\documentclass[a4paper,11pt,reqno]{amsart}
\usepackage{amssymb}
\title{Math 304 final examination}
\author{David Pierce}
\address{Mathematics Dept., Middle East Technical University, Ankara
06531} 
\email{dpierce@metu.edu.tr}
\urladdr{http://metu.edu.tr/~dpierce/Courses/303/}
\date{Saturday, June 12, 2010}
\usepackage{layout}

%\setlength{\topmargin}{-1in}
%\setlength{\textheight}{10.5in}

\usepackage{hfoldsty}
%\usepackage[oneside]{typearea}
\usepackage{pstricks,pst-plot}
\usepackage{paralist}

\theoremstyle{definition}
\newtheorem{problem}{Problem}
\newtheorem*{bonus}{Bonus}
%\renewcommand{\theenumi}{\alph{enumi}}
%\renewcommand{\labelenumi}{(\theenumi).}
\renewcommand{\theenumi}{\textbf{\Alph{enumi}}}
\renewcommand{\labelenumi}{\textbf{\theenumi.}}
\renewcommand{\theequation}{\fnsymbol{equation}}

\renewcommand{\triangle}{\mathord{\vartriangle}}

\pagestyle{empty}

\begin{document}

\maketitle
\thispagestyle{empty}

%\center{\textsc{Math 304 final examination, Saturday, June 12, 2010}}

Write your solutions on separate sheets; you may keep \emph{the}
problem sheets.  There are two numbered problems (with several
lettered parts each) and a bonus.
\emph{\.Iyi \c cal\i\c smalar; kolay gelsin.} 

\begin{problem}
This problem is about the cubic equations
\begin{gather}\label{eqn:x}
  x^3+3x^2=6x+17,\\\label{eqn:t}
  t^3=9t+9.
\end{gather}
  \begin{asparaenum}
\item
Explain the relation between the solutions of~\eqref{eqn:x} and 
%the solutions of
\eqref{eqn:t}. 
\item
For one of~\eqref{eqn:x} and~\eqref{eqn:t}, find a solution geometrically,
by intersecting conic sections (as Omar Khayyam does).
\item
Find \emph{three} solutions in the same way (some might be negative).
\item
Find a solution of~\eqref{eqn:x} or~\eqref{eqn:t} numerically (in the
manner suggested by Cardano); your steps should be justifiable.  Your
answer will involve square roots of negative numbers. 
  \end{asparaenum}
\end{problem}

\begin{problem}
This problem shows that every line through the
center of an ellipse is a diameter with certain properties.  The
method is based on Apollonius; but the algebraic geometry of Descartes
makes some simplifications possible. 
\begin{center}
\psset{xunit=2.5cm,yunit=3.5cm,plotpoints=360}
\begin{pspicture}(-1.2,-1.2)(1.2,1.4)
%\psgrid
  \parametricplot{0}{360}{t cos t sin}
\psline(-0.6,-0.8)(0.75,1)(0,1)
\psline(0.6,0.8)(0,1.25)(0,-1)
\psline(0.6,0.8)(0,0.8)
\psline[linestyle=dashed](0.15,0.2)(-0.98,0.2)(0,-0.53)
\uput[ul](0,1){$A$}
\uput[d](0,-1){$B$}
\uput[ul](0,0.2){$C$}
\uput[l](-0.98,0.2){$D$}
%\uput[r](0.98,0.2){$D'$}
\uput[r](0.6,0.8){$E$}
\uput[l](0,0.8){$F$}
\uput[u](0,1.25){$G$}
\uput[dr](0,0){$H$}
\uput[u](0.75,1){$K$}
\uput[r](0,-0.53){$L$}
\uput[dr](0.15,0.2){$M$}
\uput[dl](-0.6,-0.8){$N$}
\uput[l](-0.26,-0.34){$P$}
\end{pspicture}
\hfill
\begin{pspicture}(-1.2,-1.2)(1.2,1.4)
%\psgrid
  \parametricplot{0}{360}{t cos t sin}
\psline(-0.68,-0.9)(0.75,1)(0,1)
\psline(0.6,0.8)(0,1.25)(0,-1)
\psline(0.6,0.8)(0,0.8)
\psline[linestyle=dashed](-0.28,-0.37)(0.44,-0.9)(-0.68,-0.9)
\uput[ul](0,1){$A$}
\uput[d](0,-1){$B$}
\uput[ul](0,-0.9){$C$}
\uput[dr](0.44,-0.9){$D$}
\uput[r](0.6,0.8){$E$}
\uput[l](0,0.8){$F$}
\uput[u](0,1.25){$G$}
\uput[dr](0,0){$H$}
\uput[u](0.75,1){$K$}
\uput[ur](0,-0.57){$L$}
\uput[d](-0.68,-0.9){$M$}
\uput[l](-0.6,-0.8){$N$}
\uput[ul](-0.28,-0.37){$P$}
\end{pspicture}
\end{center}

