\chapter{Transformations} 
   
\Citation{Beauty is the first test: there is no permanent place in the world for ugly mathematics.\\ 
  \Auteur{Godfrey Harold}{Hardy}{1877}{1947}{English mathematician}{A Mathematician's Apology}{1940}} 
   
A \nuvo{transformation} \pron{£btr£ansf£e£hm£q£sn} of the plane is a one-to-one mapping of points of the plane onto 
points of the plane. It is a kind of process which associates to each point $M$ in the plane one and only point $M'$ in 
the plane. $M'$ is the \nuvo{image} of $M$ under the transformation; $M$ is the \nuvo{pre-image} \pron{pr£I£h£im£i£d} of 
$M'$. Transformations are one-to-one because every point $M'$ has one and only one pre-image under a transformation. 
   
\section{Translations and reflections}   
   
\subsection{Translations}   
   
\begin{Def}   
  The \nuvo{translation} \pron{tr£ans£hl£q£sn} by \nuvo{vector} \pron{£hvekt£e£Z} \vect{u} is the transformation which 
  to each point $M$ in the plane associates the point $M'$ such that 
  $$\vecteur{MM'} = \vect{u}.$$ 
\end{Def}   
   
The translation by \vect{u} is denoted by $\dps t_{\vect{u}}$ (read `$t$ sub $u$') or by $t$ whenever there is no 
ambiguity. The image of the point $M$ under $\dps t_{\vect{u}}$ is denoted by $\dps t_{\vect{u}}\parent{M}$ (read `$t$ 
sub $u$ of $M$'). 
   
\begin{rem}   
  The image of the figure \Set under $\dps t_{\vect{u}}$ is denoted by $\dps t_{\vect{u}}\parent{\Set}$, whereas the 
  image of the line $(AB)$ under $\dps t_{\vect{u}}$ is denoted by $\dps t_{\vect{u}}\parent{AB}$ to avoid the more 
  cumbersome, even if more correct, $\dps t_{\vect{u}}\parent{(AB)}$. 
\end{rem}   

\begin{fgre}
  \includegraphics[totalheight=3cm,keepaspectratio=true]{figures/transf01} % translation
\end{fgre}

\begin{prop}   
  Let $A$ and $B$ be two points in the plane. Let \vect{u} be a vector and $t$ be the translation by \vect{u}.  Let 
  $A'=t(A)$ and $B'=t(B)$. 
\begin{itemize}   
\item $\dps \vecteur{A'B'}=\vecteur{AB}$ so $A'B' = AB$; 
\item $\dps t(AB)=(A'B')$ and $(A'B')$ is parallel to $(AB)$; 
\item if $\vect{u} \neq \vect{0}$ then there is no \nuvo{invariant} \pron{£in£hv£Pr£Qnt} point under $t$ \idest\ for all 
  point $M$ in the plane $t(M) \neq M$. 
\end{itemize}\end{prop}   
   
\begin{rem}   
  To each point $M$ the translation by \vect{0} (null vector) associates $M$ itself. All the points in the plane are 
  invariant. This translation is said to be the \nuvo{identity} \pron{£p£hdent£et£i} of the plane and usually denoted by 
  \Id. 
\end{rem}\index{$\Id$}   
   
\subsection{Reflection in a line}   
   
\begin{Def}   
  Let \Li be a line. The \emph{reflection} \pron{r£i£hflek£sn} \emph{in} (or \emph{about} or \emph{through}) \emph{line} 
  \Li is the transformation which to each point $M$ in the plane associates the point $M'$ such that \Li is the 
  midperpendicular of $[MM']$ if $M$ is not on \Li and such that $M'=M$ if $M$ is on \Li. 
\end{Def}\index{reflection!in a line}\index{reflection!about a line}\index{reflection!through a line}   
   
The reflection in \Li is usually denoted by $s_{\Li}$. 
   
\begin{fgre}
  \includegraphics[totalheight=3cm,keepaspectratio=true]{figures/transf02}  % symétrie axiale
\end{fgre}

\begin{prop}   
  Let $A$ and $B$ be two points in the plane. Let \Li be a line and $s$ be the reflection in \Li.  Let $A'=s(A)$ and 
  $B'=s(B)$. 
\begin{itemize}   
\item $A'B' = AB$; 
\item if $(AB)$ is perpendicular to \Li, then $(A'B') = (AB)$; 
\item if $(AB)$ is parallel to \Li, then $(A'B')$ is also parallel to \Li. 
\end{itemize}\end{prop}   
   
\begin{prop}   
  Let $A$ and $B$ be two points in the plane. Let \Li be a line and $s$ be the reflection in \Li.  Let $A'=s(A)$ and 
  $B'=s(B)$. 
   
