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

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enter image description here

  • $\begingroup$ 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
    Commented Apr 29, 2017 at 17:37
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    $\begingroup$ Does that really fit together? $\endgroup$ Commented Apr 29, 2017 at 17:45
  • $\begingroup$ @Mariano, that I wonder too? $\endgroup$ Commented Apr 29, 2017 at 17:47
  • $\begingroup$ He constructed this facebook.com/… $\endgroup$ Commented Apr 29, 2017 at 17:52
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    $\begingroup$ 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. $\endgroup$ Commented Apr 29, 2017 at 19:48

4 Answers 4


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 $$ \sqrt{\frac{\sqrt 5 + 1}{2}} - \sqrt{\frac{\sqrt 5 - 1}{2}} = \sqrt \phi - \sqrt {\phi - 1} \approx 0.485868 $$

  • $\begingroup$ We can prove the height difference is less than one-half without the numerical calculation. Note that $\sqrt\phi+\sqrt{\phi-1}$ and $\sqrt\phi-\sqrt{\phi-1}$ are reciprocals summing to $2\sqrt\phi$. This sum is greater than $5/2$ because $(5/4)^2[(5/4)^2-1]=225/256<1$, so the smaller term $\sqrt\phi-\sqrt{\phi-1}$ must be less than $1/2$. $\endgroup$ Commented May 7, 2021 at 9:51

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

  • $\begingroup$ 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. $\endgroup$ Commented Apr 29, 2017 at 18:05
  • $\begingroup$ Johnson's solid is a complete list. Does that mean there is no other solid out of the list? right? $\endgroup$ Commented Apr 29, 2017 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
    Commented Apr 29, 2017 at 19:37
  • $\begingroup$ So, it really seems to be new. $\endgroup$ Commented Apr 29, 2017 at 19:44
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    $\begingroup$ Is it strictly convex? If not, it wouldn't be a Johnson solid. $\endgroup$
    – Théophile
    Commented Apr 29, 2017 at 19:48

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.


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.


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