# Regular Polyhedra

This wiki has been disputed, which means that it may be unreliable.

**Regular polyhedra** generalize the notion of regular polygons to three dimensions. They are three-dimensional geometric solids which are defined and classified by their faces, vertices, and edges.

A **regular polyhedron** has the following properties:

- faces are made up of congruent regular polygons;
- the same number of faces meet at each vertex.

There are nine **regular polyhedra** all together:

- five convex polyhedra or Platonic solids
- four "star" polyhedra or Kepler-Poinsot polyhedra.

Regular polyhedra (particularly the Platonic solids) are commonly seen in nature. For example, the icosahedral crystalline structure of iron pyrite and the tetrahedral structure of the methane molecule are shaped like Platonic solids.

These solids have also been the inspiration of many of man's creations. Besides the obvious relationship of cubes to bricks, the other Platonic solids have inspired many man-made structures and objects as well, from the geodesic dome structure seen in biodomes to a standard set of Dungeons & Dragons dice.

#### Contents

## Platonic Solids - Basic Properties

A convex solid is defined as a solid for which joining any two points on the solid surface forms a line segment that lies completely inside the solid. The five convex regular polyhedra are known collectively as the **Platonic solids**.

A proof that that there are only five is outlined below.

The following table highlights some of the fundamental properties of the five Platonic solids including the face shape and the numbers of vertices, edges, and faces:

$\begin{aligned} {\color{Blue} \textbf{Regular Polyhedron}} &&{\color{Blue} \textbf{Face Shape}} &&{\color{Blue} \textbf{Vertices}} &&{\color{Blue} \textbf{Edges}} &&{\color{Blue} \textbf{Faces}} \\ \text{Tetrahedron} &&\text{Triangle} &&4 &&6 &&4 \\ \text{Cube} &&\text{Square} &&8 &&12 &&6 \\ \text{Octahedron} &&\text{Triangle} &&6 &&12 &&8 \\ \text{Dodecahedron} &&\text{Pentagon} &&20 &&30 &&12 \\ \text{Icosahedron} &&\text{Triangle} &&12 &&30 &&20 \end{aligned}$

How many faces does a regular tetrahedron have?

From the summary above, a regular tetrahedron has 4 vertices, 6 edges, and 4 faces. In fact, it is called a "tetrahedron" because it has 4 faces. $_ \square$

What is the sum of the number of faces of all types of convex regular polyhedra?

There are five types of convex regular polyhedra--the regular tetrahedron, cube, regular octahedron, regular dodecahedron, and regular icosahedron.

Since the numbers of faces of the regular polyhedra are 4, 6, 8, 12, and 20, respectively, the answer is

$4 + 6 + 8 + 12 + 20 = 50.\ _\square$

## Platonic Solids - Additional Properties

The Platonic solids have many properties discussed below:

- They all satisfy Euler's formula.
- They have a number of important symmetries.
- They can be uniquely identified by their Schläfli symbols.
- Three of them can be constructed from multiple smaller Platonic solids.

## Euler's Formula

Since they are convex polyhedra, for each of the Platonic solids, the number of vertices $V$, the number of edges $E$, and the number of faces $F$ satisfy Euler's formula:

$V - E + F = 2.$

For example, for the octahedron (see table above), $V =6, E = 12,$ and $F = 8,$ so

$V - E + F = 6 - 12 + 8 = 2.$

If a convex regular polyhedron has 12 vertices and 30 edges, then how many faces does it have?

Let $V$ be the number of vertices, $E$ the number of edges, and $F$ the number of faces. Then from Euler's formula,

$\begin{aligned} V - E + F &= 2 \\ F & = 2 - V + E \\ & = 2 - 12 + 30 \\ &= 20. \end{aligned}$

Thus, a regular polyhedron that has 12 vertices and 30 edges has 20 faces. $_ \square$

## Symmetries

Note that each of the sides is a regular polygon, and if you rotate any Platonic solid by an edge you have two-fold symmetry. If you rotate it around opposite vertices or the centers of opposite faces, you have symmetry of order 3, 4 or 5.

