General Trema Construction Details for Three Rotations
Suppose the fixed points are located at the points $F_1$, $F_2$, and $F_3$, where all three are fixed points for the rotation by 180° and $F_2F_3$ form a horizontal segment. (Any set of three fixed points can be rotated to such an orientation.)
The functions corresponding to the chaos game moves for these fixed points are $$ \begin{aligned} f_1(P) &= P + \tfrac32(F_1-P) = -\tfrac12P + \tfrac32F_1 \\ \\ f_2(P) &= P + \tfrac32(F_2-P) = -\tfrac12P + \tfrac32F_2 \\ \\ f_3(P) &= P + \tfrac32(F_3-P) = -\tfrac12P + \tfrac32F_3 \end{aligned} $$ Now consider what happens if the chaos game is played with points along line through $F_2$ and $F_3$ starting with $F_2$. Let $P_1 = F_2$ and let $P_2 = f_3(P_1)$ be the point when $P_1$ is moved halfway towards $F_3$ and rotated 180° around $F_3$. Continue constructing a sequence $P_n$ of points by alternating chaos game moves with $F_2$ and $F_3$, where $P_{2n} = f_3(P_{2n-1})$ and $P_{2n+1} = f_2(P_{2n})$ for $n \ge 1$. See the following diagram for the first 6 points.
This gives $$ \begin{aligned} P_{2n} &= f_3(P_{2n-1}) = -\tfrac12P_{2n-1} + \tfrac32F_3 \\ \\ P_{2n+1} &= f_2(P_{2n}) = -\tfrac12P_{2n} + \tfrac32F_2 = \tfrac14P_{2n-1} - \tfrac34F_3 + \tfrac32F_2 \\ \\ P_{2n+2} &= f_3(P_{2n+1}) = -\tfrac12P_{2n+1} + \tfrac32F_3 = \tfrac14P_{2n} - \tfrac34F_2 + \tfrac32F_3 \end{aligned} $$ Define new sequences $L_n = P_{2n-1}$ and $R_n = P_{2n}$ for $n \ge 1$. Then from the result above we have: $$ \begin{aligned} L_{n+1} &= \tfrac14L_n - \tfrac34F_3 + \tfrac32F_2, \quad L_1 = F_2 \\ \\ R_{n+1} &= \tfrac14R_n - \tfrac34F_2 + \tfrac32F_3, \quad R_1 = P_2 \end{aligned} $$
These are first-order linear recurrence relations whose solutions are $$ \begin{aligned} L_n &= 2F_2-F_3 + (F_3-F_2)\left(\frac14\right)^{n-1} \\ R_n &= 2F_3-F_2 - \frac12(F_3-F_2)\left(\frac14\right)^{n-1} \end{aligned} $$
Take the limit of both sequences to define $F$ and $C$ as $$ \begin{aligned} F &= \lim_{n \rightarrow \infty} L_n = 2F_2 - F_3 \\ C &= \lim_{n \rightarrow \infty} R_n = 2F_3 - F_2 \end{aligned} $$
Notice that $F_2$ and $F_3$ divide segment $FC$ into thirds since $|FC| = 3|F_3-F_2|$.
Playing the chaos game along the line through $F_1$ and $F_2$ and the line through $F_1$ and $F_3$ would give similar results for those fixed points.
We can use this to define the other four points as
$$
\begin{aligned}
A & = 2F_2-F_1 \\
D & = 2F_1-F_2 \\
& ----\\
B & = 2F_3-F_1 \\
E &= 2F_1-F_3
\end{aligned}
$$
Now we get
$$
\begin{aligned}
\overrightarrow{AB} &= 2F_3-2F_2\\
\overrightarrow{ED} &=F_3-F_2 \\
\overrightarrow{FC} &=3F_3-3F_2 \\
\end{aligned}
$$
So segments $AB$, $ED$, and $FC$ are parallel. In addition $|FC| = \tfrac32|AB|$ and $|ED| = \tfrac12|AB|$.
The points $ABCDEF$ thus form a hexagon whose width is 3/2 the length of the base, and whose top is half the length of the base.
Moveover, the image below shows a perpendicular line from $D$ to the base. Triangles $\triangle DQF_2$ and $\triangle DPA$ are similar. Since $|DA| = 3|F_2A|$, it must also hold by similarity that $DP = 3|QP|$, i.e. the height of the hexagon is three times the distance from the base to the width.
The following GeoGebra app shows three such fixed points and the hexagon formed by the calculations above. You can move the point $P$ around the fice edges of the hexagon, either manually by moving the point on the slider, or automatically by clicking on the Play button. As $P$ moves around the hexagon, the green points $f_1(P)$, red points $f_2(P)$, and blue points $f_3(P)$ will trace where these points would occur while playing the Sierpinski gasket chaos game with the point $P$.
Notice that as $P$ moves counterclockwise around the edges of the hexagon, all three points $f_1(P)$, $f_2(P)$, and $f_3(P)$ move clockwise. The points trace out three copies of the hexagon each scaled by 1/2, with all three hexagons rotated by 180°. Their boundaries form the lines used in the trema construction.
The following image summarizes the coordinates of the vertices of the hexagon in terms of the coordinates of the fixed points, as well as the trema construction for the hexagon. Points $M$ and $N$ are the quarter points of the bottom edge, and $G$ and $H$ are quarter points of segments $CD$ and $EF$.
We use these coordinates to determine the following: $$ \begin{aligned} \overrightarrow{CD} &= 2(F_1-F_3) \\ \overrightarrow{NH} &= \tfrac52(F_1-F_3) \\ \overrightarrow{AF} &= F_1-F_3 \end{aligned} $$ Therefore $CD$, $NH$, and $AF$ are parallel, and $|CD| = 2|AF|$.
A similar calculation shows that $FE$, $MG$, and $BC$ are parallel, and $|FE| = 2|BC|$. These observations are what were used in the proofs for an isosceles hexagon, and can be used here in the same way to show that the trema calculation for the hexagon divides the hexagon into three self-similar pieces scaled by 1/2 and rotated by 180°.
Use the following GeoGebra app to experiment with changing the locations of the three fixed points. As the points are moved, the three smaller pentagons will continue to maintain the self-similarity scaling of 1/2 determined by the three red dashed lines in the trema construction.
