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\newtheorem{theorem}{Theorem}
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\title{Archimedes Dream, Notes}
\author{Burton Rosenberg}
\begin{document}
\maketitle

\section*{Elementary notes}

Assume a circle of unit radius, that is, of area $\pi$.

\begin{enumerate}
\item A square circumscribed, that is, tightly drawn outside the circle so that
the middle of each side touch the circle, will
have side length 2, and therefore area 4.
\item A square inscribed, that is, tightly drawn inside the circle
so that its four corners touch the circle, has area 2.
This can be seen by adjusting the inscribed square to the circumscribed square so
that the inscribed square is the adjoining of the midpoints of the circumscribed square's
edges. Consider the outside square as a piece of paper. The inside square is
realized by folding the outside square's four corners to the middle. This means that
the area of the outside square is enough to cover the inside square twice.
From area 2, we conclude the inside square has side length $\sqrt{2}$.
\item An inscribed regular octagon is realized from an inscribed square by pushing
out from the mid-point of each side to touch the circle. This adds to the area of the
square 4 triangles, of base the square's width, $\sqrt{2}$. The height of the triangle
is calculated by drawing a radial line from the center of the
circle to the new point of contact. The height is the length of this radial line, 1, minus
the distance to the triangle base, which is half the width of the square, $\sqrt{2}/2$.
The area of one triangle is therefore,
\[
       (1/2) \sqrt{2}( 1 - \sqrt{2}/2) = (1/2)(\sqrt{2} - 1)
\]
and of all four triangles, plus the square's area of 2,
\[
    4 (1/2)(\sqrt{2}-1)+2 = 2\sqrt{2} -2 + 2 = 2\sqrt{2}.
\]
\item A circumscribed regular octagon is realized from a circumscribed square by
cutting off four triangles from the points of the square. Drawing a radial line from the
center of the circle to the corner of the outside square, the height of a cut triangle is
the length of this line minus the distance to where the line cuts the circle. The length
of this line is one half the diagonal of a 2 by 2 square, $\sqrt{2}$, and the length to
the circle is 1, so the triangle's height is $\sqrt{2}-1$. To calculate the base, one must
recognize that this triangle has two equal angles, and hence these angles are 45 degrees.
The radial line bisects the 90 degree angle of the square, making a 45 degree angle,
hence the height cuts the triangle into two smaller isoceles triangles. This gives that
the base is twice the height, $2(\sqrt{2}-1)$. The area of one triangle is therefore,
\[
    (1/2) (\sqrt{2}-1) 2 (\sqrt{2}-1) = 2 - 2\sqrt{2}+1 = 3-2\sqrt{2}
\]
and subtracting four such triangles from the original area of 4,
\[
    4 - 4(3-2\sqrt{2}) = 4 - 12 + 8\sqrt{2} = 8\sqrt{2} - 8 = 8(\sqrt{2}-1).
\]
\end{enumerate}
 
This completes the heavy math for Archimedes Dream's parts 1 and 2.

We can continue with continued fraction expansions of $\sqrt{2}$ to get bounds as
rational numbers, I would not like to use decimal approximations since this is not
beautiful.

\section*{Further explorations}

How so we continue from 8 to 16, 16 to 32, and
so on, to regular $2^k$-gon's, circumscribed and inscribed, which give an infinite 
sequence of ever improving upper and lower bounds to $pi$?
To this end I would like to quickly jot down the following observations and proofs,
before I throw out the scrap paper.
I acknowledge 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.

