\documentclass[a4paper,12pt,twoside]{amsart}
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\title{Math 303 examination solutions (draft)}
\author{David Pierce}
\email{dpierce@metu.edu.tr}
\address{Mathematics Dept., Middle East Technical University, Ankara 06531, Turkey}
\urladdr{http://metu.edu.tr/~dpierce/Courses/303/}
 
\date{Friday, November 6, 2009}

\usepackage[polutonikogreek,english]{babel}
\usepackage[oxonia]{psgreek}

\usepackage{hfoldsty}
\usepackage{typearea}
%\usepackage{parskip}
\usepackage{paralist}
\newcommand{\Gk}[1]{\selectlanguage{polutonikogreek}#1\selectlanguage{english}}

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

\theoremstyle{remark}
\newtheorem*{remark}{Remark}

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\begin{document}
\maketitle

\emph{Note:}  There are no new diagrams in these solutions, simply
because creating them electronically is too time-consuming.


\begin{problem}

  What is wrong with the following proof that all triangles are isosceles?---:
  \begin{enumerate}
  \item
  Let a triangle be given, namely $ABC$.
  \item
  Let $BC$ be bisected at $D$.
  \item
  Let a straight line, $DE$, be drawn at right angles to $BC$.
  \item
  Let also the straight line $AE$ bisect the angle $BAC$.
  \item
  Let the straight lines $BE$ and $CE$ be drawn.
  \item
  $BE=CE$.
  \item
  Let the straight line $EF$ be drawn perpendicular to $AB$.
  \item
  Let the straight line $EG$ be drawn perpendicular to $AC$.
   \item
  $AF=AG$ and
  $EF=EG$.
  \item
  $BF=CG$.
  \item
  $AF+FB=AG+GC$.
  \item
  $AF+FB=AB$ and $AG+GC=AC$.
  \item
  $AB=AC$; in particular, $ABC$ is isosceles.
  \end{enumerate}
\mbox{}\hfill  
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  \put(-1,0){\line(1,2){1}}
   \put(1,0){\line(-1,2){1}}
   \put(0,0){\line(0,1){2}}
   \put(-1,0){\line(2,1){1.6}}
   \put(1,0){\line(-2,1){1.6}}
   \put(0,2.05){\makebox(0,0)[b]{$A$}}
   \put(1,0){\makebox(0,0)[tl]{$C$}}
   \put(-1,0){\makebox(0,0)[tr]{$B$}}
   \put(0,-0.05){\makebox(0,0)[t]{$D$}}
   \put(0,0.6){$E$}
   \put(0.6,0.8){\makebox(0,0)[bl]{$G$}}
   \put(-0.6,0.8){\makebox(0,0)[br]{$F$}}
  \end{picture}
\end{problem}

\begin{solution}
Step 12 is not justified.  In fact, if $AB>AC$, then $AF+FB=AB$, but $AC+GC=AG$.
\end{solution}

\begin{remark}
\begin{asparaenum}
\item
The diagram is misleading; but (contrary to what some people seemed to
think) the proof never assumes that $AED$ or $BEG$ or $CEF$ is a
straight line.    
\item
Step 4 may \emph{appear} unjustified; however, steps 2, 3, and 4
together say simply that the bisector of angle $BAC$ and the
perpendicular bisector of $BC$ meet at $E$.  This style of writing can
be seen for example in Euclid's Proposition I.44.   
\item
The proof does wrongly assume that $E$ lies within the triangle; but the
proof can easily be adjusted to the case where $E$ lies outside the
triangle.  Euclid usually does not bother to consider all possible
cases: we noted this for example in Proposition I.7.  The real problem
is the assumption that either both $F$ and $G$ lie on the triangle, or
both lie below the triangle. 
\end{asparaenum}
\end{remark}


\begin{problem}
Write English translations of the following words:
\begin{center}
\begin{inparaenum}[(a)]
\item
\Gk{je'wrhma},
\item
\Gk{pr'oblhma},
\item
\Gk{>an'alusic},
\item
\Gk{sunj'esic},
\item
\Gk{pol'ugwnon},
\item
\Gk{tr'igwnon}.
\end{inparaenum}
\end{center}
\end{problem}

