# Circle measurement of Archimedes

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.

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 $$F_{2n}=(1-l_{2n}^2/4)^{-1}f_{2n}$$: $$F_{2n}=\left(1-{l_{2n}^2\over4}\right)^{-1}f_{2n}= \left(1-{2\big(1-\sqrt{1-l_n^2/4}\big)\over4}\right)^{-1}{n\over2}l_n= \left({1\over2}+{1\over2}\sqrt{1-l_n^2/4}\right)^{-1}{n\over2}l_n= \left({1+\sqrt{1-l_n^2/4}\over2}\right)^{-1}{n\over2}l_n= {2\over 1+\sqrt{1-l_n^2/4}}\cdot{n\over2}l_n.$$

• Can you explain the $1+...$ in the fraction ${2\over1+\sqrt{1-l_n^2/4}}\cdot{n\over2}l_n$? – RM777 Feb 23 at 13:47
• @RM777 In ${1\over1-l_{2n}^2/4}$ I substituted $l_{2n}$ with its expression taken from the preceding equality. – Aretino Feb 23 at 14:52
• 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 …. – RM777 Feb 23 at 18:01
• See my edited answer. – Aretino Feb 23 at 18:22
• I used the first formula, previously proved, because I wanted to express everything as a function of $l_n$. – Aretino Feb 23 at 18:29

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 $$\pi$$.