Trema Construction Details for Two Rotations
For the trema construction for the Sierpinski gasket chaos game with two rotations, start with an isosceles pentagon whose width is twice the length of the base and for which the height is three times the distance from the base to the width.
Subdivide the pentagon into 7 regions by drawing lines from the vertices $A$ and $B$ of the bottom edge to the midpoints $F$ and $G$ of the opposite sides, and a horizontal line $KL$ that bisects the axis of symmetry. This creates 3 pentagons and 4 triangles.
Place the axis of symmetry along the $y$-axis. Let the half-width of the base $AB$ be $w > 0$, and let the height of vertices $C$ and $E$ above the base be $h > 0$. Then the constraints determine the coordinates of every vertex. Click the image to toggle a view with coordinates.
Using these coordinates shows that $$ \begin{aligned} \overrightarrow{ED} &= \langle 0-(-2w), 3h-h \rangle = \langle 2w,2h \rangle \\ \overrightarrow{AF} &= \langle w-(-w), 2h-0\rangle = \langle2w,2h\rangle \\ \overrightarrow{BC} &= \langle 2w-w,h-0\rangle = \langle w,h\rangle \end{aligned} $$
Hence $\overrightarrow{ED} = \overrightarrow{AF} = 2\overrightarrow{BC}$, so $ED$, $AF$, and $BC$ are parallel, and $|ED| = |AF| = 2|BC|$.
A similar calculation (or symmetry) shows that $CD$, $BG$, and $AE$ are parallel, and $|CD| = |BG| = 2|AE|$.
Experiments
You can experiment with how the shape changes as the base length and height change. Click the base dot in the GeoGebra app below and move it horizontally to change the base length, or click on the other dot and move it vertically to change the height.
The following video example uses an animated demonstration to show that the three smaller pentagons are a scaled copy of the original isosceles pentagon, with two of the copies rotated by 180°.
Proofs
We now show that the 3 pentagonal regions will always be a copy of the original isosceles pentagon scaled by 1/2 under the given constraints for any choice of the parameters $w$ and $h$, with two pentagons rotated by 180°.
The image below shows the isosceles pentagon and one of the pentagon regions in green. The table indicates how corresponding angles in the two pentagon are equal using a parallelogram and a transversal to two parallel lines.
| Pentagon $JIFDG$ (green) | Pentagon $ABCDE$ (gray) | Reason Equal |
|---|---|---|
| Angle $J$ | Angle $A$ | J equal to opposite angle parallelogram $ABJM$ |
| Angle $I$ | Angle $B$ | I equal to opposite angle parallelogram $ABNI$ |
| Angle $F$ | Angle $C$ | $HF \parallel BC$ Corresponding angles |
| Angle $D$ | Angle $D$ | Same angles |
| Angle $G$ | Angle $E$ | $AE \parallel HG$ Corresponding angles |
We have thus shown that the corresponding interior angles of pentagon $JIFDG$ and pentagon $ABCDE$ are the same. Moreover, you can use similar triangles in the figure above (e.g. $\triangle DGJ \sim \triangle DEA$) to show that each side of pentagon $JIFDG$ has length 1/2 of the corresponding side in pentagon $ABCDE$. Therefore, pentagon $JIFDG$ is a copy of pentagon $ABCDE$ scaled by 1/2.
The next image shows the green pentagon region from above and another pentagon region in brown. The table indicates how corresponding angles in the two trapezoids are equal using a parallelogram and a transversal to two parallel lines.
| Pentagon $JIFDG$ (green) | Pentagon $JKEAH$ (brown) | Reason Equal |
|---|---|---|
| Angle $J$ | Angle $J$ | Opposite angles |
| Angle $I$ | Angle $K$ | $AF \parallel ED$ Alternate interior angles |
| Angle $F$ | Angle $E$ | Opposite angles in parallelogram $AFDE$ |
| Angle $D$ | Angle $A$ | Opposite angles in parallelogram $AFDE$ |
| Angle $G$ | Angle $H$ | $AF \parallel ED$ Alternate interior angles |
We have thus shown that the corresponding interior angles of pentagon $JIFDG$ and pentagon $JKEAH$ are the same. Moreover, the corresponding sides have the same length. Therefore pentagon $JKEAH$ is also a copy of pentagon $ABCDE$ scaled by 1/2 and rotated by 180°.