\begin{solution}
Theorem, problem, analysis, synthesis, polygon, triangle.
\end{solution}

\begin{remark}
\begin{asparaenum}
\item
A \emph{transliteration} of the words into English (or Latin) letters
would be \emph{theor\^ema, probl\^ema, analysis, synthesis,
  polyg\^onon, trig\^onon,}
but this is not what was asked. 
\item
The first two words on the list have been discussed in class; these
(along with the next two) are also discussed in some notes that I put
on the web.  
\item
The last two words on the list derive from \Gk{gwn'ia} \emph{angle,}
which is apparently related to \Gk{g'onu}; this word shares its
meaning, and an Indo-European ancestor, with the English \emph{knee.}
(Here is a point where English spelling is useful; if \emph{knee} were
spelled phonetically, then its relation with \Gk{g'onu} could not be
seen.)  
\item
As a translation of \Gk{tr'igwnon}, I do find the word \emph{trigon}
in the Oxford English 
Dictionary; but the more usual word is of course \emph{triangle.} 
\end{asparaenum}
\end{remark}

\begin{problem}
Write the letters of the Greek alphabet in the standard order.  Write only the capital letters \emph{or} only the minuscule letters.
\end{problem}

\begin{solution}
\mbox{}
\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}
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{problem}
Proclus writes:
\begin{quote}
  Every problem and every theorem that is furnished with all its parts
  should contain the following elements: 
  \begin{enumerate}\renewcommand{\labelenumi}{(\theenumi)}
  \item
  an \emph{enunciation} (\Gk{pr'otasic}), 
  \item
  an \emph{exposition} (or \emph{setting out:} \Gk{>'ekjesic}), 
  \item
  a \emph{specification} (or \emph{definition of goal:} \Gk{diorism'oc}), 
  \item
  a \emph{construction} (\Gk{kataskeu'h}), 
  \item
  a \emph{proof} (\Gk{>ap'o\-deixic}), and 
  \item
  a \emph{conclusion} (\Gk{sump'erasma}). 
  \end{enumerate}
  \end{quote}
  Below is the enunciation (in Heath's translation) of Proposition I.6 of Euclid's \emph{Elements.}  Supply the remaining parts (in your own words, which may or may not be Euclid's).
  \begin{quote}
  \emph{If in a triangle two angles be equal to one another, the sides
    which subtend the equal angles will also be equal to one another.} 
  \end{quote}
 \end{problem}

\begin{solution}
\begin{asparaenum}
\item
(As above, namely:)
If in a triangle two angles be equal to one another, the sides
    which subtend the equal angles will also be equal to one another.
\item
Let $ABC$ be a triangle in which angles $ABC$ and $ACB$ are equal.
\item
We shall show that $AB=AC$.
\item
On $BA$, extended if necessary, let $BD$ be cut off equal to $CA$.
\item
Then triangle $DBC$ is equal to $ACB$, and therefore $D$ must coincide with $A$.  Consequently, $BA=CA$.
\item
Thus we have shown that, if in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another.
\end{asparaenum}
\end{solution}

\begin{remark}
Euclid's proof is a \emph{reductio ad absurdum,} that is, a proof by contradiction.  In particular, Euclid first assumes $AB\neq AC$ and \emph{then} finds $D$.  In this case, to which of Proclus's six parts does the hypothesis $AB\neq AC$ belong?  I don't know whether Proclus considers this question.
\end{remark}

  \begin{problem}
  \emph{Without} using Euclid's method of `application', prove Proposition I.8 of the \emph{Elements,} whose enunciation is,
  \begin{quote}
  \emph{If two triangles have two sides equal to two sides respectively, and have also the base equal to the base, they will also have the angles equal whch are contained by the equal straight lines.}
  \end{quote}
  \end{problem}
 
