Nested n-gons

Let $C_n$ denote the point about which all the regular $n$ -gons are rotationally symmetric (Exercise: Prove that this point is well-defined), let $V_{j,n}$ denote a vertex of $P_{j,n}$ , and let $V_{j+1,n}$ denote one of the adjacent sides' midpoints (which by definition is one of the vertices of $P_{j+1,n}$ ). Then $\overline{C_nV_{j+1,n}} \perp \overline{V_{j,n}V_{j+1,n}}$ , so by basic trigonometry, $\frac{C_nV_{j+1,n}}{C_nV_{j,n}} = \cos\left(\angle V_{j,n}C_nV_{j+1,n}\right) = \cos\frac{\pi}{n}$ so that since all the $n$ -gons are similar, $\frac{A_{j+1,n}}{A_{j,n}} = \left(\frac{C_nV_{j+1,n}}{C_nV_{j,n}}\right)^2 = \cos^2\left(\frac{\pi}{n}\right).$ This ratio doesn't depend on $j$ , so $A_{j,n}$ forms a geometric sequence with initial value $A_{0,n}$ and common ratio $r = \cos^2\left(\frac{\pi}{n}\right)$ . Since $|r|<1$ , $\mathcal{A}_n = \sum_{j=0}^\infty A_{j,n} = \frac{A_{0,n}}{1-\cos^2\left(\frac{\pi}{n}\right)} = \frac{A_{0,n}}{\sin^2\left(\frac{\pi}{n}\right)} = A_{0,n}\csc^2\left(\frac{\pi}{n}\right)$

That is, $\frac{\mathcal{A}_n}{A_{0,n}} = \csc^2\left(\frac{\pi}{n}\right) \implies \frac{\mathcal{A}_{100}}{A_{0,100}} = \csc^2\left(\frac{\pi}{100}\right) \approx \boxed{1013.545235564382986485}$

Let $h_{0}$ and $x_{0}$ be the height and a side of the initial $n-gon$

then

$x_{0} = 2r_{0}\sin(\dfrac{\pi}{n})$ and $h_{0} = r_{0}\cos(\dfrac{\pi}{n})$

$\implies h_{1} = h_{0}\cos(\dfrac{\pi}{n}) = \cos^{2}(\dfrac{\pi}{n})r_{0}, \:\ x_{1} = 2h_{0}\sin(\dfrac{\pi}{n}) = 2\cos(\dfrac{\pi}{n})\sin(\dfrac{\pi}{n})r_{0} = \sin(\dfrac{2\pi}{n})r_{0}$

$\implies h_{2} = h_{1}\cos(\dfrac{\pi}{n})r_{0} = \cos^{3}(\dfrac{\pi}{n})r_{0}, \:\ x_{2} = 2h_{1}\sin(\dfrac{\pi}{n}) = 2\cos^{2}(\dfrac{\pi}{n})\sin(\dfrac{\pi}{n})r_{0} = \sin(\dfrac{2\pi}{n}) \cos(\dfrac{\pi}{n})r_{0}$

$\implies h_{3} = h_{2}\cos(\dfrac{\pi}{n}) = \cos^{4}(\dfrac{\pi}{n})r_{0}, \:\ x_{3} = 2h_{2}\sin(\dfrac{\pi}{n}) = \sin(\dfrac{2\pi}{n}) \cos^{2}(\dfrac{\pi}{n})r_{0}$

and,

$h_{j - 1} = \cos^{j}(\dfrac{\pi}{n})r_{0}, \:\, x_{j - 1} = \sin(\dfrac{2\pi}{n}) \cos^{j - 2}(\dfrac{\pi}{n})r_{0}$

$\implies h_{j,n} = \cos^{j + 1}(\dfrac{\pi}{n})r_{0}, \:\, x_{j,n} = \sin(\dfrac{2\pi}{n}) \cos^{j - 1}(\dfrac{\pi}{n})r_{0}$

$\implies A_{j,n} = \dfrac{n}{2}\cos^{2j}(\dfrac{\pi}{n})\sin(\dfrac{2\pi}{n})r_{0}^2 \implies$

$A_{n} = \sum_{j = 0}^{\infty} A_{j,n} = \dfrac{n}{2}\sin(\dfrac{2\pi}{n})r_{0}^2\sum_{j = 0}^{\infty} \:\ (\cos^{2}(\dfrac{\pi}{n}))^{j} = \dfrac{\dfrac{n}{2}\sin(\dfrac{2\pi}{n})r_{0}^2}{\sin^{2}(\dfrac{\pi}{n})} = \dfrac{A_{0,n}}{\sin^{2}(\dfrac{\pi}{n})} \implies \boxed{\dfrac{A_{n}}{A_{0,n}} = \dfrac{1}{\sin^{2}(\dfrac{\pi}{n})}}$ $\implies \dfrac{A_{100}}{A_{0,100}} \approx \boxed{1013.54523557}$ .