Star polygons

By Martin McBride, 2022-10-23
Tags: pentagram hexagram heptagram complex polygon schläfli symbol star polygon
Categories: gcse geometry

We looked at star polygons in a previous article. We will now look at them in more detail.

We saw that some polygons are self-intersecting, that is some of their sides cross over each other, like the shape on the right, here:

Convex heptagon and complex heptagon

Self-intersecting polygons are called complex polygons. A star polygon is a type of complex polygon that creates a star shape. Here is an example of a 5-point star:

Regular pentogram

This is called a pentagram (a pentagon is a 5-sided convex polygon, a pentagram is a 5-sided start polygon). It is formed from the same 5 points that would be used to draw a convex polygon, but we connect the points in a different order to create the star shape.

In this case, the vertices we are using are from a regular pentagon (all 5 sides and all 5 angles are equal). The star shape is therefore a regular pentagram (which again has 5 equal sides and 5 equal angles, but they are different to the pentagon).

We will look at some other regular star polygons and their properties below.

Regular pentagrams

The pentagram above is formed by taking the 5 points of a regular pentagon, but connecting them in the order 1, 3, 5, 2, 4 (then back to 1).

In each case starting at vertex 1, we connect to the second vertex along in the clockwise direction:

  • Vertex 1 connects to vertex 3
  • Vertex 3 connects to vertex 5
  • Vertex 5 connects to vertex 2 (because we have gone all the way round the points and the count starts at 1 again)
  • Vertex 2 connects to vertex 4
  • Vertex 4 connects back to vertex 1, so the cycle is complete.

We can use a special notation for this star polygon. We call it {5/2}, meaning there are 5 vertices and we will form the star by stepping on 2 vertices each time.

You might guess that there are other possibilities:

  • {5/1} 5 vertices, stepping on by 1 each time.
  • {5/2} we have just seen.
  • {5/3} stepping on by 3 each time.
  • {5/4} stepping on by 4 each time.

Those are the only possibilities. {5/5} won't work because if we start at point 1 and step round 5 points we will be back at point 1. {5/6} means stepping on by 6 each time, which is a full circle plus 1, so it creates the same result as {5/1}, and so on. There are only 4 possible options for a 5-pointed star. Here is what they look like:

Regular pentogram

The {5/1} case steps through the points in the order 1, 2, 3, 4, 5 - in other words, it just creates a normal regular pentagon.

The {5/3} case looks identical to the {5/2} case. Why should that be? Well making 3 steps clockwise is the same as making 2 steps counterclockwise, so {5/3} is the mirror image of {5/2}. Since {5/2} is symmetrical that makes the results identical.

Following the same logic, {5/4} creates the same result as {5/1}.

This means that there is only one type of regular star pentagram.

General rule

For a regular polygon with s sides:

  • Type {s/1} will be a convex polygon (that is, an ordinary regular polygon). Example {5/1} is a regular pentagon.
  • Type {s/n} will be identical to type {s/(s-n)}. Example {5/2} is same shape as {5/3}.
  • Type {s/(s-1)} will be a convex polygon (from the second rule). Example {5/4} is a regular pentagon.

Regular hexagrams

We can do the same thing with hexagons. Here are the 5 possibilities for {6/1} to {6/5}:

Regular hexogram

First of all, as we would expect from the rules, {6/1} and {6/5} are regular hexagons. And {6/2} is identical to {6/4}.

The only unique cases we need to consider are {6/2} and {6/3}.

But something strange is happening here. The {6/2} case doesn't visit all the points. It creates a triangle of points 1, 3, and 5. The reason for this is that 6 and 2 contain a common factor - they are both divisible by 2. Starting from point 1 there is no way to reach an even-numbered point by repeatedly adding 2.

The {6/3} case is even worse. It creates a line between points 1 and 4. The reason again is that 6 and 3 contain a common factor - they are both divisible by 3. Starting from 1 and adding 3 we get to 4, and then back to 1.

So, there are no star polygons with 6 sides.

We can create a kind of star, like this:

Regular hexogram

There are two possible triangles we can create with a {6/2} shape. Starting at point 1 we visit points 1, 3, and 5. Starting at point 2 we visit 2, 4, and 6. If we draw both triangles together we get a star shape. However, this is not a star polygon because it is formed from two separate triangles.

The fancy name for this is a stellation. This is a star shape created by one or more polygons that visit every vertex. {6/2} is a stellation made up of 2 triangles. {5/2} is a stellation made up of a single pentagram.

We can do a similar trick in the {6/3} case, this time including 3 lines starting at points 1, 2 and 3. Here is the result:

Regular hexogram

Regular heptagrams

What about a 7-point star polygon?

Let's use what we have learned already to reduce the number of cases we have to look at:

  • {7/1} and {7/6} will create a normal convex heptagon.
  • {7/2} should create a star because 7 and 2 have no common factors (no number divides into both 7 and 2).
  • {7/3} should create a star because 7 and 3 have no common factors.
  • {7/4} will be identical to {7/3}, and {7/5} will be identical to {7/2}.

So there are only 2 cases we need to look at:

Regular hexogram

As predicted, both of these are proper star polygons.

Larger numbers of sides

The same logic can be applied to any number of sides. In many cases, the shape will have one or more types of star, but also some cases that don't create a star.

See also

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