% Exam Template for UMTYMP and Math Department courses
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% Using Philip Hirschhorn's exam.cls: http://www-math.mit.edu/~psh/#ExamCls
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% run pdflatex on a finished exam at least three times to do the grading table on front page.
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\documentclass[12pt]{exam}
\RequirePackage{amssymb, amsfonts, amsmath, latexsym, verbatim, xspace, setspace}
\RequirePackage{tikz, pgflibraryplotmarks}

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\usepackage[margin=1in]{geometry}


% Here's where you edit the Class, Exam, Date, etc.
\newcommand{\class}{University of California Davis}
\newcommand{\term}{Euclidean Geometry MAT 141}
\newcommand{\examnum}{\color{red}{Sample Midterm Examination III}}
\newcommand{\examdate}{May 1 2026}
\newcommand{\timelimit}{50 Minutes}

\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\sse}{\subseteq}
\newcommand{\lr}{\longrightarrow}

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\begin{document} 

% These commands set up the running header on the top of the exam pages
\pagestyle{head}
\firstpageheader{}{}{}
\runningheader{}{\examnum\ - Page \thepage\ of \numpages}{\examdate}
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\begin{flushright}
\begin{tabular}{p{2.8in} r l}
\textbf{\class} & \textbf{Name (Print):} & \makebox[2in]{\hrulefill}\\
\textbf{\term} & \textbf{Student ID (Print):} &
\makebox[2in]{\hrulefill}\\ \\
\textbf{\examnum} && \textbf{\examdate}\\
\textbf{Time Limit: \timelimit} &
\end{tabular}\\
\vspace{1cm}
\end{flushright}
\rule[1ex]{\textwidth}{.1pt}

\vspace{1cm}
This examination document contains \numpages\ pages, including this cover page, and \numquestions\ problems.  You must verify whether there any pages missing, in which case you should let the instructor know. {\bf Fill in} all the requested information on the top of this page, and put your initials on the top of every page, in case the pages become separated.\\

You may \textit{not} use your books, notes, or any calculator on this exam.\\

You are required to show your work on each problem on this exam.  The following rules apply:\\

\begin{minipage}[t]{3.7in}
\vspace{0pt}
\begin{itemize}

\item[(A)] \textbf{If you use a lemma, proposition or theorem which we have seen in the class or in the book, you must indicate this} and explain why the theorem may be applied.

\item[(B)] \textbf{Organize your work}, in a reasonably neat and coherent way, in the space provided. Work scattered all over the page without a clear ordering will receive little credit.  

\item[(C)] \textbf{Mysterious or unsupported answers will not receive full
credit}.  A correct answer, unsupported by calculations, explanation,
or algebraic work will receive little credit; an incorrect answer supported
by substantially correct calculations and explanations will receive
partial credit.


\item[(D)] If you need more space, use the back of the pages; clearly indicate when you have done this.
\end{itemize}

Do not write in the table to the right.
\end{minipage}
\hfill
\begin{minipage}[t]{2.3in}
\vspace{0pt}
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\vqword{Problem}
\addpoints % required here by exam.cls, even though questions haven't started yet.	
\gradetable[v]%[pages]  % Use [pages] to have grading table by page instead of question

\end{minipage}
\newpage % End of cover page

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
% See http://www-math.mit.edu/~psh/#ExamCls for full documentation, but the questions
% below give an idea of how to write questions [with parts] and have the points
% tracked automatically on the cover page.
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%http://ksuweb.kennesaw.edu/~plaval/math4381/seqlimthm.pdf
\begin{questions}

% If you want the total number of points for a question displayed at the top,
% as well as the number of points for each part, then you must turn off the point-counter
% or they will be double counted.
\newpage
\addpoints
\question[20] ({\bf Properties of $\Gamma\sse\mbox{Iso}(\R^2)$}) Solve the following parts:


\noaddpoints % If you remove this line, the grading table will show 20 points for this problem.

\begin{parts}



\part[5] Let $L=\{(x,y)\in\R^2:x=y\}$ and $M=\{(x,y)\in\R^2:x=6\}$. Find an element $g\in\Gamma=\langle \overline{r}_{L},\overline{r}_{M}\rangle$ which has a {\it unique} fixed point.\\
\textcolor{blue}{
Note that $L,M$ intersect at a single point $(6,6)$, so by a Theorem in class which shows the product of two reflections across intersecting lines is a rotation, $g:=\overline{r}_L\overline{r}_M$ is a rotation, and therefore fixes a single point.
}

