Intuition why a combined inscribed and circumscribed polygon converge faster to $\pi$? I was playing around a little with approximating $\pi$ by calculating the perimeter of a regular polygon both inscribing and circumscribing a circle.
When using trigonometry then it can be shown that $\pi$ can be approximated using the perimeters of these polygons with $n$ sides as follows
$$
\pi_i(n) = n\,\sin\left(\frac{180^\circ}{n}\right),
$$
$$
\pi_c(n) = n\,\tan\left(\frac{180^\circ}{n}\right),
$$
with $\pi_i(n)$ and $\pi_c(n)$ the approximation of $\pi$ from an inscribing and circumscribing polygon respectively.
But $\pi_i(n)$ and $\pi_c(n)$ are always a lower and upper bound respectively for $\pi$. Therefore I thought that taking the average should give a better result, which indeed reduced the error. However the rate at which the error gets smaller as $n$ increases was the same as the previous two. When I started using different weights I found the following expression gives the best result
$$
\pi_w(n) = \frac23\,\pi_i(n) + \frac13\,\pi_c(n).
$$
Graphing the errors of these approximations gives:

The rate at which the error of $\pi_i(n)$ and $\pi_c(n)$ goes down seems to be proportional to $n^{-2}$, while the rate at which the error of $\pi_w(n)$ goes down seems to be proportional to $n^{-4}$.
These converges rates can be explained by looking at the Taylor series of each expression. However I am curious if this could also be explained with pure geometry.
 A: 
By symmetry, we can discuss the approximation of $\pi$ on an angular sector  $\theta=\tfrac{2\pi}{2n}$ of the inscribing and circumscribing polygons (see the figure). 
Denoting $H_i$ and $H_c$ the half-side of these polygons, 
\begin{align}
H_i&=R\sin\theta\\
H_c&=R\tan\theta
\end{align}
Their length are related to the lower and upper approximations of $\pi$ by
\begin{align}
\pi_i(n)&=2n\frac{H_i}{R}\\
\pi_c(n)&=2n\frac{H_c}{R}
\end{align}
They approximate the length arc of circle $IC$, noted $S$, which would give the exact value of $\pi$. The question is then to find the value of the weight $\alpha$ which
minimize the difference between the lengths of the arc and the weighted approximation: $\varepsilon=\alpha  H_i+(1-\alpha)H_c-S$ for $n\to\infty$.
For small angles $\theta$, the arc $IC$ can be approximated by a parabolic shape, with a radius of curvature $R$, its length being
\begin{align}
S&=\int_0^{H_I}\sqrt{1+\frac{x^2}{R^2}}\,dx\\
&\sim H_I+\frac{1}{2}\int_0^{H_I}\frac{x^2}{R^2}\,dx\\
&\sim R\sin\theta\left( 1+\frac{1}{6}\sin^2\theta \right)
\end{align}
The error is then
\begin{equation}
\varepsilon\sim R\left( \alpha\sin\theta+\left( 1-\alpha \right)\tan\theta- \sin\theta\left( 1+\frac{1}{6}\sin^2\theta \right)\right)
\end{equation} 
For $\theta\to 0$ (or $n\to\infty$), 
\begin{equation}
\varepsilon\sim \left( \frac{1}{3}-\frac{\alpha}{2} \right)\theta^3+\left( \frac{5}{24}-\frac{\alpha}{8} \right)\theta^5+O(\theta^7)
\end{equation} 
Taking $\alpha=\frac{2}{3}$ reduces the error: $\varepsilon\sim \frac{\theta^5}{8}+O(\theta^7)$.
\begin{align}
\frac{2}{3}\pi_i(n)+\frac{1}{3}\pi_c(n)-\pi\sim 2n\frac{1}{8}\left( \frac{\pi}{n} \right)^5=\frac{\pi^5}{4}n^{-4}
\end{align}
A more exact calculation, taking into account the exact equation of the circle ($H=R\left( 1-\sqrt{1-\frac{x^2}{R^2}} \right)$) shows that the error due to the parabolic approximation is also $\sim \theta^5$. The overall error for the $\pi$ approximation when $\alpha=2/3$ is thus $\sim n^{-4}$.
