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\begin{document}
\title{Apollonian Proof}
\author{David Pierce}
\date{Logic Colloquium 2019, Prague, August}
\publishers{Mimar Sinan G\"uzel Sanatlar \"Universitesi\\
  \url{mat.msgsu.edu.tr/~dpierce/}\\
\url{polytropy.com}}

{\relscale{0.2}\maketitle}

%\setcounter{page}0
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    \begin{center}
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      \newcommand{\myxrad}{6}\newcommand{\myyrad}{4.5}
      \begin{pspicture}(-5,-3)(5.5,3.5)
        \parametricplot[linewidth=2.4pt]0{360}
              {t cos \myxrad\space mul t sin \myyrad\space mul}
           \pstGeonode[PosAngle={180,-12,78,35}]
%              PointSymbol={default,default,default,default,o}]
        (0,0){\Koc}
        (! -12 cos \myxrad\space mul -12 sin \myyrad\space mul){\Ver}
(! 90 -12 add cos \myxrad\space mul 90 -12 add sin \myyrad\space mul)
                   {\Lk}
     (!  35 cos \myxrad\space mul  35 sin \myyrad\space mul){\Vasn}
%           (! -85 cos \myxrad\space mul -85 sin \myyrad\space mul){\Pnt}
                   %                   \ncline{\Koc}{\Ver}\ncline{\Koc}{\Vasn}
                   \uput[r](\Koc){${\Koc}
                     \begin{pmatrix}
                       0\\0
                     \end{pmatrix}$}
                   \uput[r](\Ver){${\Ver}
                     \begin{pmatrix}
                       1\\0
                     \end{pmatrix}$}
                   \uput[30](\Lk){${\Lk}
                     \begin{pmatrix}
                       0\\1
                     \end{pmatrix}$}
                   \uput[r](\Vasn){${\Vasn}
                     \begin{pmatrix}
                       a\\b
                     \end{pmatrix}$}
          \rput(0,2.3){\begin{minipage}{\textwidth}\centering
          In an \textbf{affine plane,} the\\
          locus of ${\Koc}+x\cdot\vec{\Koc}{\Ver}+y\cdot\vec{\Koc}{\Lk}$,
          where
          \begin{equation*}
            x^2\pm y^2=1,
          \end{equation*}
          is (for any ${\Vasn}$,
          namely
          ${\Koc}+a\cdot\vec{\Koc}{\Ver}+b\cdot\vec{\Koc}{\Lk}$,
          on the locus)\\ fixed by
          the \emph{affinity}
            fixing ${\Koc}$ and
            interchanging ${\Ver}$ and ${\Vasn}$.\\
        \end{minipage}}
          \rput(0,-2.5){\begin{minipage}{\textwidth}\centering
            \emph{Modern proof.}
            The affinity is $
            \begin{pmatrix}
              x\\y
            \end{pmatrix}\mapsto
            \begin{pmatrix}
              a&\pm b\\b&-a
            \end{pmatrix}
            \begin{pmatrix}
              x\\y
            \end{pmatrix}$. \qedsymbol\\\mbox{}\\
              1. Apollonius's proof uses \textbf{areas.}
              2. So does an\\ \textbf{axiomatization} of affine planes
              in which the\\
              \emph{theorems} of \textbf{Pappus} and \textbf{Desargues}\\
              are just that.
        \end{minipage}}
      \end{pspicture}
    \end{center}
    \vfill\mbox{}
    \newpage

    

    \textbf{Proof of Apollonius.}
    The locus of $P$ is given by
    \begin{equation*}
      {\Xp}{\Pnt}{\Yp}
      ={\Ver}{\Xp}{\Ypn}{\Endp},
    \end{equation*}
    because\hfill
    $
    \begin{aligned}[t]
      {\Xp}{\Pnt}{\Yp}
      &\propto{\Xp}{\Pnt}^2\\
    &\propto{\Xp}{\Ver}\cdot{\Xp}{\Wv}\\
      &\propto{\Ver}{\Xp}{\Ypn}{\Endp},\\
{\Mp}{\Vasn}{\Edp}&={\Ver}{\Mp}{\Vasn}{\Endp}.
    \end{aligned}$

