\documentclass[a4paper,11pt,twoside,reqno]{amsart}

\title{\emph{Math 304 examination} {solutions}}
\author{David Pierce}
\email{dpierce@metu.edu.tr}
\address{Mathematics Dept., Middle East Technical University, Ankara
06531, Turkey} 
\urladdr{metu.edu.tr/~dpierce/}
\date{Exam date: Tuesday, March 30, 2010; this document: \today}

\usepackage{hfoldsty}
\usepackage{typearea}
\usepackage{pstricks,pst-plot}
\usepackage{paralist}

\newtheorem{problem}{Problem}
\theoremstyle{definition}
\newtheorem*{solution}{Solution}
\theoremstyle{remark}
\newtheorem*{remark}{Remark}

\renewcommand{\theenumi}{\alph{enumi}}
\renewcommand{\labelenumi}{(\theenumi).}
\renewcommand{\theequation}{\fnsymbol{equation}}
\begin{document}

\maketitle
\thispagestyle{empty}

\emph{You may use modern notation in your work; but
  Problems~\ref{prob:quad} and~\ref{prob:OK} should involve diagrams.}  

\begin{problem}
A straight line is cut into equal and unequal segments.  What is the
relationship between the square on the half and the rectangle
contained by the unequal segments?
  \end{problem}

\begin{solution}
The square exceeds the rectangle by the square on the straight line
between the points of section. 
\end{solution}

\begin{remark}
This problem is based on Proposition II.5 of Euclid's \emph{Elements.}
The language follows the style of Heath's translation of Euclid (on
the course webpage).
\end{remark}

  \begin{problem}\label{prob:quad}
    A square is equal to three roots and twenty-eight dirhams.  
What is the root?
Give a geometrical justification of your answer (as Mu\d
hammad ibn M\=us\=a al-Khw\=arizm\=\i{} or Th\=abit ibn Qurra did).
  \end{problem}

\begin{solution}
In Figure~\ref{fig:1}, the root is $AB$; $AC=3$; and $D$ bisects $AC$.  Then
\begin{gather*}
		DB^2=28+DC^2
        =28+\left(\frac32\right)^2
        =\frac{121}4,\\
      DB=\frac{11}2,\\
      AB=AD+DB=\frac32+\frac{11}2=7;
\end{gather*}
so the root is $7$.
\begin{figure}[ht]
\begin{pspicture}(3.5,4)
\psset{unit=5mm}
\psline(3,0)(7,0)(7,7)(3,7)(3,0)(0,0)(0,7)(3,7)
\psline[linestyle=dashed](7,1.5)(1.5,1.5)(1.5,7)
\psline[linestyle=dashed](1.5,5.5)(3,5.5)
\rput(5,3.5){$28$}
\uput[u](0,7){$A$}
\uput[u](7,7){$B$}
\uput[u](3,7){$C$}
\uput[u](1.5,7){$D$}
\end{pspicture}
\caption{}\label{fig:1}
\end{figure}
\end{solution}

\begin{remark}
Euclid's Proposition II.6 is behind this problem.
\end{remark}

\begin{problem}\label{prob:OK}
  Suppose a cube and nine sides are equal to ten.
Find the side by taking the
intersection of two conic sections (as Omar Khayy\=am did).  It is
preferable if one of those sections is a circle.
\end{problem}

\begin{solution}
\mbox{}[\emph{Analysis:}]\vspace{-\baselineskip}
\begin{gather}\notag
	x^3+9x=10,\\\notag
	x^3=10-9x,\\\label{eqn:Kh1}
	\frac{x^2}9=\frac{10/9-x}x,\\\label{eqn:Kh2}
	\frac x3=\frac yx=\frac{10/9-x}y,\\\notag
x^2=3y\quad\&\quad y^2=x\left(\frac{10}9-x\right).
\end{gather}
[\emph{Synthesis:}] As in Figure~\ref{fig:2}, let $ABC$ be a
semicircle with diameter $10/9$, and let $AD$, perpendicular to $AB$,
be the axis of a parabola with parameter $3$.  The semicircle and
parabola intersect at a point $C$ (as well as at $A$).  Let $CE$ be
dropped perpendicular 
to $AB$; and $CD$, to $AD$.  Then $AE=CD$; either of these is the
desired ``side''.  Indeed, 
\begin{gather*}
  CD^2=3AD,\\
    CD:3::AD:CD
        ::EC:AE::EB:EC,\\
AE^2:9::CD^2:9::EB:AE
              ::\left(\frac{10}9-AE\right):AE,\\
AE^3=10-9AE,\\
AE^3+9AE=10.
\end{gather*}
\begin{figure}[ht]
\begin{pspicture}(-2.5,-0.5)(7.5,3.75)
\psset{unit=5cm}
%\parametricplot{0}{180}{20 9 div t cos mul 20 9 div add t 20 9 div t sin mul}
\parametricplot{0}{180}{t cos 1 add 5 9 div mul t sin 5 9 div mul}
\parametricplot{-0.5}{1.5}{t t t mul 3 div}
\psline(0,0.75)(0,0)(1.11,0)
\psline(1,0)(1,0.33)(0,0.33)
\uput[d](0,0){$A$}
\uput[dr](1.11,0){$B$}
\uput[r](1,0.33){$C$}
\uput[l](0,0.33){$D$}
\uput[d](1,0){$E$}
\end{pspicture}
\caption{}\label{fig:2}
\end{figure}
\end{solution}

