%\documentclass[12pt]{exam}
\documentclass[12pt,answers]{exam}
\usepackage{mathptmx}
\usepackage{amsfonts,amsmath}
\usepackage{fullpage}
\usepackage[dvipsnames,usenames]{color}
\usepackage{graphicx}
%\usepackage[dvips]{epsfig}


\usepackage{color}
\shadedsolutions
\definecolor{SolutionColor}{rgb}{0.8,0.9,1}

\newcommand{\DS}{\displaystyle}

\def\Bbb{\mathbb}
\def\R{\ensuremath{\Bbb R}}
\def\N{\ensuremath{\Bbb N}}
\def\Z{\ensuremath{\Bbb Z}}
\def\C{\ensuremath{\Bbb C}}

\newcommand{\Point}[1]{\mbox{\sffamily\slshape{#1}}}
\newcommand{\LineL}[1]{\mbox{\sffamily\slshape{#1}}}
\newcommand{\Line} [1]{\ensuremath{\overleftrightarrow{\Point{#1}}}}
\newcommand{\Ray}  [1]{\ensuremath{\overrightarrow{\Point{#1}}}}
\newcommand{\Seg} [1]{\ensuremath{\overline{\Point{#1}}}}
\newcommand{\Betw}[3]{\ensuremath{\Point{#1}\star\Point{#2}\star\Point{#3}}}
\newcommand{\Angle}[1]{\ensuremath{\angle\Point{#1}}}
\newcommand{\Triangle}[1]{\ensuremath{\triangle\Point{#1}}}
\newcommand{\Quad}[1]{\ensuremath{\square\Point{#1}}}

\newcommand{\Item}[1]{\item[{\bf{#1}:}]}

\addpoints

\newcommand{\ExamDate}{March 26, 2009}
\newcommand{\CourseDesc}{MAT360}


\ifprintanswers\newcommand{\DocDesc}{Solutions to Midterm}
\else          \newcommand{\DocDesc}{Midterm Exam}
\fi

\newenvironment{sol}%
  {\par\smallskip\nobreak\noindent%
   \parindent=0pt\parskip=10pt plus 3pt minus 3pt\everypar{\color{darkblue}}
   \color{darkblue}%
   \bf{Solution:\quad}
   \sf
  }{\par\medskip\goodbreak}
\newcommand{\SpFill}[1]{\ifprintanswers\vskip #1\else\vfill\fi}


\begin{document}
\thispagestyle{empty}

\ifprintanswers
   \centerline{\Huge \CourseDesc \quad\hfil\quad \DocDesc}
   \bigskip
\else

\begin{coverpages}
\centerline{\Huge \CourseDesc \quad\hfil\quad \DocDesc}
\smallskip
\centerline{\bf \ExamDate}

\vskip .5in
\centerline{\Large\bf Name: \underline{\hbox to 2.5in{\relax}} \quad
                      ID: \underline{\hbox to 2in{\relax}}}

\vfill
 
\centerline{\Large \gradetable[h][questions]}

\vfill

{\parindent=0pt \parskip=10pt plus 2pt
There are \numquestions\ problems in this exam.  
Make sure that you have them all. 

Do all of your work in this exam booklet, and cross out any work that
the grader should ignore.  You may use the backs of pages, but
indicate what is where if you expect someone to look at it.  {\bf
  Books, calculators, extra papers, and discussions with friends are
  not permitted.}  
This is an ``open book'' exam only if the open book contains no math and 
is in another room.  You may use any results obtained by consulting a 
talking fish, but must give the fish appropriate credit when doing so.

\bigskip
You have 80 minutes, more or less, to complete this exam.  

\bigskip\bigskip
This version has the typos fixed, unlike the one people took.
}


\vfill
\pagebreak
{\small
\centerline{\Large Geometric Axioms}

\noindent{\sc Undefined Terms:} {\it point, line, incident, between,
  congruent.}

\medskip

\noindent{\sc Incidence axioms}
\begin{enumerate}

\Item{I1} For any two distinct points \Point{P} and \Point{Q}, there is a
 unique line \LineL{l} incident with both of them.

\Item{I2} For any line \LineL{l}, there are at least two distinct points 
\Point{P} and \Point{Q} incident with \LineL{l}.

\Item{I3} There are at least three distinct, non-collinear points.

\end{enumerate}
\noindent{\sc Betweenness axioms}

\begin{enumerate}

\Item{B1} If a point \Point{B} lies between points \Point{A} and \Point{C}, 
(written $\Betw{A}{B}{C}$), then \Point{A}, \Point{B},
 and \Point{C} are distinct collinear points, and \Betw{C}{B}{A}.

