If we take a Dodecahedron and remove the constraint that the pentagonal faces have to be regular pentagons, we get a solid called a Tetartoid. If we take the dual of that, we will end up with the corresponding irregular Icosahedron.

enter image description here

Similarly, there seems to be an irregular analogue for the tetrahedron. However, I've never seen one for the cube and octahedron (which happen to be dual of each other). Is it even possible to have irregular analogues of those two? The criterion is that the solid should be convex with all faces identical and the same number of faces meeting at each vertex. The cube analogue would have six faces which were all quadrilaterals while the octahedron analogue would have eight faces which were all scalene triangles.

EDIT: Actually the Icosahedral object might not necessarily be composed of identical triangles.


For the octahedron it is easy. Take a rhombus $R$ in the $x$-$y$-plane. The diagonals divide it into four congruent scalene triangles. The diagonals intersect in $p$, so shift $p$ in positive $z$-direction and obtain the apex of a pyramid $P$ with base $R$. Reflect $P$ at the $x$-$y$-plane and obtain a pyramid $P'$. Then $T = P \cup P'$ is the desired object. It is obviously obtained by a distortion of the standard octahedron.

For the cube we can proceed similarly and obtain an object with rhombical faces. I do not know whether it is possible to find something with more irregular faces.

  • $\begingroup$ Thanks. It was very surprising to me that the dual solid of the irregular Octahedron doesn't satisfy my requirements. I don't understand how you would go about the cube analogue. What 2-D shape would you start with? $\endgroup$ – Rohit Pandey Jul 29 '18 at 17:52
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    $\begingroup$ For the cube anlaogue I would start with a rhombus in the $x$-$y$-plane and add congruent rhombi along the edges. It is inelegant to describe this with words. The result is known as the trigonal trapezohedron. See en.wikipedia.org/wiki/Rhombohedron. $\endgroup$ – Paul Frost Jul 29 '18 at 18:40

A cube could be elongated in either of its directions of face normals. this makes some of its faces to rectangles instead.

Alternatively a pair of faces (bases, if considered a square prism) could be sheered wrt each other. This makes some of its faces into parallelograms.

Finally you could squash a cube diametrally (along its body diagonal). This makes all faces into rhombs.

And, for sure, you could combine all theese operations.

--- rk

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    $\begingroup$ The OP wants all faces identical. $\endgroup$ – Ethan Bolker Jul 29 '18 at 13:18

See the Wikipedia entry on the rhombohedron for images:

Note the 2nd shape: $6$ identical rhombi.


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