Derivation of Π

The mathematical constant Π (pi) can be derived in many different ways, but here I will present my favorite. The method was first described by none other than the mathematical genius, Archimedes, who lived in the 2nd century B.C.!!
The knowledge contained herein requires none other than an understanding of basic high-school mathematics. Archimedes’ method is so genius because it is so simple. I will take you step-by-step in this derivation, using geometric figures as a reference. Here we go!

First, what is Π?

Π is a mathematical constant that seems to show up in a variety of seemingly unrelated areas of mathematics, although it is best described as the ratio between the circumference and the diameter of a circle. In other words:

Π = Circumference/diameter


How can we calculate the value of Π using Archimedes’ method?

Archimedes envisioned the fact that as the number of sides (n) of a polygon increases, the more it approaches the configuration of a circle. Imagine increasing the sides of a polygon from 4 (a square), to 5 (a pentagon), to 6 (a hexagon), to 8 (an octagon), and so on, to infinity. Therefore, a circle can be thought of as a polygon with an infinite number of sides. Even a polygon with 32,000 sides will appear to the naked eye to look like a circle, although it is not. Thus, if we can calculate the perimeter of a polygon with a large number of sides, we can therefore approximate the circumference of the corresponding circle. Knowing the circumference, calculating the value of Π is a piece of cake! Let’s try this method using a hexagon (a polygon with 6 sides), and then we can derive a general formula to approximate Π using polygons with any number (n) of sides.
Here is a hexagon:

Now, let us circumscribe a circle around this hexagon:

I suppose you can now envision that as we increase the number of sides (n) of this polygon, the perimeter of the polygon approaches that of the circumference of the circle.



Now, let us draw 2 lines, both starting from the exact center of both the circle and hexagon. These two lines will both end at two consecutive vertices of the hexagon. The two lines are equal in distance. We will call both of these lines “r”, as they also represent the radius of the circle:

Now, let us draw a line from the exact center of both the circle and hexagon to a point that bisects this side of the polygon. We will not name this line, as it will not be used in any calculations, but it will help you envision what will come next.
Notice that this new line creates 2 new right triangles which are equal in all corresponding sides, angles, and in area.
In bisecting this 1 side of the polygon, the new unnamed line also creates the right sides of the 2 triangles (i.e. the right side of the triangle is perpendicular to the bisecting line). With this bisection, we have also divided 1 side of the hexagon into two equidistant halves. We will call each of these halves “a”:

Thus, 1 side of the hexagon = 2a.



We subsequently describe the angle immediately opposite “a” as angle “x”:


Is there a way to determine the value of angle x? We can calculate the value of angle x as follows:

The larger angle 2x is simply equal to 1/6th of the complete 360° revolution of the entire hexagon, or 60°. Therefore, angle x is equal to 30°.


Now, let us assume that the radius (r) of this circle is equal to 1/2 (you will see why later). We can then describe the relationship between angle x, a, and r (1/2) using trigonometry (remember SOH-CAH-TOA!!) as follows:

This shows that r is equal to 2a, and therefore, the triangle created by the two sides r and 2a (2a is equal to 1 side of the hexagon) is an equilateral triangle, with all angles of this triangle also equal to 2x (60°).


Using the solved equation above for “a” and “2a”, we can determine the perimeter of the entire hexagon:

Remember that the perimeter of the hexagon is 3 when we assumed that the radius of the circle “r” was equal to 1/2. Note: of course this works no matter what we assume r to be, making r = 1/2 just makes Π easier to calculate. For example, if we assumed r = 1, then the perimeter of the hexagon would have been calculated as 6 instead of 3. All other proportions could have been considered, and the end result would always be Π.

Recall also that Π is defined as the ratio between the circumference and the diameter of a circle. The diameter of a circle is simply twice the radius:

Diameter of circle = 2r

Therefore, in utilizing Archimedes’ derivation of Π, an approximation of this mathematical constant can be determined by taking the ratio of the perimeter of the hexagon (which through this method approximates the circumference of our circle) to the diameter of the circle (2r):

Not a bad estimate of Π, considering we are only using a polygon with 6 sides (n = 6). As the number of sides (n) of the polygon increases, we will get closer and closer to the “true” value of Π. But can we derive a general formula that will approximate Π for a given value of n? Read on.



Let us go back to our circle and hexagon:

Remember how we calculated the angle 2x = 60°? We said that it could be easily calculated as 1/6 of the entire rotation of the 6-sided hexagon (1/6 of 360°). Let’s generalize this calculation:

Let us carry this calculation out to solve for 2a:

Dividing each side by 2:

Remember the following relationship (using trigonometry):

Nevermind what sin x is equal to, and assuming r = 1/2 again:

Remember that 2a is equivalent to one side of a polygon with n number of sides. Another way to say this is that 2a multiplied by n will give us the perimeter of a polygon with n sides. Thus, multiplying each side of the equation by n:

That’s it!! The formula to calculate the perimeter of any polygon with n number of sides (assuming r = 1/2).

Let’s try to calculate Π for a polygon with n = 10,000 sides (I’m not going to attempt to draw this polygon, but it has MANY more sides than our hexagon! To the naked eye, it would probably look like a circle):

perimeter = 3.141592601912665692979346419289

For radius = 1/2 (i.e. diameter = 1):

We have calculated Π correct to 30 decimal places!! Of course, using calculus, the “true” value of Π can be calculated with this method, assuming a polygon with an infinite numbers of sides n, which is really a circle. This sets up a limit, but that would require 11th grade mathematics, and I need to brush up on that a bit! Maybe for another time…..