\Item{B2} Given any two distinct points \Point{B} and \Point{D}, there
  exist points  \Point{A}, \Point{C}, and \Point{E} lying on \Line{BD}
  so that \Betw{A}{B}{D},\ \Betw{B}{C}{D}, and \Betw{B}{D}{E}.

\Item{B3} If \Point{A}, \Point{B}, and \Point{C} are three distinct
  points on the same line, then exactly one of the points is between
  the other two.

\Item{B4} {\bf (Plane Separation)} For every line \LineL{l} and any three
  points \Point{A}, \Point{B}, and \Point{C} not lying on \LineL{l}:
  \begin{enumerate}
    \item If \Point{A} and \Point{B} are on the same side of \LineL{l}
      and \Point{B} and \Point{C} are also on the same side of
      \LineL{l}, then \Point{A} and \Point{C} are on the same side of
      \LineL{l}. 
    \item If \Point{A} and \Point{B} are on opposite sides of \LineL{l}
      and \Point{B} and \Point{C} are also on opposite sides of
      \LineL{l}, then \Point{A} and \Point{C} are on the same side of
      \LineL{l}. 
  \end{enumerate}

\end{enumerate}
\noindent{\sc Congruence axioms}

\begin{enumerate}

\Item{C1} If \Point{A} and \Point{B} are distinct points and if
  \Point{A'} is any point, then for each ray \LineL{r} emanating from
  \Point{A'} there is a unique point \Point{B'} on \LineL{r} such that
  $\Point{B'}\ne\Point{A'}$ and $\Seg{AB} \cong \Seg{A'B'}$

\Item{C2} If $\Seg{AB}\cong\Seg{CD}$ and $\Seg{AB}\cong\Seg{EF}$, then
$\Seg{CD}\cong\Seg{EF}$.  In addition, every segment is congruent to
  itself.

\Item{C3} If \Betw{A}{B}{C},\ \Betw{A'}{B'}{C'},
  $\Seg{AB}\cong\Seg{A'B'}$, and $\Seg{BC}\cong\Seg{B'C'}$, then 
  $\Seg{AC}\cong\Seg{A'C'}$.

\Item{C4} Given any \Angle{BAC} and any ray \Ray{A'B'} emanating from
  a point \Point{A'}, then there is a unique ray \Ray{A'C'} on a given
  side of line \Line{A'B'} such that $\Angle{B'A'C'}\cong\Angle{BAC}$.

\Item{C5} If $\Angle{A}\cong\Angle{B}$ and $\Angle{A}\cong\Angle{C}$,
  then $\Angle{B}\cong\Angle{C}$.  Moreover, every angle is congruent
  to itself.

\Item{C6}{\bf{(SAS)}} If two sides and the included angle of one
  triangle are congruent, respectively, to the two sides and the
  included angle of another triangle, then the two triangles are
  congruent.

\end{enumerate}
\noindent{\sc Continuity axiom}
\begin{enumerate}

\Item{D}{\bf (Dedekind's Axiom)} The points of any line are in
  bijection with the real numbers \R.

\end{enumerate}
\noindent{\sc Parallel axiom}
\begin{enumerate}

\Item{P} Given a line \LineL{l} and a point \Point{P} not on the line,
  there is \{no, exactly one, more than one\} line \LineL{m} through
  \Point{P} which is parallel to \LineL{l}, depending on whether the
  geometry is \{elliptic, Euclidean, or hyperbolic\}.  Note that we
  are not making a choice here, although elliptic geometry is
  inconsistent with the previous axioms. In this class, you should
  only use this axiom if you explicitly say so.
\end{enumerate}
}
\end{coverpages}


\pagebreak 
\fi %%%%%%%%%%%%%%%%%% ifprintanswers

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%

\begin{questions}
\boxedpoints

\question
\newcommand{\term}[1]{{\color{BrickRed}\sf{#1}}}

Consider the following interpretation of the terms ``point'',
``line'', and ``incidence'':

\begin{minipage}{5in}
\begin{itemize}
\item A \term{point} is any pair of real numbers $(x,y)$ such 
that $x$ and $y$ are not both zero. 
That is, any point in $\R^2$ except the origin.

\item A \term{line} is the set of \term{point}s $(x,y)$
 as above for which there are 
real numbers $a$ and $b$ not both zero such that either 
$$ (x-a)^2 + (y-b)^2 = (a^2+b^2) \qquad\text{or}\qquad ax = by .$$
That is, a \term{line} is any circle or straight 
line passing through the origin in $\R^2$.

