\chapter{Euclidean geometry}\label{ch!eucl}
\index{Euclidean!geometry|(}
\markboth{\quad\thechapter\ Euclidean geometry\hfill}{}
This chapter discusses the geometry of $n$-dimensional Euclidean space
$\E^n$,\index{Euclidean!space $\E^n$} together with its
distance\index{Euclidean!distance} function\index{distance!function}. The
distance leads to other notions such as angles\index{angle} and congruent
triangles. Choosing a Euclidean coordinate\index{coordinate!frame}
frame,\index{Euclidean!frame}\index{frame!Euclidean} consisting of an origin
$O$ and an orthonormal basis of vectors out of $O$, leads to a description
of $\E^n$ by coordinates, that is, to an identification $\E^n=\R^n$.

A map of Euclidean space preserving Euclidean distance is called a {\em
motion}\index{motion} or {\em rigid body motion}. Motions are fun to study
in their own right. My aims are

\begin{enumerate} \renewcommand{\labelenumi}{(\arabic{enumi})}
\item to describe motions in terms of linear algebra and matrixes;
\item to find out how many motions there are;
\item to describe (or classify) each motion individually.
\end{enumerate}

I do this rather completely for $n=2,3$ and some of it for all $n$. For
example, the answer to (2) is that all points of $\E^n$, and all sets of
orthonormal coordinate frames at a point, are equivalent: given any two
frames, there is a unique motion\index{motion} taking one to the other. In
other words, any point can serve as the origin, and any set of orthogonal
axes\index{orthogonal!axes} as the coordinate frames. This is the geometric
and philosophic principle that space is homogeneous and isotropic (the same
viewed from every point and in every direction). The answer to (3) in $\E^2$
is that there are four types of motions: translations and rotations,
reflections and glides (Theorem~\ref{sec!motE2}). \index{translation}
\index{rotation}
\index{reflection}

The chapter concludes with some elementary sample theorems of plane
Euclidean geometry.

\section{The metric on $\protect\R^n$}\label{Rn}
\index{triangle inequality|(}\index{metric!space}
\markright{\hfill \thesection\ The metric on $\protect\R^n$\quad}

Throughout the book, I write $\R^n$ for the vector space of $n$-tuples
$(x_1,\dots,x_n)$ of real numbers. I start by discussing its metric
geometry. The familiar Euclidean distance function\index{Euclidean!distance}
on $\R^n$ is defined by
\begin{equation}
|\bx-\by|=\sqrt{\vphantom\sum}\left(\sum(x_i-y_i)^2\right),
\quad\hbox{where $\bx= \begin{pmatrix} x_1\\ \vdots \\ x_n
\end{pmatrix}$ and
$\by= \begin{pmatrix} y_1\\ \vdots \\ y_n \end{pmatrix}$.}
\label{eq!eud}
\end{equation}
The relationship between this distance function\index{distance!function} and
the Euclidean inner product\index{Euclidean!inner product} (or dot product)
$\bx\cdot\by=\sum x_i y_i$ on $\R^n$ is discussed in
Appendix~\ref{sec!euclandlor}. The more important point is that the
Euclidean distance\index{Euclidean!distance} (\ref{eq!eud}) is a metric on
$\R^n$. If you have not yet met the idea of a {\em metric} on a set $X$, see
Appendix~\ref{app!metric}; for now recall that it is a distance function
$d(\bx,\by)$ satisfying positivity, symmetry and the triangle inequality.
Both the positivity $|\bx-\by|\ge0$ and symmetry $|\bx-\by|=|\by-\bx|$ are
immediate, so the point is to prove the triangle inequality.

\begin{thm}[Triangle inequality]
\begin{equation}
|\bx-\by|\le |\bx-\bz|+|\bz-\by|,\quad\hbox{for all
$\bx,\by,\bz\in\R^n$,}
\label{eq!1}
\end{equation}
with equality if and only if $\bz=\bx+\la(\by-\bx)$ for $\la$ a real
number between $0$ and $1$.
\end{thm}

\setcounter{figure}{\value{section}}
\addtocounter{figure}{-1}
\begin{figure}
\centering
% \leavevmode\hbox{\epsfysize=4.8 true cm\epsffile{eucl_triangle_ineq.eps}}
\centering\input{eucl_trian_inR.pstex_t}
\caption{Triangle inequality}
% \label{fig!tri}
\end{figure}

\begin{pf} Set $\bx-\bz=\bu$ and $\bz-\by=\bv$ so that
$\bx-\by=\bu+\bv$; then (\ref{eq!1}) is equivalent to
\begin{equation}
\sqrt{\vphantom\sum}\sum u_i^2+\sqrt{\vphantom\sum}\sum v_i^2
\ge\sqrt{\vphantom\sum}\left(\sum(u_i+v_i)^2\right).
\label{eq!2}
\end{equation}
Note that both sides are nonnegative, so that squaring, one sees that
(\ref{eq!2}) is equivalent to
\begin{align}
\sum u_i^2+\sum v_i^2 +2\sqrt{\vphantom\sum}\left(\sum
u_i^2\cdot\sum v_i^2\right)
& \ge\sum(u_i+v_i)^2\nonumber\\
&=\sum u_i^2+\sum v_i^2+2\sum u_iv_i.
\label{eq!2u_iv_i}
\end{align}
Cancelling terms, one sees that (\ref{eq!2u_iv_i}) is equivalent to
\begin{equation}
\sqrt{\vphantom\sum}\left(\sum u_i^2\cdot\sum v_i^2\right)\ge\sum
u_iv_i.
\label{eq!3}
\end{equation}
If the right-hand side is negative then (\ref{eq!3}), hence also
(\ref{eq!1}), is true and strict. If the right-hand side of
(\ref{eq!3}) is $\ge0$ then it is again permissible to square both
sides, giving
\begin{equation}
\sum u_i^2\cdot\sum v_j^2\ge
\left(\sum u_iv_i\right) \left(\sum u_jv_j\right).
\label{eq!4}
\end{equation}
You will see at once what is going on if you write this out explicitly for
$n=2$ and expand both sides. For general $n$, the trick is to use two
different dummy indexes $i,j$ as in (\ref{eq!4}): expanding and cancelling
gives that (\ref{eq!4}) is equivalent to
\begin{equation}
\sum_{i>j}(u_iv_j-u_jv_i)^2\ge0.\label{eq!5}
\end{equation}
Now (\ref{eq!5}) is true, so retracing our steps back through the
argument gives that (\ref{eq!1}) is true. Finally, equality in
(\ref{eq!1}) holds if and only if $u_iv_j=u_jv_i$ for all $i,j$ (from
(\ref{eq!5})) and $\sum u_iv_i\ge0$ (from the right-hand side of
(\ref{eq!3})); that is, $\bu$ and $\bv$ are proportional, $\bu=\mu
\bv$ with $\mu\ge0$. Rewriting this in terms of $\bx,\by,\bz$ gives
the conclusion. \QED \end{pf}

\section{Lines and collinearity in $\R^n$}\label{sec!col}
\markright{\hfill \thesection\ Lines and collinearity in $\R^n$\quad}
There are several ways of defining a line (already in the usual
$x,y$ plane $\R^2$); I choose one definition for $\R^n$.

\begin{defn}
Let $\bu\in\R^n$ be a fixed point and $\bv\in\R^n$ a nonzero
direction vector. The {\em line} $L$ starting at $\bu\in\R^n$ with
direction vector $\bv$ is the set
\[
L:=\bigl\{\bu+\la \bv\bigm|\la\in\R\bigr\}\subset\R^n.
\]
Three distinct points $\bx,\by,\bz\in\R^n$ are {\em collinear} if
they are on a line.
\end{defn}
If I choose the starting point $\bx$, and the direction vector
$\bv=\by-\bx$ then $L=\{(1-\la)\bx+\la \by\}$. To say that distinct
points $\bx,\by,\bz$ are collinear means that $\bz=\{(1-\la)\bx+\la
\by\}$ for some $\la$. Writing
\[
[\bx,\by]=\bigl\{\bx+\la(\by-\bx)\bigm|0\le\la\le1\bigr\}
\]
for the line segment\index{line!segment} between $\bx$ and $\by$,
the possible orderings of $\bx,\by,\bz$ on the line $L$ are controlled by
\[
\renewcommand{\arraystretch}{1.2}
\left.
\begin{array}{r}
\la \le 0 \\
0\le \la \le1 \\
1\le\la
\end{array}
\right\} \iff
\left\{
\begin{array}{l}
\bx\in[\bz,\by], \\
\bz\in[\bx,\by], \\
\by\in[\bx,\bz].
\end{array}
\right.
\]
Together with the triangle inequality Theorem~\ref{Rn}, this proves the
following result:

\begin{cor} Three distinct points $\bx,\by,\bz\in\R^n$ are collinear if and
only if (after a permutation of $\bx,\by,\bz$ if necessary)
\[
|\bx-\by|+|\by-\bz|=|\bx-\bz|.
\]
In other words, collinearity is determined by the metric. \end{cor}

\section{Euclidean space $\E^n$}\label{En}
\index{Euclidean!space $\E^n$|(}
\markright{\hfill \thesection\ Euclidean space $\E^n$\quad}
After these preparations, I am ready to introduce the main object of study:
{\em Euclidean $n$-space} $(\E^n,d)$ is a\index{metric!space} metric space
(with metric $d$) for which there exists a bijective map $\E^n\to\R^n$, such
that if $P,Q\in\E^n$ are mapped to $\bx,\by\in\R^n$ then
\[
d(P,Q)=|\by-\bx|.
\]
In other words, $(\E^n,d)$ is isometric\index{isometry} to the vector space
$\R^n$ with its usual distance function, if you like this kind of language.

Since lines and collinearity in $\R^n$ are characterised purely in terms of
the Euclidean distance function,\index{Euclidean!distance} these notions
carry over to $\E^n$ without any change: three points of $\E^n$ are
collinear if they are collinear for some isometry $\E^n\to\R^n$ (hence for
all possible isometries); the lines of $\E^n$ are the lines of $\R^n$ under
any such identification. For example, for points $P,Q\in \E^n$, the line
segment\index{line} $[P,Q]\subset\E^n$ is the set
\[
[P,Q]=\bigl\{R\in\E^n\bigm|d(P,R)+d(R,Q)=d(P,Q)\bigr\}\subset\E^n.
\]

\begin{rem}
The main point of the definition of $\E^n$ is that the map $\E^n\to\R^n$
identifying the metrics is not fixed throughout the discussion; I only
insist that one such isometry should exist. A particular choice of
identification preserving the metric is referred to as a choice of {\em
(Euclidean) coordinates}. Points of $\E^n$ will always be denoted by capital
letters $P, Q$; once I choose a bijection, the points acquire {\em
coordinates} $P=(x_1,\dots,x_n)$. In particular, any coordinate system
distinguishes one point of $\E^n$ as the origin $(0,\dots, 0)$; however,
different identifications pick out different points of $\E^n$ as their
origin. If you want to have a Grand Mosque of Mecca or a Greenwich
observatory, you must either receive it by Divine Grace or make a deliberate
extra choice. The idea of space ought to make sense without a coordinate
system,\index{coordinate!system} but you can always fix one if you like.