\begin{remark}
\begin{asparaenum}[(i).]
\item
  In the solution, \emph{analysis} and \emph{synthesis} are used in
  the sense attributed to Theon (presumably Theon of Smyrna, that is,
  \.Izmir) by Vi\`ete at the beginning of Chapter 1 of 
  the \emph{Introduction to the Analytic Art.}  In his solutions of
  cubic equations, Omar Khayy\=am gives only the synthesis; we can
  only speculate whether he had some sort of analysis like ours.  
\item
In
  our analysis, equations~\eqref{eqn:Kh1} and~\eqref{eqn:Kh2} could
  have been
  \begin{gather*}
	x^2=\frac{10-9x}x,\\
	x=\frac yx=\frac{10-9x}y,
  \end{gather*}
yielding the parabola given by $y=x^2$ and the ellipse given by
$y^2=x(10-9x)$.  This is why the problem says, ``It is
preferable if one of those sections is a circle.''
\item
I think it is better to understand the circle through the equation
$y^2=x(10/9-x)$ than to convert this equation to the more usual
modern form,
\begin{equation*}
  y^2+\left(x-\frac59\right)^2=\left(\frac59\right)^2.
\end{equation*}
\end{asparaenum}
\end{remark}

\begin{problem}
Again, a cube and nine sides are equal to ten.
  \begin{enumerate}
  \item
Find the side numerically, as the difference of the cube roots of a
\emph{binomium} and an \emph{apotome,} by Cardano's method
(really Tartaglia's 
method); your steps should be clearly justifiable.  
\item
The side is in fact a whole
number; which one? 
\end{enumerate}
\end{problem}

\begin{solution}
\begin{enumerate}
\item
We have to solve $x^3+9x=10$.  We let $x=u-v$, so
\begin{equation*}
x^3=u^3-v^3-3uv(u-v)=u^3-v^3-3uvx.
\end{equation*}
So we let
\begin{align*}
u^3-v^3&=10,&uv&=3,
\end{align*}
which we can solve:
\begin{gather*}
	u^6-u^3v^3=10u^3,\\
u^6-27=10u^3,\\
u^3=\sqrt{5^2+27}+5=2\sqrt{13}+5,\\
v^3=\frac{3^3}{2\sqrt{13}+5}=2\sqrt{13}-5.
\end{gather*}
Therefore
\begin{equation*}
x=\sqrt[3]{2\sqrt{13}+5}-\sqrt[3]{2\sqrt{13}-5}.
\end{equation*}
\item $x=1$.
\end{enumerate}
\end{solution}

\begin{remark}
  \begin{asparaenum}[(i).]
    \item
Cardano does give a formula for finding $x$, without clear
explanation.  However, this problem said ``steps should be clearly
justifiable''; so for full credit, the answer should be
\emph{derived,} as above, not just obtained from a memorized formula.
Some people who tried to memorize, remembered wrongly.
\item
Of course, the solution above did rely on the (memorized) quadratic
formula.  Memory does have its uses.
\item
Note here that $u^3$ could have been $-2\sqrt{13}+5$; but
$x$ in the end would have been the same.  Two other values of $x$ can
be obtained by considering \emph{complex} cube roots; but Cardano does
not know about these.
  \end{asparaenum}
\end{remark}

\begin{problem}
  A square-square, twelve squares, and thirty-six are equal to
  seventy-two sides.  In finding the side by Cardano's method (really
  Ferrari's method), you first solve a cubic equation.
  \begin{enumerate}
  \item\label{part:1}
Obtain that cubic equation. %in the form ``cube and squares equal to
                           %number''. 
\item
Convert that cubic equation to an equation of the form ``cube equal to roots and
number''. 
\item
The cubic equation in~\eqref{part:1} should have $6$ as a root.  Use this to
find the side in the original fourth-degree equation.
  \end{enumerate}
\end{problem}

\begin{solution}
\begin{enumerate}
\item
\mbox{}\vspace{-\baselineskip}
\begin{gather*}
	x^4+12x^2+36=72x,\\
	(x^2+6)^2=72x,\\
	(x^2+6+t)^2=2tx^2+72x+t^2+12t,\\
	2t(t^2+12t)=36^2=2^43^4,\\
	t^3+12t^2=2^33^4=648.
\end{gather*}
\item
Let $t=s-4$; then
\begin{gather*}
	s^3-48s+12\cdot16-64=2^33^4,\\
	s^3-48s=2^33^4+2^6-2^63=2^3(3^4-2^4)=8\cdot65=520.
\end{gather*}
\item
\mbox{}
\vspace{-\baselineskip}
\begin{gather*}
\begin{aligned}
	(x^2+12)^2&=12x^2+72x+108,\\
	&=12(x^2+6x+9)\\
	&=12(x+3)^2,
	\end{aligned}\\
	x^2+12=2\sqrt3(x+3),\\
	x^2=2\sqrt3-6(2-\sqrt3),\\
	x=\sqrt3+\sqrt{3-6(2-\sqrt3)}=\sqrt3+\sqrt{6\sqrt3-9}.
\end{gather*}
\end{enumerate}
\end{solution}

\begin{remark}
If we believe in negative numbers, then from $(x^2+12)^2=12(x+3)^2$ we
should obtain $x^2+12=\pm2\sqrt3(x+3)$; but the negative sign here
leads to a negative value of $x$.  The problem asks for the ``side'',
which is implicitly positive. 
\end{remark}


\end{document}