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\begin{document}
\title{Apollonian Proof}
\subtitle{Slides with commentary}
\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\thispagestyle{empty}
\vfill

We shall look at a proof by Apollonius
that seems to have been overlooked,
even by scholars of Apollonius,
be they historians like Fried and Unguru
or mathematicians like Rosenfeld.

I shall put my slides and notes for this talk
on my departmental webpage;
my blog at \url{polytropy.com}
already has an article called ``Elliptical Affinity''
from April,
with animations illustrating the proof.

The proof uses \emph{areas}
and works in an \textbf{affine plane,}
namely a principle homogeneous space
of a $2$-dimensional vector space over some field
(not of characteristic $2$).
This means the space acts \emph{simply} (or \emph{sharply})
\emph{transitively} on the plane.

I am going to offer an axiomatization of affine planes
based on areas.
\vfill
\noindent
\textbf{[Slide 2]} In an affine plane,
  we choose non-collinear points ${\Koc}$, ${\Ver}$, and ${\Lk}$,
    determining a coordinate system in which, by definition,
    \begin{compactitem}
      \item
    $\vec{\Koc}{\Ver}$
        is a unit vector in the $x$-direction;
        \item
          $\vec{\Koc}{\Lk}$ in the $y$-direction.
    \end{compactitem}
    A \textbf{conic section} with center ${\Koc}$
    is given by $x^2+y^2=1$ or $x^2-y^2=1$.


\pagebreak

%    \newpage
%    \thispagestyle{empty}
%    \mbox{}\vfill
    \begin{center}
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%        \psgrid
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              {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
    In modern terms,
    Apollonius's theorem is that,
    for an arbitrary point ${\Vasn}$ on the curve,
    the affine transformation
    that fixes ${\Koc}$ and interchanges ${\Ver}$ and ${\Vasn}$
    fixes the whole curve setwise.

    The modern proof involves
    plugging and chugging with the given rule.
    This took centuries of development
    after Descartes's \emph{Geometry} (1637).

    In modernizing Euclid,
    Hilbert reduces the theory of areas
    to a theory of lengths,
    which compose a field.
    Michael Beeson continued this work on Tuesday,
    \emph{defined} equality of rectangles.
    To prove this transitive,
    he needed Hilbert's theory of proportion,
    as simplified by Bernays.

    We shall develop a theory of proportion
    \emph{from} a theory of areas.
    Apollonius's proof will use all of this.

    In \emph{Geometric Algebra} (1957),
    Artin shows how to obtain a field from an affine space,
    axiomatized by:
    \newpage

    \begin{compactenum}
    \item
      Two points determine a line.
    \item
      \textbf{Playfair's Axiom:}
      through a point not on a line,
      a single parallel passes.
    \item
      There are three non-collinear points.
      \item
    \textbf{Desargues's Theorem:}
    except in cases of collinearity,
    the following are transitive:
    \begin{compactenum}
    \item
      being opposite sides of a parallelogram
    \item
      being sides of a triangle
      cut by a line parallel to the base.
    \end{compactenum}
    We can now define \textbf{vectors,}
    and \textbf{ratios} of parallel vectors;
    the ratios compose a field or skew-field,
    acting on the plane.
    \item
      \textbf{Pappus's Hexagon Theorem,}
      to be seen later.
    \end{compactenum}
    Though Hilbert calls it Pascal's,
    the Hexagon Theorem
    is Pappus's, as its \emph{Wikipedia} article explains,
    in the section called ``Origins''---%
    which I added a few years ago.
    I forgot the ``parallel'' case that we are using now:
    Lemma VIII in the relevant part of the \emph{Collection.}
    In his \emph{History of Greek Mathematics,}
    Heath omits to discuss Lemma VIII too.
\vfill
\noindent
\textbf{[Slide 3]}
      We first establish the equation
      of triangle and trapezoid at the top.
      In the figure:
    
\vfill

\noindent    \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}$

    \noindent
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        \pstGeonode[PosAngle={105,-60,78,75,0,0,-85},
          PointName={default,default,default,default,none,none,default}]
        (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}
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        \pstInterLL[PosAngle=90]{\Pnt}{\Xp}{\Koc}{\Vasn}{\Ypn}
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        \pstInterLL[PosAngle=75]{\Pnt}w{\Koc}{\Ver}{\Yp}
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        \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}
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        \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}

      \begin{flushright}
        Adding ${\Xp}{\Yp}{\Xpn}{\Ypn}$ yields\\
        ${\Ypn}{\Pnt}{\Xpn}
        ={\Ver}{\Yp}{\Xpn}{\Endp}$, then
        \begin{equation*}
          {\Ypn}{\Pnt}{\Xpn}
          ={\Edp}{\Yp}{\Xpn}{\Vasn}.
        \end{equation*}
      \end{flushright}

