Trema Construction Details for Three Rotations
For the construction of the Sierpinski gasket chaos game with three rotations, start with an isosceles hexagon for which the width is 3/2 the length of the base, the top is half the length of the base, and the height is three times the distance from the base to the width.
Subdivide the hexagon into 7 regions by drawing lines from the quarter points $M$ and $N$ of the bottom edge to the quarter points $H$ and $G$ from the top of the opposite sides, and a horizontal line $IL$ that bisects the axis of symmetry.
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 $F$ and $C$ 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 $$ \begin{aligned} \overrightarrow{CD} &= \langle \tfrac12 w-\tfrac32 w, 3h-h \rangle = \langle -w,2h \rangle \\ \overrightarrow{NH} &= \langle -\tfrac{3}{4}w-\tfrac12 w, \tfrac{5}{2}h-0\rangle = \langle -\tfrac{5}{4}w,\tfrac{5}{2}h\rangle = \tfrac{5}{4} \langle -w,2h \rangle\\ \overrightarrow{AF} &= \langle -\tfrac32 w-(-2w),h-0\rangle = \langle -\tfrac12 w,h\rangle = \tfrac12 \langle -w,2h \rangle \end{aligned} $$
Therefore $CD$, $NH$, and $AF$ are parallel, and $|CD| = 2|AF|$.
A similar calculation (or symmetry) shows that $FE$, $MG$, and $BC$ are parallel, and $|FE| = 2|BC|$.
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 hexagons are a scaled copy of the original isosceles hexagon, with all three copies rotated by 180°.
Proofs
We now show that the 3 hexagonal regions will always be a copy of the original isosceles hexagon scaled by 1/2 under the given constraints for any choice of the parameters $w$ and $h$, with all three hexagons rotated by 180°.
The image below shows the isosceles hexagon and one of the hexagonal regions in green. Dotted line segments have been added to form several parallelograms. The table indicates how opposite angles in these parallelograms can be used to verify that corresponding angles in the two hexagons are equal. Click on the "Show" button to see the parallelograms used for each angle pair.
| Hexagon $AMPJIF$ (green) | Hexagon $ABCDEF$ (gray) | Reason Equal |
|---|---|---|
| Angle $A$ | Angle $D$ | parallelogram |
| Angle $M$ | Angle $E$ | parallelogram |
| Angle $P$ | Angle $F$ | parallelogram |
| Angle $J$ | Angle $A$ | parallelogram |
| Angle $I$ | Angle $B$ | parallelogram |
| Angle $F$ | Angle $C$ | parallelogram |
So the corresponding interior angles of each hexagon are equal. In addition, the corresponding sides are proportional as shown by the following calculations: $$ \begin{aligned} |AF| &= \tfrac{1}{2}|DC| \\ |IF| &= |\langle \tfrac{1}{2}w, \tfrac{1}{2}h \rangle| = \tfrac{1}{2}|CB| \\ |IJ| &= 2w = \tfrac{1}{2}|BA| \\ |JP| &= |LC| = |IF| = \tfrac{1}{2}|CB| = \tfrac{1}{2}|AF| \\ |PM| &= |CB| = \tfrac{1}{2}|FE| \\ |MA| &= w = \tfrac{1}{2}|ED| \end{aligned} $$ This shows that hexagon $AMPJIF$ is a copy of hexagon $ABCDEF$ scaled by 1/2 and rotated by 180°.
The image below shows the green hexagonal region from above and another hexagonal region in brown. The first four rows of the table indicates how corresponding angles in the two hexagons are equal using the transversal to two parallel lines. The last two rows use angle results from above.
| Hexagon $AMPJIF$ (green) | Hexagon $JKGDEH$ (brown) | Reason Equal |
|---|---|---|
| Angle $M$ | Angle $K$ | $AB \parallel IL$ |
| Angle $P$ | Angle $G$ | $DC \parallel HN$ |
| Angle $I$ | Angle $E$ | $DE \parallel IL$ |
| Angle $F$ | Angle $H$ | $AF \parallel HN$ |
| Angle $A$ | Angle $J$ | $\text{green } J = A$, but $\text{green } J= \text{ brown } J$ |
| Angle $J$ | Angle $D$ | $A=D$ and $A=\text{green } J$ $\Rightarrow \text{green } J = D$ |
So the corresponding interior angles of each hexagon are equal. In addition, the corresponding sides are equal in length, so the two hexagons are congruent. Therefore hexagon $JKGDEH$ is also a copy of hexagon $ABCDEF$ scaled by 1/2 and rotated by 180°.
