%\documentclass[a4paper,twoside,12pt,reqno,draft]{amsbook}
\documentclass[a4paper,twoside,12pt,reqno,draft]{amsart}
%\documentclass[a4paper,twoside,12pt,draft]{article}
\usepackage[headings]{fullpage}
\usepackage[polutonikogreek,english]{babel}
\usepackage[oxonia]{psgreek}
%\usepackage{psgreek}
\newcommand{\Gk}[1]{\selectlanguage{polutonikogreek}#1\selectlanguage{english}}
\newcommand{\Eng}[1]{\emph{#1}}
%\newcommand{\Tur}[1]{\textsf{#1}}
%\newcommand{\letter}[1]{\textsf{#1}}  %for letters as such
\usepackage{wrapfig}

%\usepackage{showlabels}  % This doesn't show all labels; there must
                          % be a conflict with another package 

%\usepackage{showkeys}
%\usepackage[integers]{layout}

\title{Parabolas (\emph{draft})}
\author{David Pierce}
\date{April 14, 2008}

%\usepackage{verbatim}

%\begin{comment}
\address{Mathematics Department\\
Middle East Technical University\\
Ankara 06531, Turkey}

\email{dpierce@metu.edu.tr}
\urladdr{http://metu.edu.tr/~dpierce/}
%\end{comment}

\usepackage{amssymb,amsmath,amsthm}
\usepackage{amscd}     % commutative diagram
\usepackage[mathscr]{euscript}
\usepackage{url}

%\usepackage{wrapfig}

\usepackage{pstricks}
%\usepackage{pst-3d}
\usepackage{pst-plot}

\usepackage{amssymb,amsmath,amsthm}
%\usepackage{amscd}     % commutative diagram
%\usepackage[mathscr]{euscript}
\usepackage{url}
\usepackage{verbatim}  % allows a comment environment:

\usepackage{hfoldsty}  % gives old-style numerals

\newcommand{\defn}[2]{\textbf{#1#2}}
\newcommand{\tech}[2]{\textsl{#1#2}}
\renewcommand{\leq}{\leqslant}
\renewcommand{\geq}{\geqslant}

%\newcommand{\measures}{\mathrel{\mid}}
\newcommand{\ratio}[2]{#1:#2}
\newcommand{\psign}{\mathrel{:\,:}}
\newcommand{\eqr}[2]{#1\psign#2}
\newcommand{\prop}[4]{\eqr{\ratio{#1}{#2}}{\ratio{#3}{#4}}}
%\newcommand{\equim}[4]{\eqr{#1\measures#2}{#3\measures#4}}

\newcommand{\lr}{upright side}

\renewcommand{\theenumi}{\roman{enumi}}
\renewcommand{\labelenumi}{(\theenumi)}

\renewcommand{\land}{\mathrel{\&}}

\theoremstyle{plain}

\newtheorem{theorem}{Theorem}
\newtheorem{axiom}{Axiom}

\newtheorem*{enunciation}{Enunciation}

\theoremstyle{definition}

\newtheorem{proposition}{Proposition}

\theoremstyle{remark}

\newtheorem*{exposition}{Exposition}
\newtheorem*{specification}{Specification}
\newtheorem*{construction}{Construction}
%\newtheorem*{proof}{Proof}
\newtheorem*{conclusion}{Conclusion}

\begin{document}
  \maketitle
%\tableofcontents

%\setcounter{section}{-1}
\setcounter{part}{-1}

We work with a cone with distinguished axial triangle.  The cone is
cut by a plane whose intersection with the plane of the base of the
cone is perpendicular to the base of the axial triangle (extended as
necessary).  The intersection of the cutting plane with the conic
surface is a \defn{conic section}.  The intersection of the
cutting plane with the axial triangle is the \defn{diameter}{} of the
section; the diameter meets the section at a \defn{vertex}.  Straight
lines in the plane of the section and parallel to its intersection
with the base of the cone are \defn{ordinate}{} lines.  An ordinate
line from the section to the diameter may be called an
\defn{ordinate}; it cuts off from the diameter (on the side of the
vertex) an \defn{abscissa}.

We assume further that the diameter is parallel to a side of the axial
triangle.  The
intersection of the cutting plane and the conic surface is the conic
section called a \defn{parabola}{} (\Gk{parabol'h}
\Eng{application}), for reasons to be established
presently. 

\section{Propositions 11, 20}

The \defn{\lr}{} of
the parabola is a certain straight line that
\begin{enumerate}
  \item
in position is
perpendicular to the plane of the parabola at the vertex of the
parabola,
\item
in length has to the segment of the side of the axial triangle between
the vertex of the parabola and 
the apex of the triangle the same ratio that the square on the base of
the axial triangle has to the rectangle on the other two sides. 
\end{enumerate}

\begin{wrapfigure}{r}{0pt}
  \begin{pspicture}(-3,-1.4)(3,4)
%\psgrid
    \psline(-2,0)(2,0)(1,3)(-2,0)
\uput[u](1,3){$A$}
\uput[l](-2,0){$B$}
\uput[r](2,0){$C$}
\parametricplot{0}{360}
{t cos 2 mul t sin}
\psline(0.6,0)(-0.05,1.95)(-1.421,1.516)
\uput[ul](-0.05,1.95){$D$}
\uput[l](-1.421,1.516){$E$}
\psline(0.6,0)(0.6,-0.954)
\uput[ur](0.6,0){$G$}
\uput[d](0.6,-0.954){$F$}
  \end{pspicture}
\end{wrapfigure}
Suppose in particular that the axial triangle is $ABC$, with apex $A$.
Let the
vertex of the parabola be $D$ on $AB$.  Then the \lr\ is $DE$, where
\begin{equation*}
  \prop{DE}{DA}{BC^2}{BA\cdot AC}.
\end{equation*}

