Larry Riddle, Agnes Scott College

Suppose the initial polygon has **n** sides, each of length 1.
The polygon must be scaled so that **n** copies of the scaled
polygon exactly fit inside the initial polygon. The following figure
illustrates one side of the initial polygon and two copies of the
polygon after it is scaled by **r** and and the copies are
translated.

Two sides of the scaled polygons would eventually just meet at a
vertex (except for when **n** is a multiple of 4, in which case
the two polygons meet along an entire edge.) Notice that the sides of
the polygon on the left form the hypotenuses of a sequence of right
triangles, with each hypothenuse of length **r**. We need to
determine the angles **A1**, **A2**, **A3**, etc. so that we
can compute the lengths **d1**, **d2**, **d3**, etc. of the
bases of each triangle.

The interior angle **B** of a regular **n**-sided polygon is
(**n**-2)/**n***pi. Working our way around the edges of
the polygon, we get

\[\begin{array}{l} A1 = \pi - \frac{{n - 2}}{n}\pi = \frac{{2\pi }}{n} \\ A2 = 2\pi - B - \frac{\pi }{2} - \left( {\frac{\pi }{2} - A1} \right) = \frac{{2\pi }}{n} + A1 = \frac{{2\pi \cdot 2}}{n} \\ A3 = 2\pi - B - \frac{\pi }{2} - \left( {\frac{\pi }{2} - A2} \right) = \frac{{2\pi }}{n} + A2 = \frac{{2\pi \cdot 3}}{n} \\ \end{array}\]

There seems to be a pattern, and indeed, you can prove by
induction that \(Ak = \frac{{2\pi \cdot k}}{n}\) for each angle **Ak** along the edges until the
polygon meets the next polygon. We are ok in constructing these
angles and corresponding triangles as long as the angle is still
acute. This means we must have

\[Ak = \frac{{2\pi \cdot k}}{n} \le \frac{\pi }{2} \quad \Rightarrow \quad k \le \frac{n}{4}\]

Now that we know the angles, we can compute the lengths **dk**
= **r***cos(**Ak**) for **k** = 1 to
\(\left\lfloor {n/4} \right\rfloor \). By symmetry,
we have the same lengths for the polygon on the right. Therefore we
have

\[{2 \; \left( {r + \sum\limits_{k = 1}^{\left\lfloor {n/4} \right\rfloor } {r \cos \left( {\frac{{2\pi k}}{n}} \right)} } \right)} = 1 \quad \Rightarrow \quad r = \frac{1}{{2 \; \left( {1 + \sum\limits_{k = 1}^{\left\lfloor {n/4} \right\rfloor } {\cos \left( {\frac{{2\pi k}}{n}} \right)} } \right)}}\]

Note that if **n** is a multiple of 4, then **n**/4 will be
an integer. For that value of **k**, we have **Ak** = pi/2, so
that the corresponding edge is vertical. In this case, the left and
right polygons meet along this edge. See the figure.

Because the affine transformations for this iterated function
system involve only scaling by **r** and translations, each
vertex of the initial polygon is a fixed point for the transformation
that translates the scaled version back to that vertex.

If we look at a particular vertex **V** for the corresponding
transformation **f**(**x**)=**r*****x** + **b**, then

\[V = f(V) = rV + b \Rightarrow b = (1 - r)V\]

So we just need to pick the vertices in a convenient way. The easiest way to do this is to place the initial polygon with its center at the origin and with one vertex on the x-axis.

Each central angle is equal to 2***pi/n**, so if **w**
denotes the distance from the center to the vertex, then the vertices
are at the points (**w***cos(2 **pi*****k/n**),
**w***sin(2 **pi*****k/n**) for **k** = 1 to **n**.
Therefore, the translation for the **k**th function is

\[(1 - r) \; \left[ {\begin{array}{*{20}{c}} {w \cdot \cos \left( {\frac{{2\pi k}}{n}} \right)} \\ {w \cdot \sin \left( {\frac{{2\pi k}}{n}} \right)} \\ \end{array}} \right]\]