You can also look at this process from the opposite point of view. Going
from $\R^n$ to $\E^n$, I forget the distinguished origin $0\in\R^n$, the
standard coordinate system, and the vector space structure of $\R^n$,
remembering only the distance\index{distance!function} and properties that
can be derived from it.
\end{rem}

\section{Digression: shortest distance}\label{sec!shortest}
\index{shortest distance}
\index{Euclidean!space $\E^n$}
\markright{\hfill \thesection\ Digression: shortest distance\quad}
As just shown, the metric of Euclidean space $\E^n$ determines the
lines.\index{Euclidean!line} This section digresses to discuss the idea
summarised in the well known clich\'e ``a straight line is the shortest
distance between two points''; while logically not absolutely essential in
this chapter, this idea is important in the philosophy of Euclidean geometry
\index{Euclidean!geometry} (as well as spherical and hyperbolic geometry).
\index{hyperbolic!geometry}
\index{spherical!geometry}

\begin{principle}
The distance\index{distance!function} $d(P,Q)$ between two points
$P,Q\in\E^n$ is the length of the shortest curve joining $P$ and $Q$.
The line\index{line} segment\/ $[P,Q]$ is the unique shortest curve
joining $P,Q$.
\end{principle}

\begin{skpf} This looks obvious: if a curve $C$ strays off the
straight and narrow to some point $R\notin[P,Q]$, its length is at
least
\[
d(P,R)+d(R,Q)>d(P,Q).
\]

The statement is, however, more subtle: for instance, it clearly does not
make sense without a definition of a curve $C$ and its length. A curve $C$
in $\E^n$ from $P$ to $Q$ is a family of points $R_t\in\E^n$, depending on a
``time variable'' $t$ such that $R_0=P$ and $R_1=Q$. Clearly $R_t$ should at
least be a {\em continuous}\index{continuous} function of $t$ -- if you
allow instantaneous ``teleporting'' between far away points, you can
obviously get arbitrarily short paths.\index{length of path}

The proper definition of curves and lengths of curves belongs to
differential geometry or analysis. Given a ``sufficiently smooth'' curve,
you can define its length as the integral $\int_C\dd s$ along $C$ of the
infinitesimal arc length $\dd s$, given by $\dd s^2=\sum_{i=1}^n\dd x_i^2$.
Alternatively, you can mark out successive points
$P=R_0,R_1,\dots,R_{N+1}=Q$ along the curve, view the sum $\sum_{i=0}^N
d(R_i,R_{i+1})$ as an approximation to the length of $C$, and define the
length of $C$ to be the supremum taken over all such piecewise linear
approximations. To avoid the analytic details (which are not at all
trivial!), I argue under the following weak assumption: under any reasonable
definition of the length of $C$,
\begin{quote}
for any $\ep>0$, the curve $C$ can be closely approximated by a
piecewise linear path\index{length of path} made up of short
intervals $[P,R_1]$, $[R_1,R_2]$, etc., such that
\[\mbox{ length of $C$ }\geq\mbox{ sum of the lengths of the intervals}-\ep.\]
\end{quote}
However, by the triangle inequality $d(P,R_2)\le
d(P,R_1)+d(R_1,R_2)$, so that the piecewise linear path can only get
shorter\index{minimum over paths} if I omit $R_1$. Dealing likewise
with $R_2$, $R_3$, etc., it follows that the length of $C$ is $\ge
d(P,Q)-\ep$. Since this is true for any $\ep>0$, it follows that the
length of $C$ is $\ge d(P,Q)$. Thus the line interval $[P,Q]$ joining
$P,Q$ is the shortest path\index{shortest distance} between them, and
its length is $d(P,Q)$ by definition. \QED \end{skpf}
\index{triangle inequality|)}

\section{Angles}\label{angles}
\index{angle|(}
\markright{\hfill \thesection\ Angles\quad}
The geometric significance of the Euclidean inner
product\index{Euclidean!inner product}
$\bx\cdot\by={\vphantom\sum}\sum_{i=1}^n x_iy_i$ on $\R^n$
(Section~\ref{sec!euclandlor}) is that the inner product measures the size of
the {\em angle} $\angle\bx\by\bz$ based at $\by$ for $\bx, \by,\bz\in \R^n$:
\begin{equation}\label{anglemeasure}
\cos(\angle\bx\by\bz)
=\frac{(\bx-\by)\cdot(\bz-\by)}{|\bx-\by||\bz-\by|}\,.
\end{equation}
By convention, I usually choose the angle to be between $0$ and $\pi$.
In particular, the vectors $\bx-\by$, $\bz-\by$ are {\em orthogonal}
if $(\bx-\by)\cdot(\bz-\by)=0$ .\index{orthogonal!vector}

The notion of angle is easily transported to Euclidean space
$\E^n$.\index{Euclidean!space $\E^n$} Namely, the angle spanned by three
points of $\E^n$ is defined to be the corresponding angle in $\R^n$ under a
choice of coordinates. The angle is independent of this choice, because the
inner product\index{Euclidean!inner product} in $\R^n$ is determined by the
quadratic form\index{quadratic form} (Proposition~\ref{bilin}), and so
ultimately by the metric of $\E^n$. In other words, the notion of angle is
intrinsic to the geometry of $\E^n$.

There is one final issue to discuss regarding angles that is specific to the
Euclidean plane $\E^2$.\index{Euclidean!plane $\E^2$} Namely, {\em once I
fix a specific coordinate system} in $\E^2$, angles $\angle PQR$ acquire a
\begin{figure}[hb]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}}
\centering
\input{eucl_angle_dir.pstex_t}
\caption{Angle with direction}
\label{signed_angles}
\index{angle!signed}
\end{figure}
{\em direction} as well as a size, once we agree (as we usually do)
that an anticlockwise angle counts as positive, and a clockwise angle as
negative. In Figure~\ref{signed_angles},
\[
\angle PQR = -\angle RQP = \theta.
\]

Under this convention, angles lie between $-\pi$ and $\pi$. Of course
formula~(\ref{anglemeasure}) does not reveal the sign as
$\cos\theta=\cos(-\theta)$. It is important to realise that the
direction of the angle is not intrinsic to $\E^2$, since a different
choice of coordinates may reverse the sign.
\index{angle|)}

\section{Motions}\label{sec!Emot}
\index{motion}
\index{preserves distances|(}
\index{distance!function}
\markright{\hfill \thesection\ Motions\quad}

A {\em motion} $T\colon \E^n\to\E^n$ is a transformation that
preserves distances; that is, $T$ is bijective, and
\[
d(T(P),T(Q))=d(P,Q)\quad\hbox{for all $P,Q\in\E^n$.}
\]
The word motion\index{motion} is short for {\em rigid body motion};
it is alternatively called {\em iso\-metry}\index{isometry} or {\em
congruence}. To say that $T$ preserves distances\index{preserves
distances} means that there is ``no squashing or bending'', hence the
term rigid body motion. See Figure~\ref{rigid_motion}.
\setcounter{figure}{\value{section}}
\addtocounter{figure}{-1}
\begin{figure}[ht]
\centering
\leavevmode\hbox{\epsfysize=5.5 true cm\epsffile{eucl_motion.eps}}
\caption{Rigid body motion}
\label{rigid_motion}
\end{figure}

I study motions\index{motion} in terms of coordinates. After a choice of
coordinates $\E^n\to\R^n$, a motion $T$ gives rise to a map
$T\colon\R^n\to\R^n$, its coordinate expression, which satisfies
\[
|T(\bx)-T(\by)|=|\bx-\by|\quad\hbox{for all $\bx,\by\in\R^n$.}
\]
The first thing I set out to do is to get from the abstract ``preserves
distance''\index{preserves distances} definition of a motion\index{motion}
to the concrete coordinate expression $T(\bx)=A\bx+\bb$ with $A$ an
orthogonal matrix.\index{orthogonal!matrix} In the case of the Euclidean
plane $\E^2$, the result is even more concrete; $A$ is either a rotation
matrix\index{rotation!matrix} or a reflection
matrix:\index{reflection!matrix}
\[
\begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta &
\cos\theta \end{pmatrix}
\quad\hbox{or}\quad
\begin{pmatrix} \cos\theta &
\sin\theta\\ \sin\theta & -\cos\theta \end{pmatrix}.
\]

\section{Motions and collinearity}\label{sec!mcol}
\index{motion}
\markright{\hfill \thesection\ Motions and collinearity\quad}
\begin{prop}
A motion $T\colon \E^n\to\E^n$ preserves collinearity of points, so
it takes lines to lines.
\end{prop}

\begin{pf} $P,Q,R\in E^n$ are collinear if and only if, possibly
after a permutation of $P,Q,R$,
\[
d(P,R)+d(R,Q)=d(P,Q).
\]
But $T$ preserves the distance function,\index{preserves distances|)}
so this happens if and only if, possibly after a permutation,
\[
d(T(P),T(R))+d(T(R),T(Q))=d(T(P),T(Q))
\]
which is equivalent to $T(P),T(Q),T(R)$ collinear. \QED
\end{pf}

The point is of course that, as we saw in~\ref{En},
collinearity can be defined purely in
terms of distance;\index{distance!function} since a
motion\index{motion} $T$ preserves distance, it preserves
collinearity.

\section{A motion is affine linear on lines}\label{sec!affl}\index{motion}
\markright{\hfill \thesection\ A motion is affine linear on lines\quad}

\begin{prop}
If\/ $T\colon\R^n\to\R^n$ is a motion expressed in coordinates, then
\[
T((1-\la)\bx+\la \by)=(1-\la)T(\bx)+\la T(\by)
\]
for all $\bx,\by\in\R^n$ and all $\la\in\R$.
\end{prop}

\begin{pf} A calculation based on the same idea as the previous
proof: let $\bz=(1-\la)\bx+\la \by$. If $\bx=\by$ there is nothing to
prove; set $d=|\bx-\by|$. Assume first that $\la\in[0,1]$, so that
$\bz\in[\bx,\by]$. Then, as in the previous proposition,
$T(\bz)\in[T(\bx),T(\by)]$, so $T(\bz)=(1-\mu)T(\bx)+\mu T(\by)$ for
some $\mu$. But $|\bz-\bx|=\la d$, so $T(\bz)$ is the point at
distance $(1-\la)d$ from $T(\by)$ and $\la d$ from $T(\bx)$, that is,
$\mu=\la$.

