\documentclass[a4paper,11pt,twoside,reqno]{amsart}
%\documentclass[a4paper,12pt,twoside]{article}
\title{Math 303, final examination \emph{solutions}}
\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}
\usepackage[OT2,T1]{fontenc}
\usepackage[russian,polutonikogreek,english]{babel}
\usepackage[oxonia]{psgreek}
\usepackage{wrapfig}
\usepackage{hfoldsty}
\usepackage{typearea}
%\usepackage{parskip}
\usepackage{paralist}
\usepackage{pstricks}
\usepackage{pst-plot}
\newcommand{\Ru}[1]{\foreignlanguage{russian}{#1}}
\newcommand{\Gk}[1]{\foreignlanguage{polutonikogreek}{#1}}
\newtheorem{problem}{Problem}
\newtheorem*{bonus}{Bonus}
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\newtheorem*{solution}{Solution}
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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}
\textup{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?
\end{problem}
\begin{solution}
\Gk{A B G D E Z H J I K L M N X O P R S T U F Q Y W},
or
\begin{center}
\Gk{a b g d e z h j i k l m n x o p r s t u f q y w}.
\end{center}
\end{solution}
\begin{remark}
Most people seem to have learned the alphabet for this exam. If this had
been so on the first exam, I may not have asked for the
alphabet on \emph{this} exam.
\end{remark}
%\newpage
\begin{problem}
Does a square have a ratio to its side? Explain.
\end{problem}
\begin{solution}
No, since no multiple of the side can exceed the square.
\end{solution}
\begin{remark}
This problem alludes to Definition 4 of Book V of the \emph{Elements:}
\begin{quote}
Magnitudes are said to \emph{have a ratio} to one another which are
capable, when multiplied, of exceeding one another.
\end{quote}
Euclid does not seem to \emph{refer} to this definition later; but (as we
discussed in class) he \emph{uses} the definition implicitly, in
Proposition V.16 for example, where there is an unstated assumption
that $A$ and $C$ have a ratio, and (therefore) $B$ and $D$ have a
ratio. In his `quadrature of the parabola,' discussed on the last day
of class, Archimedes assumes that, if two areas are unequal, then their
difference \emph{has a ratio} (in the sense of Euclid) to either of the areas.
\end{remark}
\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}
\begin{solution}
If equimultiples $mA$ and $mC$ of $A$ and $C$ be taken, and other
equimultiples $nB$ and $nD$ of $B$ and $D$ be taken, then
\begin{gather*}
mA>nB \text{ if and only if } mC>nD,\\
mA=nB \text{ if and only if } mC=nD,\\
mA