Archimedes' Approximation of Pi

It is usually said that the calculus was invented by Isaac Newton (1642-1727) and Gottfried Wilhelm Leibniz (1646-1716) during the decade 1665-1675, but this is not entirely accurate. The ideas of calculus evolved over many centuries. The central concept behind calculus to use a limiting process to derive arithmetic results goes back to the Greeks. Archimedes of Syracuse (ca. 287-212 B.C.) was one of the first to use the limit concept to find the area and volume of various planar shapes and solids.

Elementary geometry was already used long before to find the perimeter and area of any triangle and therefore also of any polygon. A polygon is a closed planar shape consisting of straight lines. But for curved figures other concepts are needed.

The problem of determining the area of a circle was once considered a great mathematical challenge. This problem was called "squaring the circle": i.e. trying to find the square that has the same enclosed area as a circle of a given radius.

The perimeter $P(d)$ of a square is $P(d)=4d$ where $d$ is its diameter and $4$ a constant ratio. The perimeter of a circle, which is normally called its circumference is $C(d)=\pi d$ with the number $\pi$ a constant ratio. For a unit circle ($d=1$): $C(d)=\pi$. The area of a circle could be thought of as the area of infinitely small triangles each with a height $r=\frac{1}{2}d$ and with the total base equal tot the circumference of a circle. Then we could calculate the area of a circle as the sum of the areas of these many triangles as follows $A=\frac{1}{2} b h = \frac{1}{2}\pi d r = \pi r^2$. For a unit circle this gives: $A=\pi$. Finding this number $\pi$ was a great challenge and is mathematically the same thing as finding the circumference or area of a unit circle.

An Egyptian text dating to 1650 B.C. known as the Rhind Papyrus (named after the Scottish Egyptologist A. Henry Rhind) contains the statement that a circle has the same area as a square whose side is $\frac{8}{9}$ the diameter of the circle. That means that $\pi=\frac{256}{81}\approx 3.16049$, which is a fairly accurate estimation to its true value $\pi=3.14159$ rounded to 5 decimal places.

Area of circle from Rhind Papyrus

All these acient methods were based on measuring. Archimedes however took a different approach. He proposed a method that could give the value of $\pi$ to any desired accuracy by an iterative algorithm-rather than by measurement. Archimedes determined that the true value of $\pi$ lies between $3\frac{10}{71}$ and $3\frac{1}{7}$. The Works of Archimedes have been translated from Greek by T. L. Heath and published in 1897 and can be found here. The approach of Archimedes was based on algebra and geometry. An explanation can be found here. We will follow here an approach based on trigonometry which is less elaborate.

The method developed by Archimedes consists of:

The essential step in the above method is to find the iterative formulas for circumscribed and inscribed polygons. For this Archimedes used a result from geometry described by Euclid in Proposition 3 of Book VI of Euclid's Elements.


If an angle of a triangle be bisected and the straight line cutting the angle cut the base also, the segments of the base will have the same ratio as the remaining sides of the triangle; and, if the segments of the base have the same ratio as the remaining sides of the triangle, the straight line joined from the vertex to the point of section will bisect the angle of the triangle


We now proceed with how Archimedes used this result to find an iterative relationship. First he analyzed the case of circumscribed polygons. He drew a circle with center $O$ and radus $AO$ with a line perpendicular to $AO$ and through the point $C$. The $\angle AOC$ was set to $30^\circ$. The line $AC$ represents half the lenght of the side of an hexagon. Bisection of $\angle AOC$ results in a twelve sided polygon of which $AD$ is half the lenght of one side. See the picture below.