If $\la<0$, say, then $\bx\in[\by,\bz]$ with $\bx=(1-\la')\by+\la'
\bz$ and the same argument gives $T(\bx)=(1-\la')T(\by)+\la'T(\bz)$,
and you can derive the statement as an easy exercise. (The point is
to write $\la'$ as a function of $\la$; see \exref{ex!col}.) \QED
\end{pf}

\section{Motions are affine transformations} \label{sec!afflinE}
\index{affine!transformation|(}
\index{affine!linear!map|(}
\markright{\hfill \thesection\ Motions are affine transformations\quad}

\paragraph{Definition} A map $T\colon \E^n\to\E^n$ is an {\em affine
transformation} if it is given in a coordinate system by $T(\bx)=A\bx+\bb$,
where $A=(a_{ij})$ is an $n\times n$ matrix with nonzero determinant and
$\bb=(b_i)$ a vector; in more detail,
\begin{equation}
\bx=(x_i)\mapsto \by=\Bigl(\sum_{j=1}^na_{ij}x_j+b_i\Bigr),
\quad\hbox{or}\quad
\begin{pmatrix}
x_1\\
\vdots\\
x_n
\end{pmatrix}
\mapsto
A
\begin{pmatrix}
x_1\\
\vdots\\
x_n
\end{pmatrix}
+
\begin{pmatrix}
b_1\\
\vdots\\
b_n
\end{pmatrix}.
\end{equation}

% The subsection lifted from aff.tex BEGINS HERE

\begin{prop}
Let $T\colon \E^n\to \E^n$ be any map. Equivalent conditions:

\begin{enumerate}
\renewcommand{\labelenumi}{(\arabic{enumi})}
\item $T$ is given in some coordinate system by $T(\bx)=A\bx+\bb$ for $A$ an
$n\times n$ matrix.

\item For all vectors $\bx,\by\in \R^n$ and all $\la,\mu\in\R$ we have
\[
T\bigl(\la \bx+\mu \by\bigr)-T(0)=
\la\bigl(T(\bx)-T(0)\bigr)+\mu\bigl(T(\by)-T(0)\bigr).
\]

\item For all $\bx,\by\in \R^n$ and all $\la\in\R$
\[
T\bigl((1-\la)\bx+\la \by\bigr)=(1-\la)T(\bx)+\la T(\by).
\]
that is, $T$ is affine linear when restricted to any line.
\end{enumerate}
\end{prop}

\paragraph{Discussion} The point of the proposition is that condition $(3)$
is a priori much weaker than the other two; it only requires that the map
$T$ is affine when restricted to lines. Note also that using the origin $0$
in (2) seems to go against my expressed wisdom that there is no
distinguished origin in the geometry of $\E^n$. However, recall that any
point $P\in\E^n$ can serve as origin after a suitable translation.

\begin{pf} $(1)\implies(2)$ is an easy exercise. (2) means exactly that if
after performing $T$ we translate by minus the vector $\bb=T(0)$ to take
$T(0)$ back to $0$, then $T$ becomes a linear map of vector spaces. Thus
$(2)\implies(1)$ comes from the standard result of linear algebra expressing
a linear map as a matrix.

(3) is just the particular case $\la+\mu=1$ of (2). Thus the point of the
proposition is to prove $(3)\implies(2)$.

\setcounter{figure}{\value{section}}
\addtocounter{figure}{-1}
\begin{figure}
\centering
% \leavevmode\hbox{\epsfysize=6.7 true cm\epsffile{aff_span_trick.eps}}
\centering\input{aff_sumdiag.pstex_t}
\caption{Affine linear construction of $\protect\la \bx+\protect\mu \by$}
\label{fig!aff}
\end{figure}

Statement (2) concerns only the $2$-dimensional vector subspace spanned by
$\bx,\by\in V$. We use statement (3) on the two lines $0\bx$ and $0\by$ (see
Figure~\ref{fig!aff}), to get
\[
T(2\la \bx)=(1-2\la)T(0)+2\la T(\bx)
\]
and
\[
T(2\mu \by)=(1-2\mu)T(0)+2\mu T(\by).
\]
Now apply (3) again to the line spanned by $2\la \bx$ and $2\mu \by$:
\begin{align*}
T\bigl(\la \bx+\mu \by\bigr)
&=\half T(2\la \bx)+\half T(2\mu \by)\\
&=\half \bigl((1-2\la)T(0)+2\la T(\bx)\bigr)
+\half \bigl((1-2\mu)T(0)+2\mu T(\by)\bigr)\\
&=T(0)+\la\bigl(T(\bx)-T(0)\bigr)+\mu\bigl(T(\by)-T(0)\bigr),
\end{align*}
as required. \QED \end{pf}

\begin{rem} Dividing by $2$ here is just for the sake of an easy life:
$\{\half,\half\}$ is a convenient solution of $\la+\mu=1$. The point is just
that $\la \bx+\mu \by$ lies on a line containing chosen points of
$0\bx$ and $0\by$. The argument for $(3)\implies(2)$ can be made to work
provided every line has $\ge3$ points, that is, over any field with $>2$
elements. \end{rem}
\index{affine!linear!map|)}
\index{affine!transformation|)}

% The subsection lifted from aff.tex ENDS HERE

%Recall that a map $T\colon \R^n\to\R^n$ is an affine transformation
%if it is given by $T(\bx)=A\bx+\bb$, where $A=(a_{ij})$ is an
%invertible $n\times n$ matrix and $\bb=(b_i)$ a vector.

\begin{cor}
\index{Euclidean!motion}
\index{motion}
A Euclidean motion $T\colon \E^n\to\E^n$ is an affine
transformation, given in any choice of coordinates $\E^n\to\R^n$ by
$T(\bx)=A\bx+\bb$.
\end{cor}

This follows at once from Proposition~\ref{sec!mcol}, the implication
$(3)\implies(1)$ in the previous proposition, and the fact that $T$ is
bijective, so the matrix $A$ must be invertible.

\section[Euclidean motions and orthogonal transformations]{Euclidean
motions and orthogonal\\ transformations}\label{sec!orth}
\index{orthogonal!transformation}
\markright{\hfill \thesection\ Euclidean motions and orthogonal
transformations\quad}
This section makes a brief use of the relationship between the standard
quadratic form\index{quadratic form} $|\bx|^2=\sum x_i^2$ on $\R^n$ and
the associated inner product\index{Euclidean!inner product}
$\bx\cdot\by=\sum x_iy_i$. If this is not familiar to you, I refer you once
again to Appendix~\ref{app!q} for a general discussion.

\begin{prop} Let $A$ be an $n\times n$ matrix and $T\colon
\R^n\to\R^n$ the map defined by $\bx\mapsto A\bx$. Then the following are
equivalent conditions:

\begin{enumerate} \renewcommand{\labelenumi}{(\arabic{enumi})}

\item $T$ is a motion\index{motion} $T\colon\E^n\to\E^n$.

\item $A$ preserves the quadratic form; that is,
$|A\bx|=|\bx|$ for all\/ $\bx\in\R^n$.

\item $A$ is an {\em orthogonal matrix};\index{orthogonal!matrix}
that is, it satisfies $\tr AA=I_n$.
\end{enumerate}
\end{prop}

\begin{pf} $(1)\implies(2)$ is trivial. Conversely,
\[
|A\bx-A\by|^2=|A(\bx-\by)|^2=|\bx-\by|^2,
\]
where the first equality is linearity, and the second follows from (2).
Thus $T$ preserves length, so it is a motion.
$(2)\iff(3)$ is proved in Proposition~\ref{sec!fiesymm}, where you
can also read more about orthogonal matrixes\index{orthogonal!matrix} if
you wish to. \QED \end{pf}

Together with Corollary~\ref{sec!mcol}, this proves the
following very important statement:

\begin{cor}
A Euclidean motion\index{Euclidean!motion}\index{motion}
$T\colon\E^n\to\E^n$ is expressed in coordinates as
\[
T(\bx)=A\bx+\bb
\]
with $A$ an orthogonal matrix, and $\bb\in\R^n$ a vector.
\index{orthogonal!matrix|(} \end{cor}

An immediate check shows that an orthogonal
matrix\index{orthogonal!matrix} $A$ has determinant $\det A=\pm1$
(see Lemma~\ref{sec!fiesymm}).

\begin{defn}
\index{direct motion}
\index{opposite motion}
Let $T\colon\E^n\to\E^n$ be a motion\index{motion} expressed in coordinates
as $T(\bx)=A\bx+\bb$. I call $T$ {\em direct} (or {\em orientation
preserving}) if $\det A=1$ and {\em opposite} (or {\em orientation
reversing}) if $\det A=-1$.\index{Euclidean!space $\E^n$|)}
\end{defn}

The meaning of this notion in $\E^2$ and $\E^3$ is familiar in terms
of left-right orientation, and it may seem pretty intuitive that it
does not depend on the choice of coordinates. However, I leave the
proof to Ex.~\ref{ch!gp}.\ref{ex!or}.

\section{Normal form of an orthogonal matrix}
\label{sec!norm}
\index{normal form of a matrix|(}
\markright{\hfill \thesection\ Normal form of an orthogonal
matrix\quad}

The point of this section is to express an orthogonal map
$\al\colon\R^n\to \R^n$ in a simple form in a suitable orthonormal basis of
$\R^n$. This section may seem an obscure digression into linear algebra, but the
result is central to understanding motions of Euclidean space.

