Going through Archimedes' approximation to π using the method of inscribed and circumscribed polygons, one comes upon his recursive formulae for their perimeters. Representing the perimeter of the circumscribed and inscribed polygons with $n$ sides as $P_n$ and $p_n$, respectively, the recursion is:
$$
P_{2n} = \dfrac{2 p_n P_n}{p_n + P_n} \qquad
p_{2n} = \sqrt{p_n P_{2n}}
$$
It is relatively straightforward to prove these by algebraic manipulation of the double angle formulae into the following forms:
$$
\tan\alpha = \frac{\tan{2\alpha} \sin{2\alpha}}{\tan{2\alpha} + \sin{2\alpha}} \qquad
\sin\alpha = \frac{1}{\sqrt 2} \sqrt{\tan\alpha \sin{2\alpha}}
$$
However, each of these formulae is a statement about means of certain quantities:
- $P_{2n}$ is the harmonic mean of $p_n$ and $P_n$.
- $\tan\alpha$ is half the harmonic mean of $\tan{2\alpha}$ and $\sin{2\alpha}$.
- $p_{2n}$ is the geometric mean of $p_n$ and $P_{2n}$.
- $\sin\alpha$ is $\frac{1}{\sqrt 2}$ times the geometric mean of $\tan\alpha$ and $\sin{2\alpha}$.
Given that geometric reasoning was much more important is Archimedes' time – and algebraic manipulation much more rudimentary – I suspect there's a geometric way of directly seeing these facts, and this is how Archimedes would have proven these formulae. What is a diagram we could draw which would make this clear? In particular, is there a diagram which would make the appearance of the geometric and harmonic means seem natural, for example through an application of the geometric mean theorem or one of the many appearances of the harmonic mean.
Aside: there is a very similar recursive formula for the areas of the circumscribed and inscribed polygons:
$$
A_{2n} = \dfrac{2 a_{2n} A_n}{a_{2n} + A_n} \qquad
a_{2n} = \sqrt{a_n A_n}
$$
I imagine that a diagram demonstrating the formulae for the perimeter would be fairly easily adapted to show those for the areas, but if it's more convenient you are welcome to demonstrate this one instead.
Best Answer
Define $Q_n := \frac{1}{2n}P_n$ and $q_n := \frac{1}{2n}p_n$ as the half-sides of circumscribed and inscribed $n$-gons. We'll demonstrate the target relation translated to these values: $$Q_{2n}=\frac{Q_n q_n}{Q_n+q_n} \qquad\qquad 2q_{2n}^2= Q_n q_n \tag{1}$$
Let $A$, $B$, $C$ be consecutive vertices of the circumscribed $n$-gon; then midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{BC}$ (the points of tangency with the circle) are consecutive vertices on the inscribed $n$-gon; thus, $|BM|=|BN|=Q_n$. Let $L$ be the midpoint of $\overline{MN}$, so that $|ML|=|NL|=q_n$. Let $D$ be the point where $\overline{BL}$ meets the circle, $E$ be the point where the tangent at $D$ meets $\overline{BN}$, and $F$ be the midpoint of $DN$ (necessarily the altitude of $\triangle DEN$), so that $|DN|=Q_{2n}$ and $|DF|=q_{2n}$.
For the $Q_{2n}$ relation, first observe that $\angle NMD$ and $\angle BMD$ are inscribed angles (the latter being somewhat honorarily "inscribed" as the angle between a tangent and a chord) subtending congruent arcs $\overset{\frown}{MD}$ and $\overset{\frown}{DN}$, so that they themselves are congruent (by the Inscribed Angle Theorem); therefore, $\overline{MD}$ bisects $\angle BMN$, and so it cuts $\overline{BL}$ in proportion $|MB|:|ML|$ (by the Angle Bisector Theorem). Also, $\overline{DE}\parallel \overline{LN}$, so that $\triangle BDE\sim \triangle BLN$. From these facts, we can write $$\frac{|DE|}{|LN|} = \frac{|BD|}{|BD|+|DL|} = \frac{|MB|}{|MB|+|ML|} \quad\to\quad \frac{Q_{2n}}{q_n}=\frac{Q_n}{Q_n+q_n} \tag{2}$$ For the $q_{2n}$ relation, simply observe that $\triangle NDL \sim \triangle DEF$, giving $$\frac{|ND|}{|NL|}=\frac{|DE|}{|DF|} \quad\to\quad \frac{2q_{2n}}{q_n}=\frac{Q_{2n}}{q_{2n}} \tag{3}$$ Clearly, $(2)$ and $(3)$ are equivalent to $(1)$. $\square$
For now, I'll leave the area results as an exercise to the reader.