An A(PI)roximation

Topic:Circle, Constructions, Geometry, Polygons, Unit Circle

Easy as pie

Even from the earliest geometry classes, students learn about a never ending number called pie, no wait, pi. Most people know pi as 3.14, some may even know a few more digits. A good mathematician will say pi is the proportion of circumference to diameter of a circle.

But how do we know what pi is? If we need pi to calculate circumference and need circumference to calculate pi, how do we get 3.14?

Well, we can start by trying to approximate the circumference of the unit circle. But how can we do it. Let's start with a perimeter of a shape we can calculate easily: an equilateral triangle. First, lets inscribe an equilateral triangle into the unit circle

Equilateral Triangle Construction

Construction of Triangle CDE

The length from A to B is 1
We first swing the two circles centered at A and B passing through B and A respectively, giving us 2 unit circles.
That gives us two points C and D that are equidistant from the line AB.
Then, we extend the line AB to intersect with the opposite side of the unit circle around A. This gives us a new intersection point E. We also know that EA must be length 1 since it is a radius of the unit circle.
Now if we connect the points ECD we get a triangle.

Claim Triangle CDE is an equilateral triangle:

Equilateral Triangle Proof

We know that the triangle ABC is equilateral since each side must be 1 by construction.
Therefore bisection created by the line CF creates two 30-60-90 special triangles. With the angle AFC being a 90 degree angle.
That tells us that CF must be by the 30-60-90 special triangle. Furthermore, CD
We also know that the line AF must be by the same special triangle.
Therefore EF must be
Therefore we know by Pythagorean's theorem we have EC
By symmetry ED
Therefore since all 3 sides are equivalent, the triangle is equilateral.

Now that we have the side lengths of the triangle, we can calculate the perimeter

If we use this perimeter as an approximation for the circumference of the circle we will get

Our approximation is 17% inaccurate. Let's try adding an extra side and make a inscribed square.

Square Construction

Construction of Square CGBH

The length from A to B is 1
We first swing the two circles centered at A and B passing through B and A respectively, giving us 2 unit circles.
Then, we extend the line AB to intersect with the opposite side of the unit circle around A. This gives us a new intersection point C. We can note that the line AC is also of length 1.
Now swing another circle around C through A. Since the line segment AC is length 1 the orange circle is also a unit circle.
Now mark the intersection of the circle around A with the circles around C and B, call these points D and E.
Now swing a circle around B that passes through E and a circle around C that passes through D. Note that CD is equal to BE by symmetry. Or in other words, the two red circles have the same radius.
This creates an intersection point above the circle, F
Now draw a line from F to A. Note that the line FA is perpendicular to the line CB. This is true by the symmetry across the line FA and the symmetry across CB.
Then mark the intersection points of the circle around B with the line AF, call them G and H.
Then connect the points C, G, B, and H

Claim quadrilateral CGBH is a square:

Square Proof

We can note that AG and AB are both radii of the unit circle around A.
We also know by symmetry, the angle GAB is a 90 degree angle.
That makes the triangle AGB is a 45-45-90 triangle.
Then by symmetry both the triangle CAG and ABH must both be 45-45-90 triangles.
Lastly, this forces the triangle CAH is a 45-45-90 triangle.
Therefore each of the sides of the quadrilateral are equal to and each of the angles of the quadrilateral are 90 degrees
Therefore the quadrilateral is a square.

Since we know that the inscribed square has side length

, we have a perimeter equal to

This gives us a pi approximation equal to

Better, but still a ways off from 3.14... If we continue adding more sides to our inscribed polygon we can see that the approximated circumference gets closer and closer to being the circle itself. Or in other words, the perimeter of the polygon will become the circumference.

Here's The Point

So...

As the number of sides in the inscribed polygon increases the closer the perimeter will get to

. Archimedes created this proof in 200BC. To be able to calculate the perimeter of an n-gon, he calculated the length of a single side using the following geometric principles

This sketch illustrates one side of an arbitrary n-gon in green. So, we are trying to find a general solution for the length BC. Let's call this length

1. By connecting the center to the points B and C we create a isosceles triangle ABC with the two radii. 2. Then we bisect the line BC to create two symmetric triangles ABD and ACD. 3. Now, we draw the symmetric lines BE and CE in red to create two new triangles. An important characteristic that Archimedes noticed was that for every one green length we have two red lengths. In other words, if the green segments make an n-gon, the red lengths make a 2n-gon. That tells us that the length

is related to the n-gon and the 2n-gon.This directly results from the bisection of the inscribed angle. We 'divide' the n sides of the n-gon into 2n sides. So, Archimedes hoped to find a recurrence relation in which we can express the length of the 2n-gon in terms of the length of the n-gon. First, lets find an expression for the side length of an n-gon. Recall we called this length

. Using Pythagoras' Theorem, we get:

(Expr 1)

We also have:

(Expr 2)

So if we substitute Expr 2 into Expr 1 and rearrange terms we get the formula:

(Expr 3)

If we notice that the AE is a radius, we can generate the expression:

(Expr 4)

Now, if we substitute Expr 3 into Expr 4, we have:

(Expr 5)

Next, let's find an expression for the side length of the 2n-gon. Let's call this length

. Using Pythagoras' Theorem, we get:

(Expr 6)

Let's substitute more terms:

(plug in Expr 2) (plug in Expr 5) (by definition)

So now if we simplify the above expression we get:

So now we have an expression of

in terms of

. Or in other words, we have a formula that relates the length of the n-gon to the length of the 2n-gon which is what we were looking for!

A few iterations

Success!

We can see just after 8 iterations we have an approximation of pi with an accuracy of 0.0001%. You can play with how many points you want to plot using the applet above. See for yourself that the perimeter approximation will never cross the red-dotted line (y=pi). As an extra challenge, you can try to prove that the recurrence relation indeed converges to pi. The trick is to use trig substitution.