\subsection{The $2\times 2$ rotation and reflection matrixes}
\label{sec!refl}
\index{reflection!matrix}
\index{rotation!matrix}
As a prelude to an attack on the general problem, consider the
instructive case $n=2$. The conditions for a $2\times2$
matrix $A=\left(\begin{smallmatrix} a&b\\c&d
\end{smallmatrix}\right)$ to be orthogonal\index{orthogonal!matrix}
are:
\[
\tr AA=1\iff
\begin{pmatrix} a&c\\b&d \end{pmatrix}
\begin{pmatrix} a&b\\c&d \end{pmatrix}=
\begin{pmatrix} 1&0\\0&1 \end{pmatrix}
\iff
\renewcommand{\arraystretch}{1.1}
\renewcommand{\arraycolsep}{0.25em}
\left\{
\begin{array}{rcl} a^2+c^2&=&1,\\ ab+cd&=&0,\\ b^2+d^2&=&1.
\end{array}
\right.
\]
Now $(a,c)\in\R^2$ is a point of the unit circle, so I can write
\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}a}
\centering \input{eucl_new_rot.pstex_t}
\caption{A rotation in coordinates}
\end{figure}
$a=\cos\theta$, $c=\sin\theta$ for some $\theta\in[0,2\pi)$. Then
there are just two possibilities for $b,d$, giving
\[
A=
\begin{pmatrix}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{pmatrix}
\quad \hbox{or}\quad
\begin{pmatrix}
\cos\theta & \sin\theta\\
\sin\theta & -\cos\theta
\end{pmatrix}.
\]

The first of these corresponds to a direct motion\index{motion} (because
$\det A=1$), and you recognise it as a rotation\index{rotation} around the
origin through $\theta$. In fact it takes
\addtocounter{figure}{-1}
\[
\begin{pmatrix} 1\\0 \end{pmatrix}
\mapsto
\begin{pmatrix} \cos\theta\\ \sin\theta \end{pmatrix}
\quad\hbox{and}\quad
\begin{pmatrix} 0\\1 \end{pmatrix}
\mapsto
\begin{pmatrix} -\sin\theta\\ \cos\theta \end{pmatrix}.
\]
The second matrix gives an opposite motion ($\det A=-1$), and you can
understand it in several ways; for example, write
\[
A= \begin{pmatrix}
\cos\theta & \sin\theta\\
\sin\theta & -\cos\theta
\end{pmatrix}=
\begin{pmatrix}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{pmatrix}
\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}.
\]
\addtocounter{figure}{-1}
This says: first reflect in the $x$-axis, then rotate through $\theta$. It
is easy to see geometrically that this is the reflection\index{reflection}
in the line $L$ through the origin $0$ at angle $\theta/2$ to the $x$-axis.
Indeed, every point on $L$ is fixed, and the line perpendicular to $L$ is
reversed, as in Figure~\ref{fig!rot&refl}.
\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}b}
\centering
\leavevmode\hbox{\epsfysize=5.3 true cm\epsffile{eucl_rotrefl_matrix.eps}}
\caption{The rotation and the reflection}
\label{fig!rot&refl}
\end{figure}

In coordinates, this says that $\bbf_1=(\cos(\theta/2),\sin(\theta/2))$ is
an eigenvector of $A$ with eigenvalue 1, and
$\bbf_2=(\sin(\theta/2),-\cos(\theta/2))$ an eigenvector with eigenvalue
$-1$. The pair $(\bbf_1, \bbf_2)$ gives a vector space basis of $\R^2$, and
in this new basis the map is given by the matrix $\left(\begin{smallmatrix}1
& 0\\ 0 & -1 \end{smallmatrix}\right)$. You can readily check these
statements by matrix multiplication and the rules of trig, but the geometric
argument is simpler and more convincing.

\subsection{The general case}\label{ssec!normf}

In the general case I control orthogonal matrixes\index{orthogonal!matrix}
using a slightly more involved argument. \clearpage

\begin{thm}[normal form of orthogonal matrix]
Let $\al\colon \R^n\to \R^n$ be a linear map given by an orthogonal
matrix\index{orthogonal!matrix} $A$. Then there exists an
ortho\-normal basis of\/ $\R^n$ in which the matrix of $\al$ is
\[
%\renewcommand{\arraystretch}{1.4}
B=
\left( \begin{array}{ccccc}
I_{k^+}\\[4pt]
&-I_{k^-}\\[3pt]
&& B_1\\
&&&\ddots\\
&&&& B_l
\end{array}\right)
\quad\hbox{where}\quad
B_i=
\begin{pmatrix}
\cos\theta_i&-\sin\theta_i\\
\sin\theta_i&\cos\theta_i
\end{pmatrix}.
\]
Here $k^++k^-+2l=n$, and $I_{k^\pm}$ is the $k^\pm\times k^\pm$
identity matrix.
\end{thm}

\paragraph{Discussion} The rotation matrix $\left(\begin{smallmatrix}
\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta
\end{smallmatrix}\right)$ has two special cases $\theta=0$ (giving
the identity) and $\theta=\pi$:
\index{half-turn}
\[
\begin{pmatrix} -1&0\\0&-1 \end{pmatrix} =
\begin{pmatrix} \cos\pi&-\sin\pi\\
\sin\pi&\cos\pi \end{pmatrix}=180^\circ\ \hbox{rotation}.
\]
These trivial cases introduce a minor ambiguity in the normal form.
The most natural convention seems to be to disallow $\theta=0$, thus
taking $k^+$ as big as possible, but to use $\theta=\pi$ wherever
possible, so that $k^-=0$ or 1.

\begin{pf} In sketch form, this holds because $A$ is orthogonal, so its
eigenvalues have absolute value $1$. Therefore they are either $\pm1$, or
come in complex conjugate pairs $\{\la,\labar\}=\exp(\pm i\theta)$; after
this, it is enough simply to build up a basis of $\R^n$ consisting either of
real eigenvectors of $A$, or of real and imaginary parts of complex
eigenvectors.

Now I say the same thing again in more detail in $5$ steps; the sketch proof
just given already reveals that complex numbers\index{complex number} are
closely involved, so I may as well extend the action of $A$ to the complex
vector space $\C^n$, which I can do without any problems.

\paragraph{Step 1} If $\la$ is a real eigenvalue of $A$ then
$\la=\pm1$, because
\[
\hbox{$A\bx=\la \bx$ and $A$ orthogonal}\implies
|\bx|^2=|A\bx|^2=\la^2|\bx|^2.
\]

\paragraph{Step 2} If $\la$ is a complex eigenvalue of $A$ then $|\la|=1$
and $\labar=\la\1$ is also an eigenvalue (the bar denotes complex
conjugate). Indeed, given $0\ne \bz\in\C^n$ such that $A\bz=\la \bz$ (recall
I write $\bz=\tr(z_1,\dots,z_n)$ a column vector), write
$\boldzbar=\tr(\zbar_1,\dots,\zbar_n)$. Because $A$ is a real matrix,
\[
A\boldzbar=\overline{A\bz}=\overline{\la\bz}=\labar\boldzbar.
\]
Now write $z_i=x_i+iy_i$, so that $\trboldzbar
\bz=\sum|z_i|^2=\sum(x_i^2+y_i^2)>0$. Using the fact that $A$ is orthogonal,
\[
\labar\la\trboldzbar\bz=\tr (A\boldzbar)A\bz
=\trboldzbar\,^tAA\bz=\trboldzbar\bz, \quad \hbox{and thus} \quad \labar\la=1.
\]

\paragraph{Step 3} If $\la=\cos\theta+i\sin\theta$ is a complex
eigenvalue of $A$ (with $\theta\ne0,\pi$) and $\bz=\bx+i\by\in\C^n$ a
complex eigenvector then taking real and imaginary part in
$A(\bx+i\by)=A\bz=\la\bz=(\cos\theta+i\sin\theta)(\bx+i\by)$ gives
\begin{equation}
A\bx=\cos\theta \bx-\sin\theta \by,\quad
A\by=\sin\theta \bx+\cos\theta \by.\label{rotn}
\end{equation}
Now I claim that $|\bx|^2=|\by|^2$ and $\bx\cdot \by=0$, so that
scaling makes
$\bx,\by\in\R^n$ into a pair of orthonormal vectors. This is an
exercise for the reader. \begin{hint} write out the condition for
(\ref{rotn}) (with
$\theta\ne0,\pi$) to preserve $|\bx|^2,|\by|^2$ and $\bx\cdot \by$.
See \exref{ex!xy_orth}--\ref{z zbar orth ex}.\end{hint}

\paragraph{Step 4} If $\al$ preserves a subspace $W$ of $\R^n$, then
it preserves its orthogonal complement under the inner product
(compare~\ref{sec!bases})\index{orthogonal!complement $V^\perp$}
\[
W^\perp=\bigl\{\bx\in \R^n\bigm| \bx\cdot\bw=0 \quad
\hbox{for all $\bw\in W$}\bigr\}.
\]
In symbols,
\[
\al(W)=W\implies \al(W^\perp)=W^\perp.
\]
This is obvious from the definition of $W^\perp$. Look at
Figure~\ref{grid} for an example: if a motion preserves the
horizontal plane $W$ and its translates, then it will also
preserve the orthogonal complement $W^\perp$, the vertical lines.

\paragraph{Step 5. Proof of the theorem} Eigenvalues of $A$ come
from the poly\-nomial equation $p(\la)=\det(A-\la1)=0$, so that at
least one real or complex eigenvalue $\la$ exists. Step~1 or
Steps~2--3 as appropriate gives a $1$- or $2$-dimensional subspace
$W$ with $AW=W$ on which the action of $A$ is as indicated. By
induction on the dimension, I can assume that the action of $A$ on
$W^\perp$ is OK; the induction starts with $\dim W=0$ or $1$.
\index{orthogonal!matrix|)} \QED \end{pf}
\index{normal form of a matrix|)}

Complex numbers make their first incursion into real geometry during
the above proof, and it is worth pondering why; quaternions also
appear in a similar context in~\ref{sec!topson} below.

\section[Euclidean frames and motions]{Euclidean frames and motions}
\label{Eframe}
\index{motion}\index{Euclidean!motion}
\index{Euclidean!frame}\index{frame!Euclidean|(}
\markright{\hfill \thesection\ Euclidean frames and motions\quad}

\begin{defn} A {\em Euclidean frame} of $\E^n$ is a set of $n+1$
points $Q_0,Q_1,\dots,Q_n$ of $\E^n$ such that $d(Q_0, Q_i)=1$ and
the lines $Q_0Q_i$ are pairwise orthogonal for $1\le i\le n$.
\end{defn}

\begin{rem} The point of the definition is that if $Q_0,\dots,Q_n$
is a Euclidean frame\index{Euclidean!frame} then it is possible to
choose coordinates so that $Q_0$ becomes the origin $0\in\R^n$ and
the $n$ vectors $\bbe_i=\vec{Q_0Q_i}$ form an orthonormal basis of
$\R^n$.
\end{rem}

\begin{thm} If we fix one Euclidean frame\index{Euclidean!frame}
$P_0,P_1,\dots,P_n$, then Euclidean motions are in one-to-one
correspondence with Euclidean frames.
\end{thm}

\begin{pf} The correspondence is given by $T\mapsto
T(P_0),T(P_1),\dots,T(P_n)$. It is clear that the image of the Euclidean
frame\index{Euclidean!frame} $P_0,P_1,\dots,P_n$ under a
motion\index{Euclidean!motion}\index{motion} is again a Euclidean frame. The
converse, that is, the fact that two Euclidean frames are mapped to each
other by a unique motion,\index{motion} follows from the previous Remark and
Appendix~\ref{app!q}, Proposition~\ref{sec!o&lm}. \QED \end{pf}
\index{frame!Euclidean|)}

% Note that everything in this section is a direct analogue
% of~\ref{aff_frames}.

