\documentclass[a4paper,12pt,twoside]{amsart}

\title{Math 304, 2009/10, final examination \emph{solutions}}
\author{David Pierce}
\address{Mathematics Dept., Middle East Technical University, Ankara
06531, Turkey} 
\email{dpierce@metu.edu.tr}
\urladdr{http://metu.edu.tr/~dpierce/}
\date{\today}

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\begin{document}

\maketitle
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\begin{solution}
  \begin{enumerate}
  \item 
The substitution $x=t-1$ converts ($*$) into ($\dagger$); so $x$ is a
solution to ($*$) if and only if $x+1$ is a solution to ($\dagger$)).   
\item
From ($\dagger$) we have
\begin{align*}
  \frac{t^2}9&=\frac{t+1}t,&
\frac t3&=\frac yt=\frac{t+1}y
\end{align*}
for some $y$; that is, we can solve ($\dagger$) by simultaneously
solving $t/3=y/t$ and $y/t=(t+1)/y$, that is,
\begin{align*}
  t^2&=3y,&y^2&=t(t+1).
\end{align*}
These equations define a parabola and a hyperbola, respectively, as
below.  Then $AB$ is a solution to ($\dagger$).
\begin{center}
  \begin{pspicture}(-3,-1)(5,4.5)
%\psgrid
    \psplot{-3}{3.6}{x x mul 3 div}
    \psplot{0}{4}{x x 1 add mul sqrt}
    \psplot{0}{0.6}{x x 1 add mul sqrt neg}
    \psplot{-3}{-1}{x x 1 add mul sqrt}
    \psplot{-1.6}{-1}{x x 1 add mul sqrt neg}
\psline(-3,0)(5,0)
\psline(0,-1)(0,4.5)
\psline(0,3.88)(3.42,3.88)
\psline(0,1.64)(-2.20,1.64)
\psline(0,0.48)(-1.20,0.48)
\uput[l](0,3.88){$A$}
\uput[dr](3.42,3.88){$B$}
\uput[r](0,1.64){$C$}
\uput[dl](-2.20,1.64){$D$}
\uput[ul](0,0.48){$E$}
\uput[dl](-1.20,0.48){$F$}
\psdots(0,3.88)(3.42,3.88)(0,1.64)(-2.20,1.64)(0,0.48)(-1.20,0.48)(0,0)
\uput[dl](0,0){$G$}
\end{pspicture}
\end{center}
\item
The negative solutions of ($\dagger$) are $CD$ and $EF$.  (The parabola and
hyperbola intersect also at $G$, but no solution to ($\dagger$)
corresponds to this, since the corresponding value of $y$ is $0$.)
\item
Let $t=u+v$; then
\begin{equation*}
  t^3=3uvt+u^3+v^3.
\end{equation*}
Then ($\dagger$) holds, provided
$uv=3$ and $u^3+v^3=9$.
Solving these, we have
\begin{align*}
  u^6+u^3v^3&=9u^3,&
u^6+27&=9u^3,&
u^3=\frac92\pm\sqrt{\frac{81}4-27}=\frac{9\pm3\sqrt{-3}}2.
\end{align*}
So if $u$ is a cube root of $(9+3\sqrt{-3})/2$, then
 one solution to ($\dagger$) is $u+3/u$.
  \end{enumerate}
\end{solution}

\begin{remark}
  Cardano could not give a meaning to the solution we found in the
  last part; today we can, and the three choices of the cube root give
  the three solutions found geometrically earlier.
\end{remark}

\begin{solution}
  \begin{enumerate}
  \item 
If ($\ddag$) holds, then in particular it holds when $C$ is $F$.
Therefore ($\ddag$) is equivalent to
\begin{align*}
  \frac{CD^2}{AC\times CB}&=\frac{EF^2}{AF\times FB}=\frac{a^2}{b^2},&
\frac{x^2}{b^2-y^2}&=\frac{a^2}{b^2},&
b^2x^2&=a^2b^2-a^2y^2,
\end{align*}
which is equivalent to ($\parallel$).
\item
Let $HG=e$.  Then ($\mathsection$) becomes
\begin{equation*}
  \frac{e-b}{e+b}=\frac{b-d}{b+d},
\end{equation*}
which yields $e=b^2/d$.
\item
Since $CDL\sim FEG$, and
\begin{equation*}
  FEG=\frac12\Bigl(\frac{b^2}d-d\Bigr)c,
\end{equation*}
we have
\begin{equation*}
  CDL=\frac{x^2}{c^2}FEG=\frac{x^2}{2c}\Bigl(\frac{b^2}d-d\Bigr).
\end{equation*}
We assume angle $BAK$ is right; otherwise, we can just
multiply throughout by its sine.)
Also $AHK$ and $CHM$ are both similar to $FHE$, which is $cd/2$; so
\begin{equation*}
  AKH-CHM=\frac{cd}2\Bigl(\frac{b^2}{d^2}-\frac{y^2}{d^2}\Bigr).
\end{equation*}
So ($**$) holds if and only if
\begin{gather*}
  \frac{x^2}c\Bigl(\frac{b^2}d-d\Bigr)
=c\frac{b^2-y^2}d,\\
x^2(b^2-d^2)=c^2(b^2-y^2),\\
b^2x^2=a^2(b^2-y^2),
\end{gather*}
which is equivalent to ($\parallel$).
\item
In ($**$), let $C$ be $F$; then the equation becomes
\begin{equation*}
  FEG=AHK-FHE,
\end{equation*}
so $AHK=FEG+FHE=GHE$.
\item
By part \textbf C, it is enough to show
\begin{equation*}
  PDM=EHG-PHL.
\end{equation*}
We have
\begin{align*}
  PDM
&=CDL+CHM-PHL&&\\
&=AHK-PHL&&\text{ [by ($**$)]}\\
&=EHG-PHL&&\text{ [by \textbf D]}.
\end{align*}
  \end{enumerate}
\end{solution}
\vfill
\end{document}