      \newpage

\noindent
      \begin{compactitem}
      \item
        $P$ is a random point on the conic section.
      \item
        ${\Pnt}{\Xp}$, ${\Vasn}{\Mp}$,
        and ${\Ver}{\Endp}$ are parallel to ${\Koc}{\Lk}$.
          \item
      ${\Koc}{\Edp}$ is a third proportional to ${\Koc}{\Mp}$
            and ${\Koc}{\Ver}$.
          \item
            Hence the given equation 
      ${\Mp}{\Vasn}{\Edp}={\Ver}{\Mp}{\Vasn}{\Endp}$ holds.
      \item
        ${\Pnt}{\Yp}\parallel{\Edp}{\Vasn}$.
      \end{compactitem}
      We conclude:
      \begin{compactitem}
        \item
      ${\Xp}{\Pnt}{\Yp}$
          varies as the square on its side ${\Xp}{\Pnt}$.
          \item
      By the property that Apollonius derives
      from the cone itself,
      the square on the \emph{ordinate} ${\Xp}{\Pnt}$
      varies jointly as the two \emph{abscissas}
      ${\Xp}{\Wv}$ and ${\Xp}{\Ver}$.
\item
      The trapezoid ${\Ver}{\Xp}{\Ypn}{\Endp}$ also
      varies jointly as the abscissas,
      since in particular
      \begin{equation*}
        {\Xp}{\Wv}={\Xp}{\Koc}+{\Ver}{\Koc}
        \propto{\Xp}{\Ypn}+{\Ver}{\Endp}.
      \end{equation*}
      \end{compactitem}
      \vfill
      
      \noindent
      \textbf{[Slide 4]}
Michael Beeson reviewed the proof of Euclid's
Proposition \textsc i.35,
which yields as a consequence 37 and its converse, 39:
triangles on the same base are equal
if and only if the line joining their apices
is parallel to the common base.

\vfill
      


\begin{multicols}{2}
  \raggedright
  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
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%    \psset{linestyle=dashed}
    \ncline BD\ncline CE
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  \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


I analyze the proof of 37 and 39 into six parts,
corresponding to six axioms for an affine plane,
consisting, as a structure, of
\begin{compactitem}
\item
  a sort for points,
\item
  a sort for polygons,
\item
  for each $n$ greater than $2$,
  a map sending an $n$-tuple of points
  to the polygon with those points as vertices,
\item
  a relation of equality of polygons,
\item
  operations of an abelian group on the sort of polygons.
\end{compactitem}
\vfill


    \begin{multicols}{2}\raggedright
      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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          %          \ncline DA\ncline DE
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          \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}
    \vfill
    \noindent
    \textbf{[Slide 5]}
    In Axiom 1,
    the polygons shown as equal can be taken as identical;
    but equality as in Axioms 2, 5, and 6
    will be a \emph{congruence}
    with respect to the abelian-group operations
    on polygons.

    In Axiom 2,
    we use equations $ABC=0$
    to express that the set of points,
    each collinear with two points,
    is determined in this way by any two of its points.

    In Axiom 4, the common order, if finite, is automatically prime.
\vfill
    \newpage

    \begin{minipage}{5cm}
      \begin{pspicture}(5,4)
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%              \psgrid
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        (0.6,0.2)D(4.4,0.2)C(2,3.8)F(4,3.8)A
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        \pstInterLL F{F'}ACE
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        \psset{linestyle=dotted}
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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.\vphantom{\frac11}
      \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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        \pstHomO[HomCoef=0.35,PosAngle=180]OB[A]
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        \pstTranslation[PosAngle=0] ODA[M]
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    \end{minipage}
    \vfill
    \noindent
\textbf{[Slide 6]}
    By design, our six axioms yield Euclid \textsc i.37 and 39,
    and these in turn give us the Hexagon Theorem,
    by Pappus's own proof
    (except he uses an intersection point of the bounding lines
    $DF$ and $CA$).

    It remains to prove Desargues's Theorem.

    Michael Beeson used the diagram of \textsc i.43 to define
    $\alpha=\beta$ in the rectangular case
    when $G$ lies on $OL$.

    We strengthen \textsc i.43 with its converse and more,
    in order to establish first a special case of Desargues.
    
\vfill
    \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
    \begin{minipage}{4cm}
      \begin{pspicture}(1.5,0)(5.5,8.5)
        %    \psgrid
        \psset{PointSymbol=none}
        \pstGeonode[PosAngle={0,180,180,0}]
        (5,8.3)A(0.6,8)O(0.6,2.5)C
        \pstHomO[HomCoef=0.35,PosAngle=90]OA[D]
        \pstTranslation[PointName=none]ACD[x]
        \pstInterLL[PosAngle=180] DxOCF
        \pstTranslation[PointName=none]OCA[P]
        \pstHomO[HomCoef=0.48]AP[B]
        \pstTranslation[PointName=none]OBC[Q]
        \pstTranslation[PointName=none]OCD[K]
        \pstInterLL[PosAngle=-65] DKOBE
        \pstInterLL[PosAngle=115] ACOBG
        \pstInterLL[PosAngle=-170] DFOBH
        {\psset{PointName=none}
        \pstInterLL DKCQL
        \pstTranslation CPF[M]
        \pstTranslation CQF[N]
        \pstTranslation AEH[x]}
        \pstInterLL[PosAngle=90] HxOAK
        \ncline CP\ncline BQ\ncline QC\ncline EL
        \ncline FM\ncline FN
                {\psset{linestyle=dotted}
                  \ncline HK\ncline KG\ncline AE}
        \psset{linewidth=2.4pt}
        \ncline OA\ncline OC\ncline OB
        \ncline AB\ncline DE
        \psset{linestyle=dashed}
        \ncline AC\ncline DF
        \psset{linestyle=dotted}
        \ncline CB\ncline FE
      \end{pspicture}
    \end{minipage}