  Assume\,\footnote{To assume: take or accept as being true for the purpose of argument.} \pron{£e£hsj£Um} that $(AB)$ 
  is not parallel to \Li. Let $I$ be their common point. Then $I$ is \nuvo{invariant} under $s$ \idest\  $s(I) = I$. 
  Moreover, $I$ is the only invariant point on $(AB)$. 
\end{prop}   
   
\subsection{Reflection in a point}   
   
\begin{Def}   
  Let $O$ be a point. The \emph{reflection in} (or \emph{about} or \emph{through}) \emph{point} $O$ is the 
  transformation which to each point $M$ in the plane associates the point $M'$ such that $O$ is the midpoint of $[MM']$. 
\end{Def}\index{reflection!in a point}\index{reflection!about a point}\index{reflection!through a point}   
   
The reflection in $O$ is usually denoted by $S_{O}$. 
   
\begin{fgre}
  \includegraphics{figures/transf03}  % symétrie centrale
\end{fgre}

\begin{prop}   
  Let $A$ and $B$ be two points in the plane. Let $O$ be a point and $S$ be the reflection in $O$.  Let $A'=S(A)$ and 
  $B'=S(B)$. 
\begin{itemize}   
\item $\vecteur{A'B'} = - \vecteur{AB}$ and consequently, 
\item $A'B' = AB$, 
\item $(A'B')$ is parallel to $(AB)$; 
\item the only invariant point under $S$ is $O$. 
\end{itemize}\end{prop}   
   
\section{Orientation of the plane, rotations}   
   
\subsection{Orientation of the plane}   
   
There are two directions on a circle: the clockwise and the anticlockwise direction. The anticlockwise direction is 
chosen to be the \emph{positive direction}. The other one is the \emph{negative direction}. 
\index{positive!direction}\index{negative!direction}\index{direction!positive}\index{direction!negative} 
   
To give a direction to the plane is to choose the same direction for all the circles in the plane.  The plane is then 
said to be \nuvo{directed}. The plane is generally assumed to be positively directed. 
   
\begin{fgre}
  \includegraphics{figures/transf04}  % orientations du plan
\end{fgre}
   
\begin{Def}   
  The triangle $ABC$ is said to be \emph{positive} if on the circumcircle of $ABC$ one goes from $A$ to $C$ through $B$ 
  in the positive direction. If $ABC$ is not positive then it is said to be \emph{negative}. 
\end{Def}\index{positive!triangle}\index{negative!triangle}   
   
\begin{fgre}
  \includegraphics{figures/transf05}  % triangle positif
\end{fgre}
   
\begin{rem}   
  If $ABC$ is positive then $BCA$ and $CAB$ are also positive whereas $CBA$, $BAC$ and $ACB$ are negative. 
\end{rem}   
   
\begin{Def}   
  The triangles $ABC$ and $DEF$ are said to have the \nuvo{same sense} if they are both positive or both negative. 
\end{Def}\index{sense}   
   
\subsection{Rotation}   
   
\begin{Def}   
  The \emph{rotation of angle §a anticlockwise about $O$} is the transformation which to each point $M\neq O$ in the 
  plane associates the point $M'$ such that: 
\begin{itemize}   
\item if $M\neq O$ then $OM = OM'$, $OMM'$ is positive and $\angle MOM' = §a$; 
\item if $M=O$ then $M'=O$. 
\end{itemize}   
\end{Def}   
   
\begin{rem}   
  The rotation of $§a^{\circ}$ anticlockwise about $O$ is equal to the rotation of $(360-§a)^{\circ}$ clockwise about 
  $O$. Therefore, the rotation of $180^{\circ}$ anticlockwise about $O$ is equal to the rotation of $180^{\circ}$ 
  clockwise about $O$ and to the reflection in $O$. 
\end{rem} % remplacé \degre par une construction du style $^{\circ}$ mais pb de distance ?   
   
\begin{prop}   
  Let $A$ and $B$ be two points in the plane. Let $O$ be a point and $R$ be the reflection of $§a^{\circ}$ anticlockwise 
  about $O$. Let $A'=R(A)$ and $B'=R(B)$. Let $C$ be such that $\vecteur{AC} = \vecteur{A'B'}$. Then: 
\begin{itemize}   
\item $A'B' = AB$; 
\item $\angle BAC = §a$; 
\item If the angle of $R$ is not null then $O$ is the only invariant point under $R$. 
\end{itemize}\end{prop}   
   
\begin{prop}   
  Translations and rotations --- and therefore reflections in a point --- map a triangle onto a triangle with the same 
  sense.  Reflections in a line map a triangle onto a triangle with the opposite sense. 
\end{prop}   
   
\section{Isometries}   
\begin{Def}   
  An \nuvo{isometry}\,\footnotemark{} \pron{£p£hs£om£itr£i} is any transformation from the plane onto itself that 
  preserves distance. 
\end{Def}   
\footnotetext{from Greek `isometria': equality of measure. The prefix `iso' means `equal' as in `isosceles': having two 
  equal legs. And `metria' is the measure as in `geometry' which is --- originally --- the science of the measure of 
  Earth.} 
   
In other words, if $h$ is an isometry, $A$ and $B$ two points in the plane, $A'$ and $B'$ their images under $h$ 
respectively, then $A'B' = AB$. 

