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Let $f_n$ or $F_n$ be areas of the regular $n-$ polygon described to the unit circle or circumscribed. Show

$f_{2n}=\sqrt{f_nF_n}$ and $F_{2n}=\frac{2f_{2n}F_{n}}{f_{2n}+F_n}$

In the solutions there is only the fact that this follows from elementary geometric considerations.

I tried to explain the result by elementary geometry, but I did not succeed.

I have already searched the net for approaches. For example you can construct a $2n$ corner out of a $n-$ corner which is in a circle by drawing a point between two corners of a side.

To clarify what I mean I have added a sketch using the example of a 4 corner.

Why is the geometric mean of the area content of the outer and inner 4 corners now the area content of the inner $8$ corner?

I have no approach to the formula for the outer $2n$ corner.

How can I draw the $2n$ corner from the outer $n$ and why is the harmonic mean of the area of the inner $2n$ corner and the area of the inner $n$ corner?

I hope someone can help me.

gi

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Let $l_n$ and $L_n$ be the lengths of the sides of the regular $n$-polygon inscribed in, or circumscribed to, the unit circle. The inscribed polygon with $n$ sides is formed by $n$ equal triangles, with base $l_n$ and altitude $\sqrt{1-l_n^2/4}$. We have then: $$ f_n={n\over2}l_n\sqrt{1-l_n^2/4},\quad F_n={n\over2}L_n. $$ On the other hand $L_n:l_n=1:\sqrt{1-l_n^2/4}$, that is: $$ L_n={l_n\over\sqrt{1-l_n^2/4}},\quad\text{and}\quad F_n={n\over2}{l_n\over\sqrt{1-l_n^2/4}}. $$ The area of the inscribed polygon with $2n$ sides is obtained by adding to $f_n$ the area of $2n$ small triangles: $$ f_{2n}=f_n+2n\cdot{1\over2}{l_n\over2}\big(1-\sqrt{1-l_n^2/4}\big)= {n\over2}l_n=\sqrt{f_n\cdot F_n}. $$ For $F_{2n}$ one can do a similar reasoning: $$ l_{2n}^2=l_n^2/4+\big(1-\sqrt{1-l_n^2/4}\big)^2=2\big(1-\sqrt{1-l_n^2/4}\big), $$

$$ \tag{*} F_{2n}={L_{2n}^2\over l_{2n}^2}f_{2n}={1\over1-l_{2n}^2/4}f_{2n}= {2\over1+\sqrt{1-l_n^2/4}}\cdot{n\over2}l_n $$ and finally: $$ {1\over f_{2n}}+{1\over F_{n}}={2\over nl_n}\big(1+\sqrt{1-l_n^2/4}\big) ={2\over F_{2n}}. $$ EDIT.

To obtain the last result in $(*)$, we must substitute $l_{2n}^2=2\big(1-\sqrt{1-l_n^2/4}\big)$ and $f_{2n}={n\over2}l_n$ into $F_{2n}=(1-l_{2n}^2/4)^{-1}f_{2n}$: $$ F_{2n}=\left(1-{l_{2n}^2\over4}\right)^{-1}f_{2n}= \left(1-{2\big(1-\sqrt{1-l_n^2/4}\big)\over4}\right)^{-1}{n\over2}l_n= \left({1\over2}+{1\over2}\sqrt{1-l_n^2/4}\right)^{-1}{n\over2}l_n= \left({1+\sqrt{1-l_n^2/4}\over2}\right)^{-1}{n\over2}l_n= {2\over 1+\sqrt{1-l_n^2/4}}\cdot{n\over2}l_n. $$

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  • $\begingroup$ Can you explain the $1+...$ in the fraction ${2\over1+\sqrt{1-l_n^2/4}}\cdot{n\over2}l_n$? $\endgroup$ – RM777 Feb 23 at 13:47
  • $\begingroup$ @RM777 In ${1\over1-l_{2n}^2/4}$ I substituted $l_{2n}$ with its expression taken from the preceding equality. $\endgroup$ – Aretino Feb 23 at 14:52
  • $\begingroup$ I still don't understand the equation could you please elaborate on this step please? Your Claim is that ${1\over1-l_{2n}^2/4}f_{2n}= {2\over1+\sqrt{1-l_n^2/4}}\cdot{n\over2}l_n$-------- I have tried a couple times to calculate this result but I couldn't succeed ----- First I have sepecified $f_{2n}$ which must be according to first Formula $\frac{2n}{2}l_{2n}\sqrt{1-l_{2n}^2/4}$ Therefor we first get $\rightarrow \frac{1}{1-l_{2n}^2/4}f_{2n}=\frac{1}{1-l_{2n}^2/4}\cdot \frac{2n}{2}l_{2n}\sqrt{1-l_{2n}^2/4}$ I have continiued to manipulate the term to get your result …. $\endgroup$ – RM777 Feb 23 at 18:01
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    $\begingroup$ See my edited answer. $\endgroup$ – Aretino Feb 23 at 18:22
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    $\begingroup$ I used the first formula, previously proved, because I wanted to express everything as a function of $l_n$. $\endgroup$ – Aretino Feb 23 at 18:29
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The key idea is that a regular $n$ sided polygon is composed of $n$ isoceles triangles each having a common vertex at the center of the polygon. Also, each isoceles triangle is split by an altitude into two congruent right triangles. Thus, the area of the polygon is $2n$ times $T$ the area of one of the right trangles whose vertex angle is $t:=\pi/n$. Given a circle of radius $1$ there are two cases. If the polygon is inscribed inside the circle, then $T = \frac12 \sin(t)\cos(t).\,$ If the polygon is circumscribed outside the circle, then $T = \frac12 \sin(t)/\cos(t).$

Putting this together, we have $\,f_n = n\sin(\pi/n)\cos(\pi/n)\,$ and $\,F_n = n\sin(\pi/n)/\cos(\pi/n).\,$ The relations between $\,f_n,F_n\,$ and $\,f_{2n},F_{2n}\,$ is a consequence of simple trigonometric identities such as $\,1 = \sin^2(t)+\cos^2(t),\,$ but these are based on geometric relations such as the Pythagorean theorem. You also need the doubling formulas $\,\sin(2x)=2\cos(x)\sin(x),\,\,\cos(2x)=\cos^2(x)-\sin^2(x).$

Note that Archimedes used the perimeters of polygons and not their areas to approximate $\pi$.

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