Inscribed Circles

Let's make the height of the triangles (based on the sides of the $n$ -gon) equal to 1 for simplicity. Then the sides of the $n$ -gon are $2\tan\left(\frac{\pi}{n}\right)$ and the area of the triangles is $\tan\left(\frac{\pi}{n}\right)$ . Consider a right triangle with its hypotenuse connecting the center of a circle to an adjacent vertex of the $n$ -gon to see that the radius of the circles is $\tan\left(\frac{\pi}{n}\right)\tan\left(\frac{\pi}{4}-\frac{\pi}{2n}\right)$ .The limit we seek is $\lim_{n \to \infty}\frac{n\pi \tan^2\left(\frac{\pi}{n}\right)\tan^2\left(\frac{\pi}{4}-\frac{\pi}{2n}\right)}{\tan\left(\frac{\pi}{n}\right)}=\pi\lim_{n \to \infty}n \tan\left(\frac{\pi}{n}\right)=\pi^2\approx \boxed{9.86960440}$

A regular $n$ -gon is formed from $n$ congruent central isosceles triangles where $\frac {2\pi}n$ is the measure of the angle between the equal sides. Let the side length of the equal sides be 1, then the area of the central isosceles triangle, $A_\triangle = \frac 12 \sin \frac {2\pi}n$ .

Let $M$ be the midpoint of the base of the central isosceles triangle $PQ$ ; then $\angle POM = \angle QOM = \frac \pi n$ . Let also the center of the circle inscribed in the central isosceles triangle be $I$ and its radius be $r$ , and the altitudes from $I$ to $OP$ and $OQ$ be $IL$ and $IN$ respectively. Then we have:

$\begin{aligned} OM & = OI + IM \\ AO\cos \frac \pi n & = \frac {IL}{\sin \frac \pi n} + IM & \small \color{#3D99F6} \text{Note that } AO = 1 \text{ and } \\ \cos \frac \pi n & = \frac r{\sin \frac \pi n} + r & \small \color{#3D99F6} IL = IM = r \end{aligned}$

$\begin{aligned} \implies r & = \frac {\cos \frac \pi n}{\frac 1{\sin \frac \pi n}+1} = \frac {\sin \frac {2\pi}n}{2 \left(1+\sin \frac \pi n \right)} \end{aligned}$

Then the area of the incircle is $A_\bigcirc = \dfrac {\pi \sin^2 \frac {2\pi}n}{4 \left(1+\sin \frac \pi n\right)^2}$ and the required limit:

$\begin{aligned} \lim_{n \to \infty} \frac {nA_\bigcirc}{A_\triangle} & = \lim_{n \to \infty} \frac {n \pi \sin^2 \frac {2\pi}n}{4 \left(1+\sin \frac \pi n\right)^2} \times \frac 2{\sin \frac {2\pi}n} \\ & = \lim_{n \to \infty} \frac {n \pi \sin \frac {2\pi}n}{2 \left(1+\sin \frac \pi n\right)^2} \\ & = \lim_{n \to \infty} \frac {\pi^2 \sin \frac {2\pi}n}{\frac {2\pi}n} \times \lim_{n \to \infty} \frac 1{\left(1+\sin \frac \pi n\right)^2} \\ & = \pi^2 \approx \boxed{9.86960440} \end{aligned}$

Let $OP_j=1$

Notice that $\displaystyle\lim_{n\to\infty}nT_j(n)=\pi$ (equal to a unit circle)

And $A_j(n)=r_I^2\pi$ , $r_I$ is the inradius of $\triangle OP_jP_{j-1}$

So $r_I={2T_j(n)}{OP_j+OP_{j-1}+P_jP_{j-1}}=\frac{2T_j(n)}{2+P_jP_{j-1}}$

$\displaystyle\lim_{n\to\infty}\frac{A(n)}{T_j(n)}$

$=\displaystyle\lim_{n\to\infty}\frac{nA_j(n)}{T_j(n)}$

$=\displaystyle\lim_{n\to\infty}\frac{n^2A_j(n)}{nT_j(n)}$

$=\displaystyle\lim_{n\to\infty}\frac{n^2A_j(n)}{\pi}$

$=\displaystyle\lim_{n\to\infty}\frac{n^2r_I^2\pi}{\pi}$

$=\displaystyle\lim_{n\to\infty}(nr_I)^2$

$=\displaystyle\lim_{n\to\infty}(\frac{n2T_j(n)}{2+P_jP_{j-1}})^2$

$=\displaystyle\lim_{n\to\infty}(\frac{n2T_j(n)}{2+0})^2$

$=\displaystyle\lim_{n\to\infty}({nT_j(n)})^2$

$=\pi^2$

$\underset{n\to \infty }{\text{lim}}\frac{\pi n \left(\frac{\sin \left(\frac{2 \pi }{n}\right)}{2 \left(\sin \left(\frac{\pi }{n}\right)+1\right)}\right)^2}{\frac{1}{2} \sin \left(\frac{2 \pi }{n}\right)} \Longrightarrow \pi^2$

$\dfrac{x}{2} = r\sin(\dfrac{\pi}{n}) \implies h = r\cos(\dfrac{\pi}{n}) \implies T_{j}(n) = \dfrac{1}{2}\sin(\dfrac{2\pi}{n})r^2$ for each triangle $P_{j - 1}OP_{j}$ .

and,

$T_{j}(n) = rR + \dfrac{1}{2}xR = \dfrac{R}{2}(2r + x) \implies R = \dfrac{2T_{j}(n)}{2r + x} =$ $\dfrac{\sin(\dfrac{2\pi}{n})r^2}{2r + 2r\sin(\dfrac{\pi}{n}))} = \dfrac{\sin(\dfrac{2\pi}{n})r}{2(1 + \sin(\dfrac{\pi}{n}))}$

$\implies C_{j}(n) = \dfrac{\pi\sin^2(\dfrac{2\pi}{n})r^2}{4(1 + \sin(\dfrac{\pi}{n}))^2}$

$\implies A(n) = \dfrac{\dfrac{\pi}{2}(\dfrac{n}{2}\sin^2(\dfrac{2\pi}{n})r^2)}{(1 + \sin(\dfrac{\pi}{n}))^2}$

and,

For fixed $j$ we have: $\dfrac{A(n)}{T_{j}(n)} = \dfrac{\pi(\dfrac{n}{2}\sin(\dfrac{2\pi}{n}))}{(1 + \sin(\dfrac{\pi}{n}))^2}$ .

$\lim_{n \rightarrow \infty} (1 + \sin(\dfrac{\pi}{n}))^2 = 1$

and,

For $\pi\lim_{n \rightarrow \infty} \dfrac{n}{2}\sin(\dfrac{2\pi}{n})$ :

Using the inequality: $\cos(x) < \dfrac{\sin(x)}{x} < 1 \implies \cos(\dfrac{2\pi}{n}) < \dfrac{\sin(\dfrac{2\pi}{n})}{\dfrac{2\pi}{n}} < 1 \implies\pi\cos(\dfrac{2\pi}{n}) < \dfrac{n}{2}\sin(\dfrac{2\pi}{n}) < \pi$

and $\pi\lim_{n \rightarrow \infty} \cos(\dfrac{2\pi}{n}) = \pi \implies \pi\lim_{n \rightarrow \infty} \dfrac{n}{2}\sin(\dfrac{2\pi}{n}) = \pi * \pi = \pi^2$

$\implies \lim_{n \rightarrow \infty} \dfrac{A(n)}{T_{j}(n)} = \boxed{\pi^2 \approx 9.86960440}$ .