    \vfill
    \noindent
      \textbf{[Slide 7]}
      Now we obtain Desargues's Theorem,
      that in triangles $ABC$ and $DEF$,
      where the bold solid and bold dashed sides are parallel,
      the bold dotted sides are also parallel,
      so that the triangles are \textbf{similar.}

      The converse will follow,
      that similar triangles are \emph{perspective from a point.}

      When we assume $AB\parallel OC$,
      the result follows by \textsc i.43 plus.

      To continue, by Pappus,
      $BAG\sim EDH$ yields $BAE\sim GKH$,
      where $GK\parallel BA$.

      Thus when $BG$ and $EH$ are bases of similar triangles
      with apices on $OA$,
      so are $BE$ and $GH$.

      We show that we can maintain similarity
      while moving the apices along $OA$.

      Then the special case of Desargues yields the general.
      \vfill

      \newpage

    \begin{minipage}{6cm}
      \begin{pspicture}(6,8.5)
        \psset{PointSymbol=none}
        \pstGeonode[PosAngle={180,180,0,0}]
        (0.6,6)A(0.6,0.2)D(5.4,1)H(3,8.3)E
        \pstHomO[HomCoef=0.5,PosAngle=180]AD[C]
        \pstTranslation[PointName=none] CED[x]
        \pstInterLL[PosAngle=0] DxEHF
        \pstTranslation[PointName=none] EAF[x]
        \pstInterLL[PosAngle=180] FxADB
        \pstTranslation[PointName=none] EAC[x]
        \pstInterLL[PosAngle=0,PointSymbol=default] CxEHL
        \pstTranslation[PointName=none] DHB[x]
        \pstInterLL[PosAngle=0] BxEHG
        \pstTranslation[PointName=none] HDE[x]
        \pstInterLL[PosAngle=180] ExADK
        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
        \ncline KD\ncline EH
        \ncline AE\ncline BF
        \psset{linestyle=none,hatchangle=0}
        \pspolygon[fillstyle=hlines]
        (K)(E)(A)(G)(B)(F)
        %        \psset{hatchsep=8pt}
        \pspolygon[fillstyle=vlines]
        (K)(E)(C)(H)(D)(F)
        \psset{linestyle=solid,linewidth=2.4pt}
        \ncline KF\ncline AG\ncline CH
        \psset{linestyle=dashed}
        \ncline CE\ncline DF
        \psset{linestyle=dotted}
        \ncline KE\ncline BG\ncline DH
      \end{pspicture}
    \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}
    \vfill
    \noindent
    \textbf{[Slide 8]}
    In two steps now, if $AHB\sim CKD$, then
    $AH'B\sim CK'D$.

    \begin{pspicture}(0,-1)(5,3)
      \psset{PointSymbol=none}
      \pstGeonode[PosAngle={-90,-90,180,0}]
        (1,0.6)O(4.4,0.6)D(0.6,2.8){K'}(3.2,2.8)K
        \pstTranslation[PosAngle=-90]{K'}OK[C]
        \pstHomO[HomCoef=0.4,PosAngle=135]OK[H]
        \pstTranslation[PointName=none] K{K'}H[x]
        \pstInterLL[PosAngle=180] HxO{K'}{H'}
        \pstTranslation[PointName=none] KDH[x]
        \pstInterLL[PosAngle=-90] HxODB
        \pstTranslation[PointName=none] O{K'}H[x]
        \pstInterLL[PosAngle=-90] HxODA
        \pspolygon(O)(D)(K)(K')
        \ncline OK\ncline KC
        \ncline AH\ncline HB\ncline H{H'}
        \psset{linestyle=dotted}
        \ncline A{H'}\ncline{H'}B
        \ncline C{K'}\ncline{K'}D
    \end{pspicture}
\hfill
    \begin{pspicture}(7,5)
      \psset{PointSymbol=none}
      \pstGeonode[PosAngle={180,0,0,0}]
      (0.6,1.4)O(6.4,2.3)A(6,0.6)B(5.2,4.5)C
      \pstHomO[HomCoef=0.6,PosAngle={45,-90,90}]O{A,B,C}[D,E,F]
      \psset{PosAngle=135}
      \pstInterLL EFOAH
      \pstInterLL BCOAG
      \ncline OA\ncline OB\ncline OC
      \pspolygon(A)(B)(C)
      \pspolygon(D)(E)(F)
    \end{pspicture}

    Hence if $EF\parallel BC$ and $DF\parallel CA$,
    so $HFD\sim GCA$, then $HED\sim GBA$, so $DE\parallel AB$.
    
    \vfill
\end{document}