Platonic Solids and the Euler...

Platonic Solids and the Euler Characteristic

Keith Jones

Sanford Society, SUNY Oneonta September 2013

What is a Platonic Solid?A Platonic Solid is a 3-dimensional object with flat faces andstraight edges (a.k.a a polyhedron) satisfying:

I all faces are congruent,

I all faces are regular polygons,

I the same number of faces meet at each vertex, and

I the solid is convex (no indentations).

Images from WikipediA.

The Five Platonic SolidsThere are only five solids satisfying these properties:

1. The tetrahedron has 4 triangular faces, 6 edges, and 4 vertices:

2. The cube or hexahedron has 6 square faces, 12 edges, and 8vertices:

3. The octahedron has 8 triangular faces, 12 edges, and 6 vertices:

4. The dodecahedron has 12 pentagonal faces, 30 edges, and 20vertices.

5. The icosahedron has 20 triangular faces, 30 edges, and 12vertices.

Platonic Solids are not a New Idea

http://rabbithole2.com

According to a certain website, these stones are believed to date to2000 B.C.

But this is a dubious website dedicated to conspiracy theories. Nosolid evidence these people knew of the Platonic solids.

Still, the platonic solids are named after Plato, after all.

How to count edges and vertices

Suppose a solid has F faces, each face is an p-sided polygon, andq edges meet at each vertex.

Then there are E =pF

2edges and V =

qvertices.

Example: The Dodecahedron. 12 faces,5 · 12

2= 30 edges,

2 · 30

3= 20 vertices.

If you have E edges, you have F = 2Ep faces and V = 2E

q vertices.

2edges and V =

qvertices.

2= 30 edges,

2 · 30

3= 20 vertices.

q vertices.

2edges and V =

qvertices.

2= 30 edges,

2 · 30

3= 20 vertices.

q vertices.

2edges and V =

qvertices.

2= 30 edges,

2 · 30

3= 20 vertices.

q vertices.

The Euler Characteristic

The Euler Characteristic χ for the surface a polyhedron with Ffaces, E edges, and V vertices is:

χ = F − E + V

or χ = F + V − E

I tetrahedron: F = 4, E = 6, V = 4. So χ = 4 + 4 − 6 = 2

I hexahedron: F = 6, E = 12, V = 8. So χ = 6 + 8 − 12 = 2

I octahedron: F = 8, E = 12, V = 6. So χ = 8 + 6 − 12 = 2

I dodecahedron: F = 12, E = 30, V = 20. So χ = 2

I icosahedron: F = 20, E = 30, V = 12. So χ = 2

In all cases, χ = 2! Why?

χ = F − E + V or χ = F + V − E

A Theorem for the Euler Characteristic

For the surface of any convex polyhedron, χ = F + V − E = 2.

This is known as Euler’s Polyhedron Formula.

There Can Be Only Five!

How do we know these are the only five platonic solids?

From Euler’s Characteristic Formula, we know thatF + V − E = 2 > 0.

If each face meets p edges, and each vertex meets q edges, we canwrite

pand V =

So we have2E

q− E = 2 > 0.

pand V =

So we have2E

q− E = 2 > 0.

pand V =

So we have2E

q− E = 2 > 0.

pand V =

So we have2E

q− E = 2 > 0.

Continuing the calculation...

q− E > 0

q− 1

Now, p and q must both be larger than 2.

q− E > 0

q− 1

q− E > 0

q− 1

q− E > 0

q− 1

The only possible (p, q) pairs:

I p = 3, q = 3 — 13 + 1

3 — Tetrahedron —

I p = 4, q = 3 — 14 + 1

3 — Hexahedron —

I p = 3, q = 4 — 13 + 1

4 — Octahedron —

I p = 5, q = 3 — 15 + 1

3 — Dodecahedron —

I p = 3, q = 5 — 13 + 1

5 — Icosahedron —

I p = 3, q = 3 — 13 + 1

I p = 4, q = 3 — 14 + 1

I p = 3, q = 4 — 13 + 1

I p = 5, q = 3 — 15 + 1

I p = 3, q = 5 — 13 + 1

I p = 3, q = 3 — 13 + 1

I p = 4, q = 3 — 14 + 1

I p = 3, q = 4 — 13 + 1

I p = 5, q = 3 — 15 + 1

I p = 3, q = 5 — 13 + 1

I p = 3, q = 3 — 13 + 1

I p = 4, q = 3 — 14 + 1

I p = 3, q = 4 — 13 + 1

I p = 5, q = 3 — 15 + 1

I p = 3, q = 5 — 13 + 1

I p = 3, q = 3 — 13 + 1

I p = 4, q = 3 — 14 + 1

I p = 3, q = 4 — 13 + 1

I p = 5, q = 3 — 15 + 1

I p = 3, q = 5 — 13 + 1

Spot the Non-Platonic Solid!

Visithttp://www.learner.org/interactives/geometry/platonic.html

For Print and Fold instructions to make your own Platonic Solids!

Thanks for your time!

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