The last image shows the brown pentagon region from above and another pentagon region in yellow. The table indicates how corresponding angles in the two pentagons are equal using a transversal to two parallel lines.
| Pentagon $JKEAH$ (brown) | Pentagon $LIHBC$ (yellow) | Reason Equal |
|---|---|---|
| Angle $H$ | Angle $C$ | $BG \parallel CD$ $AF \parallel BC$ sum of corresponding angles |
| Angle $J$ | Angle $L$ | $BG \parallel CD$ |
| Angle $K$ | Angle $I$ | $AF \parallel ED$ |
| Angle $E$ | Angle $H$ | $AF \parallel ED$ $AE \parallel BG$ sum of corresponding angles |
| Angle $A$ | Angle $B$ | remaining angles in pentagons |
We have thus shown that the corresponding interior angles of pentagon $JIFDG$ and pentagon $LIHBC$ are the same. Moreover, the corresponding sides have the same length. Therefore pentagon $LIHBC$ is also a copy of pentagon $ABCDE$ scaled by 1/2 and rotated by 180°
Iterated
Function
System
Position the initial isosceles pentagon so that the base of width $2w$ is on the $x$-axis and the axis of symmetry is the $y$-axis.
The dark gray pentagon is the result of scaling the initial trapezoid by 1/2 (towards the origin). This is then translated up by $\tfrac32 h$ for the green pentagon. Another copy of the dark gray pentagon is then rotated by 180° around the origin to form the red pentagon, which then has to be translated to the left by $w$ and up by $\tfrac32 h$ so that the tip of the red pentagon moves to $(-w,0)$. The blue pentagon is the dark gray pentagon rotated by 180° and then translated to the right by $w$ and up by $\tfrac32 h$.
The following animation illustrates the affine transformations for the trema construction starting with the isosceles trapezoid.
The iterated function system for this orientation of the isosceles pentagon is the following.
To find the fixed points, solve the following equations: $$ \begin{aligned} f_1:\quad \begin{bmatrix} x \\ y \end{bmatrix} &= \left[ {\begin{array}{cc} {1/2} & 0 \\ 0 & {1/2} \\ \end{array}} \right] \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} 0 \\ 3h/2 \end{bmatrix} = \begin{bmatrix} x/2 \\ y/2+3h/2 \end{bmatrix} \quad \Rightarrow \quad F_1 = (0, 3 h) \\ \\ f_2:\quad\begin{bmatrix} x \\ y \end{bmatrix} &= \left[ {\begin{array}{cc} {-1/2} & 0 \\ 0 & -1/2 \\ \end{array}} \right] \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} -w \\ 3h/2 \end{bmatrix} = \begin{bmatrix} -x/2-w \\ 0 \end{bmatrix} \quad \Rightarrow \quad F_2 = (-\tfrac23 w, h) \\ \\ f_3:\quad\begin{bmatrix} x \\ y \end{bmatrix} &= \left[ {\begin{array}{cc} {-1/2} & 0 \\ 0 & -1/2 \\ \end{array}} \right] \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} w \\ 3h/2 \end{bmatrix} = \begin{bmatrix} x/2+w \\ 0 \end{bmatrix} \quad \Rightarrow \quad F_3 = (\tfrac23 w, h) \end{aligned} $$
If the Sierpinski gasket chaos game with two rotations is played with the fixed points $(-A,B)$, $(A,B)$, and $(0,3B)$, then the corresponding trema construction would use $w = \tfrac32 A$ and $h = B$.
The 54°
Isosceles
Pentagon
Consider the isosceles pentagon where the two bottom legs make a 54° angle with the base. If the width of the base is $2w$, then the height from the base to the width of the pentagon is $h = w\tan 54^\circ$. This completely determines the geometry of the pentagon. Click on the button "Add Fractal" below to see the fractal obtained by iterating the trema construction on this isosceles pentagon.
The 54° angle was chosen because one can place five copies of this isosceles pentagon around the edges of a regular pentagon. Click on the button "5 Pentagons" to see this arrangement, and then "5 Fractals" for the image obtained by iterating the trema construction on each pentagon. If this 5-fractal image is scale by 1/2, then it will fit exactly within the star-shaped center. Scale by another 1/2 and the new image will fit within the now smaller star-shaped center. This can continue indefinitely, each scaling by 1/2. Click the "Add" buttons to see the first three steps.
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Just as the Sierpinski gasket construction can be done with any triangle, not just an equilateral triangle, so too can the trema construction for two rotations be applied to any pentagon for which the width is twice the length of the base and the height is 3 times the distance from the base to the width [General Construction Details]. An example of such a pentagon is shown below, divided into the 6 regions by the trema construction. Click the buttons to display the fractal generated by iterating this construction. The black squares are the fixed points that would be used in the chaos game with two rotations.