Another symmetry of the Platonic solids is that they are duals of each other. That is, if you take the centers of all the faces of one Platonic solid, and make those the new vertices of another convex solid, you get the Platonic solid's dual. Or, in the case of the tetrahedron, it is its own dual. The following are the Platonic duals:

- Tetrahedron $\leftrightarrow$ Tetrahedron
- Octahedronn $\leftrightarrow$ Cube
- Dodecahedron $\leftrightarrow$ Icosahedron.

## Schläfli Symbols

Also, each Platonic solid can be represented by its Schläfli symbol $\{m,n\},$ where $m$ is the number of edges for each face and $n$ is the number of faces that meet at a vertex. The Schläfli symbol for the Platonic solids are as follows:

- Tetrahedron $\rightarrow \{ 3,3\}$
- Cube $\rightarrow \{ 4,3 \}$
- Octahedron $\rightarrow \{ 3,4\}$
- Dodecahedron $\rightarrow \{ 5,3 \}$
- Icosahedron $\rightarrow\{ 3,5 \}.$

The Schläfli symbols can be used to see directly which Platonic solids are duals of each other (i.e. by reversing $m$ and $n$). One can then see very quickly that the duals correspond to those outlined above.

The following is a proof that there can't be any more than five Platonic solids:

Prove that there can only be five Platonic solids.

Consider a convex regular polyhedron with Schläfli symbol $\{m,n\}.$ This implies, based on the definition of the Schläfli symbols above, that

- the faces are made up of $m$-gons;
- $n$ faces meet at a point.
First off, since it's a three-dimensional solid,

- $m \geq 3$ (since regular polygons must have three or more faces);
- $n \geq 3$ (since at least three faces must meet at any vertex or it would be flat).
Also, since the solid is convex, the sum of all of the interior angles of the polygons that meet at a vertex must add up to less than $360^{\circ}$ because otherwise they would be planar $($if they add up to exactly $360^{\circ})$ or non-convex $($if they add up to something $\gt 360^{\circ}).$

Consider the interior angles of the first few regular $m$-gons:

- triangle: $60^{\circ}$
- square: $90^{\circ}$
- pentagon: $108^{\circ}$
- hexagon: $120^{\circ}$.
For the hexagon, the three interior angles that meet at a vertex will add up to something $\gt 360^{\circ}$.

Therefore, the faces can only be triangles, squares, or pentagons, so $3 \leq m \leq 5.$

Let's first consider $m = 3$ (regular triangular faces).

Since the interior angle of a regular triangle is $60^{\circ}$, there can only be $3, 4,$ or $5$ regular triangles meeting at a face $($since $6 \times 60^{\circ} = 360^{\circ}).$ This implies that there are only three Platonic solids with triangular faces, namely

- $\{3,3\}$
- $\{3,4\}$
- $\{3,5\}.$
For $m=4$ (square faces), since the interior angle of a square is $90^{\circ},$ there can only be $n=3$. Since for $n=4$ the four angles would add up to $360^{\circ},$ there is only one Platonic solid with square faces, namely

- $\{4,3\}.$
Finally for $m=5$ (regular pentagonal faces), since the interior angle of a regular pentagon is $108^{\circ},$ there can only be $n=3$. Since for $n=4$ the four angles would add up to something $\gt 360^{\circ},$ there is only one Platonic solid with pentagonal faces, namely

- $\{5,3\}.$
So, that leaves

only fivepossible Platonic solids:

- $\{3,3\}$ - Tetrahedron
- $\{3,4\}$ - Octahedron
- $\{3,5\}$ - Icosahedron
- $\{4,3\}$ - Cube
- $\{5,3\}$ - Dodecahedron. $_\square$

The concept of Schläfli symbols can be extended to two dimensional objects (e.g. infinite lattices) and four dimensional objects (e.g. tesseracts and hexadecachoronsas).

Here is an example of how to apply the Schläfli symbols to the dual of a two-dimensional lattice:

What is the dual of an infinite triangular lattice?

**Definitions**:

The

**dual**of a solid with Schläfli symbols $\{m,n\}$ has Schläfli symbols $\{n,m\}.$The

**Schläfli symbols**for a geometric construct is defined by $\{m,n\}$, where $m$ is the number of edges for each face and $n$ is the number of faces that meet at a vertex.

## Constructing Platonic Solids

Another feature of Platonic solids is that for three of them, namely the cube, tetrahedron and octahedron, you can construct them from smaller Platonic solids.

