Finding a way to describe the position of vertices of $n$ nested hexagons

Question

I am trying to find a way to describe precisely the points of $n$ nested regular hexagons when they are in the same position i.e. essentially there are $3$ lines (diagonals) with the total of $2n$ points on each. I want the following to hold:

1. the red lines given in the image should be parallel (even though its only shown for one side, this should symmetrically hold for other parts of the hexagon as well). Now, by this I mean that the slope created by the bottom right vertex of the innermost hexagon and the right top vertex of the outermost hexagon create a slope that is equal to the slope of the other red lines.

2. The lines with the $||$ notation would also be parallel. This should again hold for other parts of the hexagons given the symmetry.

Is there a way to describe the position of all the points in the hexagon by this logic? Or would we need to know something more. Without the loss of generality, I am assuming that in $\mathbb{R}^2$ the vertices of the innermost hexagon are $(2,0),(-2,0),(-1,\sqrt{3}),(-1,-\sqrt{3}),(1,\sqrt{3}),(1,-\sqrt{3})$.

Answer 1

Assuming the hexagons are to be "equally spaced" ...

Consider the inner-most hexagon to have center $O$ and "radius" $r$. (OP's example uses $r=2$.) Let $R$ be its bottom-right vertex and $S'$ its right-most. We extend $r$ by $n-1$ copies of length $s$, obtaining hexagons of radii $r+s$, $r+2s$, $\ldots$, $r+(n-1)s$. Let the bottom-right vertex of the $r+s$ hexagon be $R'$, and the top-right vertex of the $r+(n-1)s$ hexagon be $S$.

The desired parallelism property is merely a requirement that $\triangle OSR$ and $\triangle RS'R'$ be similar. Setting sides proportional, we have $$\frac{r+(n-1)s}{r}=\frac{r}{s} \quad \to\quad (n-1)s^2+rs-r^2=0 \tag{1}$$ Solving for $s$ (and discarding an extraneous negative root) gives $$s = r\;\frac{-1 + \sqrt{4n-3}}{2 (n-1)} \tag{2}$$ so that the outermost hexagon has radius $$r+(n-1)s = \frac{r}{2}\left(\,1 + \sqrt{4n-3}\,\right) \tag{3}$$ In the example, $n=4$ (for a total of $4$ hexagons), and $s=\frac{r}{6}(-1+\sqrt{13})\approx 0.434259 r$.

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Fun Fact. For $n=2$, we have $r=s\phi$, where $\phi:=1.618\ldots$ is the Golden Ratio.