MATHEMATICS AND ORIGAMI

Jesús de la Peña Hernández

11 STELLATE POLÍGONS
As is well known, to construct a stellate regular polygon with n sides (associated to the
division of a circumference in n parts) and species p (pitch between vertices) these conditions
are required:

• p > 1


• p and n must be prime between themselves.


• 
  

  2

  
  

n

p<

With the antecedent conditions, stellate polygons of 3, 4, 6, etc sides are impossible,

whereas there can be of 5,7,8, etc.

We are going to focus on the folding of pentagon and heptagon: there is only one stel-

late pentagon (n = 5; p = 2), but two heptagons (n = 7; p = 2) and (n = 7; p = 3).

As can be seen, the hexagon can ́t be stellate; the only that n = 6 can produce is two op-

posite overlapped equilateral triangles. Nevertheless we shall study several solutions of what

we call hexagonal stars and also those beautiful figures known as ice crystals.

Likewise convex regular polygons, the stellate can be perfectly constructed by folding,

but we can also find more or less imperfect versions. Besides, a lot of stars are offered that, in

most of the cases are not regular polygons and, some of them, are not even flat: they may have

a bit of volume.

11.1 STELLATE PENTAGON (S. FUJIMOTO)
It is a perfect solution obtained from an argentic rectangle.

Let ́s see first some relations within the convex regular pentagon and between it and the

argentic rectangle.

In fig. 1:

β 3 δ

2

1

=

for β and 3δ see the same arc in the circumference, the former from one of its points, and the

latter from its center. In like manner it is:

2

2

2

1 δ

ε= γ=γ=

Moreover:

β= 2 α ; and being β 3 δ

2

1

= , we ́ll have: β 3 2 γ

2

1

= × ; β= 3 γ

The antecedent can also be observed directly:

5

3 π

β= ;

10 5

2 π π

γ= = therefore: β= 3 γ

1