Straight line $AB$ is given, and angle $BAK$ is given.  The point $C$ moves
along $AB$, and as it moves, straight line $CD$ remains parallel to $AK$.  But $D$ moves along $DC$ as $C$ moves, so that $D$ traces out a curvilinear figure
$ADB$, as shown above with two possible positions of $DC$. 

Recall that the curvilinear figure $ADB$ is an \textbf{ellipse} with \textbf{diameter} $AB$ and \textbf{ordinates} parallel to $AK$ if and only if
\begin{equation}\label{eqn:p}
  CD^2\propto AC\times CB
\end{equation}
(that is, the square on $CD$ varies as the rectangle formed by $AC$
and $CB$).

Let $E$ be chosen at random on $ADB$, and let straight line $EF$ be
drawn parallel to $KA$, meeting $AB$ at $F$.  Let straight line $EG$
be drawn, meeting $BA$ extended at $G$ 
so that
\begin{equation}
  \frac{AG}{GB}=\frac{AF}{FB}.
\end{equation}
Let $H$ be the midpoint of $AB$, and let straight line $HE$ be drawn and extended to meet $AK$ at $K$.  Let $L$ be taken on $AB$ (extended if necessary) so that
straight line $DL$ is parallel to $GE$.  Finally, let $M$ be the point
of intersection of $DC$ and $HK$ (both extended if necessary).

For computations, let
\begin{align*}
  AH&=b,&
EF&=c,&
HF&=d,&
CD&=x,&
CH&=y.
\end{align*}
Also, let $a$ be such that
\begin{equation}
\frac{a^2}{b^2}=\frac{EF^2}{AF\times FB}=\frac{c^2}{b^2-d^2}.
\end{equation}


\begin{asparaenum}
\item\label{item:a}
Show that~\eqref{eqn:p} holds if and only if
\begin{equation}
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.
\end{equation}
\item
Find  $HG$ in terms of $b$ and $d$.
\item\label{item:tri}
Show that~\eqref{eqn:p} holds if and only if
\begin{equation}\label{eqn:ell}
  \triangle CDL=\triangle AHK-\triangle CHM.
\end{equation}
(Angle $BAK$ is not assumed to be a right
angle; but the computations can be performed as if it were.)
\item\label{item:b}
Assuming~\eqref{eqn:p} holds (and hence~\eqref{eqn:ell} holds, for \emph{all} possibilities for $C$), show
\begin{equation*}
\triangle AHK=\triangle GHE.
\end{equation*}
\item
Assume~\eqref{eqn:p} holds.  Let
$EH$ be extended to meet the ellipse again at $N$, and let $EN$ meet $DL$ (extended as necessary) at $P$.  Show that the curvilinear figure $ADB$
is an ellipse with diameter $EN$ whose ordinates are parallel to
$EG$.  (You will probably want to use part~\ref{item:tri}, translated appropriately.)
%(As far as I know, the best way to do this is by means of
       %part~\eqref{item:b}.) 
\end{asparaenum}
\end{problem}




\begin{bonus}
  What are your suggestions for improving the course?
\end{bonus}

\emph{Geldi\u giniz i\c cin te\c sekk\"urler.  \.Iyi tatiller!}


\vfill
%\layout

\end{document}