Archimedes found a relationship between the ratio $AO:AC$ of the hexagon and $AO:AD$ of the twelve sided polygon. \begin{align*} \frac{CO}{AO} &= \frac{CD}{AD} \text{ based on Euclid } \to \\ \frac{CO+AO}{AO}&= \frac{CD+AD}{AD} =\frac{AC}{AD} \\ \frac{AO}{AD} &=\frac{CO+AO}{AC}=\frac{AO}{AC}+\frac{CO}{AC} \\ &\text{applying Pythagoras gives: } \\ &=\frac{AO}{AC}+\sqrt{\frac{AO^2}{AC^2}+1} \end{align*}

Archimedes thus found that the ratio of the radius to half the length of the side of a 12-sided polygon depends on the ratio of a hexagon. Let us define the following recursive relations for the fractions: \begin{align} \xi_0&=\frac{AO}{AC} \notag \\ \xi_n &= \xi_{n-1} + \sqrt{\xi^2_{n-1} +1} \; ,n>0 \label{eq01} \end{align} If we let $L_n$ be the length of one side of a $6.2^n$-sided polygon and $P_n$ its perimeter, let $R=AO$ and $d=2AO$ then \begin{align*} \xi_n &= \frac{R}{\tfrac{1}{2}L_n}=\frac{d}{L_n} \to \\ \frac{6.2^n}{\xi_n}&=\frac{6.2^nL_n}{d}=\frac{P_n}{d}>\frac{C}{d}=\pi \end{align*}

As $P_n$ is a decreasing series Archimedes found an approximation for an upper bound of $\pi$ for $n=4$: $\pi \less 3\tfrac{1}{7}$

Next Archimedes analyzed the circumfence of the circle with inscribed polygons. Let $AB$ be the diameter of the circle and let $\angle ABC = 30^\circ$ with $C$ a point on the circumfence of the circle. Join $BC$. We have $\angle ACB$ is a right angle.

A triangle inscribed in a semicircle is a right triangle

Next let $AD$ bisect the angle $BAC$ and meet $BC$ in $K$ and the circle in $D$. Join $BD$. Note that $BC$ represents the lenght of one side of a hexagon and $BD$ of one side of a 12-sided polygon. Archimedes found a recursive relationship between the ratio $AB:BC$ and $AB:BD$. First note that $\angle BAC=\angle KAC$ because $AD$ bisects the angle $BAC$. But also $\angle KAC=\angle KBD$ because both intercept the arc $CD$. The angles at $C$ and $D$ are both right angles. It follows that the triangles $ADB$ and $BDK$ are similar and also $ADB$ is similar to $ACK$. We have therefore \begin{align*} \frac{AD}{BD} &=\frac{BD}{DK} \\ &=\frac{AC}{CK} = \frac{AB}{BK} \text{ (from Proposition 3 of Book VI of Euclid's Elements) } \\ &=\frac{AB+AC}{BK+CK} \\ &=\frac{AB+AC}{BC} \\ &=\frac{AB}{BC}+\frac{AC}{BC} \\ &=\frac{AB}{BC}+\sqrt{\left(\frac{AB}{BC}\right)^2-1} \end{align*} Now via Pythagoras theorem: \begin{align*} AB^2=AD^2+BD^2 \to \\ \frac{AB^2}{BD^2}=\frac{AD^2}{BD^2}+1 \to \\ \frac{AB}{BD}&=\sqrt{\frac{AD^2}{BD^2}+1} \to \\ &=\sqrt{2\frac{AB}{BC}\left(\frac{AB}{BC}+\sqrt{\left(\frac{AB}{BC}\right)^2-1}\right)} \end{align*}

Archimedes thus found that the ratio of the diameter to the length of the side of a 12-sided polygon depends on the ratio of a hexagon. Let us define this as the following recursive relation: \begin{align} \eta_0&=\frac{AB}{BC}\notag \\ \eta_n &= \sqrt{2\eta_{n-1}\left(\eta_{n-1}+\sqrt{\eta_{n-1}^2-1}\right)} \; ,n>0 \label{eq02} \end{align} If we let $L_n$ be the length of one side of a $6.2^n$-sided polygon inscribed in the circle and $p_n$ its perimeter, let $d=AB$ then \begin{align*} \eta_n &= \frac{d}{L_n} \to \\ \frac{6.2^n}{\eta_n}&=\frac{6.2^nL_n}{d}=\frac{p_n}{d}\less\frac{C}{d}=\pi \end{align*}

As $p_n$ is an increasing series Archimedes found an approximation for an lower bound of $\pi$ for $n=4$: $\pi > 3\tfrac{10}{71}$

