4
$\begingroup$

A friend of mine, Jose Joel Leonardo, is claiming to have discover a new "regular" polyhedron, or, at least, a new type of regular polyhedron. Can anyone tell whether he is right or not? Below there is a image with a model from which you would be able to construct the polyhedron. Thanks in advance. enter image description here

enter image description here

enter image description here

$\endgroup$
  • $\begingroup$ what is the definition of regular polyhedron? According to Wikipedia, it is a polyhedron whose symmetry group acts transitively on its flag, and there are 5 convex ones, and 4 more regular star polyhedra (non-convex ones), making nine regular polyhedra in all. The above one may be "regular" being pretty, but does it satisfy the formal definition? en.wikipedia.org/wiki/Regular_polyhedron $\endgroup$ – Mirko Apr 29 '17 at 17:37
  • 2
    $\begingroup$ Does that really fit together? $\endgroup$ – Mariano Suárez-Álvarez Apr 29 '17 at 17:45
  • $\begingroup$ @Mariano, that I wonder too? $\endgroup$ – Emmanuel José García Apr 29 '17 at 17:47
  • $\begingroup$ He constructed this facebook.com/… $\endgroup$ – Emmanuel José García Apr 29 '17 at 17:52
  • 1
    $\begingroup$ Paper is flexible, and allows for lots and lots of wiggle... Really fitting together is not something that you can prove by making a paper model. $\endgroup$ – Mariano Suárez-Álvarez Apr 29 '17 at 19:48
5
$\begingroup$

No, they are not all regular polygons.

The pictured solid works with isosceles triangles that are almost equilateral but not perfect. I found this critical fact,

Dihedral angles between tetrahedron faces from triangles' angles at the tip

This says that when a regular hexagon and two regular pentagons meet at a vertex, the dihedral angle between the hexagon and a pentagon has

$$ \cos \theta = \frac{(1 - \sqrt 5) \sqrt 3}{\sqrt{10 + 2 \sqrt 5}}, $$ $$ \sin \theta = 2 \sqrt{\frac{\sqrt 5 - 1}{5 + \sqrt 5}} $$

It follows that the first vertex of a pentagon is this distance above the plane of the hexagon, $$ \sqrt{\frac{\sqrt 5 - 1}{2}}, $$ while the highest point of the pentagon is $$ \sqrt{\frac{\sqrt 5 + 1}{2}} $$ over the hexagon plane. The height difference, which would be exactly one half for an equilateral triangle, is actually $$ \sqrt{\frac{\sqrt 5 + 1}{2}} - \sqrt{\frac{\sqrt 5 - 1}{2}} = \sqrt \phi - \sqrt {\phi - 1} \approx 0.485868 $$

$\endgroup$
3
$\begingroup$

Informally speaking, a regular polyhedron is one where all vertices, edges, and faces look alike. A cube is regular, for example. There are only five (convex) regular solids, so we can be skeptical about discoveries of new ones.

There is also such a thing as a semiregular polyhedron: the restrictions are a looser than above, as only the vertices need to look alike. A soccer ball is semiregular (but not regular, because there are two different kinds of faces).

According to these definitions, your friend's shape is therefore not regular or even semiregular. Assuming it fits together, it's just a plain old polyhedron.

$\endgroup$
3
$\begingroup$

It's a polyhedron, but not a regular one.

A polyedron is called "regular" if its symmetry group acts transitively on its "flags". Intuitively this means that all faces must be equivalent, in particular all faces must have the same shape.

There are very few regular polyhedra, you can find a (complete) list here: https://en.wikipedia.org/wiki/Regular_polyhedron#The_regular_polyhedra

There are plenty of non-regular polyhedra, you can find a lot of examples here: https://en.wikipedia.org/wiki/Polyhedron, your specific example might or might not appear in one of their lists.

$\endgroup$
3
$\begingroup$

As the other answers said, this is not a regular polyhedron. I suspect you referred to the fact that all the faces are regular polygons. Such a polyhedron would be called a Johnson solid: https://en.m.wikipedia.org/wiki/Johnson_solid

They are completely enumerated and at first glance I couldn't see your friend's among them.

Edit: As confirmed by Will Jagy's answer - the triangles are not equilateral, so this is not a Johnson solid

$\endgroup$
  • $\begingroup$ Can anyone answer the question raised by Mariano? Does that really fit together? My friend thinks it does, but he is only based on the fact that he has constructed it using paper. $\endgroup$ – Emmanuel José García Apr 29 '17 at 18:05
  • $\begingroup$ Johnson's solid is a complete list. Does that mean there is no other solid out of the list? right? $\endgroup$ – Emmanuel José García Apr 29 '17 at 18:35
  • $\begingroup$ Thats correct - though there are some other categories of polyhedra with regular polygon sides - the Archimedean solids, prisms and anti prisms, but this doesn't fit into those groups either $\endgroup$ – Quickdraw Apr 29 '17 at 19:37
  • $\begingroup$ So, it really seems to be new. $\endgroup$ – Emmanuel José García Apr 29 '17 at 19:44
  • $\begingroup$ Is it strictly convex? If not, it wouldn't be a Johnson solid. $\endgroup$ – Théophile Apr 29 '17 at 19:48

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.