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