Basel Problem - Area of $\frac 16$ of Circle with Radius $\sqrt{\pi}$.

Question

There are several proofs to the solution of the well-known Basel Problem, i.e. $$\sum_{n=1}^\infty \frac 1{n^2}=\frac {\pi^2}6$$

Is is possible to create a geometrical interpretation of this identity in the form of the area of $\frac 16$ of a circle with radius $\sqrt{\pi}$?

Answer 1

(For the best expierience- please use a compass)

I. An approximation for $\sqrt{\pi}$ : II. How to split a circle into 6 equal part:

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EDIT: Epilog and history:

I have start w/ some famous square roots (without a compass): On this way we can win by hand roots like $\sqrt{3}$; $\sqrt{10}$ since $=\sqrt{ 3^2+1^2}$

Note: $\sqrt{3}<\sqrt{\pi}$ and $\pi<\sqrt{10}$

or the golden ratio- we need add to $1$ with the compass the $\sqrt{5}$ (or vice versa) and split in the middle:

Answer 2

TOPIC: Area of $\frac{1}{6}$ of Circle with Radius $\sqrt{\pi}$

Last time I show how to get an easy approximation for radius via Pythagorean theorem: $$ (\sqrt{\pi})^2\geq1.7^2+0.5^2 $$ and have give a suggestion to do different- let me show for $\pi\leq\frac{22}{7}$ (by using intercept theorem for $\frac{\sqrt{7}}{7}$) and fully geometrical solving this time:

Answer 3

Topic: Radius $\sqrt{\pi}$

Today (and this gonna be my last post- unless I get requests) with the golden ratio:

$\phi=\frac{1+\sqrt{5}}{2}$ ; Property : $\phi^2=\phi+1$

I found this pythagorean relation: $$ (\sqrt{\pi})^2\leq\phi^2+(\frac{\phi}{4})^2+0.6^2 $$ Accuracy: $$$$ \begin{align} \frac{17}{16}\phi^2+0.36\approx3.141661\\ \pi\approx3.141593 \\ { For-comparison:} \frac{22}{7}\approx3.14\color{red}{2857} \\ \end{align}