Now we analyze the relation between the two recurrent sequences. Let us denote the sequence of lenghts of $\tfrac{1}{2}$ the lenght of the side of a circumscribed $M2^n$ sided polygon ($M>=3$) with $AA_n$ and the sequence of the lenght of the side of an inscribed $M2^n$ sided polygon with $BB_n$. The sequence $A_0/B_0,A_1/B_1,A_2/B_2,...$ replaces the sequence used by archimedes: $C,D,E,F,G,$ We further choose $A_0$ and $B_0$ such that the $\angle AOA_0=\angle ABB_0$ and it follows that $\angle AOA_n=\angle ABB_n$ for each $n>=0$. Note that the angle for the inscribed polygon is measured from the diameter whereas the angle for the circumscribed is measured from the radius. However they both describe $M2^n$ sided polygons. We use the following fact.

Given a triangle inscribed in a semicircle with diameter $AB$ and radius $AO$ and $\alpha=\angle BAC$ then $\angle BOC=2\alpha$.

Relationships between circumscibed and inscribed sequence




$0$ $\xi_0=\frac{AO}{AA_0}=\frac{AB_0}{BB_0}$ $\eta_0=\frac{AB}{BB_0}=\frac{OA_0}{AA_0}$
$1$ $\xi_1=\frac{OA_0}{AA_0)}+\frac{OA_0}{AA_0}=\xi_0+\eta_0$ \begin{align*} \eta_1 &=\sqrt{\left(\frac{AB}{BB_0}+\frac{AB_0}{BB_0}\right)^2+1} \\ &= \sqrt{\left(\xi_0+\eta_0 \right)^2+1} \\ &= \sqrt{\xi_1^2+1} \\ \end{align*}
... ... ...
$n$ $\xi_n=\frac{OA_{n-1}}{AA_{n-1}}+\frac{OA_{n-1}}{AA_{n-1}}=\xi_{n-1}+\eta_{n-1}$ $\eta_n=\sqrt{\xi_n^2+1}$

Note that $\xi_n$ and $\eta_n$ represents the trigonometric definitions of respectively the cotangent ($\cot$) and cosecant ($scs$) of the following angle: \[ \angle \alpha=\frac{\tfrac{1}{2}2\pi}{M2^n}=\frac{\pi}{M2^n} \]

We can now summarize the algorithm of Archimedes to calculate $\pi$.
Algorithm of Archimedes for calculating $\pi$
Given a unit circle with diameter=1. Let $\xi_0=\cot \pi/M$ and $\eta_0=\csc \pi/M$.

  • \[\xi_n=\xi_{n-1}+\eta_{n-1}, \; \eta_n=\sqrt{\xi_n^2+1} \text{ for } i=1,...,N\]
  • \[P_{M2^N}=\frac{M2^N}{\xi_N} > \pi > p_{M2^N}=\frac{M2^N}{\eta_N}\]

Archimedes used $M=6$, $\xi_0=\sqrt{3}$, $\eta_0=2$ and $N=4$ to find $3\tfrac{1}{7}>\pi>3\tfrac{10}{71}$

Here follows a VB.Net implementation of the Archimedean algorithm

    ' PiArchimedes: This function approximates PI with the algorithm of Archimedes
    ' INPUT:
    ' M = number of sides of the starting polygon,should be > 3
    ' N = number of iterations
    ' OUTPUT:
    ' lbnd = lower bound of Pi
    ' ubnd = upper bound of Pi
    Public Shared Sub PIArchimedes(ByVal M As Integer, ByVal N As Integer,
                                        ByRef lbnd As Double, ByRef ubnd As Double)

        Check_Condition(M >= 3 And N > 0, "Parameters of function PIArchimedes")

        Dim i As Integer = 0
        Dim x As Double = cot(PI / M)
        Do While i \less N
            x = x + Sqrt(Pow(x, 2) + 1)
            i = i + 1
        ubnd = M * Pow(2, N) / Sqrt(Pow(x, 2) + 1)  'circumscribed polygons
        lbnd = M * Pow(2, N) / x                    'inscribed polygons

    End Sub


Copyright 2012 Jacq Krol. All rights reserved. Created July 2012; last updated July 2012.