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\title{Angles in analytic geometry}
\author{David Pierce}
\date{November 3, 2006; recompiled, \today}

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\fancyhead[CO]{Angles in analytic geometry (draft)}
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  \maketitle

The words \Eng{synthetic} and \Eng{analytic} are sometimes used
as opposites or complements.  The geometry pioneered by
Rene Descartes \cite{Descartes-Geometry} is called \textbf{analytic
  geometry;} by contrast, the geometry of ancient mathematicians like
Euclid of Alexandria \cite{MR17:814b} and Apollonius of Perge
\cite{Apollonius} is called 
\textbf{synthetic geometry.} 

  The word \Eng{synthetic} comes from the Greek \Gk{sunjetik'oc} meaning
  \Eng{skilled in putting together} or \Eng{constructive.}  This Greek
  adjective derives from the verb \Gk{sunt'ijhmi} \Eng{put together,
  construct.}  The word \Eng{analytic} is the English form of
  \Gk{>analutik'oc}, which derives from the verb \Gk{>anal'uw}
  \Eng{undo, set free, dissolve.}  

What do these words mean in the context of mathematics?
Although we refer to ancient geometry as synthetic, the Ancients
evidently recognized both analytic and synthetic
methods.  Pappus of Alexandria writes:
\begin{quote}
  Now \textbf{analysis} is a method of taking that which is sought as
  though it were admitted and passing from it through its consequences
  in order to something which is admitted as a result of
  synthesis; for in analysis we suppose that which is sought to be
  already done, and we inquire what it is from which this comes about,
  and again what is the antecedent cause of the latter, and so on
  until, by retracing our steps,  we light upon something already
  known or ranking as a first principle; and such a method we call
  analysis, as being a reverse solution.

But in \textbf{synthesis,} proceeding in the opposite
  way, we suppose to be already done that which was last reached in
  the analysis, and arranging in their natural order as consequents
  what were formerly antecedents and linking them one with another, we
  finally arrive at the construction of what was sought; and this we
  call synthesis. \cite[p.~597]{MR13:419b}
\end{quote}

The main point seems to be that synthesis (and synthetic geometry in
particular) should start from first principles and build from there;
while analysis (and analytic geometry) is a kind of search for
principles from which a desired result would follow.

Euclid of Alexandria begins his \emph{Elements} with five principles:
\begin{enumerate}
  \item
any two points can be joined by a [straight] line;
\item
any [straight] line can be extended indefinitely;
\item
a circle can be drawn with any center and radius;
\item
all right angles are equal;
\item
if two angles, say $ABC$ and $BCD$, are together less than two right
angles, then lines $BA$ and $CD$, extended as necessary beyond $B$ and
$D$, must meet.
\end{enumerate}

The 47th proposition that Euclid derives from these principles is
commonly known by another name:

\begin{theorem}[Pythagoras]
In a right triangle, the square on the hypotenuse
  is equal to the squares on the legs.
\end{theorem}

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\emph{Proof.}
The proof is based on the picture at the right, where $ABC$ is a triangle
with right angle at $A$, the squares on the sides are drawn as shown, and
$AL$ is perpendicular to $BC$.

The square $ABFG$ is twice the triangle $CBF$, which is congruent to
$DBA$, which is half the rectangle $DBML$.  So $ABFG$ is equal to
$DBML$.  Likewise, $ACKH$ is equal to $ECML$.  But $DBML$ and $ECML$
are together the square $BCED$.\hfill\qed

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One way to \emph{analyze} the Pythagorean Theorem is to understand it
as `really' 
being about lengths:  If the side of $ABC$ opposite angle $A$ has
length $a$, and so forth, and the angle at $A$ is right, then 
\begin{equation}\label{eqn:P}
a^2=b^2+c^2.  
\end{equation}
We shall see how such an equation can
arise when we understand the points $A$, $B$, and $C$ as ordered pairs
(or triples) of real numbers. 

