Larry Riddle, Agnes Scott College

#### Description

Start with a pentagon with sides of length 1. Along each side, construct the basic motif for a pentadentrite. These five copies fit nicely together to form six smaller pentagons as shown in the figure below, where each motif is shown in a different color. Continue the pentadentrite construction along each side.

You can also see how five copies of the pentadentrite fit together by successively adding copies using the buttons below.

#### IteratedFunctionSystem

IFS
Animation
An alternative view of the 2nd form of McWorter's pentigree is to consider each of the six pentagons as a scaled version of the unit pentagon with an appropriate rotation and translation as shown in the figure below. Because of the symmetry of the pentagon, it is possible to rotate them in various ways and still create the same image. Here we rotate each of the scaled pentagons by the same angle A = 11.8185873°.

The scaling factor is still r = 0.381966. This leads to the following IFS [Details]:

 $${f_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {0.341} & { - 0.071} \\ {0.071} & {0.341} \\ \end{array}} \right]{\bf{x}}$$ scale by r, rotate by 11.82° $${f_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {0.341} & { - 0.071} \\ {0.071} & {0.341} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {0.649} \\ {0.136} \\ \end{array}} \right]$$ scale by r, rotate by 11.82° $${f_3}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {0.341} & { - 0.071} \\ {0.071} & {0.341} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {0.071} \\ {0.659} \\ \end{array}} \right]$$ scale by r, rotate by 11.82° $${f_4}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {0.341} & { - 0.071} \\ {0.071} & {0.341} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {-0.604} \\ {0.271} \\ \end{array}} \right]$$ scale by r, rotate by 11.82° $${f_5}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {0.341} & { - 0.071} \\ {0.071} & {0.341} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {-0.445} \\ {-0.491} \\ \end{array}} \right]$$ scale by r, rotate by 11.82° $${f_6}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {0.341} & { - 0.071} \\ {0.071} & {0.341} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {0.330} \\ { - 0.575} \\ \end{array}} \right]$$ scale by r, rotate by 11.82°

#### SimilarityDimension

We have a hyperbolic IFS with each map being a similitude of ratio r < 1. Therefore the similarity dimension, d, of the attractor of the IFS is the solution to

$\sum\limits_{k = 1}^6 {{r^d}} = 1 \quad \Rightarrow \quad d = \frac{{\log (1/6)}}{{\log (r )}} = 1.6995$

#### References

1. Edgar, Gerald A. Measure, Topology, and Fractal Geometry, Springer-Verlag, 1990.