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\centerline{MATH 392 -- Geometry Through History}
\centerline{Midterm Exam Solutions}
\centerline{March 22, 2016}
\vskip 10pt
\noindent
I.  Short Answer.  Answer any four (4) of the following briefly (two or three sentences 
will suffice for each).  Only the best 4 will be used to compute your total score.
\begin{enumerate}
\item[(A)] (5) Approximately when and where was Euclid active?  How much do we
know about his life?    
\item[] {\it Solution:}  Euclid lived around 300 BCE, in Alexandria in Egypt
for a part of his life, at least.  
Almost nothing else about his life is known besides a tradition that he 
was connected in some way with Plato's school of philosophy and a few anecdotes 
about his teaching.
(Most of this is deduced from a statement of the 
commentator Proclus from about 750 years later.)
\item[(B)] (5) State the 5 Postulates for geometry Euclid included at the start
of Book I of the {\it Elements}.  What was the function of these statements
in Euclid's development of geometry?
\item[] {\it Solution:}  Postulate I: A straight line segment can be drawn
between any two points.  Postulate II: A straight line segment can be extended
indefinitely in both directions.  Postulate III:  A circle can be constructed
given any point as the center and given any line segment as radius.  Postlate IV:
All right angles are equal.  Postulate V:  If a line falling on two other lines
makes the angles on one side less than two right angles, then the two lines if
extended will meet on the side with the angles summing to less than two right angles.
These were unproved starting assumptions, from which Euclid wanted to deduce all
known facts about geometry.  (One could also say that it was pretty clear that
Euclid thought of these as intuitively obvious facts about the geometry of physical
space.)
\item[(C)] (5) What was Proclus's opinion about Euclid's Postulate V?  How was this
train of thought continued in later work?
\item[] {\it Solution:}  Proclus thought that Postulate V was too complicated and
unobvious to be a postulate.  He wanted to prove it as a theorem from the other
Postulates and results derived from them.  This line of attack was taken up 
by many other mathematicians up until the late 18th century CE.
\item[(D)] (5) Approximately when and where was Girolamo Saccheri, S.J. active?  What
were his main contributions?
\item[] {\it Solution:}  Saccheri lived from 1667 to 1733 in Italy.  He 
attempted to prove Postulate V arguing by contradiction using the properties
of figures that we now call Saccheri quadrilaterals.  He proved many theorems
about hyperbolic geometry in the process in his book {\it Euclid Freed of Every Flaw},
but he eventually declared that he had found a contradiction to the nature of 
a straight line and ``punted.''
\item[(E)] (5) Why is Gauss now always included with Bolyai and Lobachevsky in 
discussions of the developers of hyperbolic geometry when he did not publish
any work on that subject?
\item[] {\it Solution:}  He was one of the first mathematicians to realize that
no contradiction would result from a geometry where Euclid's Postulate V
did not hold.  He developed many of the theorems of hyperbolic geometry
independently and before Bolyai and Lobachevski.  But he decided not
to publish his work because he did not think it would be understood and
he wanted to avoid controversy and preserve his reputation(!)
\item[(F)] (5) What does Pasch's Axiom say and why do we include
that statement in a complete set of axioms for Euclidean geometry?
\item[] {\it Solution:}  Pasch's Axiom says that a line that enters
a triangle through a point on one or two of the sides must also exit
the triangle at a point one of the sides not already crossed.  It is necessary
to include this as an axiom to provide complete proofs of several 
propositions.  
\end{enumerate}
\vskip 10pt
\noindent
II. (20) State and prove Proposition 29 in Book I of the {\it Elements}.  What
is especially notable about this proposition?  (Note:  If you need to, you 
may ``buy'' the statement to be proved from me for 5 of the possible points
on this problem.) 
\vskip 10pt
\noindent
{\it Solution:}  Refer to the figure above.
\begin{figure}
\begin{center}
\includegraphics[width=3in,height=3in]{EuclidI29.png}
\caption{Figure for Euclid I.29}
\end{center}
\end{figure}
The statement is:  If lines $\overleftrightarrow{AB}$ and 
$\overleftrightarrow{CD}$ are parallel 
and line $\overleftrightarrow{GH}$ is a transversal, then (a) alternate interior angles are equal, 
(b) corresponding angles are equal, and (c) the interior angles on one side of
the transversal add to two right angles.  

