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\centerline{MATH 136 -- Calculus 2}
\centerline{Discussion -- Archimedes' {\it Quadrature of the Parabola}}
\centerline{November 22, 2016}
\vskip 10pt
\noindent
{\it Background}
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One of the most interesting surviving works of the ancient Greek mathematician 
Archimedes (287 - 212 BCE) is his {\it Quadrature of the Parabola}.  The
problem he solved was that of determining the area of the region between a
a parabola and one of its secant lines -- a region called a \emph{parabolic segment}.
The result he showed (with two different proofs, each of an extremely high 
degree of ingenuity) was:
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{\it The area of the parabolic segment bounded the secant line $AB$ is
$\frac{4}{3}$ of the area of the triangle $\Delta ABC$, where $C$ is the point
on the parabola whose tangent line is parallel to $AB$.}  (See Figure 1 for a special case.)
\footnote[1]{``Quadrature'' for Archimedes meant finding a \emph{square} with the same area as another region under consideration.  Other geometric constructions known to the 
Greeks would allow a geometer to get a square with the same area as $4/3$ times
the area of the triangle.  Areas of other figures with curved boundaries--for 
instance circles--had been determined before this.  This was the first 
determination of an area quite this complicated, though.
The word ``quadrature'' later evolved into a 
near synonym for ``integral.''} 
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\noindent
We will retrace his steps today, using calculus to derive some of the
intermediate results he did via pure geometry.
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\noindent
{\it Questions}
\begin{enumerate}
\item[(A)]  Say the parabola is $y = x^2$ and the points $A,B$ are
$A = (-1,1)$ and $B = (2,4)$ respectively (see Figure 1 on the next page).
Find the equation of the 
secant line and compute the area of the region between the secant line 
and the parabola using calculus. 
\item[(B)] Find the point $C$ on the parabola where the tangent line
is parallel to the secant line through $A$ and $B$.  Determine the 
area of $\Delta ABC$ and verify that Archimedes' result holds.  
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\item[] Needless to say, though, what you did is not what Archimedes did.  Moreover he
demonstrated the statement above for all points $A$, $B$, not just
the specific ones in parts A,B.  His idea was the following:  No matter
where $A$ and $B$ are, once we have the vertex $C$, the
two secant lines $AC$ and $BC$ form two 
new parabolic segments, each with a vertex.  These points are called $C'$ and $C''$
in Figure 2.  Archimedes' key insight here was that each of the triangles
$\Delta ACC'$ and $\Delta BCC''$ had $\frac{1}{8}$ of the area of $\Delta ABC$.
Therefore 
\begin{equation}
\label{rec}
{\rm area}(\Delta ACC') + {\rm area}(\Delta BCC'')
= \frac{1}{4}\cdot {\rm area}(\Delta ABC).
\end{equation}
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\item[(C)]  With the points $C',C,C''$ we now have 4 new parabolic segments.
We repeat the process of finding the vertex and inscribing a triangle in each 
one.  Explain why \eqref{rec} shows that the area of the four new triangles
obtained this way is
$$\frac{1}{4^2}\cdot {\rm area}(\Delta ABC) = \frac{1}{16}\cdot {\rm area}(\Delta ABC).$$
\item[(D)]  If we continue creating new triangles this way indefinitely, and add the areas, 
explain why the total area is given by the following infinite series:
$${\rm area}(\Delta ABC)\cdot \sum_{n = 1}^\infty \frac{1}{4^n}$$
Determine the sum of this series.  On the other hand, it is clear that the 
triangles we are looking at eventually ``fill up'' all of the area between
the parabola and the secant line $AB$.  This was the way Archimedes proved
his result in his \emph{first proof}.  The second one is even more amazing!
\end{enumerate}

\begin{figure}
\begin{center}
\includegraphics[width=50mm]{ArchQP.png}
\caption{The setting for Archimedes' result}
\end{center}
\end{figure}

\begin{figure}
\begin{center}
\includegraphics[width=60mm]{ArchQPTwo.png}
\caption{Inscribing new triangles}
\end{center}
\end{figure}
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