\item A \term{point} is \term{incident} 
with a \term{line} if it satisfies the corresponding equation.
\end{itemize}
\end{minipage} \hfill
\begin{minipage}{1.5in}
\includegraphics[width=1.4in]{circlelines}
\end{minipage}

\begin{parts}
\part[6] Does this define an incidence geometry?  That is, do axioms
     {\bf I1}, {\bf I2}, and {\bf I3} hold?  Fully justify your
     answer.\footnote{It might be useful for you to remember that for any
       three distinct non-collinear points in $\R^2$, there is a
       unique circle on which they all lie.  
       Also, C.S.~Lewis once said the following line:
       ``The next best thing to being wise oneself is to live in a
       circle of those that are.''}

\begin{solution}
Throughout, we will use \term{line} or \term{point} to refer to the objects
as defined in the model, and line, point, or circle to refer to the
``usual'' Euclidean meanings in $\R^2$.

First, let's check {\bf I1}:  Given two \term{point}s
$\Point{P}=(x_1,y_1)$ 
and $\Point{Q}=(x_2,y_2)$, we need to confirm that they determine a unique
\term{line}.  There are two cases:  

\begin{minipage}{4.5in}
First, suppose that \Point{P} and
\Point{Q} are collinear with the origin in $\R^2$.
Either the line \Line{PQ} is vertical, 
in which case $y_1 = y_2 = 0$ and so \Point{P} and \Point{Q} lie on the line
$y=0$ (that is, $a=0$ and $b=1$), or \Line{PQ} has slope $m$, in which
case the line is $y=mx$ (that is, $a=m$ and $b=1$).  Thus, if \Line{PQ} is
of line type, it is not possible to find a circle in $\R^2$ passing through
\Point{P}, \Point{Q} and the origin.
\end{minipage}
\begin{minipage}{1in}
  \centerline{\includegraphics[width=.75in]{linePQ0}}
\end{minipage}

\begin{minipage}{1in}
  \centerline{\includegraphics[width=.75in]{circlePQ0}}
\end{minipage}
\begin{minipage}{4.5in}
In the other situation, \Point{P}, \Point{Q}, and the origin are not
collinear in $\R^2$, so there is a unique circle containing all three
(see footnote).  In either case, there is a unique \term{line} defined
by \Point{P} and \Point{Q}.
\end{minipage}

The next two are easy. Axiom {\bf I2} holds, since every line has
infinitely many points on it.  Axiom {\bf I3} also holds easily.  For
example, the \term{point}s $(1,0)$, $(-1,0)$, and $(0,1)$ do not lie
on the same \term{line}.

Since axioms {\bf I1}, {\bf I2}, and {\bf I3} hold, this does define an
incidence geometry.  In fact, with suitable definitions of \term{between}
and \term{congruence}, all of the other axioms except the continuity and
parallel axioms hold.
\end{solution}

\SpFill{3pt}
\part[4] Does this model define a Euclidian, elliptic, or hyperbolic
geometry (or none of the those)?  Again, fully justify your answer.

\begin{solution}
Certainly there are parallel \term{line}s (for example, all of the
lines of the form $ax=by$ are parallel to each other.  In addition,
any two circles which are tangent at the origin are also parallel to
one another.)  Thus, the geometry is not elliptic.

Now, given a \term{line} \LineL{l} and a \term{point} \Point{P} 
not on \LineL{l}, we must determine how many \term{line}s passing
through \Point{P} which are parallel to \LineL{l}.

\begin{minipage}{.75in}
  \centerline{\includegraphics[width=.75in]{circleLlineM}}
\end{minipage}
\begin{minipage}{4in}
If \LineL{l} is of circle type, then there are two possibilities.
First, if \Point{P} lies on the line \LineL{m} which is tangent to
\LineL{l} at the origin, \term{line}s \LineL{l} and \LineL{m} are
parallel.  There can be no other parallel \term{line} (all other
circles or lines passing through \Point{P} will intersect \LineL{l}).
In the other possibility, there is a unique circle through \Point{P}
which will be tangent to \LineL{l} at the origin.  In both cases, there
is a unique \term{line} through \Point{P} parallel to \LineL{l}.
\end{minipage}
\begin{minipage}{.75in}
  \centerline{\includegraphics[width=.75in]{circleLlineM}}
\end{minipage}

\begin{minipage}{4.75in}
In case \LineL{l} is of line type, then the line in $\R^2$ connecting
\Point{P} to the origin is a \term{line} \LineL{m} parallel to \LineL{l}.  
But also the circle which passes through \Point{P} and is tangent to
\LineL{l} at the origin is another \term{line} \LineL{n} which is parallel to
\LineL{l}.   Thus, there are two lines parallel to \LineL{l} through
\Point{P}. 
\end{minipage}
\begin{minipage}{.75in}
  \centerline{\includegraphics[width=.75in]{circlelinePmn}}
\end{minipage}