\newpage
   
\begin{prop}   
  Translations, reflections and rotations are isometries. 
\end{prop}   
   
We can state the following useful properties of isometries: 
   
\begin{prop}   
  Let $A$, $B$ and $C$ be three points in the plane. Let $h$ be any isometry. Let $A'=h(A)$, $B'=h(B)$ and $C'=h(C)$. 
\begin{itemize}   
\item the image under $h$ of a circle of center $A$ and radius $r$, is a circle of center $A'$ and radius $r$; 
\item the image under $h$ of the segment $[AB]$ is the segment $[A'B']$; 
\item if $I$ is the midpoint of $[AB]$ then $h(I)$ is the midpoint of $[A'B']$; 
\item the image under $h$ of the line $(AB)$ is the line $(A'B')$; 
\item if $\angle ABC$ is right then $\angle A'B'C'$ is right; 
\item if the lines \Li and \Lip are parallel then $h(\Li)$ and $h(\Lip)$ are parallel; 
\item $\angle ABC = \angle A'B'C'$. 
\end{itemize}\end{prop}   
   
\section{Composition}   
   
Let $h$ and $k$ be two transformations of the plane. Let $M$ be a point in the plane. We can first apply $h$ to any 
point $M$ and, in so doing, obtain point $M^{\ast}$. Then we can apply $k$ to point $M^{\ast}$ and obtain $M'$. The 
transformation which maps $M$ onto $M'$ is said to be the \nuvo{composite} \pron{£hk£omp£ez£it} of $h$ with $k$. We have 
composed \pron{k£em£hp£wzd} those two transformations to obtain a new one.\index{compose} That operation is called the 
\nuvo{composition} \pron{£bk£omp£e£hz£i£sn}. 
   
Usually, the composite of $h$ with $k$ ---~\idest\  $h$ followed by $k$~--- is denoted by $k\circ h$ (read `$k$ circle 
$h$' or `$k$ of $h$'). The first transformation to operate is written on the right hand side for it is there that we 
write the point to be transformed: $k\circ h(M)$. Here we can write $k\circ h(M) = k(M^{\ast})=M'$. 
   
\subsection{Composition of two translations}   
   
\begin{prop}   
  The composite of the translation by \vect{u} with the translation by \vect{v} is the translation by 
  $\vect{u}+\vect{v}$. 
  $$ 
  t_{\vect{v}}\circ t_{\vect{u}} = t_{\vect{u}}\circ t_{\vect{v}} = t_{\vect{u}+\vect{v}}$$ 
\end{prop}   
   
The second figure below is a schematic \pron{sk£i£hm£at£ik} representation of the composition of two translations.  The 
first to operate maps $M$ onto $M^{\ast}$ then the second maps $M^{\ast}$ onto $M'$. The short-cut from $M$ to $M'$ 
represents the composite of the two preceding translations. 
   
\begin{fgre}
  \includegraphics{figures/transf06}
\end{fgre}
   
\subsection{Composition of two reflections in a point}   
   
\begin{prop}   
  The composite of the reflection in $O$ with the reflection in $O'$ is the translation by $2\vecteur{OO'}$. 
  $$ 
  S_{O'}\circ S_O = t_{2\vecteur{OO'}} \quad \text{and} \quad S_{O}\circ S_{O'} = t_{2\vecteur{O'O}}$$ 
\end{prop}   
   
\begin{fgre}
  \includegraphics{figures/transf07}
\end{fgre}

{\slshape Try your hand at proving that, for any point $M$, \vecteur{MM'} is truly equal to $2\,\vecteur{OO'}$. Use the
  figure above and the triangle law (see chap.\ref{vect-chap} prop. \ref{prop-trianglelaw}).}

\section{Projections}   
   
\begin{Def}   
  The orthogonal projection from the plane to line \Li is the mapping which to each point $M$ in the plane associates 
  the point $M'$ such that $M'$ is the intersection point of \Li and the perpendicular to \Li passing through $M$. 
\end{Def}   
   

\begin{fgre}
  \includegraphics{figures/transf08}  % projection
\end{fgre}
   
\begin{rem}   
  A projection is not an isometry for it does not map the plane onto itself. Moreover a projection does not preserve 
  length.  (See figure above where $AB \neq CD$ and $A'B' = C'D'$. 
\end{rem}   
   
\begin{prop}   
  A projection preserves the midpoints \idest{} if $I$ is the midpoint of $[AB]$ and if $A'$, $B'$ and $I'$ are the 
  respective image under the projection on \Li then $I'$ is the midpoint of $[A'B']$. 
\end{prop}
\endinput   
   
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