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In Euclidean geometry, two distinct lines intersecting at a point
determine a plane in the following sense.  Let the lines be $OA$ and
$OB$.  If $C$ is on $OA$, and $D$ is on $OB$, then there is a unique
parallelogram $CODP$ for some point $P$.  (The parallelogram
is `degenerate' if $C$ or $D$ is $O$.)  Such points $P$ compose a
plane, and every point $P$ in this plane determines a unique such
parallelogram.  Therefore, instead of working with the points $P$, we
can work with the pairs $(x,y)$, where $x$ is the `signed' distance of
$C$ from $O$ (that is, $x$ is negative if it is on the opposite side
of $OA$ from $A$), and $y$ is the signed distance of $D$ from $O$.
Here we understand signed distances to be just real numbers; so our
plane becomes the set $\R\times\R$ or $\R^2$ of ordered pairs of real
numbers. 

So \emph{plane} analytic geometry is about the set $\R^2$; we think of
its elements as points.
We conceive of $\R^2$ as having \textbf{axes,}
called $X$ and $Y$ respectively.  The \textbf{$X$-axis} consists of points
$(x,0)$; the \textbf{$Y$-axis} consists of points $(0,y)$.  Nothing
that we have said so far requires these axes to be perpendicular;
indeed, it is not yet clear what it would \emph{mean} for the axes to
be perpendicular, since these axes are just sets of ordered pairs of
numbers.  However, Equation~\eqref{eqn:P} is a clue.

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Not everything interesting that
we can say about $\R^2$ requires us to conceive of the axes as
perpendicular.  For example, from the inequality
\begin{equation*}
  0\leq(x-a)^2=x^2-2ax+a^2,
\end{equation*}
we obtain
\begin{equation*}
  2ax-a^2\leq x^2.
\end{equation*}
This means that every point on the curve defined by $y=x^2$ is above
the point with the same $X$-coordinate on the line $y=2ax-a^2$.  As
the picture shows, this makes visual sense, even if the two axes are
not perpendicular.

The same element of $\R^2$ can be written as $\vec u$ or $(u_1,u_2)$.
Then $u_1{}^2+u_2{}^2\geq0$, so 
$\sqrt{u_1{}^2+u_2{}^2}$ is a well-defined, non-negative number: let
us call this number the \defn{norm} of $\vec u$ and denote it by
\begin{equation*}
  \size{\vec u}.
\end{equation*}
So, by definition, we have the identity
\begin{equation}\label{eqn:norm}
  \size{\vec u}=\sqrt{u_1{}^2+u_2{}^2}.
\end{equation}
The norm is intended to express a notion of \emph{distance:}
$\size{\vec u}$ should make sense as the distance between $\vec u$ and
$\vec 0$ (that is, $(0,0)$).  \emph{Does} it make sense?  Well, in
Euclidean geometry, $\vec u$ is the length of the hypotenuse of a
right triangle whose legs have lengths $u_1$ and $u_2$.  But what is a
right triangle in $\R^2$?

We can add elements of $\R^2$ coordinate-wise:
\begin{equation}
  \vec u+\vec v=(u_1+v_1,u_2+v_2).
\end{equation}
Likewise, we can multiply them by real numbers:
\begin{equation}
  a\cdot\vec u=(a\cdot u_1,a\cdot u_2).
\end{equation}
Here $a$ may be called a \defn{scalar;} the elements of $\R^2$ are
then called \defn{vectors.}  
The operations on vectors have various nice properties that follow
from the corresponding properties of operations on scalars.

Two vectors are \defn{parallel} if one of
them is a scalar multiple of the other:  if $a\cdot\vec u=\vec v$, or
$a\cdot\vec v=\vec u$, then
\begin{equation}
  \vec u\parallel\vec v.
\end{equation}
Some algebraic consequences of~\eqref{eqn:norm} follow almost
immediately:

\begin{theorem}
  For all $\vec u$ in $\R^2$ and $a$ in $\R$,
  \begin{enumerate}
    \item\label{item:1}
$0\leq\size{\vec u}$;
\item\label{item:2}
$0=\size{\vec u}\Iff \vec 0=\vec u$;
\item\label{item:3}
$\size{a\cdot\vec u}=\size a\cdot\size{\vec u}$.
  \end{enumerate}
\end{theorem}