Let an arbitrary point on the parabola be chosen, different from the
vertex.  Without loss of generality, let it be $F$, in the base of the
cone.  Let the ordinate $FG$ be dropped to the axis.
Then
\begin{equation*}
  FG^2=BG\cdot GC.
\end{equation*}
But
\begin{gather*}
  \prop{BC}{BA}{GC}{DA},\\
\prop{BC}{AC}{BG}{DG},
\end{gather*}
and therefore, compounding these ratios, we have
\begin{equation*}
  \prop{BC^2}{BA\cdot AC}{GC\cdot BG}{DA\cdot DG},
\end{equation*}
that is,
\begin{equation*}
  \prop{BC^2}{BA\cdot AC}{FG^2}{DA\cdot DG},
\end{equation*}
so that
\begin{gather*}
  \prop{DE}{DA}{FG^2}{DA\cdot DG},\\
  \prop{DE\cdot DG}{DA\cdot DG}{FG^2}{DA\cdot DG},\\
DE\cdot DG=FG^2.
\end{gather*}

In particular, squares on ordinates are to each other in the ratio of
the corresponding abscissas.

\section{Proposition 17, 18, 19}

\begin{wrapfigure}{R}{0pt}
  \begin{pspicture}(0,0)(4,4)
%\psgrid
    \parabola(0,0)(2,4)
\psline(1,3)(1,0)
\psline(0,1)(1.5,4)
\psline(1,3)(2,1)
  \end{pspicture}
\end{wrapfigure}
Straight lines drawn ordinatewise at the vertex fall outside the
section; the section need not be parabola.  For, suppose a straight
line drawn from the vertex passes 
within the section, but is not the diameter.  Apollonius seems to slip
here by assuming that this straight line must meet the section.  If it
does meet the section, then a chord is formed that is not bisected by
the diameter: therefore that chord is not drawn ordinatewise.

Suppose this straight line does not meet the section.
Then it must meet the chord of the section that is in the base of the
cone.  This chord is ordinate, and therefore the other straight line
is not ordinate.

\begin{wrapfigure}{l}{0pt}
  \begin{pspicture}(0,0)(4,4)
%\psgrid
    \parabola(0,0)(2,4)    
\psline(0.6,2.04)(3,3)
  \end{pspicture}
\end{wrapfigure}
Straight lines within a section that are parallel to a chord meet the
section in both directions; again, the section need not be a
parabola.  For, the chord cuts off a piece of the section.  If the
straight line is within the region bounded by the chord and this
cut-off piece, then it must meet the section, as it cannot meet the
chord. Suppose the straight line is on the other side of the chord.
From one end of the chord, draw another chord to the other unbounded
piece of the section.  The original straight line meets this chord
either inside or outside the section; in either case, it must meet the
section.  By symmetry, it meets the section in the other direction too.

Hence, in particular, ordinate lines within the section must meet the
section in both directions.

\section{Propositions 22, 24, 26}

\begin{wrapfigure}{R}{0pt}
  \begin{pspicture}*(0,-1)(4.4,5)
%\psgrid
    \parabola(-1,-5)(2,4)    
\psline(1,-5)(1,5)
\psline(3,3)(1,5)
\psline(1,2)(2,4)
\psline(-1,-5)(3,3)
  \end{pspicture}
\end{wrapfigure}
Every chord of the parabola meets the diameter (extended as
necessary).  For, from the ends of the chord, let ordinates be
dropped.  If these are equal, then there is only one corresponding
abscissa, and the chord must consist of these two ordinates.  If the
ordinates are unequal, then the chord is not parallel to the diameter,
so these two straight lines must meet.

Similarly, a straight line outside the parabola, but meeting it at a
point, must meet the diameter.  For, drop the ordinate from the
meeting point, and drop another with a greater abscissa.  (We can do
this by 19.)  The latter ordinate is longer; extended to meet the
original straight line (as it must), it becomes longer still.
Therefore the original straight line is not parallel to the diameter;
so the two straight lines meet.

Hence a straight line parallel to the diameter meets the parabola at
most once (and then it must pass within the parabola).  And it meets
at least once:  For, if the line is within the section, the claim is
obvious.  Suppose it is outside the section.  From a point $A$ of the
straight line, drop a
straight line $AB$ ordinatewise to the diameter.  Within the parabola,
find an abscissa $CD$ that, with the upright side, makes a rectangle
equal to $AB^2$.  Let the corresponding ordinate (on the same side of
the diameter as $AB$) be $DE$.  Then $AB$ and $DE$ are equal and
parallel; so $AE$ is parallel to the diameter and is an extension of
the original straight line.

\vfill

%Testing: what happens if I type a long line?  Does it still wrap too
%soon? 

\mbox{}\hfill

%\bibliographystyle{plain}
%\bibliography{../../../TeX/references}

\end{document}
