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\title{Archimedes and Pi}
\author{Burton Rosenberg}
\begin{document}
\maketitle

\section*{Introduction}

Proposition 3 of Archimedes' {\em Measurement of a Circle} states that $\pi$ 
is less than $22/7$ and greater than $223/71$. 
The approximation $\pi_a \approx 22/7$ is referred to as Archimedes Approximation
and is very good. 
It has been reported that a 2000 B.C. Babylonian approximation is $\pi_b \approx 25/8$.
We will compare these two approximations. The author, in the spirit of 
idiot's advocate, will venture his own approximation of $\pi_c \approx 19/6$. The
Babylonian approximation is good to one part in 189, the author's, one part in 125,
and Archimedes an astonishing one part in 2484.

Archimedes' approach is to circumscribe and inscribe regular $n$-gons around
a unit circle. He begins with a hexagon and repeatedly subdivides the side
to get 12, 24, 48 and 96-gons.
The semi-circumference of these polygons converge on $\pi$ from above and
below.

In modern terms, Archimede's derives and uses the cotangent half-angle formula,
\[
\cot x/2 = \cot x + \csc x.
\]
In application, the cosecant will be calculated from the cotangent according
to the (modern) identity,
\[
\csc^2 x = 1 + \cot^2 x
\]
Greek mathematics dealt with ratio's more than with numbers. Among the
often used ratios are the proportions among the sides of a triangle.
Although Greek mathematics is said to not know trigonometric functions, we
shall see how conversant it was with these ratios 
and the formal manipulation of ratios, resulting in a theory essentially equivalent to
that of trigonometry.

\section*{For the circumscribed polygon}

We use the notation of the Dijksterhuis translation of Archimedes.
Circumscribe a hexagon around a circle centered at $E$. 
Let $\Gamma$ be a point of tangency between the circle and the hexagon.
Let $Z$ be a vertex of the face bisected by $\Gamma$.
Consider the right triangle $\triangle\,Z\Gamma E$. Let $H$ be the point
on $\Gamma Z$ such that $HE$ bisects the angle $\angle\,ZE\Gamma$.
The segment $H\Gamma$ is the half side of a circumscribed 12-gon.
Continuing, let $\Theta$ be the point on $\Gamma H$ such that $\Theta E$ bisects
$\angle\, HE\Gamma$, $K$ be the point on $\Theta\Gamma$ such that $KE$
bisects $\angle\,\Theta E\Gamma$, and $\Lambda$ be the point of $K\Gamma$
such that $\Lambda E$ bisects $KE\Gamma$. The segments $\Theta\Gamma$,
$K\Gamma$ and $\Lambda\Gamma$ are half sides of circumscribed
24, 48 and 96-gons, respectively.


Archimedes states the following ratio's for the hexagon,
\[\begin{array}{rcl}
      (E \Gamma,\Gamma Z )& >& (265,153)\\
  (ZE,\Gamma Z)& =& (306,153)
\end{array}\]
These two ratios approximate $\cot(\pi/6)$ and $\csc(\pi/6)$, respectively.
We do not how he arrived at this excellent approximation of $\cot(\pi/6)=\sqrt{3}$ by a rational.

Archimedes then derives the cotangent half angle formula starting from Proposition
3 of Euclid's {\em Elements}, Book VI.

\begin{lemma}[Euclid VI(3)]
Let $\triangle\, Z\Gamma E$ be a
right triangle, with the right angle at $\Gamma$. Extend from $E$ a bisector
meeting $Z\Gamma$ at $H$. Then,
\[
(ZH,\Gamma H) = (ZE,\Gamma E).
\]
where the notation $(A,B)$ is the ratio $A$ to $B$. 
\end{lemma}

{\bf Proof:} A proof does not appear in the original. 

Let $E\Gamma$ be the radius of a circle with center $E$, and Z$\Gamma$
the tangent at $\Gamma$. Let $X$ be the intersection of $EZ$ with the circle.
Construct a tangent to the circle at $X$ and let $H$ be the intersection of this tangent
with $Z\Gamma$. 
Triangle $\triangle\, Z\Gamma E$ is congruent to $\triangle\, ZXH$. Therefore
$(ZE, \Gamma E)=(ZH,HX)$ and $HX=\Gamma H$.
\hfill $\Box$

\begin{lemma}[Cotangent Half Angle Formula]
In the notation of the previous lemma, 
\[
  (\Gamma E, \Gamma H) =(ZE+E\Gamma, \Gamma Z).
\]
\end{lemma}