\begin{proof}
  We observed~\eqref{item:1} while defining $\size{\vec u}$.
  For~\eqref{item:2}, the direction $\Leftarrow$ follows by
  computation: $\size{\vec0}=\sqrt{0^2+0^2}=0$; for the direction
  $\Rightarrow$, we prove the \emph{contrapositive:}  If
  $\vec0\neq\vec u$, then one of $u_1$ and $u_2$ is not $0$; without
  loss of generality, we may assume $u_1\neq0$.  Then
  \begin{equation*}
    \size{\vec u}^2=u_1{}^2+u_2{}^2\geq u_1{}^2>0,
  \end{equation*}
so $\size{\vec u}>0$, and in particular $\size{\vec u}\neq0$.
Finally, for~\eqref{item:3}, just compute: 
\begin{align*}
\size{a\cdot\vec u}
&=\size{(a\cdot u_1,a\cdot u_2)}\\
&=\sqrt{(a\cdot u_1)^2+(a\cdot u_2)^2}\\
&=\sqrt{a^2\cdot(u_1{}^2+u_2{}^2)}\\
&=\size a\cdot\sqrt{u_1{}^2+u_2{}^2},
\end{align*}
which is $\size a\cdot\size{\vec u}$.
\end{proof}

The theorem does not violate any notion of $\size{\vec u}$ as the
distance between $\vec u$ and $\vec 0$.  For example, if $a\cdot\vec
u=\vec v$, then the distance from $\vec 0$ to $\vec v$ ought to be
$\size a$ times the distance to $\vec u$; but this is
what~\eqref{item:3} expresses.

But we should like
$\size{\vec u+\vec v}$ to make sense as the length of the side of a
triangle whose other two sides have lengths $\size{\vec u}$ and
$\size{\vec v}$.  In particular, we want
\begin{equation}\label{eqn:t}
  \size{\vec u+\vec v}\leq\size{\vec u}+\size{\vec v}.
\end{equation}
We cannot \emph{assume} that this is true; it is
already either true or not, since it is stating a possible property of
$\R$.  In fact, we shall prove that it is true.
Towards doing so, we note first that (since norms are
non-negative,)~\eqref{eqn:t} is logically equivalent to
\begin{align*}
  \size{\vec u+\vec v}^2
&\leq(\size{\vec u}+\size{\vec v})^2\\
&=\size{\vec u}^2+2\cdot\size{\vec u}\cdot\size{\vec v}+\size{\vec v}^2,
\end{align*}
which is equivalent to
\begin{equation}
  \frac{\size{\vec u+\vec v}^2-\size{\vec u}^2-\size{\vec v}^2}2
  \leq\size{\vec u}\cdot\size{\vec v}.
\end{equation}
It turns out to be convenient to give the left-hand member of this
inequality an abbreviation and a name: it is the \defn{scalar product}
or \defn{dot-product} of $\vec u$ and $\vec v$, and it is denoted
\begin{equation*}
  \vec u\cdot\vec v.
\end{equation*}
So, by definition, we have the identity
\begin{equation}\label{eqn:dot}
  \vec u\cdot\vec v=\frac{\size{\vec u+\vec v}^2-\size{\vec
  u}^2-\size{\vec v}^2}2.
\end{equation}
Presently we shall see an alternative expression for $\vec u\cdot\vec
v$; but let us first note that some basic properties of the scalar
product follow directly from~\eqref{eqn:dot}:

\begin{theorem}\label{thm:b}
For all $\vec u$ and $\vec v$ in $\R^2$:
\begin{enumerate}
  \item
$\vec u\cdot\vec v=\vec v\cdot \vec u$;
\item
$\vec u\cdot\vec 0=0$;
\item
$\vec u\cdot\vec u=\size{\vec u}^2$.
\end{enumerate}
\end{theorem}

\begin{proof}
  Left to the reader.
\end{proof}

To be able to say much more, we need:

\begin{lemma}
For all $\vec u$ and $\vec v$ in $\R^2$,
\begin{equation}
  \vec u\cdot\vec v=u_1\cdot v_1+u_2\cdot v_2.
\end{equation}
\end{lemma}

\begin{proof}
  Just compute.
\end{proof}

The lemma allows us to show:

\begin{theorem}\label{thm:linear}
  For all $\vec u$, $\vec v$, and $\vec w$ in $\R^2$, and $a$ in $\R$, 
  \begin{enumerate}
    \item
$\vec u\cdot(\vec v+\vec w)=\vec u\cdot\vec v+\vec u\cdot\vec w$;
\item\label{item:a2}
$\vec u\cdot(a\cdot\vec v)=a\cdot(\vec u\cdot\vec v)$.
  \end{enumerate}
\end{theorem}

\begin{proof}
  Computation.
\end{proof}

A special case of~\eqref{item:a2} is
\begin{equation*}
  \vec u\cdot(-\vec v)=-\vec u\cdot\vec v.
\end{equation*}
Using this, in~\eqref{eqn:dot}, we can replace $\vec v$ with $-\vec v$
and rearrange to get
\begin{equation}\label{eqn:lc}
  \size{\vec u-\vec v}^2=\size{\vec u}^2+\size{\vec v}^2-2\cdot\vec
  u\cdot\vec v.
\end{equation}
Note the similarity to the Law of Cosines.