Since we sometimes have two parallels and sometimes have a unique one,
the geometry is not hyperbolic, Euclidean, or elliptic.
\end{solution}
\SpFill{3pt}
\end{parts}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\ifprintanswers\bigskip\else\pagebreak\fi
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newcommand\CrossOffMessage{
\ifprintanswers{\relax}\else
\centerline{%
\sf You may skip {\bf one} of problems 2, 3, 4, 5, or 6
 by crossing off the page.}
\bigskip\fi}

\CrossOffMessage

\question[10]
Rewrite the following proof, supplying valid reasons for each of the steps.

\begin{minipage}{3.75in}
In any triangle \Triangle{ABC}, if the perpendicular bisectors of two 
of the sides of \Triangle{ABC} meet, all three perpendicular bisectors of the
three sides meet in a common point \Point{X}, called the {\it circumcenter}.

\medskip{\bf Proof:}
Let \Point{M}, \Point{N}, and \Point{P} be the midpoints of sides
\Seg{AB}, \Seg{BC}, and \Seg{CA} respectively.  Let the perpendicular
bisector of \Seg{AB} meet that of \Seg{BC} at a point \Point{X}. We
must show that \Seg{PX} is perpendicular to \Seg{AC}.

Now, $\Triangle{MXB}\cong\Triangle{MXA}$ and 
$\Triangle{NXB}\cong\Triangle{NXC}$.  Thus $\Seg{CX}\cong\Seg{AX}$,
and so $\Triangle{CPX}\cong\Triangle{APX}$.  Therefore
$\Line{PX}\perp\Seg{AC}$.
\end{minipage} \qquad
\begin{minipage}{2.5in}
\includegraphics[width=2.5in]{circumcenter}
\end{minipage}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\ifprintanswers\bigskip\else\pagebreak\fi
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\CrossOffMessage

\question[10]
Let \Quad{ABCD} be a Saccheri quadrilateral, with right angles at
\Point{A} and \Point{B}, and $\Seg{AD}\cong\Seg{BC}$ as usual.
Also, let \Point{E} and \Point{F} be the midpoints of \Seg{AD} and
\Seg{BC} respectively, and let \Point{G} be the point of intersection
of \Seg{EC} and \Seg{DF}.  Prove that $\Seg{EG}\cong\Seg{GF}$.

\rightline{\includegraphics[width=3in]{saccheri}}
\SpFill{10pt}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\ifprintanswers\bigskip\else\pagebreak\fi
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\CrossOffMessage

\question[10]
Prove the Hypotenuse-Leg congruence condition.  That is, suppose
\Triangle{ABC} and \Triangle{PQR} are right triangles with right
angles at \Angle{B} and \Angle{Q}.  Furthermore, suppose that 
$\Seg{AB}\cong\Seg{PQ}$ and $\Seg{AC}\cong\Seg{PR}$.  Show that 
$\Triangle{ABC}\cong\Triangle{PQR}$.  (Hint: an isosceles triangle
could be helpful.)

\SpFill{10pt}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\ifprintanswers\bigskip\else\pagebreak\fi
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\CrossOffMessage

\question[10]
Prove that the Euclidean parallel postulate holds if and only if the
following statement holds:

Let lines \LineL{l} and \LineL{m} be parallel, and let line \LineL{t}
be perpendicular to \LineL{l}.  Then \LineL{t} is also perpendicular
to \LineL{m}.

\SpFill{10pt}

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\ifprintanswers\bigskip\else\pagebreak\fi
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\noaddpoints
\CrossOffMessage

\question[10]
%Let $\gamma$ be a circle with center \Point{O}, and let \Point{A} and
%\Point{B} be two points on $\gamma$.  Let \Point{M} be the midpoint of
%\Seg{AB}. Prove that if $\Point{O}\ne\Point{M}$, then \Line{OM} is
%perpendicular to \Line{AB}. 
%
%\rightline{\includegraphics[width=2.5in]{circlechord}}
Let \Point{D} be any point interior to triangle \Triangle{ABC}. Prove
that the measure of \Angle{BAC} is less than the measure of \Angle{BDC}.
\SpFill{10pt}
\addpoints

\end{questions}
\end{document}