\section{Frames and motions of\/ $\protect\E^2$}\label{sec!E2frame}
\markright{\hfill \thesection\ Frames and motions of\/
$\protect\E^2$\quad}
It is worth noting two useful consequences of Theorem~\ref{Eframe}, whose
proofs are left as easy exercises (see Figure~\ref{fig!E2frame} and
\exref{ex!E2frame}):

\setcounter{figure}{\value{section}}
\addtocounter{figure}{-1}
\begin{figure}[ht]
\centering
\leavevmode\hbox{\epsfysize=5 true cm\epsffile{eucl_frames.eps}}
\caption{The Euclidean frames $P_0,P_1,P_2$ and $P'_0,P'_1,P'_2$}
\label{fig!E2frame}
\end{figure}

\begin{cor}
\begin{enumerate}

\item Suppose that $[P,Q]$ and $[P',Q']$ are two line segments in $\E^2$ of
the same length $d(P,Q)=d(P',Q')>0$. Then there exist {\em exactly two}
motions\index{motion} $T\colon\E^2\to\E^2$ such that $T(P)=P'$, and
$T(Q)=Q'$.

\item Let $\triangle PQR$ and $\triangle P'Q'R'$ be two triangles in
$\E^2$ with all sides equal:
\[
d(P,Q)=d(P',Q'),\quad d(P,R)=d(P',R'),\quad d(Q,R)=d(Q',R').
\]
(I assume that the three vertexes of each triangle are distinct and
noncollinear.) Then there is a unique motion $T\colon\E^2\to\E^2$ such that
$T(P)=P',T(Q)=Q',T(R)=R'$.
\end{enumerate}
\end{cor}

\section[Classification of motions of\/ $\protect\E^2$]{Every motion
of $\E^2$ is a translation,\\ rotation, reflection or glide}
\label{sec!motE2}
\index{motion}
\index{reflection}
\index{rotation}
\index{translation}
\index{glide|(}
\markright{\hfill \thesection\ Classification of motions of\/
$\protect\E^2$\quad}

Let us list the motions of $\E^2$ we know, expressed in coordinates (see
Figure~\ref{rot_and_glide}).

\begin{enumerate}

\item The translation\index{translation} $\Trans(\bb)\colon \bx\mapsto
\bx+\bb$ for $\bb\in\R^2$.

\item The rotation\index{rotation} through angle $\theta$ about a point
$O\in\E^2$; if $O$ is the origin of the coordinate system, this is written
\[
\Rot(O,\theta)\colon \begin{pmatrix} x_1\\ x_2\\ \end{pmatrix}
\mapsto
\begin{pmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta
\end{pmatrix}
\begin{pmatrix} x_1\\ x_2\\ \end{pmatrix}.
\]

\begin{figure}[p]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}a}
\centering
\leavevmode\hbox{\epsfysize=2.7 true cm\epsffile{eucl_rotglide.eps}}
\caption{$\protect\Rot(O,\theta)$ and $\protect\Glide(L,\bv)$}
\label{rot_and_glide}
\end{figure}

\item The reflection\index{reflection} in a line $L$; if $L$ is the
$x_1$-axis $(x_2=0)$ then
\[
\Refl(L)\colon \begin{pmatrix} x_1\\ x_2 \end{pmatrix} \mapsto
\begin{pmatrix} x_1\\ -x_2 \end{pmatrix}.
\]

\item The {\em glide} (or glide reflection)\index{glide} in a line $L$
through a vector $\bv$ along $L$. Reflect in $L$ and translate in $\bv$. If
$L$ is the $x_1$-axis $(x_2=0)$ and $\bv=(a,0)$ then this is given by:
\[
\Glide(L,\bv)\colon \begin{pmatrix} x_1\\ x_2\\ \end{pmatrix} \mapsto
\begin{pmatrix} x_1+a\\ -x_2\\ \end{pmatrix}.
\]
Here $\bv$ is parallel\index{parallel!lines} to $L$, and the
reflection\index{reflection} and translation\index{translation}
commute\index{commutative law}.

\end{enumerate}

I use self-documenting notation such as $\Rot(O,\theta)$ and $\Glide(L,\bv)$
for these motions. In each case, I have chosen coordinates in an obvious way
to make the formula as simple as possible. Obviously (1) and (2) are direct
motions, and (3) and (4) opposite.\index{direct motion}\index{opposite
motion} Note that (3) is a particular case of (4) (where the translation
vector\index{translation!vector} is 0). It is sometimes convenient to view
(1) as a limiting case of (2), when the centre of rotation\index{centre of
rotation} is very far away and the angle of rotation\index{angle!of
rotation} correspondingly small.

\begin{thm} That's all, folks! \end{thm}

\begin{pf} There are several ways of proving this. (Why not devise your own?
See \exref{ex!rot} and \exref{ex!gli} for an argument in terms of
$\bx\mapsto A\bx+\bb$, and Ex.~\ref{ch!comp}.\ref{comp of reflns ex} for an
argument in terms of composing reflections.)\index{composite!of reflections}

The proof given here is based on the following geometric idea taken from
Nikulin and Shafarevich \cite{NSh}: let $P,Q$ and $P',Q'$ be two pairs of
distinct points with $d(P,Q)=d(P',Q')\ne0$. By Corollary~\ref{sec!E2frame},
we know that there are {\em exactly two} motions of $\E^2$ such that
$T(P)=P'$ and $T(Q)=Q'$. In Step~1 below, I construct a
reflection\index{reflection} or glide, and in Step~2 a
rotation\index{rotation} or translation.\index{translation} Now if $T$ is
any motion,\index{motion} pick any two distinct points $P\ne Q$, and set
$P'=T(P)$, $Q'=T(Q)$. Then $T$ must be one of the two motions constructed in
Steps~1--2, both of which are in my list.

\paragraph{Step 1} I first find a reflection\index{reflection} or glide.
Write $\bu=\vec{PQ}$ and $\bu'=\vec{P'Q'}$. First I need to find the line of
reflection $L$. The direction of $L$ and of $\bv$ is the vector bisecting
the angle between $\bu$ and $\bu'$ (that is, $\half(\bu+\bu')$ if the
vectors are not opposite).
\begin{figure}[p]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}b}
\centering
\leavevmode\hbox{\epsfysize=3.9 true cm\epsffile{eucl_direction_glide.eps}}
\caption{Construction of glide}
\label{const_glide}
\end{figure}
Doing this arranges that the reflection\index{reflection} or glide
reflection in any line parallel\index{parallel!lines} to $L$ takes
$\vec{PQ}$ into a vector\index{parallel!vector} {\em parallel to}
$\vec{P'Q'}$. Now choose $L$ among lines with the given direction so that
$d(L,P)=d(L,P')$, and write $A$ and $A'$ for the feet of the respective
perpendiculars from $P$ and $P'$ to $L$ and $\bv=\vec{AA'}$ (see
Figure~\ref{const_glide}). Since reflection\index{reflection} in $L$ takes
$\bu$ into a vector parallel to $\bu'$ by construction, and
$d(P,Q)=d(P',Q')$, it is clear that $\Glide(L,\bv)$ does what I want.

\paragraph{Step 2} There exists a rotation or translation\index{translation}
$T\colon \E^2\to\E^2$ such that $P\mapsto P'$ and $Q\mapsto Q'$. I suppose
first that $P\ne P'$, and that the lines $PQ$ and $P'Q'$ intersect at a
single point in an angle $\theta$.
\begin{figure}[p]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}c}
\centering
\leavevmode\hbox{\epsfysize=5.9 true cm\epsffile{eucl_construct_rotation.eps}}
\caption{Construction of rotation}\index{rotation}
\label{const_rot}
\end{figure}

Then the (signed) angle of rotation\index{angle!of rotation} must be $\theta$;
the centre must be the point $O$ of the perpendicular
bisector\index{perpendicular bisector} of the line $PP'$ determined
by $POP'=\theta$ (see Figure~\ref{const_rot}). Then by construction
$\Rot(O,\theta)$ takes $P\mapsto P'$, and the interval $[P,Q]$ to an
interval out of $P'$ of the same length with $d(P,Q)=d(P',Q')$ and the
same direction as $[P'Q']$; hence it takes $Q\mapsto Q'$.

The proof just given does not work if $P=P'$, or if the lines $PQ$
and $P'Q'$ are parallel,\index{parallel!lines} but these special
cases are easy to deal with, and I leave them as exercises (see
\exref{ex!sp}). \QED \end{pf}

\section{Classification of motions of\/ $\E^3$}
\label{sec!motE3}
\index{motion}
\markright{\hfill \thesection\ Classification of motions of\/
$\E^3$\quad}
\begin{thm} A motion $T\colon \E^3\to \E^3$ is one of the following:
\begin{enumerate}

\item {\rm Translation}\index{translation} by a vector $\bv$.

\item {\rm Rotation}\index{rotation} about a directed line $L$ as
axis through an angle $\theta$.\index{angle!of rotation|(}

\item {\rm Twist}: the same followed by a translation along $L$.

\item {\rm Reflection}\index{reflection} in a plane.

\item {\rm Glide}: a reflection in a plane followed by a translation by a
vector in the plane.\index{glide|)}

\item {\rm Rotary reflection}: the rotation through $\theta$ about a
directed axis $L$ followed by a reflection in a plane $\Pi$ perpendicular to
$L$.