\begin{theorem}[Cauchy--Schwartz]
  For all $\vec u$ and $\vec v$ in $\R^2$,
  \begin{equation}\label{eqn:cs}
  \size{\vec u\cdot\vec v}\leq\size{\vec u}\cdot\size{\vec v},
\end{equation}
with equality if and only if $\vec u\parallel\vec v$.
\end{theorem}

\begin{proof}
  Let $x$ be a scalar.  Now matter how $x$ changes, we must have
  \begin{equation*}
    0\leq\size{\vec u-x\cdot\vec v},
  \end{equation*}
equivalently,
  \begin{equation}\label{eqn:non-neg}
    0\leq\size{\vec u-x\cdot\vec v}^2.
  \end{equation}
Now compute:
\begin{align*}
  \size{\vec u-x\cdot\vec v}^2
&=(\vec u-x\cdot\vec v)\cdot(\vec u-x\cdot\vec v)&&\\
&=\vec u\cdot\vec u-2x\cdot\vec u\cdot\vec v+x^2\cdot\vec v\cdot\vec
  v&&\text{[by Theorem~\ref{thm:linear}]}\\
&=\size{\vec u}^2-2x\cdot\vec u\cdot\vec v+x^2\cdot\size{\vec v}^2
  &&\text{[by Theorem~\ref{thm:b}]} 
\end{align*}
This is a quadratic polynomial in $x$; it may be written in the more
usual fashion as
\begin{equation}\label{eqn:poly}
\size{\vec v}^2\cdot x^2
-2\cdot(\vec u\cdot\vec v)\cdot x+
\size{\vec u}^2.
\end{equation}
By the general theory of such
things, the polynomial $ax^2+bx+c$ takes an extreme value at
$-b/(2a)$, and this extreme value is $c-b^2/(4a)$; this is a maximum
value if $a>0$.  In particular, our polynomial~\eqref{eqn:poly} has minimum
value 
\begin{equation*}
  \size{\vec u}^2-\frac{(\vec u\cdot\vec v)^2}{\size{\vec v}^2}.
\end{equation*}
This cannot be negative, by~\eqref{eqn:non-neg}.  That is,
\begin{gather*}
0\leq  \size{\vec u}^2-\frac{(\vec u\cdot\vec v)^2}{\size{\vec
    v}^2},\\
\frac{(\vec u\cdot\vec v)^2}{\size{\vec v}^2}\leq\size{\vec u}^2,\\
(\vec u\cdot\vec v)^2\leq\size{\vec u}^2\cdot\size{\vec v}^2,
\end{gather*}
and therefore
\begin{equation*}
\size{\vec u\cdot\vec v}\leq\size{\vec u}\cdot\size{\vec v}.  
\end{equation*}
Finally, this inequality is an equation if and only if it is possible
for $\vec u-x\cdot\vec v$ to be $\vec0$; but this is possible if and
only if $\vec u$ and $\vec v$ are parallel.
\end{proof}

If we accept that there is a function $\cos$ on $\R$ that takes on
every value in the interval $[-1,1]$, then, by the Cauchy--Schwartz
Inequality~\eqref{eqn:cs}, 
Theorem, if $\vec u$ and $\vec v$ 
are non-zero, there is $\theta$ such that
\begin{equation}\label{eqn:cos}
  \cos \theta=\frac{\vec u\cdot\vec v}{\size{\vec u}\cdot\size{\vec v}};
\end{equation}
in particular, $\cos\theta=1$ if and only if $\vec u\parallel\vec v$.
Rewriting~\eqref{eqn:cos} as
\begin{equation}
\size{\vec u}\cdot\size{\vec v}\cdot\cos \theta=\vec u\cdot\vec v
\end{equation}
and substituting into~\eqref{eqn:lc}, we get a more familiar form of
the Law of Cosines.


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