\vfill
\part[5] Draw the $\Gamma$-orbit of the point $(0,0)$ where $\Gamma:=\langle t_{(2,3)},t_{(-1,0)}\rangle$.\\
\textcolor{blue}{
$\Gamma(0,0)$ consists of the intersection points of the black and red lines. The black lines have slope $\frac32$, translated around by powers of $t_{(-1,0)}$, while the red lines have slope $\frac{0}{-1}=0$, translated around by powers of $t_{(2,3)}$. (Side note: if we were to consider the quotient $\R^2/\Gamma$, then each of the parallelograms would be a fundamental domain). You may also just simply plot all the points instead of drawing the lines.
\begin{figure}[htp]
        \centering
        \includegraphics[width=0.5\linewidth]{Midterm/points on gamma.png}
    \end{figure}
}
\vfill

\part[5] Give an instance of a line $L\in\R^2$ such that its image under the quotient $\R^2\lr\R^2/\Gamma$ is finite, where $\Gamma:=\langle t_{(2,3)},t_{(-1,0)}\rangle$ as in $(b)$.\\
\textcolor{blue}{
The image of the line $L=\{y=0\}\subseteq \R^2$ is a line segment $\{(x,0):0\leq x\leq 1\}$ with length $1$.
}
\vfill

\part[5] Show that any non-trivial isometry in $\Gamma:=\langle t_{(2,3)},\overline{r}\circ t_{(1,0)}\rangle$ cannot have fixed points.\\
\textcolor{blue}{
We want to show that if $f\in\Gamma$ is not the identity, then $f(x,y)\neq (x,y)$ for all $(x,y)\in\R^2$. By definition, $\Gamma$ consists of isometries which are compositions of $g:=t_{(2,3)}$ and $h:=\overline{r}\circ t_{(1,0)}$. First, we note that for any translation $t_{(a,b)}$, we have $\overline{r}\circ t_{(a,b)}(x,y)=\overline{r}(x+a,y+b)=(x+a,-y-b)=t_{(a,-b)}\circ \overline{r}(x,y)$. In particular, $t_{(a,b)}$ and $\overline{r}$ commute at the cost of a sign change to $t_{(a,-b)}$. Now, let $f\in\Gamma$ be nontrivial. By definition, $f$ is a word of the form $\displaystyle f=\prod_{i=1}^k g^{n_i}h^{m_i}=g^{n_k}h^{m_k}g^{n_{k-1}}h^{m_{k-1}}\cdots g^{n_1}h^{m_1}$. This is a product of translations and $\overline{r}$, which by the commutation formula $\overline{r}\circ t_{(a,b)}=t_{(a,-b)}\circ \overline{r}$, we may move all the $\overline{r}$ to the right, which gives us a single glide reflection if the total number of $h$ is odd, or a single translation if the total number of $h$ is even. Since $f\neq$ Id, we know that in the latter case where $f$ is a single translation, that it must not be the trivial translation, hence $f$ fixes no points. If $f$ is a single glide reflection, we need to show that the translation part is not trivial (otherwise, $f$ is a reflection, which has fixed points). To that end, it suffices to show the $x$-coordinate is not fixed. Since the number of $h$ is odd, there must be a translation $t_{(1,0)}$ somewhere. But since $h^2=t_{(2,0)}$ and $g^n=t_{(2n,3n)}$, we must have $f=t_{(2m,3n)}t_{(1,0)}\overline{r}$ for some $m,n\in\Z$, but clearly $2m+1\neq 0$, hence the $x$-coordinate is not fixed. Thus, $f$ has no fixed points.
}


%\vfill
%\part[5] Find two elements $g_1,g_2$ in the group
%$$\Gamma:=\langle t_{(-4,6)},t_{(-3,9)},t_{(5,-15)},t_{(2,-3)},t_{(-1,3)},t_{(1,-1.5)}\rangle$$ which generate $\Gamma$, i.e. such that $\Gamma=\langle g_1,g_2\rangle$.



\vfill

\end{parts}

%\newpage
%\addpoints
%\question[20] ({\bf Properties of $\Gamma\sse\mbox{Iso}(\R^2)$}) Solve the following parts:


%\noaddpoints % If you remove this line, the grading table will show 20 points for this problem.