In the case of a cube, it's straightforward to see, for example, that if you take $n^3$ unit cubes, you can make a larger $n \times n \times n$ cube:

However, it isn't so obvious that you can construct a tetrahedron from smaller tetrahedra and octahedra, or that you can construct an octahedron from smaller tetrahedra and octahedra.

Here are some examples of (1) a tetrahedron (left) built from four tetrahedra and one octahedron, and (2) an octahedron (right) built from six octahedra and eight tetrahedra:

If you remove the octahedron from the structure on the left you get a Sierpinski pyramid with linear dimension two. Further similarly constructed subdivisions of the smaller tetrahedra result in more and more intricate Sierpinski pyramids with higher linear dimensions. The process, of course, can be repeated ad infinitum.

Try out these problems for a bit more on Sierpinski pyramids.

This shelf is displaying several Sierpinski tetrahedra:

The linear dimensions of the five big ones from left to right are 1, 2, 4, 8, and 16.

You can fill the void in the second one perfectly with one octahedron.

The third one can be filled with five octahedra, a big one and four smaller ones.

How many octahedra (of different sizes) would it take to fill all the voids in a Sierpinski tetrahedra of linear dimension 1024?

**Image credit:** http://www.fractalnature.com/

To conclude this discussion on constructing Platonic solids, here is a more challenging question which will require deriving the general equations for how many smaller octahedra and tetrahedra with the same linear dimension it would take to construct a tetrahedron of larger linear dimension:

For this problem, we will define a "unit polyhedron" to be a polyhedron of edge length one.

The picture to the right shows how a larger tetrahedron can be built from four unit tetrahedra and one unit octahedron.

Amanda wishes to construct a much larger tetrahedron with many more unit tetrahedra and octahedra. In the end, she builds one using exactly 364 unit octahedra and some unit tetrahedra.

How many unit tetrahedra did she use?

**Assumption:** The final construction is one solid tetrahedron, with nothing extra sticking out, and nothing missing (no holes). No unit octahedra or tetrahedra are cut in any way.

**Image credit:** http://www.matematicasvisuales.com/

## Kepler-Poinsot (or "Star") Polyhedra

The Platonic solids above are all examples of **convex** regular polyhedra. However, not all regular polyhedra are convex. In fact, there are exactly four **non-convex** regular polyhedra, also called Kepler-Poinsot polyhedra.

These four polyhedra have the same properties of the convex polyhedra above, but aren't convex. That is,

- faces can pass through the interior of the solid;
- faces don't necessarily need to be convex polygons.

The four Kepler-Poinsot polyhedra pictured above are the

- small stellated dodecahedron
- great dodecahedron
- great stellated dodecahedron
- great icosahedron.

Two of these--the two stellated dodecahedrons--are not only **not** convex themselves but also have faces that are **not** convex. Specifically, they have pentagrams, or "five-pointed stars" for faces. The great dodecahedron and great icosahedron, however, share the same convex face types as their Platonic solid counterparts, namely the pentagon and triangle, respectively. The lighter shade of red in each image highlights these faces.

The Kepler-Poinsot (or "Star") polyhedra have the following basic properties:

$\begin{aligned} {\color{Blue} \textbf{Regular Polyhedron}} &&{\color{Blue} \textbf{Face Shape}} &&{\color{Blue} \textbf{Vertices}} &&{\color{Blue} \textbf{Edges}} &&{\color{Blue} \textbf{Faces}} \\ \text{Small stellated dodecahedron} &&\text{Pentagram} &&12 &&30 &&12 \\ \text{Great dodecahedron} &&\text{Pentagon} &&12 &&30 &&12 \\ \text{Great stellated dodecahedron} &&\text{Pentagram} &&20 &&30 &&12 \\ \text{Great icosahedron} &&\text{Triangle} &&12 &&30 &&20 \end{aligned}$

Note that they don't necessarily follow Euler's equation. For example, for the great dodecahedron,

$V - E + F = 12 - 30 + 12 = -6 \neq 2.$

However, like the Platonic solids, the Kepler-Poinsot polyhedra are also duals of each other:

- $\text{Small stellated dodecahedron} \leftrightarrow \text{Great dodecahedron}$
- $\text{Great stellated dodecahedron} \leftrightarrow \text{Great icosahedron}$.

## See Also

**Cite as:**Regular Polyhedra.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/regular-polyhedra/