        \vfill
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                          %\psgrid
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          PointName={default,default,default,default,none,none,default}]
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        (! 90 -12 add cos \myxrad\space mul 90 -12 add sin \myyrad\space mul){\Lk}
        (!  35 cos \myxrad\space mul  35 sin \myyrad\space mul){\Vasn}
        (! 0 47 cos)x
        (0,1)t
        (! -85 cos \myxrad\space mul -85 sin \myyrad\space mul){\Pnt}
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        \pstInterLL[PosAngle=-105]{x'}x{\Koc}{\Ver}{\Mp}
        \pstTranslation[PointName=none] x{\Ver}t
        \pstInterLL[PosAngle=-60]{t'}t{\Koc}{\Ver}{\Edp}
        \pstTranslation[PointName=none]{\Mp}{\Vasn}{\Ver,\Pnt}[u,v]
        \pstInterLL[PosAngle=90]{\Ver}u{\Koc}{\Vasn}{\Endp}
        \pstInterLL[PosAngle=-60]{\Pnt}v{\Koc}{\Ver}{\Xp}
        \pstInterLL[PosAngle=90]{\Pnt}{\Xp}{\Koc}{\Vasn}{\Ypn}
        \pstTranslation[PointName=none]{\Edp}{\Vasn}{\Pnt}[w]
        \pstInterLL[PosAngle=75]{\Pnt}w{\Koc}{\Ver}{\Yp}
        \pstInterLL[PosAngle=-105]{\Pnt}{\Yp}{\Koc}{\Vasn}{\Xpn}
        \pstTranslation[PosAngle=150]{\Ver}{\Koc}{\Koc}[\Wv]
        \pstTranslation[PosAngle=-150,PointName=none]{\Vasn}{\Koc}{\Koc}[\Wnv]
%        \pstInterLL[PosAngle=10,PointName=none]{\Vasn}{\Edp}{\Ver}{\Endp}{\Fp}
        \ncline[linewidth=2.4pt]{\Wv}{\Edp}
        \ncline{\Mp}{\Vasn}
        \ncline{\Ver}{\Endp}
        \ncline{\Pnt}{\Ypn}
        \ncline{\Koc}{\Lk}
        \psset{linestyle=dashed}
        \ncline{\Vasn}{\Edp}
        \ncline{\Pnt}{\Yp}
        \psset{linestyle=dotted}        
        \ncline{\Wnv}{\Endp}
      \end{pspicture}
      \vfill\mbox{}
\hfill
        Adding ${\Xp}{\Yp}{\Xpn}{\Ypn}$
        yields\\
\hfill                  ${\Ypn}{\Pnt}{\Xpn}
          ={\Ver}{\Yp}{\Xpn}{\Endp}$, then
          \begin{equation*}
            {\Ypn}{\Pnt}{\Xpn}
          ={\Edp}{\Yp}{\Xpn}{\Vasn}.
          \end{equation*}
      
  
\newpage

\begin{multicols}{2}
  Fundamental to the geometry of areas is
%Each \textbf{axiom} will justify a step in the proof of
\textbf{Euclid \textsc i.37, 39.}

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               {\psset{linewidth=2.4pt}
                 \ncline AD\ncline DC\ncline CA
                 \ncline CF\ncline FD\ncline DC}
    \ncline AF
%    \psset{linestyle=dashed}
    \ncline BD\ncline CE
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    \psset{linestyle=dotted}
    \ncline AH\ncline HC
    %        \pstHomO[HomCoef=1.8,PosAngle=-90]CD[K]
    %        \ncline[linestyle=dashed]DK
  \end{pspicture}
  \begin{enumerate}
  \item
    Assuming $AF\parallel CD$,
    let
    \begin{align*}
      AC&\parallel BD,
      &CE&\parallel DF.
    \end{align*}
  \item
    By translation,
    \begin{equation*}
    ACE=BDF.      
    \end{equation*}
  \item
    By polygon algebra,
    \begin{equation*}
      ACDB=ECDF.
    \end{equation*}
    \begin{comment}
      \begin{multline*}
        ACDB
        =ACGB+CDG\\
        =ACE-BGE+CDG\\
        =BDF-BGE+CDG\\
        =ECDF.
      \end{multline*}
    \end{comment}
  \item
    By bisection,
    \begin{gather*}
      ACDB=2ACD,\\
      ECDF=2FCD.
    \end{gather*}
  \item
    By halving,
    \begin{equation}\label{eqn:ACD}
      ACD=FCD.
    \end{equation}
  \item
    If $AF\nparallel CD$,
    let $AH\parallel CD$.
    \begin{align*}
      ACD&=HCD,\\
      FCD&=HCD+FCH,
    \end{align*}
    so \eqref{eqn:ACD} fails.
  \end{enumerate}
\end{multicols}
    \newpage

    \begin{multicols}{2}
      In order 3, 6, 1, 5, 4, 2,
      the steps
      are justified by:
      
      \begin{asparadesc}
      \item[Axiom 1.]
        The polygons compose an abelian group $\Pi$ where,
        $*$ and $\dag$ being strings of vertices,
        \begin{gather*}
          A*{}= A* A={}* A,\\
          A* B+B\mathrel{\dag} A= A* B\mathrel{\dag},\\
          -ABC\cdots= \cdots CBA.
        \end{gather*}
      \item[Axiom 2.]
        $ABC=0$ means that $A$, $B$, and $C$ are collinear.
      \item[Axiom 3.]
        Playfair's Axiom.
      \item[Axiom 4.]
        All nonzero elements of $\Pi$
        have the same order,
        not $2$.
        
