\documentclass[12pt]{article}
\usepackage{graphicx,amssymb,amsmath}
\newtheorem{thm}{Theorem}
\newtheorem{prop}{Proposition}
\newtheorem{lem}{Lemma}
\newtheorem{cor}{Corollary}
\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\topmargin}{0in}
\setlength{\textheight}{8.5in}
\begin{document}
\centerline{MONT 107Q -- Thinking about Mathematics}
\centerline{Discussion -- Descartes and Coordinate, or ``Analytic'' Geometry}
\centerline{April 19 and 21, 2017}
\vskip 10pt
\noindent
{\it Background}
\vskip 10pt
In 1637, the French philosopher and mathematician Ren\'e Descartes published 
a pamphlet called {\it La G\'eometrie} as one of a series of discussions about
particular sciences intended apparently as illustrations of his general
ideas expressed in his work {\it Discours de la m\'ethode pour bien conduire 
sa raison, et chercher la v\'erit\'e dans les sciences} (in English: Discourse
on the method of rightly directing one's reason and searching for truth in the 
sciences).  This work introduced other mathematicians to a new way of dealing
with geometry.  Via the introduction (at first only 
in a rather indirect way) of numerical coordinates to describe points
in the plane, Descartes 
showed that it was possible to \emph{define geometric objects by means
of algebraic equations} and to apply techniques from \emph{algebra} to deduce
geometric properties of those objects.  

He illustrated his new methods first by considering a problem first studied
by the ancient Greeks after the time of Euclid:
\vskip 10pt
\noindent
\emph{Given three (or four) lines in the plane, find the locus of points $P$
that satisfy the relation that the square of the distance from $P$ to the 
first line (or the product of the distances from $P$ to the first two of the 
lines) is equal to the product of the distances from $P$ to the other two
lines.}
\vskip 10pt
The resulting curves are called \emph{three-line} or \emph{four-line} loci
depending on which case we are considering.  


For instance, the locus of points satisfying the condition above for 
the four lines 
\begin{eqnarray*}
x + 2 &=& 0,\\
x - 1 &=& 0,\\
y + 1 &=& 0,\\ 
y - 1 &=& 0
\end{eqnarray*}
is shown at the top on the back of this sheet.
The point shown in black is $P = (1,1)$.  Note that it satisfies the defining condition
\emph{the product of the distances from $P$ to the first two of the 
lines is equal to the product of the distances from $P$ to the other two
lines} since it lies on the line $x - 1 = 0$ so the first product is zero, 
and it lies on the line $y - 1 = 0$, so the second product is also zero.


The Greek mathematician Apollonius (ca. 262 BCE - ca. 190 BCE) considered
this problem in Book III of his masterwork called the {\it Conics}.  He showed via
\emph{extremely elaborate} ``synthetic'' (i.e. Euclid-style) proofs that
\emph{both the three- and four-line loci are conic sections -- the
ellipses, hyperbolas, and parabolas that are obtained as plane sections
of a cone.}  This lines in Figure 1 give a \emph{hyperbola}; the other conics can be obtained by varying the 
positions of the lines, making them meet at non-right angles, etc.  

\begin{figure}
\begin{centering}
\label{expformfig}
\includegraphics[width=3in,height=3in]{FourLine.png}
\caption{A four-line locus}
\end{centering}
\end{figure}

Descartes actually learned about this work by reading an account given
by the later Greek mathematician Pappus (ca. 290 - 350 CE), whose work we
have encountered previously.  Pappus' 
{\it Mathematical Collection} preserved much of the earlier work of Greek 
mathematicians and earlier thinking about the complementary
roles of {\it analysis} and {\it synthesis} in mathematics.  This work
by Pappus was translated into Latin in the 16th century
and reintroduced much Greek advanced mathematics to Europe around the 
time of F. Vi\`ete, and slightly later, R. Descartes.  With his ``analytic'' 
geometry in the plane, Descartes was able to derive Apollonius's results 
in a much easier way, and he also showed how to solve analogous problems when 
the locus was described by any number $\ge 3$ of lines.  

Today, using our knowledge of coordinate geometry, we want to understand
what Descartes did, why it was such an advance, and why it essentially
(re)-united algebra and geometry.
\vskip 10pt
\noindent
{\it Questions}
\vskip 10pt
\begin{itemize}
\item[(A)]  A line $L$ in the coordinate plane is given by the equation 
$Ax + By + C = 0$.  Let $Q = (x_0,y_0)$ be a point.  The 
(perpendicular) distance from $L$ to $Q$ is given by 
$$d(L,Q) = \frac{|A x_0 + B y_0 + C|}{\sqrt{A^2 + B^2}}$$
Using this formula, determine the distance from the point 
$$Q = \left(\frac{-1+\sqrt{5}}{2},0\right)$$
to each of the lines given above (the ones that produce the plot at the top of the 
page).  Deduce that $Q$ lies on the four-line locus for these lines.
\item[(B)] Use part A to prove Apollonius's result that starting 
from any three or four lines 
$$L_i:  A_i x + B_i y + C_i = 0, i = 1,2,3,4$$
the locus of points $P = (x,y)$ such that {\it the square of the distance
from $P$ to line $L_1$ is equal to the product of the distances from $P$
to $L_2$ and $L_3$} is defined by an algebraic equation
of total degree 2 in the coordinates of the point $P$ -- that is, 
an equation of the form:
\begin{equation}
\label{conic}
D x^2 + E xy + F y^2 + G x + Hy + I = 0.
\end{equation}
for some real numbers $D,E,F,G,H,I$ (after possibly ignoring the absolute values).
Do the same for the locus of points $P = (x,y)$ such that {\it the product of the
distances from $P$ to $L_1$ and to $L_2$ is equal to the product of the distances
from $P$ to $L_3$ and to $L_4$}.  
\item[(C)]  Give examples of equation of the form (\ref{conic}) in part (B) defining each of the
three conic sections -- ellipse, parabola, hyperbola.  Also, what is the 
locus of points defined by the equation $x^2 - y^2 = 0$?  What about $x^2 + y^2 = 0$?
What about $x^2 + y^2 + 1 = 0$? 
\end{itemize}
\end{document}





\end{document}