Let $A_n$ by the area of a regular circumscribed $n$-gon, $a_n$ the same for
the inscribed $n$-gon, $P_n$ be the semi-perimeter of a regular circumscribed $n$-gon,
and $p_n$ the same for the inscribed $n$-gon.

\begin{lemma}
A regular $n$-gon circumscribed around a unit circle has area equal to its
semi-perimeter, $P_n=A_n$.
\end{lemma}

{\bf Proof:} Consider a single triangular sector of the circumscribed $n$-gon.
It has height 1 and base $x$. Therefore both the area and
the semi-perimeter are $nx/2$.


\begin{lemma} 
A regular $n$-gon inscribed in a unit circle has semi-permeter equal to the
area of a regular inscribed $2n$-gon,
$p_n=a_{2n}$.
\end{lemma}

{\bf Proof:} Consider a single triangular sector of the inscribed $n$-gon. It has height
1 and base $x$. Therefore the semi-permeter is $p_n=nx/2$. Consider the 
two triangular sectors of the $2n$-gon which fit neatly inside this one sector
of the $n$-gon. They are two triangles back to back, of common base 1 and 
each of height $x/2$. This gives area $a_{2n}=nx/2=p_n$.

I consider this very interesting. Using trigonometric functions, we can quickly
get formulae for $a_n$ and such, and show the simple relationships between
these quantities used by Archimedes.

\begin{lemma}
$a_n = n \sin (\pi/n) \cos (\pi/n)$.
\end{lemma}

{\bf Proof:} 
Consider a single triangular sector of the inscribed $n$-gon, and bisect with a radial
line from the circle's center. 
This forms two right triangles with hypotenuse 1 and angle is $\pi/n$,
so $\sin( \pi/n)$ and $\cos (\pi/n)$ are the height and base, respectively.
Add up the area of these $2n$ triangles.

\begin{lemma}
$A_n = n \tan (\pi/n)$.
\end{lemma}

{\bf Proof:} 
 Consider a single triangular sector of the circumscribed $n$-gon, and bisect with a radial
line from the circle's center. This forms two right triangles of base 1 and angle $\pi/n$.
Hence $\tan (\pi/n)$ is the triangle's height. Add up the area of these $2n$ triangles.

\begin{lemma}
The area of an regular inscribed $2n$-gon is the geometric mean of the areas of
regular inscribed and circumscribed $n$-gons, $a_{2n} = \sqrt{a_n A_n}$.
\end{lemma}

{\em Proof:} 
Verify the identity,
\[
      \sqrt{a_n A_n} = \sqrt{ n \sin (\pi/n) \cos(\pi/n) n \tan(\pi/n) } = n\sin(\pi/n).
\]
Now apply the double angle formula,
\[
    n\sin(2\pi/(2n)) = n \sin(\pi/(2n))\cos(\pi/(2n)) = a_{2n}.
\]

\begin{lemma}
The semi-perimeter of a regular circumscribed $2n$-gon is half the harmonic mean
of the semi-perimeters of regular inscribed and circumscrbied $n$-gons,
\[
1/P_{2n} = (1/2) (1/p_n + 1/P_n).
\]
The same is true of perimeters.
\end{lemma} 

{\em Proof:} Given the relationships between semi-perimeters and areas, we show,
\[
     1/A_{2n} = (1/2) (1/a_{2n} + 1/A_n).
\]
Using a form of the half-angle formula for tangent,
\[
    \tan \theta = \pm \sin (2\theta) / (1+\cos (2\theta))
\]
we have,
\[
   1/A_{2n}=
   \frac{1}{2n \tan (\pi/(2n))} = \pm\left(\frac{1}{2n\sin( \pi/n) } +
   \frac{1}{2n\tan(\pi/n)}\right)
     = (1/2) (1/a_{2n} + 1/A_n).
\]
To get the statement for perimeters, multiply both sides by $1/2$ and rename variables.

