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diagram shows a bisected equilateral triangle of side length = radius

The idea was to use an infinite series of triangles. The red then green then the... to get the area of this sector then the area of the circle is 16 times this. If it is a unit circle than area should equal Pi.

Here is the series I got using Pythagorean’s theorem , is it correct?

$$\begin{align} A&=3r^{2} + 12\sum_{ n=0}^{\infty}2^{n-1}x_{n}\left(1-\sqrt{r^{2}-\frac{x{_{n}}^{2}}{4}}\right),\\ x_{0}&=r\sqrt{2-\sqrt{3}} ,\\x_{n+1}&=\sqrt{2r^{2}-2r\sqrt{r^{2}-\frac{x{_{n}}^{2}}{4}}} \end{align}$$

So-for-a-unit-circle

$$\begin{align} \pi&=3 + 12\sum_{ n=0}^{\infty}2^{n-1}x_{n}\left(1-\sqrt{1-\frac{x{_{n}}^{2}}{4}}\right),\\ x_{0}&=\sqrt{2-\sqrt{3}} ,\\x_{n+1}&=\sqrt{2-2\sqrt{1-\frac{x{_{n}}^{2}}{4}}} \end{align}$$

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    $\begingroup$ Looks like Viete's formula, except you are starting with a triangle instead of a square. $\endgroup$ – Wojowu Jun 16 at 14:25
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    $\begingroup$ Welcome to MSE. Please type your posts using MathJax to format the math. $\endgroup$ – saulspatz Jun 16 at 14:48
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    $\begingroup$ Archimedes did it first! $\endgroup$ – herb steinberg Jun 16 at 16:15
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    $\begingroup$ "There was more imagination in Arquimedes' head than in Homer's" (Voltaire) $\endgroup$ – Piquito Jun 16 at 17:01
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    $\begingroup$ This is called the Method of Exhaustion, and as noted, it is thousands of years old. $\endgroup$ – Eric Lippert Jun 16 at 22:35
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Numerical calculation suggests that you have made an error (or that I have, of course.)

from math import sqrt, pi

xs = [sqrt(2-sqrt(3))]

def f(x):
    return sqrt(2-2*sqrt(1-x*x/4))

def a(x):
    return 1 - sqrt(1-x*x/4)

for _ in range(50):
    xs.append(f(xs[-1]))

answer = 3+sum(2**(n-1)*xs[n]*a(xs[n]) for n in range(50))

print("answer=", answer)
print((pi-3)/(answer-3))

produces the output

answer= 3.0117993877991496
11.999999999999849

Have you dropped a factor of $12$ before the sum, perhaps?

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  • $\begingroup$ Yes it seems that is the case $\endgroup$ – Bodi Osman Jun 16 at 17:13
  • $\begingroup$ Fixed it thank you so much $\endgroup$ – Bodi Osman Jun 16 at 17:14
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Note that, in $x_{n+1} =\sqrt{2-2\sqrt{1-\frac{x{_{n}}^{2}}{4}}} $ if $x_n = 2\sin(t) $ then

$\begin{array}\\ x_{n+1} &=\sqrt{2-2\sqrt{1-\frac{x{_{n}}^{2}}{4}}}\\ &=\sqrt{2-2\sqrt{1-\sin^2(t)}}\\ &=\sqrt{2-2\cos(t)}\\ &=2\sqrt{\dfrac{1-\cos(t)}{2}}\\ &=2\sin(\dfrac{t}{2})\\ \end{array} $

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    $\begingroup$ Excuse my ignorance, but what do you mean I should notethis? $\endgroup$ – Bodi Osman Jun 16 at 23:02
  • $\begingroup$ Just that your iteration divides the angle by 2, so it goes from an n-gon to a 2n-gon. $\endgroup$ – marty cohen Jun 17 at 18:27
  • $\begingroup$ Ah yes thank you $\endgroup$ – Bodi Osman Jun 17 at 19:34

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