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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.

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3 Answers 3

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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.

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  • $\begingroup$ Is this figure convex? $\endgroup$ Commented Aug 16, 2018 at 23:30
  • $\begingroup$ Also, how did you make the animations? $\endgroup$ Commented Aug 16, 2018 at 23:31
  • $\begingroup$ No. Geogebra.$\,$ $\endgroup$ Commented Aug 16, 2018 at 23:37
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    $\begingroup$ You don't have to distort the triangle as much as I did. With smaller distortion it will be convex all right. $\endgroup$ Commented Aug 17, 2018 at 3:29
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    $\begingroup$ Using a SAT solver, I've checked all possible colorings here - I believe there is only one solution up to isometries of the icosahedron and permutations of the colors. $\endgroup$ Commented Dec 26, 2023 at 22:44
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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

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Here, a graphical solution is presented for the icosahedron. As suggested in the comments and in the discussion of Ivan Neretin's answer, the icosahedron is not the dual of the tetartoid, nor could it be except for the Platonic case; and unlike the tetrahedron or octahedron there are enough triangular faces at each vertex to allow concave solutions.

Below left is a graphical representation of a triangular icosahedron. On the right the edges have call been labeled $a,b,c$ representing the distinct lengths of the sides of each triangle. The reader can verify that each face in the graph properly has one side of each length.

enter image description here

The existence of the graph does not fully prove the existence of the polyhedron. We must verify that there are ranges of the side lengths that allow the polyhedron to be assembled in real space. For instance, if we try to specify $a=9, b=6, c=4$ we will find that the icosahedron fails to come together.

Let the angles of the triangles be $\alpha$ opposite side $a$, $\beta$ opposite side $b$, and $\gamma$ opposite side $c$. Without loss of generality we may assume $a>b>c$ from which $\alpha>\beta>\gamma$. From the graph we can then identify four vertices having three $\alpha$ angles, one $\beta$ and one $\gamma$; and cyclic permutations thereof for two remaining groups of four vertices.

One criterion for existence is then the largest face angle at any vertex must be less than the sum of the other angles. With $\alpha>\beta>\gamma$ this implies

$\alpha<\beta+3\gamma.$

Since the three angles of any triangle sum to $180°$, we may render this as

$\alpha<135°-\beta/2.$

Equivalently,

$\alpha<90°+(\alpha-\beta)/2.$

The latter form, with $\alpha$ being the larger angle, implies that the icosahedron must exist when this angle is less than $90°$, thus always for acute triangles. In terms of side lengths, this translates to $a^2<b^2+c^2$. For instance, with $a=6,b=5,c=4$ we have correct angles to form an icosahedron with the graphical structure pictured above.

The second necessary condition is that the polygonal angles at each vertex should sum to less than $360°$. With $\alpha>\beta>\gamma$ this implies

$3\alpha+\beta+\gamma<360°.$

Again $\alpha+\beta+\gamma=180°$, so this criterion is also equivalent to the maximum angle $\alpha$ measuring less than $90°$.

Therefore twenty congruent, acute triangles may indeed be assembled to form the faces of an icosahedron using the graphical scheme pictured above.

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