Polyhedra

Definition (Polyhedra)

A Polyhedron is a $3$ dimensional figure bounded by finitely many flat faces. Examples are

• Cube
• Various Prisms
• Various Pyramids Faces meet at an edge and edges meet at vertices.

Definition (Convex)

A polyhedron is convex if for any two points in the polyhedron, if the line segment connecting them is also contained in the polyhedron.

Lemma (Planar Graphs from Convex Polyhedra)

If we have a convex polyhedron, then we have a Planar Graph of its edges and vertices. We can essentially make a wireframe diagram of a polytope. For example, given a cube, we can draw

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source code

which is obviously planar. Note that the polytope face “behind” the cube was “ripped open” to become the ambient space, which is our face.

Definition (Regular Polyhedra / Platonic Solid)

A regular polyhedron or platonic solid is a polyhedron where all

• the edges are the same length
• the faces are regular polygons
• all faces have the same number of sides
• the same number of faces at each vertex

Lemma (Characterization of Platonic Solids)

What are the platonic solids? We can answer this by turning polytopes into planar graphs.

Let’s assume we have $s$ sides per face and we are $d-$regular (or $d$ edges at each vertex). Or equivalently, $d$ faces at each vertex. By Euler’s Formula we have

$$|F|-|E|+|V|=2$$

and by Handshake Lemma, we have

$$|V|\cdot d=2|E|$$

In particular,

$$|V|=\frac{2|E|}{d}$$

So, by the Face Handshake Lemma where $|S|$ is the number of sides of a platonic solid,

$$\begin{array}{rl}2& =|V|-|E|+|F|\\ & =|E|\cdot \left(\frac{2}{d}+\frac{2}{|S|}-1\right)\end{array}$$

and that for any polyhedron, each vertex has at least degree $3$, we have that

• $d,|S|\ge 3$
• $2/d+2/|S|>1$
• $|E|=2\cdot (2/d+2/|S|-1{)}^{-1}$
• $|V|=2|E|/d$
• $|F|=2|E|/|S|$

giving us the following cases:

1. If $d=s=3$, then $|V|=4,|E|=6,|F|=4$, a regular tetrahedron
2. If $d=3,s=4,$ then $|V|=8,|E|=12,|F|=6$, a cube
3. If $d=3,s=5$, then $|V|=20,|E|=30,|F|=12$, a regular dodecahedron
4. If $d=3,s\ge 6$, then it does not work. We would get $3$ hexagons at a vertex.
5. If $d=4,s=3$ then $|V|=6,|E|=12,|F|=8$, a regular octahedron
6. If $d=4,s\ge 4$, this does not work.
7. If $d=5,s=3$, then $|V|=12,|E|=30,|F|=20$, a regular icosahedron
8. If $d=5,s\ge 4$, this does not work.
9. If $d\ge 6,s\ge 3$, then this does not work. We have only $5$ platonic solids.

See Wikipedia for images.

Lemma (Faces Have At Least 3 Sides)

Every face has at least $3$ sides, such that $2|E|\ge 3|F|$. More generally, if $G$ only has faces with at least $k$ sides, then

$$|E|\ge \frac{k|F|}{2}$$

Lemma (Faces Have Maximum Edges)

If each face has at least $k$ sides, then the maximum number of edges is

Proof: By Euler’s Formula,

$$\begin{array}{rl}2& =|V|-|E|+|F|\\ & \le |V|-|E|+\frac{2|E|}{k}\\ & =|V|-|E|\cdot \left(1-\frac{2}{k}\right)\end{array}$$

such that

$$|E|\le \frac{|V|-2}{1-2/k}$$

and we are done.

Corollary (Faces Have Maximum Sides)

Every polyhedron has a face with at most $5$ sides. This is from Theorem (Planar Graphs have Bounded Minimum).