{\bf Proof:}
To Euclid's equation, Archimedes applies the operation of {\em componendo}
(Euclid V, Proposition 18),
\[
(ZH,\Gamma H)=(ZE,\Gamma E)
 \Rightarrow 
(ZH+\Gamma H, \Gamma H)=(ZE+\Gamma E,\Gamma E)
\]
and $ZH+\Gamma H = Z\Gamma$; then the operation of {\em permutando}
(Euclid V, Proposition 16),
\[
(Z\Gamma, \Gamma H) = (ZE+\Gamma E,\Gamma E)
\Rightarrow
(\Gamma E,\Gamma H) = (ZE+\Gamma E,Z\Gamma).
\]
\hfill$\Box$

{\bf N.B.:}
Let the angle $\angle\, ZE\Gamma$ be $\theta$. We 
transcribe the above lemma to modern notation, thus seeing how
it is a half angle formula.
\[
(\Gamma E,\Gamma H) =
{\Gamma E}/{\Gamma H} = \cot \theta/2
\]
and
\[
(ZE+\Gamma E, Z\Gamma) 
= \frac {ZE+\Gamma E} {Z\Gamma} 
=  \csc \theta + \cot \theta
\]
%\tan \theta\ 2 =\frac{\tan \theta}{1+\sec \theta}

\begin{lemma}[Pythagorean cosecant formula]
In the notation of the above two lemmas,
\[
   ((HE)^2,(\Gamma H)^2) =  (  (\Gamma E)^2+(\Gamma H)^2, (\Gamma H)^2) 
\]
\end{lemma}

{\bf Proof:}
$HE$ is the hypotenuse of the right triangle $\triangle\,HE\Gamma$. Apply
the Pythagorean formula, Euclid book 1, Proposition 47.
\hfill $\Box$

{\bf N.B.:}
Replacing the ratios of the previous lemma with modern notation,
\[ \csc^2\theta= 1 + \cot^2 \theta. \] 

The above lemma does not appear explicitly in the proof. Rather it is implicit
in the next three lines of calculation,
\[\begin{array}{rcl}
 (\Gamma E, \Gamma H ) &>& (571,153)   \\
( (EH)^2,(\Gamma H)^2) &>& (349450,23409)   \\
  ( EH,\Gamma H) &>& (591 \,\,1/8,153)\\
\end{array}\]
These are, respectively, approximations for $\cot(\pi/12), \csc^2(\pi/12)$ and
$ \csc(\pi/12)$.
In the orignal,
Archimedes sometimes used equality when inequality is correct.
Although his calculation of the square root were properly rounded, 
the notation did not always highlight this.

The proof continues by iteratively bisecting three more times the 
angle defining the polygon, defining points
on $\Theta, K$ and $\Lambda$, the vertices adjacent to the tangency 
at $\Gamma$ on a 24, 48 and 96-gon, respectively. 
In the original, the following ratio are given. We have added an annotation column
giving the trignometric interpretation.
\[\begin{array}{rcl}
  (E\Gamma, \Gamma \Theta) &>& ( 1162 \,\,1/2, 153) \\
 (E\Theta, \Theta\Gamma) &>& (1172\, \,1/8,153) \\
  (E\Gamma, \Gamma K) &>& ( 2334\,\,1/4, 153)\\
 (EK, K\Gamma) &>& (2339\,\,1/4,153) \\
 (E\Lambda, \Lambda\Gamma) &>& (4673\,\,1/2,153) \\
\end{array}\]
The proof concludes noting that half the circumference of the circle is less than
half the perimeter of the circumscribed 96-gon. The ratio is then rounded,
\[
   \pi < 96 \cdot (153, 4673\,\,1/2) < 22/7
\]

\section*{For the inscribed polygon}

Let $\Gamma B$ be a face of the inscribed polygon. Construct a diameter of
the circle it one end at $\Gamma$ and the other call $A$. Then $\triangle \,\Gamma B A$
is a right triangle. 
Call $H$ the point at which the bisector of angle  $\angle\, BA\Gamma$ crosses the circle
on the arc between $\Gamma$ and $B$.
Segment $\Gamma H$ is the face of an inscribed polygon of twice the number of
faces as the original polygon. Continuing, bisect $\angle\,HA\Gamma$ to locate point
$\Theta$, bisect $\angle\,\Theta A\Gamma$ to locate point $K$, and finally
$\angle\,KA\Gamma$ to locate point $\Lambda$. If $\Gamma B$ is the face of
a hexagon, then $H\Gamma$, $\Theta\Gamma$, $K\Gamma$, and $\Lambda\Gamma$
are faces of 12, 24, 48 and 96-gons.