%\begin{parts}
%\part[5] Show that $\Gamma=\langle t_{(2,3)},\overline{r}\circ t_{(1,0)}\rangle$ is fixed point free.\\
%\textcolor{blue}{
%We want to show that if $f\in\Gamma$ is not the identity, then $f(x,y)\neq (x,y)$ for all $(x,y)\in\R^2$. By definition, $\Gamma$ consists of isometries which are compositions of $g:=t_{(2,3)}$ and $h:=\overline{r}\circ t_{(1,0)}$. First, we note that for any translation $t_{(a,b)}$, we have $\overline{r}\circ t_{(a,b)}(x,y)=\overline{r}(x+q,y+b)=(x+a,-y-b)=t_{(a,-b)}\circ \overline{r}(x,y)$. In particular, $t_{(a,b)}$ and $\overline{r}$ commute at the cost of a sign change to $t_{(a,-b)}$. Now, let $f\in\Gamma$ be nontrivial. By definition, $f$ is a word of the form $\displaystyle f=\prod_{i=1}^k g^{n_i}h^{m_i}=g^{n_k}h^{m_k}g^{n_{k-1}}h^{m_{k-1}}\cdots g^{n_1}h^{m_1}$. This is a product of translations and $\overline{r}$, which by the commutation formula $\overline{r}\circ t_{(a,b)}=t_{(a,-b)}\circ \overline{r}$, we may move all the $\overline{r}$ to the right, which gives us a single glide reflection if the total number of $h$ is odd, or a single translation if the total number of $h$ is even. Since $f\neq$ Id, we know that in the latter case where $f$ is a single translation, that it must not be the trivial translation, hence $f$ fixes no points. If $f$ is a single glide reflection, we need to show that the translation is not trivial (otherwise, $f$ is a reflection, which has fixed points). To that end, it suffices to show the $x$-coordinate is not fixed. Since the number of $h$ is odd, there must be a translation $t_{(1,0)}$ somewhere. But since $h^2=t_{(2,0)}$ and $g^n=t_{(2n,3n)}$, we must have $f=t_{(2m,3n)}t_{(1,0)}\overline{r}$ for some $m,n\in\Z$, but clearly $2m+1\neq 0$, hence the $x$-coordinate is not fixed. Thus, $f$ has no fixed points.
%}

%\vfill
%\part[5] Let $L=\{(x,y)\in\R^2:x=y\}$ and $M=\{(x,y)\in\R^2:x=6\}$. Find an element $g\in\Gamma=\langle \overline{r}_{L},\overline{r}_{M}\rangle$ which has a {\it unique} fixed point.\\
%\textcolor{blue}{
%Note that $L,M$ intersect at a single point $(6,6)$, so by a Theorem in class which shows the product of two reflections across intersecting lines is a rotation, $g:=\overline{r}_L\overline{r}_M$ is a rotation, and therefore fixes a single point.
%}

%\vfill
%\newpage
%\part[5] Draw the $\Gamma$-orbit of the point $(0,0)$ where $\Gamma=\langle t_{(2,3)},t_{(-3,9)}\rangle$.\\
%\textcolor{blue}{
%$\Gamma(0,0)$ consists of the intersection points of the black and red lines. The black lines have slope $\frac32$, translated around by $t_{(-3,9)}$, while the red lines have slope $\frac{9}{-3}=-3$, translated around by $t_{(2,3)}$. (Side note: if we were to consider the quotient $\R^2/\Gamma$, then each of the parallelograms would be a fundamental domain).
%\begin{figure}[htp]
%        \centering
%        \includegraphics[width=0.3\linewidth]{Midterm/points on gamma.png}
%    \end{figure}
%}


%\vfill
%\part[5] Find two elements $g_1,g_2$ in the group
%$$\Gamma:=\langle t_{(-4,6)},t_{(-3,9)},t_{(5,-15)},t_{(2,-3)},t_{(-1,3)},t_{(1,-1.5)}\rangle$$ which generate $\Gamma$, i.e. such that $\Gamma=\langle g_1,g_2\rangle$.\\
%\textcolor{blue}{
%Let $g_1=t_{(1,-1.5)},g_2=t_{(-1,3)}$. We show that all the other listed translations are powers of $g_1,g_2$. (In general, you would have to check for compositions of $g_1,g_2$ as well, since some of the listed translations may be a power of $g_1$ composed with a power of $g_2$ or something like that, so just looking at powers doesn't cover all possible cases. But for this problem, it's sufficient.) We observe: $g_1^2=t_{(2,-3)}, g_1^{-4}=t_{(-4,6)},g_2^{3}=t_{(-3,9)},g_2^{-5}=t_{(5,-15)}$. Thus, $\Gamma=\langle g_1,g_2\rangle$.
%}

%\vfill


%\end{parts}

\newpage
\addpoints
\question[20] ({\bf Reflections in $\R^2$}) Consider the two lines $L_0=\{y=0\},L_1=\{x=y\}\sse\R^2$ and the two lines $M_0=\{x=y+1\},M_1=\{x=-y+1\}\sse\R^2$.