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          \pstTranslation[PosAngle=-45] DAE[B]
          \ncline AD\ncline CF\ncline BE
          \psset{linestyle=dotted}
          \psset{linestyle=dashed}
          %          \ncline FG\ncline BG
          %          \ncline DA\ncline DE
          \psset{linestyle=solid,linewidth=2.4pt}
          \ncline AC\ncline CB\ncline BA
          \ncline DF\ncline FE\ncline ED
          \ncline AG\ncline CG
        \end{pspicture}      \item[Axiom 5, 6.]
        If
        $ABCG$,
        $ABED$, and $BCFE$ are parallelograms,
        then
        \begin{equation*}
          CGA=ABC=DEF.
        \end{equation*}
      \end{asparadesc}

    \end{multicols}

    \newpage

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        \psset{linewidth=2pt,linestyle=solid}
        \ncline AB\ncline BC\ncline DE\ncline EF
        \psset{linestyle=dashed}
        \ncline CD
        \ncline AF
        %    \psset{linestyle=dotted}
        %    \ncline AC\ncline BE\ncline DA
      \end{pspicture}
    \end{minipage}\hfill
    \begin{minipage}{7cm}
\centerline{\emph{From Euclid \textsc i.37, 39:}}
      
$\bullet$ \textbf{Pappus's Hexagon Theorem,}
by his proof:
      In hexagon  $ABCDEF$, if
      \begin{align*}
        AB&\parallel DE,&
        BC&\parallel EF,
      \end{align*}
      then $FAD=FAEB=FAC$, so
      \begin{equation*}
        CD\parallel FA.
      \end{equation*}
    \end{minipage}

    \begin{minipage}{6cm}
      $\bullet$ \textbf{Euclid \textsc i.43 plus.}
      \begin{align*}
        OGL=0
        &\iff \alpha=\beta\\
        &\iff BD\parallel AC.
      \end{align*}
$\bullet$ \emph{Desargues's Theorem.}
    \end{minipage}\hfill
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    \newpage
    \begin{minipage}{8cm}
      \textbf{Desargues's Theorem.}
      If
      \begin{equation*}
        AB\parallel DE\And AC\parallel DF,
      \end{equation*}
      then
      $BC\parallel EF$,
      so $ABC\sim DEF$.
      
      \emph{Proof.}
        \begin{compactitem}
        \item
          True when
          $AB\parallel OC$,
          by \textsc i.43+.
        \item
          Enough now that, since
          \begin{equation*}
            BAG\sim EDH,
          \end{equation*}
          for all $X$ (not shown) on $OA$,
          \begin{equation*}
            BXG\sim EYH
          \end{equation*}
          for some $Y$ on $OA$.
        \end{compactitem}
\makebox[0pt][l]{Note  $BAE\sim GKH$
      by Pappus,
where $BA\parallel GK$.}
    \end{minipage}
    \hfill
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        \psset{linestyle=dashed}
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        \psset{linestyle=dotted}
        \ncline CB\ncline FE
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    \end{minipage}

    \newpage

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        \pstInterLL[PosAngle=0] BxEHG
        \pstTranslation[PointName=none] HDE[x]
        \pstInterLL[PosAngle=180] ExADK
        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
        \ncline KD\ncline EH
        \ncline AE\ncline BF
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    \end{minipage}
    \hfill
    \begin{minipage}{6cm}
      \textbf{Lemma.}
      Given
      \begin{equation*}
        AEC\sim BFD,
      \end{equation*}
      we noted
      \begin{equation*}
        AEB\sim CLD
      \end{equation*}
            for some $L$ on $EF$.
      Now let
      \begin{equation*}
        KF\parallel AG\parallel CH.
      \end{equation*}
      By Pappus twice,
      \begin{equation*}
        BG\parallel KE\parallel DH,
      \end{equation*}
      whence
      \begin{equation*}
        AGB\sim CHD.
      \end{equation*}
    \end{minipage}

    \begin{comment}
      \newpage
      \noindent
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        \psset{linestyle=dashed}
        \psset{linestyle=dotted}
        \psset{linestyle=none,hatchangle=0}
        \pspolygon[fillstyle=hlines]
        (H)(M)(N)(L)(D)(F)
        \pspolygon[fillstyle=vlines]
        (G)(M)(N)(L)(A)(C)
        \uput*[30](E){$E$}
        \uput*[60](K){$K$}
      \end{pspicture}

      \begin{math}
        \begin{gathered}[b]
          AB\parallel DE,\\
          OC\parallel EH\parallel DK\parallel BG,\\
          OA\parallel EM\parallel KL,\\
          NM\parallel DF\parallel AC
        \end{gathered}
      \end{math}

      implies
      \begin{math}
        \begin{aligned}[t]
          HM&\parallel DL&&\text{[Desargues case]}\\
          GE&\parallel AK&&\text{[Pappus]}\\
          MG&\parallel LA&&\text{[Desargues case]}\\
          HF&\parallel NL\parallel GC&&\text{[Pappus]}\\
          FE&\parallel CB&&\text{[Desargues case]}
        \end{aligned}
      \end{math}
    \end{comment}

\end{document}
