Larry Riddle, Agnes Scott College

## McWorter's Pentigree

#### Description

Starting from the origin, draw a line segment of length r at an angle of 36°. Turn left by 72° and draw another line segment of length r. Next turn right by 144° and draw a third line segment of length r. The remaining three line segments, each of length r, are formed by making a right turn of 72° followed by two left turns of 72°. The end of the last line segment will be at the point (1,0).

The six red line segments in the figure give the basic motif (first iteration) for this IFS. Each subsequent iteration is done by replacing each line segment with a scaled version of the basic motif.

These six transformations are given by rotations of 36°, 108°, -36°, -108°, -36°, and 36° respectively. Thus the endpoints of the fourth segment and the last segment lie on the x-axis. The value of r must be chosen so that the endpoint of the last segment is at the point (1,0). From the figure we see that we must have

$\begin{array}{l} r + 2r\cos {36^ \circ } = 1 \\ \Rightarrow r = \frac{1}{{1 + 2\cos {{36}^ \circ }}} = \frac{1}{{1 + 2\left( {\frac{{1 + \sqrt 5 }}{4}} \right)}} = \frac{{3 - \sqrt 5 }}{2} = 0.381966 \\ \end{array}$

#### IteratedFunctionSystem

IFS
Animation
It is now possible to determine how each of the six rotated segments must be translated. For example, the second segment must be translated to (r cos36°, r sin36°) after a rotation by 108° and a scaling by a factor of r = 0.381966. Similar calculations with the other segments led to the following IFS [Details]:

 $${f_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {0.309} & { - 0.255} \\ {0.255} & {0.309} \\ \end{array}} \right]{\bf{x}}$$ scale by r, rotate by 36° $${f_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { - 0.118} & { - 0.363} \\ {0.363} & { - 0.118} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {0.309} \\ {0.225} \\ \end{array}} \right]$$ scale by r, rotate by 108° $${f_3}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {0.309} & {0.225} \\ { - 0.225} & {0.309} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {0.191} \\ {0.588} \\ \end{array}} \right]$$ scale by r, rotate by −36° $${f_4}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { - 0.118} & {0.363} \\ { - 0.363} & { - 0.118} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {0.500} \\ {0.363} \\ \end{array}} \right]$$ scale by r, rotate by −108° $${f_5}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {0.309} & {0.225} \\ { - 0.225} & {0.309} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {0.382} \\ 0 \\ \end{array}} \right]$$ scale by r, rotate by −36° $${f_6}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {0.309} & { - 0.225} \\ {0.225} & {0.309} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {0.691} \\ { - 0.225} \\ \end{array}} \right]$$ scale by r, rotate by 36°

#### L-System

L-system
Animation
Angle 36
Axiom F
F —> +F++F−−−−F−−F++F++F−

First three iterations of the L-system

#### SimilarityDimension

The pentigree is self-similar with 6 non-overlapping copies of itself, each scaled by the factor r < 1. Therefore the similarity dimension, d, of the attractor of the IFS is the solution to

$\sum\limits_{k = 1}^6 {{r^d}} = 1 \quad \Rightarrow \quad d = \frac{{\log (1/6)}}{{\log (r )}} = 1.86172$

#### SpecialProperties

Five copies of the pentigree fit together to form a set with five-fold rotational symmetry as illustrated in the following figure. Edgar calls this the "2nd form of McWorter's pentigree". [Details]