# What’s the easiest way to explain why only 5 polyhedrons exist?

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What’s the easiest way to explain why only 5 polyhedrons exist?

In: Mathematics I think the easiest way to explain is by looking at the corners of the shapes. In order to get a regular polyhedron, we need to construct it entirely out of regular polygons. Looking at the regular polygons, only 3 of them (triangle, square, and pentagon) can actually form a corner, where 3 or more polygons meet. With a hexagon, 3 regular hexagons connected by their corners form a plane and can’t actually fold into a corner. And anything higher than that will overlap.

Likewise, while we can form a corner with 3 squares and regular pentagons, if we try to connect 4 regular pentagons or squares, we will run into the same problem. On the other hand, we can connect up to 5 triangles and still form a corner.

In total, we have 5 regular polyhedrons. 3 formed by triangles with a 3 triangle corner, 4 triangle corner, and 5 triangle corner, one formed by a 3 square corner, and one formed by a 3 pentagon corner. You can basically go through every regular polygon and show that everything above 5 sides has too wide of an angle to connect with itself, the same way that a sphere doesn’t take up all space.

Then, you go through triangles, squares, and pentagons, and you can find that triangles have lots of wiggle room, they can connect with three different dihedral angles before it starts hitting itself. These three angles form the tetrahedron, octahedron, and icosahedron. The cube has less freedom, and can only connect to form the cube before it hits itself. Similarly, the pentagon is only able to get the dodecahedron. The faces must be regular polygons but it’s easy to see that the only possibilities are triangles, squares and pentagons. Hexagons just make a flat surface and beyond that doesn’t work at all.

A polyhedron needs at least three faces meeting at each vertex, and:

* for triangles, 3, 4 and 5 per vertex give the tetrahedron, octahedron and icosahedron; 6 triangles make a flat surface and beyond that can’t work;
* for squares, 3 per vertex give a cube; 4 squares make a flat surface and beyond that can’t work; and
* for pentagons, 3 per vertex given a dodecahedron and beyond that can’t work.