Question 3

3. Given a circle of radius one, find the length of a side of an inscribed, regular 24-gon by using the length of a side of an inscribed regular 12-gon (like we did in class). Use this length to approximate π by providing a fraction (square roots are okay, but decimals are not) that is an under-estimate of π.

Other than using calculus, the value of π can be approximated using polygons. Traditionally, Archimedes, who established this method of approximation used polygons inscribed and circumscribed around a circle to establish an lower and upper bounds on the value of pi. But, in this post I will only approximate pi using inscribed polygons. So, how does this approximation work? First, let's find the approximation of π using hexagon and then a 12-gon before we get to a 24-gon.

First, make a circle:

 Then I draw a hexagon inside the circle.

 

After the hexagon is drawn, I can draw a line across the circle that cuts through the hexagon and connects two opposite corners. This line that cuts through the circle is the diameter.

I can then finish "cutting" the hexagon by connecting the last two pair of opposites like this picture. From the previous picture, we can see that all of these "cuts" is the circle's diameter. It can be anything, but in this problem, I will assume that the diameter is 2 and because  radius = diameter / 2, r = 2/2 = 1. Because the triangles created are equilateral triangles we know that all the sides of the triangles and the hexagon is equal to the radius which is equal to 1.
We can then approximate pi using this:

C = 2πr

π = C / 2r

But, we can't find the area without using pi, so we will approximate it using the perimeter of the hexagon.

The perimeter of the hexagon is: the number of the sides x the length of the sides, in this case it's:

Perimeter = 6 x 1 = 6.

So,

π is approximately = 6 / 2(1) = 6/2 = 3

Now, to improve this approximation let's add more lines and cut each of the triangles in half to prepare and make a 12-gon(like the picture below).

Then, connect the line in the middle to the corners to make this picture below. I've made the 12-gon.

 
Here, in this picture, I take 1 slice of the equilateral triangle from before out so that it is easier to see the length of the sides to this 12-gon.This picture shows the variables of all the things we are missing. N is a length of one side of the 12-gon, this is what I ultimately want to find because with this length I can find the perimeter of the hexagon and use it. However, first, I'd have to find the length of m, and the only way to do that is find the length of L.

Because L + m = 1 (it is the radius), once I find L then 1 - L = m.

 
So, to make this problem even easier, I've split the two triangles and made them larger so it is easier to see. This triangle is the one in which we will use to determine L. We know that r is the radius and so it is 1. We then know that h = 1/2r (because the equilateral triangle from the hexagon was cut into 2) so, in this case, h = ½.

Then we use the pythagorean theorem:

½² + L² = 1²

¼ + L² = 1

L² = 1 – ¼

L² = ¾

L = √3/2

Then, to find n:

First, we find m: L + m = 1, so m = 1 -  √3/2 = (2-√3)/2

I know h = ½

So, let's use the pythagoren theorem:

½² + (2-√3/2)² = n²

¼ + (4-4√3+3)/4 = n²

                                           (8-4√3)/4 = n²

                                           2-√3 = n²

                                           √(2-√3) = n

Now that I have n, which is the length of one side of the 12-gon, I can find the perimeter of the 12-gon to approximate pi:

Perimeter = 12 x √(2-√3) = 12√(2-√3)

I can then approximate pi using this:

C = 2πr
Assuming C is approximately the perimeter of the 12-gon:

π = Perimeter / 2r = 12√(2-√3)/2 = 6√(2-√3) = 3.10582


Then, to approximate pi from a 24-gon, cut the triangles of the 12-gon in half and repeat the procedure again. The result will look like this:

Then I split the triangles like before, so that I can determine the length of one side:
After a similar math process to the 12-gon, I determine that one side of the 24-gon is:
√(2-(√2+(√3)))
So, to approximate pi, I then need to find the perimeter of the 24-gon
Perimeter = 24 x √(2-(√2+(√3))) = 24√(2-(√2+(√3)))
I can then approximate pi using this:

C = 2πr
Assuming C is approximately the perimeter of the 24-gon:

π = Perimeter / 2r = 24√√(2-(√2+(√3)))/2 = 12√(2-(√2+(√3))) = 3.1326286

Bonus - Go all the way to 96. This makes you current with Archimedes.

The 24-gon is then cut again to create a 48-gon, and I can use the previous knowledge of the previous length to figure out that the length of one side of it is:

 √(2-(√2+(√2+(√3))))

Then, knowing this, I can "cut"the triangles in the 48-gon in half again to create a 96-gon. And, using the information from the previous polygon, I determined the length of one side of the 96-gon to be:

 √(2-(√2+(√2+(√2+(√3)))))

Then, to approximate pi, I have to first find the perimeter of this polygon:

Perimeter = 96 x  √(2-(√2+(√2+(√2+(√3))))) = 96 √(2-(√2+(√2+(√2+(√3)))))

I can then approximate pi using this:

C = 2πr
Assuming C is approximately the perimeter of the 24-gon:

π = Perimeter / 2r = 96 √(2-(√2+(√2+(√2+(√3)))))/2 = 48√(2-(√2+(√2+(√2+(√3))))) = 3.14103195