The image below shows the green hexagonal region from above and another hexagonal region in yellow. The rows of the table indicates how corresponding angles in the two hexagons are equal using the transversal to two parallel lines.
| Hexagon $AMPJIF$ (green) | Hexagon $NBCLKP$ (yellow) | Reason Equal |
|---|---|---|
| Angle $A$ | Angle $N$ | $AF \parallel NP$ |
| Angle $M$ | Angle $B$ | $MP \parallel BC$ |
| Angle $P$ | Angle $C$ | $HP \parallel DC$ $MP \parallel BC$ sum of corresponding angles |
| Angle $J$ | Angle $L$ | $HP \parallel DC$ |
| Angle $I$ | Angle $K$ | $FE \parallel PG$ |
| Angle $F$ | Angle $P$ | $FE \parallel PG$ $AF \parallel NP$ sum of corresponding angles |
So the corresponding interior angles of each hexagon are equal. In addition, the corresponding sides are equal in length, so the two hexagons are congruent. Therefore hexagon $NBCLKP$ is also a copy of hexagon $ABCDEF$ scaled by 1/2 and rotated by 180°.
Iterated
Function
System
Position the initial isosceles hexagon so that the base of width $2w$ is on the $x$-axis and the axis of symmetry is the $y$-axis.
The dark gray hexagon is the result of scaling the initial hexagon by 1/2 (towards the origin). This is then rotated by 180° and translated up by $\tfrac32 3h$ for the green hexagon. Another copy of the dark gray pentagon is then rotated by 180° around the origin to form the red hexagon, which then has to be translated to the left by $\tfrac34 w$ and up by $\tfrac32 h$ so that the lower left vertex of the red hexagon moves to $(-w,0)$. The blue hexagon is the dark gray hexagon rotated by 180° and then translated to the right by $\tfrac34 w$ 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 hexagon is the following.
To find the fixed points, solve the following equations: $$ \begin{aligned} f_1:\quad \begin{bmatrix} x \\ y \end{bmatrix} &= \left[ {\begin{array}{cc} {-1/2} & 0 \\ 0 & {-1/2} \\ \end{array}} \right] \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} 0 \\ 3h \end{bmatrix} = \begin{bmatrix} -x/2 \\ -y/2+3h \end{bmatrix} \quad \Rightarrow \quad F_1 = (0, 2h) \\ \\ f_2:\quad\begin{bmatrix} x \\ y \end{bmatrix} &= \left[ {\begin{array}{cc} {-1/2} & 0 \\ 0 & -1/2 \\ \end{array}} \right] \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} -3w/4 \\ 3h/2 \end{bmatrix} = \begin{bmatrix} -x/2-3w/4 \\ 0 \end{bmatrix} \quad \Rightarrow \quad F_2 = (-\tfrac12 w, h) \\ \\ f_3:\quad\begin{bmatrix} x \\ y \end{bmatrix} &= \left[ {\begin{array}{cc} {-1/2} & 0 \\ 0 & -1/2 \\ \end{array}} \right] \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} 3w/4 \\ 3h/2 \end{bmatrix} = \begin{bmatrix} x/2+3w/4 \\ 0 \end{bmatrix} \quad \Rightarrow \quad F_3 = (\tfrac12 w, h) \end{aligned} $$
If the Sierpinski gasket chaos game with three rotations is played with the fixed points $(-A,B)$, $(A,B)$, and $(0,2B)$, then the corresponding trema construction would use $w = 2A$ and $h = B$.
General Case
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 hexagon for which the width is 3/2 the length of the base, the top is half 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 hexagon 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.