\vskip 10pt
Proof:  We prove (c) first and deduce the other two statements from that one.
Suppose $\angle EFD + \angle FEB < $ two right angles.  Then Postulate V implies
that the lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$ extended must meed on 
that side of the transversal $\overleftrightarrow{GH}$.  But this cannot be true since
we assumed $\overleftrightarrow{AB}$ and 
$\overleftrightarrow{CD}$ are parallel.  Similarly, if
$\angle EFD + \angle FEB > $ two right angles, then $\angle CFE + \angle AEF$
must be less than two right angles and we do the same argument on that side.
It follows that $\angle EFD + \angle FEB = $ two right angles.
Now for (b), corresponding angles are ones like the pair $\angle GEB$
and $\angle EFD$.  By a previous proposition (I.13 to be exact), 
$\angle GEB + \angle FEB = $ two right angles since $G,E,F$ are collinear.
But then from (c) we also have $\angle EFD + \angle FEB = $ two right angles.
Hence $\angle GEB = \angle EFD$ by Common Notion 3 (``if equals are subtracted
from equals, the remainders are equal'').  Finally for (a), opposite interior
angles are angles like the pair $\angle AEF$ and $\angle EFD$.  
We have $\angle GEB = \angle AEF$ by I.15 (equal vertical angles).  Then 
from (b) and Common Notion 1, we have the desired statement $\angle AEF = \angle EFD$.  
\vskip 10pt
This was the first use made of Postulate V in Book I of the {\it Elements}.
\vskip 10pt
\noindent
III.  Definitions and related facts.
\begin{enumerate} 
\item[(A)] (10) What is the \emph{angle of parallelism} for a segment $AP$
in hyperbolic geometry?  What does the Bolyai-Lobachevsky theorem tell us
about the angle of parallelism?  
\item[] {\it Solution:}  Let $\overline{AP} \perp \overline{PQ}$.  The ray
line $\overrightarrow{AB}$ from $A$ is said to be parallel to $\overrightarrow{PQ}$
in the direction of $Q,B$ if for any point $S$ in the quadrilateral $PQBA$, 
the ray $\overrightarrow{AS}$ intersects $\overrightarrow{PQ}$, but $\overrightarrow{AB}$
does not meet $\overrightarrow{PQ}$. 
The angle of parallelism $\Pi(AP)$ is the angle $\angle PAQ$.  The Bolyai-Lobachevsky
theorem says that this angle, which depends only on the length $x = AP$ can be
computed by 
$$\tan(\Pi(x)/2) = e^{-x/k}$$
for some constant $k$.  
\item[(B)] (10) What is a \emph{horocycle} in hyperbolic geometry? 
What is the closest Euclidean analog to this curve?
\item[] {\it Solution:}  Consider a pencil of parallel lines.  If $P$
is a given point, then the horocycle through $P$ is the locus of points
$Q$ such that $Q$ corresponds to $P$ with respect to the pencil.  (In other
words, the line of the pencil passing through the midpoint of the
line segment $\overline{PQ}$ is perpendicular to $\overline{PQ}$.  The 
closest Euclidean analog is probably the circle, but the center of the 
circle would be a point ``at infinity.'' 
\end{enumerate}
\vskip 10pt
\noindent
IV. Refer to the figure at the top of the next page.  A \emph{median} of a triangle is the 
line segment joining a vertex to the midpoint of the opposite side.  
In the figure, $\overline{AA'}$, $\overline{BB'}$, and the dotted line
$\overline{CC'}$ are medians.  In this problem, from the parts below you will 
find a proof of the theorem that 
\vskip 10pt
\centerline{\emph{The three medians of a triangle meet in a single point (called
the centroid of the triangle).}}
\vskip 10pt
\noindent  

\begin{figure}
\begin{center}
\includegraphics[trim=.9in 1in 1in 1in,clip,width=3in,height=3in]{Medians.png}
\caption{$\Delta ABC$ with medians $AA'$ and $BB'$ in solid black}
\end{center}
\end{figure}

