Five Regular Solids, Part One


Observe the five colourful objects that head my blog.

The tetrahedron

Leftmost is a tetrahedron — a pyramid with a triangular base.

Notice how regular and symmetric it is. It has four corners — mathematicians call them vertices (singular vertex) — and each vertex is host to exactly three edges.  Each vertex is a corner of exactly three of the four faces of the tetrahedron. And each face has exactly three edges. There are six edges all together.

This regularity means that you get the same picture of the tetrahedron no matter which vertex is on top.

Compare that with the the standard Egyptian pyramid (or square pyramid), which has a square base and four triangular sides. Four edges meet at the peak vertex of the Egyptian pyramid, but the four base vertices play host to only three edges. Pick up an Egyptian pyramid and turn it about: you get a different picture depending on which vertex is uppermost. The Egyptian pyramid is not regular.

Notice that regularity does not depend on the lengths of the edges or the area of the faces. We can shorten one of the edges to create a distorted tetrahedron that is still regular, although somewhat twisted. Regularity means each of the faces has the same number of sides and the same number of edges meet at each vertex.

If you feel particularly geeky, you might like to learn a new technical word that will lose you friends when you mention it at parties. Both the tetrahedron and the Egyptian pyramid are three-dimensional polytopes, objects with flat faces bounded by edges that meet in vertices. A cut diamond is a polytope; a beach ball is not.

A regular polytope is one in which each vertex has the same number of edges meeting there and each face is bounded by the same number of edges.

If you are like me, you will prefer to call polytopes “solids” — much more descriptive.

The tetrahedron is the simplest regular solid. The other pictures in my header show four others: the cube, the octahedron, the dodecahedron and the icosahedron.

The Cube

The cube has six faces, each with four sides. There are eight vertices and 12 edges. Three edges meet at each vertex.

The Octahedron

Think of the octahedron as two Egyptian pyramids, one inverted, stuck together on their square bases. It has eight triangular faces, six vertices and 12 edges. Four edges meet at each vertex.

The Dodecahedron

Now things get more difficult to visualize unless you have a model. The dodecahedron has 12 five-sided faces and 20 vertices.  Three edges meet at each vertex.

We can use those last facts to compute the number of edges without counting them on the diagram or a model.

Each face has five edges and there are 12 faces. So there appear to be 5 x 12 = 60 edges. But we have counted each edge twice; once from the point of view of one face and once from the point of view of the adjacent face. So there are 60/2 = 30 edges.

The Icosahedron

The last regular solid in my header is the icosahedron. Its 20 faces are all triangular. There are 12 vertices. Five edges meet at each vertex. How many edges?

Let’s use the same method of computing edges that we used for the dodecahedron. Three edges per face and 20 faces gives us an initial count of 3 x 20 = 60 edges. But each one has been counted twice. So there are 60/2 = 30 edges. Same as the dodecahedron.

We can also count edges from the point of view of the vertices. There are 12 vertices and five edges meet at each. An initial count of the edges is 12 x 5 = 60. But we have counted each edge twice — once at each end. So the number of edges is 60/2 = 30.

We’ve developed two rules for counting edges, rules that apply to all regular solids.

Let F be the number of faces, V be the number of vertices and E be the number of edges. Also suppose each face has m sides or edges to it and let n be the number of edges meeting at each vertex. (m and n are the same for each face or vertex exactly because we are talking about regular solids.) Then the number of edges can be expressed in two different ways.

E = \frac {Fm} {2}

and

E = \frac {Vn} {2}

A bit of high school algebra gives us the following.

F = \frac {nV} {m} \qquad V = \frac {mF} {n}

These rules are true for all five of the regular solids.

Now — are they true for all regular solids? Can you think of any more regular solids? Remember each face has to have the same number of edges and each vertex must play host to the same number of edges.

More to come.

Advertisements

About aharmlessdrudge

Way back during the late Bronze age -- actually it was the 1950s -- all of us in high school had to take a vocational test to determine our interests and, supposedly, our future careers. I cannot remember the outcome, but I do recall one question that gave me pause. "If you were to win a Nobel prize, would it be in literature or in physics?" I hesitated over the question: although I enjoyed mathematics and science more than English class, I did have a couple of unfinished (and very bad) novels hidden away at home. I cannot remember what I chose back then, but the dilemma followed me to university, where I switched from mathematics to English and -- after a five-year stint in journalism -- back to mathematics. I recently retired as a professor of statistics. Retirement. What a good chance to revive my literary ambitions. I have finished a novel -- more about that in good time -- and a rubble of drafts of articles about mathematics and statistics is taking up space on my hard disk.
This entry was posted in Uncategorized and tagged , , , , , , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s