramid on each $k$-gon (the tip of the pyramids are at $(0,0,s)$ and $(0,0,h -s)$. The centre $C$ is at $(0,0,\tfrac{h}{2})$.

For the triangles to be congruent one can write $h$ as a function of $s$ (it's $h = \tfrac{ 2\cos(\pi/k)-1+s^2}{2s}$). If $k>3$, requiring each face to be at the same distance from $C$ (i.e. $C$ will be the centre of an insphere) will fix a value of $s$ (it's $=\sqrt{2\cos(\pi/k)+1}$). The point of the faces which minimises the distance to $C$ are [rather, seem to be] the circumcenter of the triangles (only checked this for $k=4,6$ and $7$ [I was too lazy to do the algebra for general $k$]).

From there it follows that these solids are pseudo-Catalan (they cannot be Catalan [if $k \neq 5$] since the vertices at the tip of the pyramids have degree $k$ while the other vertices have degree 5. Hence there is no global symmetry which sends a face from the pyramids to the antiprism.

I would tend to believe that these solids are in a larger family with scalene triangles. A similar construct based on trapezohedra (instead of dipyramids) would be fun (but I have no idea how to do this at the moment).

EDIT: the case $k=3$ is singular: if you force the planes of the faces to touch the insphere, you get a trapezohedron (whose faces are rhombi; i.e. the triangles of the pyramid align perfectly with those of the antiprism). If you further use the remaining parameter so that the closest point to $C$ is the same on each [triangular] face, it actually gives the cube (!).

Answer 2

Here is another (and hopefully simpler) example (though definitively not a complete list of possible solids). Take a $k$-dipyramid (the equatorial vertices have $xy$-coordinate which are $k^\text{th}$-roots of unity and $z=0$). Let the tips of the pyramids be at $(0,0,\pm 1)$. When $k$ is even (so $k \geq 4$), one can cut this pyramid along the plane which goes through the tips and the roots of unity $\pm 1$. This cuts the dipyramid along a square. Now rotate one of the two pieces by 90° and paste them back together. The resulting solids (which should, I assume, be called gyrate dipyramids) satisfy the required conditions.

To see that these are not Catalan solids (unless $k=4$, which is just taking the octaeder, cutting it and putting it back together) just observe that there are two types of faces: those which touch the square where the glueing occurred and the others.

Here are some pictures for $k=6$ and $k=8$.