\noindent
Let $O$ be the point where two medians meet as in the figure\footnote{{\it Comment}:  Such
a point $O$ must exist by Pasch's Axiom, applied to the triangle $\Delta AA'B$ and 
the line $\overleftrightarrow{BB'}$.} 
The idea of this proof is to show that 
\vskip 10pt
\centerline{(*)  $O$ is a point such that $AO = 2\cdot OA'$ and $BO = 2\cdot OB'$.}
\vskip 10pt\noindent
In other words, (any) two medians \emph{cut each other} at a point $O$ in two line segments
in the ratio $1:2$, with the shorter distance being the distance from $O$
to the midpoint of the opposite side.  
\begin{enumerate}
\item[(A)] (5) Explain why establishing (*) is enough to show that the 
three medians all intersect at $O$.  (Don't think too hard about this -- it's very easy!)
\item[] {\it Solution:}  If (*) is true then the third median $\overline{CC'}$ must
also pass through the point $O$ because there is only one point on each median that
cuts the median into two line segments in the ratio $1:2$.
\item[(B)] (5) Show that ${\rm area}(\Delta ABA') = {\rm area}(\Delta ACA')$, 
${\rm area}(\Delta OBA') = {\rm area}(\Delta OCA')$, and similarly
${\rm area}(\Delta BAB') = {\rm area}(\Delta BCB')$ and 
${\rm area}(\Delta OAB') = {\rm area}(\Delta OCB')$.
\item[] {\it Solution:}  These all follow from the statement of I.38 (triangles
on equal bases and in the same parallels have equal area).  A more modern way
to say this is that each pair of triangles has equal bases and equal altitudes,
so the area formula $\frac{1}{2}bh$ shows the areas are the same.  
\item[(C)] (5) Deduce that 
$${\rm area}(\Delta OBA') = {\rm area}(\Delta OCA') = {\rm area}(\Delta OCB') 
= {\rm area}(\Delta OAB').$$
\item[] {\it Solution:}  We already know that the first two of these are 
equal and the last two of them are equal by (B).  So we need to show that
one of the first pair is equal to one of the second pair.   Note
that since ${\rm area}(\Delta ABA') = {\rm area}(\Delta ACA')$, each of these
triangles has half the area of $\Delta ABC$.  Since 
${\rm area}(\Delta BAB') = {\rm area}(\Delta BCB')$, the same is true
for those two triangles.  It follows that 
$${\rm area}(\Delta BAB') = {\rm area}(\Delta BOA) + {\rm area}(\Delta OAB') = 
{\rm area}(\Delta ABA') = {\rm area}(\Delta BOA) + {\rm area}(\Delta OBA').$$
Hence ${\rm area}(\Delta OAB') = {\rm area}(\Delta OBA')$.  Then all four of those
triangles are equal in area. 
\item[(D)] (5) Finally, show that $AO = 2\cdot OA'$ and $BO = 2\cdot OB'$, which
is the desired statement (*) above.
\item[] {\it Solution:}  Look at parts (B) and (C).  The triangle $\Delta ACA'$ is 
split into three equal smaller triangles.  The triangle $\Delta ABA'$ has the 
same area and is split into $\Delta OBA'$ (which is equal in area to each of 
the three smaller triangles) and $\Delta OAB$.  This says 
$${\rm area}(\Delta OAB) = 2\cdot {\rm area} (OBA').$$  
But now those two triangles have bases along the same line and a common vertex at $B$,
so the bases satisfy $AO = 2\cdot OA'$.  The statement $BO = 2\cdot OB'$ follows similarly 
by considering the triangles $\Delta BAB'$ and $\Delta BCB'$.  
\end{enumerate}
\vskip 10pt
\noindent  
V. Consider the figure at the top of the next page.  
\begin{enumerate}
\item[(A)] (15) Show that in the hyperbolic plane
it is possible to construct three pairwise parallel
hyperbolic lines as in the figure -- that is show
how to construct three lines $\ell_1, \ell_2$, and $\ell_3$ and points $A,A'$ on
$\ell_1$, $B,B'$ on $\ell_2$ and $C,C'$ on $\ell_3$ such that  
$\overrightarrow{AA'}$ is parallel to $\overrightarrow{BB'}$
in the direction of $A'$ and $B'$, and (at the same time) 
$\overrightarrow{B'B}$ is parallel to $\overrightarrow{CC'}$ in the direction of $B$ and $C'$, 
and (at the same time) $\overrightarrow{A'A}$ is parallel to $\overrightarrow{C'C}$
in the direction of $A$ and $C$.  
\item[] {\it Solution:}  Recall from Problem Set 3 that given any angle $\angle AOB$, less
than two right angles, there is a line $\overleftrightarrow{PQ}$ 
that such that $\overrightarrow{QP}$ is 
parallel to $\overrightarrow{OA}$ in the direction of $P$ and $A$ and 
also $\overrightarrow{PQ}$ is parallel to $\overrightarrow{OB}$ in the direction of $Q$ and $B$.  The construction
was as follows.  Let $\overrightarrow{OD}$ be the angle bisector of $\angle AOB$
and let $D$ be a point along that bisector such that the angle of parallelism
$\Pi(OD) = \angle AOD = \angle BOD$.  Then the line perpendicular to $\overrightarrow{OD}$
through $D$ is the line.  To get the lines as in the figure, take three rays meeting
at a point $O$ at pairwise angles of $120^\circ$ (for instance -- any three angles 
summing to $360^\circ$ with none $\ge 180^\circ$ would also work).  Apply the 
result from the problem set to each pair of rays to get three lines.  The only remaining
thing to show is that the lines are pairwise parallel.  But that follows since for
instance once we know $\overrightarrow{AA'}$ and $\overrightarrow{BB'}$ are both 
parallel to the ray shown as dotted black in the figure, then they must also be
parallel to each other.  
\item[(B)] (5) Generalize your argument to show that there
are similar ``asymptotic $n$-gons'' of parallel lines for any $n \ge 3$.
\item[] {\it Solution:}  Apply the same result from the Problem Set 3 to 
$n$ rays meeting at angles of $\frac{360^\circ}{n}$.  
\end{enumerate}

\begin{figure}
\begin{center}
\includegraphics[width=3in,height=3in]{HypPars.png}
\caption{An ``asymptotic triangle'' in the hyperbolic plane}
\end{center}
\end{figure}
\end{document}


{\it Format}
\vskip 10pt
The exam will cover the material on Euclidean and hyperbolic geometry that
we have studied since the start of the semester (i.e. Problem Sets 1 - 3 and
the first quiz).  There will be roughly equal coverage of the Euclidean and
hyperbolic cases.  