 \begin{solution}
 Suppose $ABC$ and $DEF$ are triangles such that $AB=DE$, $BC=EF$, and $AC=DF$.  We shall show that angles $ABC$ and $DEF$ are equal.  To this end, let $AG$ be dropped perpendicular to $BC$, extended if necessary [by I.12].  On $EF$, extended if necessary, cut off $EH$ equal to $BG$ [by I.3].  Erect $HK$ perpendicular to $EF$ [by I.11] and equal to $AG$ [by I.3 again].  Then $EK= AB$ and angles $KEF$ and $ABC$ are equal [by I.4], and similarly, since $HF=GC$, we have $FK=CA$.  Hence $EK=ED$ and $FK=FD$.  Therefore $K$ and $D$ coincide [by I.7], and in particular, angles $DEF$ and $ABC$ are equal.
 
Now, we have used two propositions [namely I.11 and 12] that Euclid proves by means of I.8.  However, alternative proofs are as follows.  

If $A$ does not lie on the straight line $BC$, then by drawing a circle with center $A$ that cuts the line, we may assume $B$ and $C$ have been chosen so that $AB=AC$.  Draw an equilateral triangle $BCD$ (on the opposite side of $BC$ from $A$) [by I.1].  Draw the straight line $AD$, which cuts $BC$ at a point $E$.  Then angles $BAD$ and $CAD$ are equal [by I.5 and 4], and therefore angles $AEB$ and $AEC$ are equal [again by I.4], so the latter angles are right.  Therefore $AE$ has been dropped perpendicular to $AB$.

If $A$ does lie on $BC$, we may still assume $AB=AC$.  Draw an equilateral triangle $BCD$ and straight line $AD$.  Then angles $BAD$ and $CAD$ are equal [by I.5 and 4], so they are right.  Thus $AD$ has been erected perpendicular to $BC$.
 \end{solution}
 
 \begin{remark}
 \begin{asparaenum}
 \item
 It is not necessary to name the propositions used.
 \item
 Some people argued by contradiction that if (in the notation above) angle $ABC$ is greater than $DEF$, then $BC$ must be greater than $EF$.  This is Proposition I.24; but I.24 relies on I.23, which in turn relies on I.8.  It is not clear to me that there is a way to prove I.24 without first proving I.8.
 \item
 One person suggested an interesting argument that I understand as follows.  If angle $ABC$ is greater than $DEF$, then inside the former angle, there must be an angle $ABG$ equal to $DEF$.  We may then assume $BG=BC=EF$.  But then $GA=FA$ [by I.4], so we have violated I.7, which is absurd; therefore $ABC=DEF$.  Now, if this argument is valid, then what is the point of I.3?  If straight line $AB$ is greater than straight line $C$, why does Euclid not declare that there must be a part of $AB$, namely $AE$, that is equal to $C$?  Why does Euclid feel the need to \emph{construct} $AE$?
 \end{asparaenum}
 \end{remark}
 
  \begin{problem}
  In triangle $ABC$, suppose $BC$ is bisected at $D$, and straight line $AD$ is drawn.  Assuming $AB$ is greater than $AC$, prove that angle $BAD$ is less than $DAC$.
  \end{problem}
 
 \begin{solution}
 Extend $AD$ to $E$ so that $DE=DA$.  Then angles $DEC$ and $DAB$ are equal, and $CE=BA$ [by I.4].  But then angle $CAE$ is greater than $CEA$ [by I.18], so $CAD>DAB$.
 \end{solution}
 
 \begin{remark}
 I think the argument just given is the best of several variants that were found by different people.  The argument I had thought originally of was more complicated:  Since angle $BDA$ must be greater than $ADC$, inside angle $BDA$ we can construct angle $ADE$ equal to $ADC$, with $DE=DC$.  Then $BE$ is parallel to $AD$ [why?], so $E$ lies outside triangle $ABD$.  Therefore angle $BAD$ is less than $EAD$; but the latter is equal to $DAC$.
 \end{remark}
 
\end{document}
