\documentclass[a4paper,12pt,twoside]{amsart}
%\documentclass[a4paper,12pt,twoside]{article}
\title{Math 303, final examination}
\author{David Pierce}
%\email{dpierce@metu.edu.tr}
%\address{Mathematics Dept., Middle East Technical University, Ankara 06531, Turkey}
%\urladdr{metu.edu.tr/~dpierce/}
\usepackage{url}
\date{Tuesday, January 12, 2010}
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\usepackage[oxonia]{psgreek}
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\theoremstyle{definition}
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\newtheorem*{bonus}{Bonus}
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\begin{document}
\maketitle\thispagestyle{empty}
\begin{problem}
In the 8th century \textsc{b.c.e.,} the colony of Cumae (\Gk{K'umh})
was founded, near what is now Naples, by settlers from Euboea (E\u
griboz), and also from Cyme (\Gk{K'umh}) in western Anatolia near what
is now Alia\u ga.\footnote{Paul Harvey, \emph{The Oxford Companion to
Classical Literature} (1980); Bilge Umar, \emph{T\"urkiye'deki
Tarihsel Adlar} (\.Istanbul: \.Inkil\^ap, 1993).} From the Greek
alphabet as used in Cumae, the Latin alphabet was ultimately derived;
this came to have 23 letters:
\begin{center}
A B C D E F G H I K L M N O P Q R S T V X Y Z.
\end{center}
In the year 863 \textsc{c.e.}, a monk from Salonica named Cyril invented
the so-called Glagolitic alphabet in order to translate holy scripture
from Greek into Old Bulgarian. Soon after that, the simpler Cyrillic
alphabet was invented.\footnote{S. H. Gould, \emph{Russian for the
Mathematician}
(Springer-Verlag, Berlin--Heidelberg--New York, 1972). Many
alphabets can be seen in Carl Faulmann, \emph{Yaz\i\ Kitab\i}
(T\"urkiye \.I\c s Bankas\i\ K\"ult\"ur Yay\i nlar\i, 2001).} After
some changes (such as the abolition of a few letters by the Soviet
government in 1918), the Cyrillic alphabet became the 33-letter
Russian alphabet of today:
\newcommand{\gap}{\hfill}
\begin{center}
\Ru{A\gap B\gap V\gap G\gap D\gap E\gap \"E\gap Zh\gap Z\gap I\gap \CYRISHRT\gap K\gap L\gap M\gap N\gap O\gap P\gap R\gap S\gap T\gap U\gap F\gap H\gap C\gap Q\gap X\gap W\gap \CYRHRDSN\gap Y\gap \CYRSFTSN\gap \CYREREV\gap Yu\gap Ya.}
\end{center}
This alphabet retains 19 of the 24 letters of the Greek alphabet, in
their original order, though not always in the original form. What
\emph{are} the 24 letters of the Greek alphabet?
\begin{center}
\newcommand{\letter}{\makebox[3mm]{}}
\begin{tabular}{|*{24}{c|}}\hline
\letter&
\letter&
\letter&
\letter&
\letter&
\letter&
\letter&
\letter&
\letter&
\letter&
\letter&
\letter&%\\\hline\hline
\letter&
\letter&
\letter&
\letter&
\letter&
\letter&
\letter&
\letter&
\letter&
\letter&
\letter&
\letter\\\hline
\end{tabular}
\end{center}
\end{problem}
%\newpage
\begin{problem}
Does a square have a ratio to its side? Explain.
\end{problem}
\vfill
\begin{problem}
Suppose a magnitude $A$ has a ratio to a magnitude $B$, and a
magnitude $C$ has a ratio to a magnitude $D$. What does it mean to
say that $A$ has the \emph{same} ratio to $B$ that $C$ has to $D$
(according to Definition 5 of Book V of Euclid's \emph{Elements})?
\end{problem}
\vfill\vfill
\enlargethispage{2\baselineskip}
\newpage
\begin{problem}
Suppose a straight line $AB$ is bisected at $C$, and another point,
$D$, is chosen on $AB$. What is the relation between the squares on
$AC$ and $CD$ and the rectangle contained by $AD$ and $DB$?
