Agnes Scott College
Larry Riddle, Agnes Scott College

Details for McWorter's Pentigree IFS


Description

The translations used in the IFS are determined by the coordinates of the starting endpoints P1, P2, P3, P4, and P5 of the line segments as shown in the figure below. Each line segment is obtained by scaling the unit horizontal line by r, rotating by the appropriate angle, then translating the line so that the origin is moved to P1, P2, P3, P4, or P5. The scaling factor is \(r = \frac{{3 - \sqrt 5 }}{2} = 0.381966\).

translations

The coordinates of the five points can be determined using trigonometry by starting at (0,0) and moving along the red path from point to point. See the "Trig Value Details" page for the exact values of sine and cosine at 36° and 72°.

Maple
calculations

\( P1 = \left( {\begin{array}{*{20}{c}} {r\cos {{36}^ \circ }} \\ {r\sin {{36}^ \circ }} \\ \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { \frac{1}{4}\sqrt 5 - \frac{1}{4}} \\ {\frac{3}{8}\sqrt {10 - 2\sqrt 5 } - \frac{1}{8}\sqrt {50 - 10\sqrt 5 } } \\ \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {0.309017} \\ {0.224514} \\ \end{array}} \right) \)
 

\( P2 = P1 + \left( {\begin{array}{*{20}{c}} { - r\cos {{72}^ \circ }} \\ {r\sin {{72}^ \circ }} \\ \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - \frac{1}{4}\sqrt 5 + \frac{3}{4}} \\ {\frac{1}{4}\sqrt {10 - 2\sqrt 5 } } \\ \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {0.190983} \\ {0.587785} \\ \end{array}} \right) \)
 

\( P3 = P2 + \left( {\begin{array}{*{20}{c}} {r\cos {{36}^ \circ }} \\ { - r\sin {{36}^ \circ }} \\ \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\frac{1}{2}} \\ { - \frac{1}{8}\sqrt {10 - 2\sqrt 5 } + \frac{1}{8}\sqrt {50 - 10\sqrt 5 } } \\ \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {0.5} \\ {0.363271} \\ \end{array}} \right) \)
 

\( P4 = P3 + \left( {\begin{array}{*{20}{c}} { - r\cos {{72}^ \circ }} \\ { - r\sin {{72}^ \circ }} \\ \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\frac{3}{2} - \frac{1}{2}\sqrt 5 } \\ 0 \\ \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {0.381967} \\ 0 \\ \end{array}} \right) \)
 

\( P5 = P4 + \left( {\begin{array}{*{20}{c}} {r\cos {{36}^ \circ }} \\ { - r\sin {{36}^ \circ }} \\ \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - \frac{1}{4}\sqrt 5 + \frac{5}{4}} \\ { - \frac{3}{8}\sqrt {10 - 2\sqrt 5 } + \frac{1}{8}\sqrt {50 - 10\sqrt 5 } } \\ \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {0.690983} \\ { - 0.224514} \\ \end{array}} \right) \)