\end{enumerate}
\end{thm}

\begin{figure}[b]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}a}
\centering
\leavevmode\hbox{\epsfysize=4.8 true cm\epsffile{eucl_twist_rotrefl.eps}}
\caption{$\protect\Twist(L,\theta,\bv)$ and
$\protect\RotRefl(L,\theta,\Pi)$}
\label{twist_and_rot}
\end{figure}

\begin{figure}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}b}
\centering
\leavevmode\hbox{\epsfysize=4.8 true cm\epsffile{eucl_3space_grid.eps}}
\caption{A grid of parallel planes and their orthogonal lines}
\index{parallel!hyperplanes}
\label{grid}
\end{figure}

(2) is a special case of (3), and (4) a special case of (5). In all cases
where a motion is defined as a composite of two others, these two
commute\index{commutative law}. (6) is also called a {\em rotary inversion},
because it is also the rotation around the directed axis $L$ through
$\pi+\theta$, followed by a point reflection in $L\cap\Pi$. Clearly (1--3)
are direct motions\index{direct motion}\index{opposite motion} and (4--6)
opposite. Notice that any motion\index{motion} leaves invariant a grid of
parallel planes and their orthogonal lines (Figure~\ref{grid}).

\begin{pf} See, for example, \exref{ex!motE3} or Rees \cite{Rees}, p.~16,
Theorem~17 for a geometric proof. I give a coordinate geometry proof based
on the use of the normal form \index{normal form of a matrix} of
Theorem~\ref{sec!norm}. Let $T\colon \E^3\to\E^3$ be a motion expressed in
coordinates as $T\colon \bx\mapsto A\bx+\bb$; write $T=T_1\circ T_2$ where
$T_i$ are given (in the same coordinate system) by
\[
T_2\colon \bx\mapsto A\bx\quad\hbox{and}\quad
T_1\colon \by\mapsto \by+\bb.
\]
Then by Theorem~\ref{sec!norm}, there exists an orthogonal
coordinate system such that
\[
A= \begin{pmatrix}
\pm1\\
&B\\
\end{pmatrix},
\quad\hbox{where}\quad
B=
\begin{pmatrix}
\cos\theta&-\sin\theta\\
\sin\theta&\cos\theta\\
\end{pmatrix}.
\]
In these coordinates, $T$ has the form
\begin{equation}
T\colon
\begin{pmatrix}
x_1\\x_2\\x_3\\
\end{pmatrix}
\mapsto
\begin{pmatrix}
\pm x_1\\ \vspace{5pt}
\begin{pmatrix}
\cos\theta&-\sin\theta\\
\sin\theta&\cos\theta\\
\end{pmatrix}
\begin{pmatrix}
x_2\\x_3\\
\end{pmatrix}
\end{pmatrix}+
\begin{pmatrix}
b_1\\b_2\\b_3\\
\end{pmatrix}.\label{sepn of vars}
\end{equation}
For the proof, I have to verify that this map is a motion\index{motion} of
one of types \hbox{(1--6)}. This can be done, for example, by a direct
coordinate calculation. It is better to argue using the following {\em
separation of variables}: (\ref{sepn of vars}) breaks $T$ up as a {\em
product} (not composite) of two motions\index{motion} $T=t'\times t''\colon
\E^1\times\E^2 \to\E^1\times\E^2$, where $T'\colon \E^1\to\E^1$ and
$T''\colon \E^2\to\E^2$ are given in coordinates by
\[
T'\colon x_1\mapsto\pm x_1+b_1
\quad
\hbox{and}
\quad
T''\colon \begin{pmatrix}
x_2\\x_3\\
\end{pmatrix}
\mapsto \begin{pmatrix}
\cos\theta&-\sin\theta\\
\sin\theta&\cos\theta\\
\end{pmatrix}
\begin{pmatrix}
x_2\\x_3\\
\end{pmatrix}
+ \begin{pmatrix}
b_2\\b_3\\
\end{pmatrix}.
\]
In other words, (\ref{sepn of vars}) separates the $3$ variables in
such a way that $T(\bx)=\by$ with $\by=(y_1,y_2,y_3)$, where $y_1$ is
a function of $x_1$ only, and $y_2,y_3$ functions of $x_2,x_3$ only.
Now both $T'$ and $T''$ are motions\index{motion} in their own right.
This is the real point of the theorem. (It is easy to guess how to
generalise the result to all dimensions.)

$T''$ is a direct motion,\index{motion} and is a
translation\index{translation} if $\theta=0$ or rotation\index{rotation} if
$\theta\ne0$; this follows by Theorem~\ref{sec!motE2}, or by direct
observation. In terms of coordinates $(x_2,x_3)$ of $\E^2$, it is the
rotation through an angle $\theta$\index{angle!of rotation|)} about the
point determined by
\[
\begin{pmatrix}
x_2\\x_3\\
\end{pmatrix}
=
\begin{pmatrix}
\cos\theta&-\sin\theta\\
\sin\theta&\cos\theta\\
\end{pmatrix}
\begin{pmatrix}
x_2\\x_3\\
\end{pmatrix}
+
\begin{pmatrix}
b_2\\b_3\\
\end{pmatrix},
\]
that is, solving for $x_2,x_3$ by inverting a $2\times2$ matrix:
\[
\begin{pmatrix}
x_2\\x_3\\
\end{pmatrix}
=
\frac{-1}{2-2\cos\theta}
\begin{pmatrix}
\cos\theta-1&-\sin\theta\\
\sin\theta&\cos\theta-1\\
\end{pmatrix} \begin{pmatrix}
b_2\\b_3\\
\end{pmatrix}.
\]

The theorem follows easily on sorting out the cases. \QED \end{pf}

\section{Sample theorems of Euclidean geometry}\label{sec!sample}
\index{Euclidean!geometry}
\markright{\hfill \thesection\ Sample theorems of Euclidean
geometry\quad}

This chapter has mainly been concerned with the foundations of Euclidean
geometry and a description of Euclidean motions.\index{motion} I do not have
time to give many results of substance from Euclidean geometry, either the
theory of Euclid's elements, or the much more extensive 19th century
subject, but I do not want to omit to mention it altogether. Coxeter
\cite{Cx} is very entertaining on this subject.

\subsection{Pons asinorum}\label{sec!pons}

\begin{prop} {\rm Pons asinorum}, ``Bridge of asses''. Equivalent
conditions on a triangle $\triangle ABC$:
\begin{enumerate}

\item $d(A,B)=d(A,C)$;

\item $\theta=\angle ABC=\theta'=\angle ACB$;

\item there exists a motion\index{motion} $T\colon \triangle
ABC\mapsto\triangle ACB$.
\end{enumerate}

\begin{figure}[htb]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}a}
\centering
\leavevmode\hbox{\epsfysize=4.6 true cm\epsffile{eucl_pons.eps}}
\caption{Pons asinorum}
\label{asses}
\end{figure}
\end{prop}

\begin{pf} $(1)\iff(2)$ is an easy consequence of trigonometry,
because in Figure~\ref{asses},
\[
d(A,O)= d(A,B) \sin\theta= d(A,C)\sin\theta'.
\]
From our point of view, $(3)\implies(1)$ or (2) is obvious, and (1) or
$(2)\implies(3)$ follows by Corollary~\ref{sec!E2frame}. Euclid talked
directly about the motion\index{motion} of the plane consisting of picking
up the triangle and laying it down over itself so that $A,B,C$ match up with
$A,C,B$ in order; alternatively, you can drop a perpendicular $AO$ from $A$
to $BC$, and argue on congruent triangles.\index{congruent triangles} \QED
\end{pf}

\subsection{The angle sum of triangles}
\index{angle!sum|(}
\begin{thm} The sum of angles in a triangle $=\pi$.
\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}b}
\centering
\leavevmode\hbox{\epsfysize=4 true cm\epsffile{eucl_angle_sum.eps}}
\caption{Sum of angles in a triangle $=\protect\pi$}
\label{angles_in_triangle}
\end{figure}
\end{thm}
\begin{pf}
Let $\triangle ABC$ be a given triangle. Consider the motion\index{motion}
$T=\Trans(\vec{AC})$ and set $\triangle A'B'C'=T(\triangle ABC)$ as in
Figure~\ref{angles_in_triangle}. Then because $T$ is a motion\index{motion},
I get $\triangle A'B'C'\equiv\triangle ABC$, where $\equiv$ is congruence
(see \exref{ex!congruent_triangles}). Also, since $T$ is a Euclidean
translation,\index{Euclidean!translation}\index{translation}
$d(B,B')=d(A,C)$, therefore also $\triangle ABC\equiv\triangle BA'B'$. Hence
\[
\al+\be+\ga=\angle B'CC'+\angle BCB'+\angle ACB=\pi
\]
since the angles combine to form a straight line. \QED \end{pf}

\begin{rem}
The statement {\em the sum of angles in a triangle equals $\pi$} is
equivalent to the parallel postulate\index{parallel!postulate}
(see~\ref{sec!nonE} and~\ref{sec!parallels}). The proof used translation in
$\E^2$, coming from the coordinate model. Figure~\ref{angles_in_triangle}
makes sense in spherical geo\-metry\index{spherical!geometry} (or
hyperbolic) geometry,\index{hyperbolic!geometry} but there $d(A,A')>d(B,B')$
(respectively $d(A,A')<d(B,B')$).\index{angle!sum|)}
\end{rem}

\subsection{Parallel lines and similar triangles}\label{sec!par_sim}

A distinguishing feature of Euclidean geometry is the existence of unique
parallel lines (compare~\ref{sec!parallels}). Parallel lines fall on lines
in same ratio, and conversely; they are also responsible for the existence
of similar triangles. The following proposition makes these statements
precise.

\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}c}
\centering \input{eucl_parallel_ratio.pstex_t}
\caption{Parallel lines fall on lines in the same ratio}
\label{same ratio}
\end{figure}

\index{parallel!lines|(}
\begin{prop} \begin{enumerate}
\item If $L_1,L_2,L_3$ are three parallel lines in $\E^2$, and they
meet a line $M$ in $P_1,P_2,P_3$, then the (signed) ratio of distances
$d(P_1,P_2):d(P_2,P_3)$ is independent of $M$ (Figure~\ref{same
ratio}).
%%%%%%%%%%%%%
%%%%%%%%%%%%%
% The converse is discussed in \ref{aff_rat_parallels}.
% Not written yet: L1,L2,L3 fall in same ratio on 3 lines
% M,M',M'' (not parallel) implies L1,L2,L3 are parallel.
%%%%%%%%%%%%%
%%%%%%%%%%%%%
% Conversely if $L_1,L_2,L_3$ meet two distinct lines $M,M'$ in
% $P_1,P_2,P_3$ and $P'_1,P'_2,P'_3$, then
% \[
% d(P_1,P_2):d(P_2,P_3)=d(P'_1,P'_2):d(P'_2,P'_3)\implies
% \hbox{$L_1,L_2,L_3$ are parallel.}
% \]

\item Consider the two triangles $\De ABC$ and $\De AB'C'$ of
Figure~\ref{sim}. The following are equivalent:
\begin{enumerate}

\item $BC$ is parallel to $B'C'$.

\item Equality of ratios: $d(A,B):d(A,B')=d(A,C):d(A,C')$.

