3
$\begingroup$

Scouring through Wikipedia, I've found the following analogs to platonic solids that are composed of irregular faces.

Cube = Trigonal Trapezohedron

Dodecahedron = Tetartoid

Tetrahedron = Disphenoid

I couldn't find the analogs for the Octahedron and Icosahedron. Does that mean they don't exist? How do I prove either way?

EDIT: My bad, Octahedron exists as well. That leaves Icosahedron.

$\endgroup$
2
$\begingroup$

An icosahedron can be viewed as a pentagonal antiprism with two pyramids built on its bases, hence it can also be made with isosceles triangles (see image below).

And I suspect that other more exotic topologies are possible.

enter image description here

$\endgroup$
2
$\begingroup$

Why, of course it can be done. Start with the net, assign the lengths so that all triangles will be equal and scalene:

2D net

Then fold it and glue everything together.

3D

True, the figure looks god-awfully ugly. This is a consequence of the fact that it can't be made face-transitive, unlike your three examples. (To make 20 faces symmetrically equivalent, we'd have to keep at least one 5-fold symmetry axis, which would keep passing through a vertex, which would make its five edges equal, which in turn doesn't sit well with the triangles being scalene.)

If you agree on isosceles faces, then you may keep a 5-fold axis and arrive at Aretino's example.

I suspect there are multiple ways to connect the scalene triangles (this is the case with octahedron, after all), but this post is getting too long, so let's call it a day.

$\endgroup$
  • $\begingroup$ Is this figure convex? $\endgroup$ – Rohit Pandey Aug 16 '18 at 23:30
  • $\begingroup$ Also, how did you make the animations? $\endgroup$ – Rohit Pandey Aug 16 '18 at 23:31
  • $\begingroup$ No. Geogebra.$\,$ $\endgroup$ – Ivan Neretin Aug 16 '18 at 23:37
  • $\begingroup$ Ok, then it doesn't quite fit my requirement "platonic solids with irregular faces". I meant they should have all other properties of platonic solids. Still cool shape :) $\endgroup$ – Rohit Pandey Aug 17 '18 at 1:13
  • 1
    $\begingroup$ You don't have to distort the triangle as much as I did. With smaller distortion it will be convex all right. $\endgroup$ – Ivan Neretin Aug 17 '18 at 3:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.