# I found the value of PI on my own!

This note is an extension to my previous note. In that note, I asked the community to find the mistake in my solution yielding 0=1. In that note, this mistake was the multiplication of inifinty and 0 which is basically an indeterminate form. This note is a correct version of the previous note and my formula seems to give a good approximation for $$\pi$$.

For any $$n$$ sided regular polygon inscribed in a circle of radius $$r$$, our objective is to calculate the perimeter of the polygon.

The figure represents the given situation. $$O$$ is the center of the circle and $$AB$$,$$BC$$ are edges of a regular polygon. $$OL$$ and $$OM$$ are perpendiculars drawn on line segments $$AB$$ and $$BC$$ respectively. Since perpendiculars from center bisect the chord, thus we have,

$$BL = BM$$

$$OB = OB$$

$$\angle OLB = \angle OMB$$

Therfore by RHS Criteration of congruency, $$\Delta OLB \cong \Delta OMB$$.

Thus by CPCT, we have

$$\angle LBO = \angle OBM$$

For any regular polygon of $$n$$ sides, each angle is given by $$\frac{180(n-2)}{n}$$.

Therefore, $$\angle LBM = \frac{180(n-2)}{n}$$. Thus $$\angle LBO = \frac{180(n-2)}{2n}$$.

Clearly,

$$\frac{BL}{r} = cos(\frac{90(n-2)}{n}°)$$

$$\ BL = rcos(\frac{90(n-2)}{n}°)$$ and thus,

$$BC = 2BL = 2rcos(\frac{90(n-2)}{n}°)$$

Since it is a regular polygon,

Perimeter = $$(n)(BA) = 2nrcos(\frac{90(n-2)}{n}°)$$ $$.........................(1)$$

But as $$n$$ approaches infinity the polygon tends to coincide the circle in which it is inscribed. Thus in that case the perimeter of the polygon becomes equal to the circumference of the circle in which it is inscribed.

Thus, Comparing the perimeter of the inifinite-sided polygon with the circumference of a circle we get,

$$ncos(\frac{90(n-2)}{n}°)=nsin(\frac{180}{n}°) = \pi ......................... (2)$$

Let us take a reasonable approximation for $$n$$ which resembles infinity. Thus, let $$n=10^{18}$$

Plugging in the value of $$n$$ in the 2nd equation we get,

$$10^{18} \times sin(18 \times 10^{-17}°) = \pi$$ which yields us $$3.141592653589793238462643383279$$ using a scientific calculator which is actually $$\pi$$ upto 30 decimal places!

Basically, if you want a better approximation just increase the power of 10.

Note by Akshat Joshi
2 years ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

How did you determine a value for sin?

- 11 months, 2 weeks ago

This is the Archimedes's method of calculating $$π$$.

- 2 years ago

I seriously did not know that Archimedes used this method. You can check my previous notes, on how I started to find out Pi. In the previous note the mistake I did was multiplying infinity and zero and wrote it as zero giving me the wrong answer. Today, I realized that instead of taking an infinite value, I can actually use a large finite value.

- 2 years ago

- 2 years ago

I just found that Archimedes did not exactly use this method. Rather he had put polygons inside and outside the circle and compared the outside and inside perimeters to get a range in which Pi lies ( 3.1408 < π < 3.1429 ). And at that time, I feel that the concept of inifinity was not properly developed.

- 2 years ago

In fact you should read the comments of the video, which proves that this is not Archimedes' method.

- 2 years ago