\end{problem}
\vfill
\begin{problem}
In the diagram, $BAC$ is the diameter of a circle, $A$ is the
center, and $AD$ is at right angles to $BC$. Straight line $DC$ is
drawn. From a point $E$ on the circumference between $B$ and $D$,
the straight line $EF$ is drawn at right angles to $AD$, and $EA$ and $ED$ are drawn.
Show that the square on $DE$ has the same ratio to the square on
$DC$ that the straight line $DF$ has to $DA$. (\emph{Suggestion:}
express $DE^2$ and $DC^2$ in terms of $DF$, $FA$, and $DA$.)
\end{problem}
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\uput[ur](0,-1.2){$F$}
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\vfill\vfill\vfill
\newpage
\begin{problem}
In the diagram on the left below, $ABC$ is an axial triangle of a cone
whose base is the circle $CDEBFG$, and $DKG$ and $EMF$ are at right
angles to $BC$. Planes through $DKG$ and $EMF$ cut the cone, making
sections $DHG$ and $ELF$, with diameters $HK$ and $LM$, respectively;
and these diameters are parallel to $AC$. The \textbf{parameters} (the
`upright sides' or \emph{latera recta}) of the sections are not shown;
but let them be $HN$ and
$LP$. What is the ratio of $HN$ to
$LP$ (in terms of straight lines that \emph{are} shown in the diagram)?
\end{problem}
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\begin{problem}
We know that an ellipse or an hyperbola has two `conjugate' diameters,
each diameter being situated ordinatewise with respect to the other.
A parabola cannot have conjugate diameters in this sense.
Nonetheless, suppose, in the diagram on the right above, $AB$ is the
diameter of a parabola, and $AC$ is drawn ordinatewise, and $AC$ is
also the diameter of another parabola, and $AB$ is situated
ordinatewise with respect to $AC$. Suppose the two parabolas meet at
$D$ (as well as at $A$). Let the respective ordinates $DB$ and $DC$
be dropped. Finally, suppose the parabola with diameter $AB$ has
parameter $E$ (not shown), and the parabola with diameter $AC$ has parameter $F$.
Show that
\begin{align*}
E:AC&::AC:AB,&AC:AB&::AB:F.
\end{align*}
(\emph{Remark.} It follows then that $E$ is to $F$ as the \emph{cube}
on $AC$ is to
the cube on $AB$. In particular, if $E$ is twice $F$, then the cube
on $AC$ is double the cube on $AB$.
According to Eutocius in his
\emph{Commentary on Archimedes's Sphere and Cylinder,} Menaechmus
discovered this method of `duplicating' the cube, along with another
method involving a parabola and a hyperbola. This work is the earliest
known use of conic sections.
For Menaechmus however, the angle $BAC$ would have been
right.)
\end{problem}
\newpage
\begin{problem}
In the triangle $ABC$ below, $FG$ is parallel to $DC$, and $DE$ is
parallel to $AG$. Show that $AC$ is parallel to $FE$. (You may use
the theory of proportion developed in Books V and VI of the
\emph{Elements.} In that case, you will probably want to use
\emph{alternation:} if $A:B::C:D$, then $A:C::B:D$. You may use also
that if $A:B::E:F$ and $B:C::D:E$, then $A:C::D:F$. Alternatively, it
is possible to
avoid the theory of proportion by showing, as a lemma, that, in the
diagram, $FE$ is
parallel to $AC$ if and only if the parallelogram bounded by $BF$ and
$BC$, in the angle $B$, is equal to the parallelogram bounded by $BE$
and $BA$. Or maybe you can find another method.
In modern
terms, this problem can be set in a two-dimensional
vector-space; but if the scalar field of that space is non-commutative,
then the claim is false.)
\end{problem}
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\vfill
\begin{bonus}
How can this exam and this course be improved? (Responses may be
submitted also by
email in the next few days: \url{dpierce@metu.edu.tr}.
Meanwhile, \emph{iyi \c cal\i\c smalar; ondan sonra, iyi tatiller!})
\end{bonus}
\end{document}