As stated, there are some very nice symmetries in these formulas, adding to the
curious interrelationships of these quantities. Other forms of the these
equations are:

\begin{corollary}
$ 1/A_{2n} = (1/2) (1/a_{2n} + 1/A_n)$ and $p_{2n}=\sqrt{P_{2n}p_n}$.
\end{corollary}

As a note, if one attempts to construct relationships working
with the other version of the half-angle formula,
\[
    \tan \theta = \pm (1-\cos 2\theta)/\sin 2\theta
\]
This gives,
\begin{lemma}
$A_{2n}/(2n^2)= 1/A_n - 1/a_{2n}$.
\end{lemma}
 
Note that this formula is numerically unstable due to a subtraction and then
a large amplification of the truncated result.
Not only did Archimedes derive his formulas without aid of a developed
theory of trigonometry, but also he found numerically well-conditioned 
recurrence relations and avoided the badly behaved relations.

Here are some values for $a_n$ and $A_n$,

\begin{figure}[ht]
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
\hline
$n$ & $a_n$ & $A_n$ & $a_n/A_n$ \\
\hline
\hline
4 & 2 & 4  & $1/2$ \\
\hline
8 & $2\sqrt{2}$ & $8(\sqrt{2}-1)$ & $(2+\sqrt{2})/4$ \\
\hline
16 & $4\sqrt{2-\sqrt{2}}$ &  
$8\sqrt{2}(2-\sqrt{2})/(2+\sqrt{2+\sqrt{2}})$ & \\
\hline
\end{tabular}
\end{center}
\caption{Values for $n=2^i$}
\end{figure}

There is another approach to recurrence, which might be easier
for hand calculation. Here is a simple consequence of our other
formulas.

\begin{corollary} $a_n/A_n = \cos^2 \pi/n$.
\end{corollary}

\begin{lemma}
\[
\frac{a_{2n}}{A_{2n}} = \frac{ \sqrt{a_n/A_n}+1}{2}
\]
\end{lemma}

{\em Proof:} 
Follows from half-angle formula,
\[
   \cos^2 (\theta/2) = (\cos \theta + 1)/2
\]
and the previous corollary.

The result of this formula is the following sequence
for $a_n/A_n$:
\[
  \frac{2}{4},
  \frac{2+\sqrt{2}}{4}, 
  \frac{2 + \sqrt{2 + \sqrt{2}}}{4},
  \frac{2 + \sqrt{2 + \sqrt{2 + 
                      \sqrt{2}}}}{4},
  \ldots
\]

Finally, it is easy enough to derive,
\[
   a_{2n} = a_n \sqrt{A_n/a_n}
\]
giving the sequence for $a_n$,
\[
    2, \frac{4}{\sqrt{2}}, \frac{8}{\sqrt{2}\sqrt{2+\sqrt{2}}},
    \frac{16}{\sqrt{2}\sqrt{2+\sqrt{2}}\sqrt{2+\sqrt{2+\sqrt{2}}}}, \ldots
\]

Note that now we can continue with continued fractions or with
a geometric construction (straight-edge and compass) to construct
convergents to $\pi$.

Beging with the special property of a hexagon, that it
is six equalateral triangles, and if inscribed  in a unit
circle then the side length is one, we have the follow
tabled derived from our recurrences or by direct
reasoning.
The direct reasoning begins with the hexagon and, to
get an inscribed triangle, connects alternate vertices.
The side length is twice the height of a side length one
equilateral triangle. Further, the circumscribed triangle
is seen to be equilateral with side length twice that of
the inscribed triangle. To go the 12-gon, subdivide the
6-sector to get two triangles of height one and base 1/2.
Immediately it follows that the area is one.

\begin{figure}[ht]
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
\hline
$n$ & $a_n$ & $A_n$ & $p_n$  & $P_n$ \\
\hline
\hline
$3$ & $3\sqrt{3}/4$ & $3\sqrt{3}$  & $3\sqrt{3}/2$ & $3\sqrt{3}$ \\
\hline
$6$ & $3\sqrt{3}/2$ &  & $3$ & \\
\hline
$12$ & $3$ & & &  \\
\hline
\end{tabular}
\end{center}
\caption{Values for $n=3 (2^i)$}
\end{figure}

\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.

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

\end{document}