\item Equality of angles: $\angle ABC=\angle AB'C'$ and
$\angle ACB=\angle AC'B'$.
\end{enumerate}\end{enumerate}\end{prop}

\begin{figure}[ht]\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}d}
\centering
\leavevmode\hbox{\epsfysize=4.6 true cm\epsffile{eucl_similar_triangles.eps}}
\caption{Similar triangles}
\label{sim}
\end{figure}
\begin{pf} All this is trivial in coordinate geometry;
see~\exref{ex!similar_triangles}.
\end{pf}

Two triangles satisfying the conditions of the second part are
called {\em similar}.\index{similar triangles} Corresponding pairs of angles
of a pair of similar triangles are equal.

\subsection{Four centres of a triangle}\label{three centres}

\begin{prop}[Centroid]\index{centroid} The three medians of a triangle
$ABC$ meet in a point~$G$.
\end{prop}

\begin{pf} (See~\ref{aff_cent} for a slightly different proof.) Let $A'$,
$B'$, $C'$ be the midpoints of $BC$, $AC$, $AB$ and let $G$ be the point on
$AA'$ with $d(A,G)=2d(G,A')$. If $L$, $M$ are the midpoints of $AG$ and
$CG$, then by similar triangles\index{similar triangles|(}
\[
LM\parallel AC\parallel A'C'\quad\hbox{and}\quad LC'\parallel
GB\parallel MA',
\]
(where $\parallel$ is parallel),\index{parallel!lines} so that $LMA'C'$
is a parallelogram, $G$ is its centre, so $MGC'$ is a straight line.
Hence $G$ lies on each of $AA'$, $BB'$, $CC'$, so it is the centroid.
\QED \end{pf}

\begin{figure}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}d}
\centering
% \leavevmode\hbox{\epsfysize=4.8 true cm\epsffile{eucl_centroid.eps}}
\input{eucl_centroid.pstex_t}
\caption{The centroid}
\label{centroid}
\end{figure}

\begin{prop}[Circumcentre]\index{circumcentre} The three perpendicular
bisector\index{perpendicular bisector}s of sides $AB$, $BC$ and $AC$ meet
in a point $O$. This is the centre of the circle circumscribed around $ABC$.
\end{prop}

\begin{pf} This is almost obvious, since the perpendicular
bisector\index{perpendicular bisector} of $AB$ is
determined as the locus of points equidistant from $A$ and $B$, so that any two
of the perpendicular bisector\index{perpendicular bisector}s intersect at
the point $O$ determined by $d(A,O)=d(B,O)=d(C,O)$. \QED \end{pf}

\begin{figure}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}e}
\centering
\input{eucl_circum.pstex_t}
% \leavevmode\hbox{\epsfysize=4.8 true cm\epsffile{eucl_circumcentre.eps}}
\caption{The circumcentre}\index{circumcentre}
\label{circumcentre}
\end{figure}

\begin{prop}[Orthocentre]\index{orthocentre|(} The three perpendiculars
dropped from a vertex onto the opposite side intersect in a point $H$.
\end{prop}

\begin{figure}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}f}
\centering
\input{eucl_ortho.pstex_t}
% \leavevmode\hbox{\epsfysize=4.5 true cm\epsffile{eucl_ortho.eps}}
\caption{The orthocentre}
\label{ortho}
\end{figure}

\begin{pf} In vector notation, $H$ is the point given by
$\vec{OH}=3\vec{OG}$, where $O$ is the circumcentre\index{circumcentre}
and $G$ the centroid. Indeed, in Figure~\ref{ortho}, $BB'$ is the median
and $OB'$ the perpendicular bisector\index{perpendicular bisector} of $AC$;
since $\vec{GB}=2\vec{B'G}$
and $\vec{GH}=2\vec{OG}$, it follows that the two triangles $\triangle
GB'O$ and $\triangle GBH$ are similar.\index{similar triangles}
Therefore the line $BH$ is perpendicular to $AC$, and $H$ lies on this
perpendicular. $H$ lies on each of the other two perpendiculars for
similar reasons. \QED \end{pf}

Note that, as a byproduct of the above proof, we also see that the centroid $G$
lies on the segment $[O,H]$ determined by the circumcentre and the orthocentre,
and divides it into ratio $(1:2)$.

\begin{prop}[Incentre]\index{incentre} The angle
bisector\index{angle!bisector}s of the three angles $\angle CAB$, $\angle ABC$ and
$\angle ACB$ meet in a point $K$. This is the centre of the circle
inscribed into $ABC$.
\end{prop}

\begin{pf} This is exactly analogous to the case of the circumcentre above
(see~Ex.~\ref{ch!eucl}.\ref{ex!incentre}). \QED \end{pf}

\subsection{The Feuerbach 9-point circle}
\label{9-point circle}
\index{Feuerbach circle|(}

\begin{thm}[The Feuerbach circle]$\!\!$\footnote{The Feuerbach circle is
alternatively called the Euler circle, because it was discovered by Poncelet
and Brianchon. The reason why the young Bavarian schoolmaster Feuerbach's
name appears in the context is his beautiful theorem that the circle touches
the inscribed circle of the triangle (indicated in Figure~\ref{9-point} with
a dotted line). Purists may prefer the noncommital name 9-point circle.} The
following 9 points lie on a circle (see Figure~\ref{9-point}):

$3$ feet $P,Q,R$ of the perpendiculars dropped from a vertex to the
opposite side;

$3$ midpoints $A',B',C'$ of the sides;

$3$ midpoints $D,E,F$ of $AH,BH,CH$, where $H$ is the
orthocentre.\index{orthocentre|)}
\end{thm}

\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}g}
\centering
\centering\input{eucl_ninepoint_incircle.pstex_t}
\caption{The Feuerbach 9-point circle}
\label{9-point}
\end{figure}

\begin{pf} The intellectual achievement here is the statement, of
course. The proof is rather easy because there are so many parallel and
perpendicular lines in Figure~\ref{9-point}. By similar
triangles,\index{similar triangles|)} the following lines are
parallel:\index{parallel!lines|)}
\[
A'B'\parallel DE\parallel AB\quad
\hbox{and}\quad A'E\parallel B'D\parallel CR.
\]
But $AB\perp CR$ by construction, hence $A'B'DE$ is a rectangle. Thus
the circle with diameter $A'D$ also has $B'E$ as diameter; arguing in
the same way one sees that $A'C'DF$ is also a rectangle, so that the
same circle with diameter $A'D$ also has $C'F$ as diameter. Finally,
$\angle A'PD=90^\circ$, which is a sufficient condition for the circle
with diameter $A'D$ to pass through $P$, so that this same circle passes
also through the feet of the perpendiculars. \QED \end{pf}
\index{Feuerbach circle|)}

\section*{Exercises to Chapter~\thechapter}
\markright{\hfill Exercises to Chapter~\thechapter\quad}
\addcontentsline{toc}{section}{Exercises to Chapter~\thechapter}
\begin{enumerate}
\renewcommand{\labelenumi}{\thechapter.\theenumi.}
\addtocounter{section}{1}
\item Redo the proof of Theorem~\ref{Rn} in detail in the cases $n=1$ and
$n=2$.

\item The angle between nonzero vectors $\bu,\bv\in\R^n$ can be defined by
\[
\cos\theta=\sum u_iv_i/|\bu||\bv|.
\]
Prove that the right-hand side is in the interval $[-1,+1]$, so that its
$\arccos$ is defined.

\item\label{ex!col} The line $L=\bx\by$ in $\R^n$ is the set
$\{(1-\la)\bx+\la \by| \la\in\R\}$. If $\bz\in L$, write $\by$ in terms of $\bx$ and
$\bz$. Complete the proof of Corollary~\ref{sec!affl}.

\item\label{ex!bijnon}
Show that the assumption that $T$ is bijective in the definition of
motion\index{Euclidean!motion}\index{motion} of Euclidean space is
superfluous; that is, a map $T\colon\E^n\to\E^n$ that preserves
distances\index{preserves distances} is bijective, therefore a motion.
\begin{hint} prove that $T$ is affine linear. Compare
Ex.~\ref{app!metric}.\ref{isom ex}.\end{hint}

%by rereading the proofs around~\ref{sec!affl}

\item\label{ex!xy_orth} Complete the proof of Step~3 in
Theorem~\ref{sec!norm} using the hint given in the text.

\item\label{z zbar orth ex} Let $A$ be a (real) orthogonal
matrix.\index{orthogonal!matrix} \begin{enumerate}

\item If $\bbe,\bbf\in\R^n$ are eigenvectors of $A$ belonging to distinct
eigenvalues $\la\ne\mu$, prove that $\bbe\cdot \bbf=0$.

\item If $\bz\in\C^n$ is a complex eigenvector with complex eigenvalue
$\la\notin\R$, prove that $\bz\cdot\bz=0$. (Here $\bx\cdot \by=\sum_j
x_jy_j$ is the usual inner product.)\index{Euclidean!inner product} Use this
to give a better proof of Step~3 in Theorem~\ref{sec!norm}. \end{enumerate}

\index{motion}\index{Euclidean!motion}

\item\label{refl ex} \begin{enumerate}

\item Let $T\colon\E^2\to\E^2$ be the motion obtained by
reflecting\index{reflection} in the $x$-axis then rotating\index{rotation}
through $\theta$ around the origin. Show that $T$ is the reflection in a
certain line (to be specified).

\item Calculate the eigenvalues and eigenvectors of the reflection
matrix\index{reflection!matrix} $A=\left( \begin{smallmatrix} \cos\theta &
\sin\theta\\ \sin\theta &-\cos\theta \end{smallmatrix} \right)$.

\item Relate (a) and (b).
\end{enumerate}

\item\label{ex!rot}
\begin{enumerate}

\item Let $\theta$ be a nonzero angle and $\bb$ a translation vector in the
plane.\index{translation!vector} Give a geometric construction for a point
$P\in \E^2$ such that
\[
\Rot(O,\theta)(P)=\Trans(-\bb)(P).
\]
\begin{hint} draw a picture, to find points $P,Q$ with $\bb=\vec{QP}$ such
that $O$ is on the perpendicular bisector\index{perpendicular bisector} of
$PQ$ and $\angle POQ=\theta$.\end{hint}

\item By solving linear equations, find $x,y$ such that
\[
A \begin{pmatrix} x_1\\x_2\end{pmatrix}+
\begin{pmatrix} b_1\\b_2 \end{pmatrix}= \begin{pmatrix} x_1\\x_2
\end{pmatrix},
\quad\hbox{where}\quad A= \begin{pmatrix} \cos\theta & \sin\theta\\
\sin\theta& -\cos\theta \end{pmatrix}.
\]

\item Express the motion\index{motion} $T\colon\E^2\to\E^2$ defined in
coordinates by $T(\bx)=A\bx+\bb$ in the form $T=\Rot(P,\theta)$.