You should know the statements of Postulates I - V from 
Euclid, Pasch's Axiom, and definitions of terms we have used (especially
the \emph{revised} notion of parallelism and the angle of parallelism for the 
hyperbolic case, pencils of parallel lines, horocycles, horospheres.)  

Know the statement of the Bolyai-Lobachevsky theorem, and be prepared to use it 
to deduce facts about angles of parallelism in the hyperbolic case.

I will ask you to construct one or two short proofs you have not seen before (comparable 
to the proof from Quiz 1; also see the Review/Practice Problems below).  Other 
questions will ask for one or two of the more substantial assigned proofs 
listed below.  

I might also include a short answer question with several parts on aspects of the 
historical development, so you should know approximately when each of the following 
key figures lived, what their main contributions were, and how they figure into what 
we have done so far:
\begin{itemize}
\item Euclid of Alexandria
\item Proclus
\item Girolamo Saccheri, S.J.
\item Adrien-Marie Legendre
\item Janos Bolyai
\item Carl Friedrich Gauss
\item Nikolai Lobachevsky
\end{itemize} 

\vskip 10pt
\noindent
{\it Assigned Proofs}
\vskip 10pt
\begin{enumerate}
\item The proof of Proposition 29 in Book I of Euclid's {\it Elements}.  Also -- what
is especially notable about this Proposition?
\item The proof of Proposition 38 in Book I of the {\it Elements}, assuming Proposition 37.   
\item The proof of the Saccheri-Legendre Theorem (Theorem 2.2 in McCleary).  Also know
that this result holds in both Euclidean and hyperbolic geometry.  Why?
\item The proof that if $ABCD$ is a Saccheri quadrilateral on the base AB, then 
\begin{itemize} 
\item if HAA holds, then $AB < CD$ and the angle sum in any triangle is less
than 2 right angles, 
\item if HOA holds, then $AB > CD$ and the angle sum in any triangle is greater
than 2 right angles.  
\end{itemize}
(This is covered in Theorem 3.12 in McCleary.)  
\item The proof that the ratio of lengths of concentric arcs of two horocycles intercepted
by two lines in the parallel pencil is an exponential function of the distance between
the arcs.  (This is Theorem 4.9 in McCleary; we did the same proof in class as well.)
\end{enumerate}

\vskip 10pt
\noindent
{\it Review/Practice Problems}
\vskip 10pt
\begin{enumerate}
\item[(A)]  Assuming Postulate V, prove that the three internal angle bisectors of a triangle meet
in a single point.  (Where does your proof use Postulate V?)  Hint:  Take two bisectors that meet 
and join the intersection point to the third vertex.  Under what conditions does
that line bisect the angle at that third vertex?
\item[(B)]  Assuming Postulate V, prove that the three perpendicular bisectors of the sides of a 
triangle meet in a single point.  (Where did you use Postulate V?)  Hint:  You can use
the same strategy as in (A).
\item[(C)]  Look at the figure in problem 4.20 on page 70 in McCleary.  Show that in the 
hyperbolic plane, 
it is possible to construct three nonintersecting 
hyperbolic lines as in the top left diagram -- that is show
how to construct three lines $\ell_1, \ell_2$, and $\ell_3$ and points $A,A'$ on
$\ell_1$, $B,B'$ on $\ell_2$ and $C,C'$ on $\ell_3$ such that  
$\overrightarrow{AA'}$ is parallel to $\overrightarrow{BB'}$
in the direction of $A'$ and $B'$, and (at the same time) 
$\overrightarrow{B'B}$ is parallel to $\overrightarrow{CC'}$ in the direction of $B$ and $C'$, 
and (at the same time) $\overrightarrow{A'A}$ is parallel to $\overrightarrow{C'C}$
in the direction of $A$ and $C$.  Then generalize your argument to show that there
are similar ``$n$-gons'' of parallel lines for any $n \ge 3$.
\item[(D)]  Determine the derivative $\frac{d}{dx}(\Pi(x))$ and show that it is always
negative.
\item[(E)]  Let $x,y > 0$.  Express the angle of parallelism $\Pi(x+y)$ 
in terms of $\Pi(x)$ and $\Pi(y)$.
\end{enumerate}
\end{document}