The approximation to $\cot (\angle\,BA\Gamma)$ is given,
\[
   (BA,B\Gamma) < (1351,780).
\]
Archimedes then derives the formula,
\[
(AH,H\Gamma) = (A\Gamma+AB,B\Gamma).
\]
Expanding this out in modern notation, we again see the half angle formula
for cotangent.
Archimedes next applies the formula giving an approximation to $\cot(\angle\,HA\Gamma)$,
\[
   (HA,H\Gamma) < (2911,780). 
\]
Using the cosecant formula, without statement, Archimedes approximates,
\[
   (A\Gamma, \Gamma H) < (3013\,\,3/4,780).
\]
Assuming now that his reader is intelligently following his iteration, successive cotangents
and cosecants are given for the remaining bisections,
\[\begin{array}{rcl}
(A\Theta,\Theta\Gamma) &<&(1823,240)\\
(A\Gamma,\Gamma\Theta) &<& (1838\,\,9/11,240)\\
(AK,K\Gamma) &<& (1007,66)\\
(A\Gamma,\Gamma K) &<& (1009\,\,1/6,66)\\
(A\Lambda,\Lambda\Gamma) &<& (2016\,\,1/6,66)\\
(A\Gamma,\Gamma\Lambda) &<& (2017\,\,1/4,66)
\end{array}
\],
Since $A\Gamma$ is the diameter of a unit radius circle, the semi-perimeter of the
inscribed 96-gon is,
\[
(96 \cdot 66, 2017\,\,1/40)=(6336,2017\,\,1/4).
\]
Hence,
\[
\pi > \frac{6336}{2017\,\,1/4} > 3 \,\, 10/71.
\]
Thus completing the demonstration.

\section*{Error and convergence}

Archimedes did not talk of convergence properties. He did consider
error, since the extraction of square roots is approximate and he
carefully rounded appropriately, the lower bound being subject
to rounding down, the upper bound to rounding up.

The semiperiemeters of circumscribed and
inscribed $n$-gons, $P_n$ and $p_n$ respectively, are given by,
\begin{eqnarray*}
  P_n &=& n/\cot(\pi/n)\\
  p_n &=& n/\csc(\pi/n)
\end{eqnarray*}
Using the cosecant identity,
\[
    1/p_n^2 = 1/n^2 + 1/P_n^2.
\]
or
\[
  P_n - p_n = \frac{(P_n p_n)^2 }{n^2(P_n+p_n)}
\]
When $n\ge 6$, we have the bounds
$3<P_n<2\sqrt{3}$ and $3<p_n<\pi$. This gives a simpler
inequality for the error, 
\[
   P_n - p_n < 2(\pi/n)^2.
\]


\section*{Appendix: Sources}

Archimedes studied in Alexandria (269--263 B.C.) and for the rest of his life
communicated his results by letter to his friends there. He generally first
sent results with encouragement to find proofs, and later full treatments
including proofs. None of this has survived. All modern and medieval
copies of  his work seem to come through a 10-th century compendium
of his work. 

In the 12-th century, this compendium was taken apart, erased and the
parchment reused. Such a practice was common and the resulting
book is called a Palimpsest. Being a religious text it became sacred
and for the next 600 years was read from and 
cared for by the Greek Orthodox adherents at the monastery of Mar Saba
in the Dead Sea. It was moved to the library of the Greek Patriarch in the Christian
quarter of Jerusalem and then in the early 1800's to the Metochion in Constantinople,
a daughter house of the Church of the Holy Sepulchre.

In 1907 Johan Heiburg discovered the manuscript by reading
and identifying the erased text.  The manuscript disappeared
in the mid 20-th century, to turn up mysteriously at auction.
It was sold at Christie's of New York on October 29, 1998 to
an anonymous collector for two million dollars. 
The collector loaned the Archimedes Palimpsest to the Walters Art
Gallery in Baltimore, Maryland where it now resides.
This history is based on Heath and the Walters Art Gallery web site.



\section*{References}

G. M. Phillips, {\em Archimedes the numerical analyst,} Amer. Math.
Monthly, 88(3), March 1981, reprinted in {\em Pi: A source book,} Lennart Berggren,
Jonathan Borwein and Peter Borwein, Springer 1997.

E. J. Dijksterhuis, {\em Archimedes}, Princeton University Press, 1987.

Sherman Stein, {\em Archimedes, What did he do besides cry Eureka?}, 
The Mathematical Assoc. of America, 1999.

The Walters Art Gallery, {\em Archimedes Palimpsest},
\verb,http://www.thewalters.org/archimedes,

Archimedes, {\em Works}, 
Sir Thomas Heath ed., Cambridge University Press, 1897.

Xavier Gourdon, Pascal Sebah, \verb,http://numbers.computation.free.fr/Constants/constants.html,.

%Continue Fractions, \verb,http://www.mcs.surrey.ack.uk/Personal/R.Knott/Fibonacci/cfINTRO.html,

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