\item Relate (a) and (b).
\end{enumerate}

\item\label{ex!gli} Let $A=\left( \begin{smallmatrix} \cos\theta &
\sin\theta\\ \sin\theta & -\cos\theta \end{smallmatrix}\right)$ be the
reflection matrix of \ref{sec!refl}, and consider the motion\index{motion}
$T(\bx)=A\bx+\bb$; give a proof in coordinates that it is a glide
reflection.\index{glide} \begin{hint} you need to turn
Figure~\ref{const_glide} into coordinates.\end{hint}

\item\label{ex!sp} In the proof of Theorem~\ref{sec!motE2}, Step~2, there are
3 special cases:
\begin{enumerate}

\item $P=P'$,

\item $PQ$ and $P'Q'$ are parallel,

\item and $PQ$ and $P'Q'$ are opposite (that is $PQ$ and $Q'P'$ parallel).
\end{enumerate}
Complete the proof of Step~2 in any of these cases by constructing a
suitable translation\index{translation} or rotation\index{rotation} taking
$P\mapsto P'$ and $Q\mapsto Q'$.

\item Find the two motions $\E^2\to\E^2$ taking $(0,0)\mapsto(1,2)$ and
$(0,\sqrt2\,)\mapsto(2,3)$. Write each as $\bx\mapsto A\bx+\bb$.
\begin{hint} the easy way: for the direct motion,\index{motion} translate
then rotate; for the opposite motion, reflect then translate then
rotate.\end{hint} Express them as rotation and glide.

\item\label{ex!E2frame} Prove Corollary~\ref{sec!E2frame}, (1). \begin{hint}
as in Figure~\ref{fig!E2frame}, make a Euclidean
frame\index{frame!Euclidean} with $P_0=P$,
$\vec{P_0P_1}=\frac{\vec{PQ}}{d(P,Q)}$ and $P_2$ a third point; if I do the
same for $P',Q'$, there are $2$ choices for $P_2'$, one on either side of
the line $P'Q'$. The statement now follows by
Corollary~\ref{Eframe}.\end{hint}

\item\label{ex!frame} Let $P_0,P_1,P_2\in\E^2$ be distinct noncollinear
points. Show that there is a unique Euclidean
frame\index{frame!Euclidean}\index{Euclidean!frame} so that $P_0=(0,0)$,
$P_1=(a,0)$ with $a>0$ and $P_2=(b,c)$ with $c>0$. Deduce that a
motion\index{Euclidean!motion} of $\E^2$ is uniquely determined by its
effect on any 3 distinct noncollinear points.

\item Let $P_0,P_1,P_2$ and $P'_0,P'_1,P'_2\in\E^2$ be two pairs of
distinct noncollinear points such that $d(P_i,P_j)=d(P'_i,P'_j)$ for
all $i,j$. Prove that there exists a unique motion\index{motion}
$T\colon\E^2\to\E^2$ taking $P_i\mapsto P'_i$ for $i=1,2,3$.
\begin{hint} you know enough motions to send $P_0\mapsto P'_0$. Then
fixing $P_0=P'_0$, to send $P_1\mapsto P'_1$ in exactly 2 different
ways. Where does this leave $P_2$?\end{hint}

\item Let $P_0,\dots,P_n$ be $n+1$ points spanning $\E^n$. Prove
that a point $Q\in\E^n$ is uniquely determined by its distances from
all of the $P_i$. \begin{hint} take $P_0$ as origin; the $n$ vectors
$\bbe_i=\vec{P_0P_i}$ are linearly independent. The vector
$\bbf=\vec{P_0Q}$ is determined by $\bbf\cdot \bbe_i$, so it is enough to
determine $f\cdot e_i$ from distances in $\triangle
P_0P_iQ$.\end{hint}

\item\label{ex!congruent_triangles}
\index{congruent triangles}
Let $\De ABC$ and $\De DEF$ be two triangles in $\E^2$. Prove that the
following 4 conditions are equivalent:
\begin{enumerate}

\item 3 sides are equal $AB=DE$, $BC=EF$, $CA=FD$;

\item equal side--angle--side: $AB=DE$, $CA=FD$ and
$\angle CAB=\angle FDE$;

\item angle--side--angle: $\angle ABC=\angle DEF$, $BC=EF$ and
$\angle BCA=\angle EFD$;

\item There exists a motion\index{motion} $T$ taking $A\mapsto D$, $B\mapsto E$, $C\mapsto F$.
\end{enumerate}
The triangles $\De ABC$ and $\De DEF$ are {\em
congruent}\index{congruent triangles} if these conditions hold; in
symbols, $\De ABC\equiv\De DEF$.

\item\label{ex!similar_triangles} Prove Proposition~\ref{sec!par_sim}
by computing in a suitably chosen coordinate system.

\item\label{ex!incentre}
By analogy with the proof of Proposition~\ref{three centres} (Circumcentre),
prove that the three angle bisectors\index{angle!bisector} of angles $\angle
CAB$, $\angle ABC$ and $\angle ACB$ meet in a point $K$. Show also that this
is the centre of the circle inscribed in $ABC$ (a circle touching all sides
of $\triangle ABC$).

\end{enumerate}
\index{Euclidean!geometry|)}

\endinput


Former text on $\E^n$

How to introduce coordinates in space, and fix up constructions and
arguments to be valid independently of the coordinate system is one of the
main foundational points of the course; hence the superfluous notation
$\E^n$, which I write to mean that I forget the distinguished origin
$0\in\R^n$, the standard coordinate system, and the vector space structure
of $\R^n$, remembering only the distance\index{distance!function} and
properties that can be derived from it. For example, a plane
$\Pi\subset\R^3$ (defined by $ax+by+cz=d$ with $d\ne0$, say) is a copy of
$\E^2$, but with no God-given origin or $x$- or $y$-axes. If you want to
have a Grand Mosque of Mecca or a Greenwich observatory in $\E^n$, you must
receive it by Divine Grace, or make a deliberate extra choice. I usually
think of a point $P\in\E^n$, and $x\in\R^n$ as its coordinates, once an
identification $\E^n=\R^n$ has been chosen. The distinction is somewhat
esoteric and not very important, so you can just ignore it and take
$\E^n=\R^n$ if you find it confusing. The idea of space ought to make sense
without a coordinate system, but you can always fix one if you like. We will
see presently that the distance function\index{distance!function} determines
an ``affine linear structure'' on $\E^n$, so that the metric geometry of
$\E^n$ is affine linear.

Lines and collinearity in $\E^n$ can be defined in terms of a coordinate
system $\E^n=\R^n$. In other words, to know if 3 points $P,Q,R\in\E^n$ are
collinear, use an identification $\E^n=\R^n$ to take them into coordinates
$\bx,\by,\bz\in\R^n$, then apply Definition~\ref{sec!col}. The point of the
corollary is that the notion of line is determined purely in terms of the
metric of $\E^n$. In fact, three points $P,Q,R\in\E^n$ are collinear if and
only if $d(P,Q)=d(P,R)+d(R,Q)$, etc., and this does not depend on the
particular identification $\E^n=\R^n$.

% Old section on lines and collinearity

For $\bx,\by\in\R^n$, the {\em line segment\/} $[\bx,\by]$ consists of
points $\bx+\la(\by-\bx)$ for $\la$ a real number between $0$ and $1$; note
that here I am using the vector space structure of $\R^n$, namely addition
and multiplication by a real number.

\begin{thm}[Triangle inequality]
\begin{equation}
|\bx-\by|\le |\bx-\bz|+|\bz-\by|,\quad\hbox{for all
$\bx,\by,\bz\in\R^n$,} \label{eq!1}
\end{equation}
with equality if and only if $\bz$ is in the line segment\/
$[\bx,\by]$; in other words, $\bz$ is {\em collinear} with $\bx$ and
$\by$ (see below) and {\em between} them.
\end{thm}

\section{Lines and collinearity in $\R^n$}\label{sec!col}
\markright{\hfill \thesection\ Lines and collinearity in $\R^n$\quad}
There are several ways of defining a line (already in the usual
$x,y$ plane $\R^2$); I choose one definition for $\R^n$.

\begin{defn}
Let $\bu\in\R^n$ be a fixed point and $\bv\in\R^n$ a nonzero
direction vector. The {\em line} $L$ starting at $\bu\in\R^n$ with
direction vector $\bv$ is the set
\[
L:=\bigl\{\bu+\la \bv\bigm|\la\in\R\bigr\}\subset\R^n.
\]
Three distinct points $\bx,\by,\bz\in\R^n$ are {\em collinear} if
they are on a line.
\end{defn}

%The main new ingredient is a
%notion of Euclidean distance\index{Euclidean!distance} between points; the
%distance function\index{distance!function} leads to other notions such as
%angles\index{angle} and congruent triangles. The connection to affine
%5geometry is established by showing that straight lines (and hence linear
%subspaces) are determined by the distance function.

% This follows at once from the previous proposition, the implication
% $(3)\implies(1)$ in Proposition~\ref{prop!lines_affinemaps}, and the
% fact that $T$ is bijective, so the matrix $A$ must be invertible.

%You can phrase the same argument without vector notation, in terms of
%the Euclidean notion of similar triangles\index{similar triangles|(}.

%The last clause of the theorem lies at the cornerstone of the interaction
%between the metric and linear geometry of $\R^n$. It shows that the
%line segment\index{line!segment}
% \[
% [\bx,\by]=\bigl\{\bx+\la(\by-\bx)\bigm|0\le\la\le1\bigr\}
% \]
%familiar from affine geometry
%(compare~\ref{sec!afl}) is determined in terms of the metric:
%it is the set of points of $\R^n$ where the triangle inequality turns
%into an equality. Hence {\em collinearity is also a metric property}.

%In particular, the geometry of $\E^n$ is affine linear as discussed in
%the previous chapter: the notion of affine linear
%subspaces,\index{affine!linear!subspace} their geometry,
%intersection properties,\index{dimension!of intersection} etc.,
%also make sense on $\E^n$.

%(just as I did when I introduced $\aff^n$),
%As I proved above, the distance function
%determines an ``affine linear structure'' on $\E^n$, so that the
%metric geometry of $\E^n$ is affine linear.

%See~\ref{sec!affl} for an application of the proposition.
%Note also that using the origin $0$ in the
%statement seems to go against my expressed wisdom that there is no
%distinguished origin in the geometry of $\aff^n$. However, recall
%that any point $P\in\aff^n$ can serve as origin
%after a suitable translation.