\noaddpoints % If you remove this line, the grading table will show 20 points for this problem.

\begin{parts}
\part[5] Show that the only fixed point of the isometry $\overline{r}_{L_1}\circ\overline{r}_{L_0}$ is $(0,0)$.\\
\textcolor{blue}{
Recall a Theorem from class states that the product of two reflections across intersecting lines is a rotation about the intersection point. Since $L_0,L_1$ intersect only at the point $(0,0)$, we know that the composition $\overline{r}_{L_1}\circ \overline{r}_{L_0}$ is a rotation about $(0,0)$, hence the only fixed point is $(0,0)$.
}

\vfill
\part[5] Prove that the isometry $\overline{r}_{M_1}\circ\overline{r}_{M_0}\circ\overline{r}_{L_1}\circ\overline{r}_{L_0}$ is a rotation.\\
\textcolor{blue}{
$M_0,M_1$ are two lines intersecting at $(1,0)$, so $\overline{r}_{M_1}\circ \overline{r}_{M_0}$ is a rotation about $(1,0)$. In particular, the angle between $L_0,L_1$ is $\frac{\pi}{4}$, and the angle between $M_0,M_1$ is $\frac{\pi}{2}$, so $\overline{r}_{L_1}\circ \overline{r}_{L_0}=R_{\pi/2,(0,0)}$, and $\overline{r}_{M_1}\circ \overline{r}_{M_0}=R_{\pi,(1,0)}$. From the proof of the theorem (saw in week 2 discussion) for closure of the set of translations and rotations, we know that $R_{\theta,P}\circ R_{\phi,Q}$ for $P\neq Q$ is a translation if $\frac{\theta}{2}+\frac{\phi}{2}\equiv \pi \pmod{2\pi}$, and a rotation otherwise. Since $\frac{\pi}{4}+\frac{\pi}{2}\not\equiv \pi \pmod{2\pi}$, $\overline{r}_{M_1}\circ\overline{r}_{M_0}\circ\overline{r}_{L_1}\circ\overline{r}_{L_0}=R_{\pi,(1,0)}\circ R_{\pi/2,(0,0)}$ is a rotation.
}

\vfill
\part[5] Show that there exist two lines $N_0,N_1\sse\R^2$ such that $$\overline{r}_{M_1}\circ\overline{r}_{M_0}\circ\overline{r}_{L_1}\circ\overline{r}_{L_0}=\overline{r}_{N_1}\circ\overline{r}_{N_0}.$$\\
\textcolor{blue}{
By Theorem from class, we know that any rotation is a product of two reflections. Therefore, since $\overline{r}_{M_1}\circ\overline{r}_{M_0}\circ\overline{r}_{L_1}\circ\overline{r}_{L_0}$ is a rotation, there does exist lines $N_0,N_1\sse\R^2$ such that $\overline{r}_{M_1}\circ\overline{r}_{M_0}\circ\overline{r}_{L_1}\circ\overline{r}_{L_0}=\overline{r}_{N_1}\circ\overline{r}_{N_0}.$ (Note the question only asks us to show existence, so you don't have to find the explicit lines, though you are certainly welcome to.)
}

\vfill
\part[5] Find the image of a point $(x,y)\in\R^2$ under the isometry $\overline{r}_{M_0}\circ\overline{r}_{L_1}$.\\
\textcolor{blue}{
Note that $L_1,M_0$ have the same slope, hence are parallel. So $\overline{r}_{M_0}\circ \overline{r}_{L_1}$ is a translation in the direction of the line $y=-x$ (since this line is perpendicular to the $L_1,M_0$) by twice the distance between $L_1,M_0$. The distance between $L_1,M_0$ is the distance between any perpendicular segment between two points on the lines, so we compute the distance between $(0,0)\in L_1,(0.5,-0.5)\in M_0$, which is $\sqrt{2}/2$. So $\overline{r}_{M_0}\circ\overline{r}_{L_1}=t_{(a,-a)}$, where $\sqrt{a^2+(-a)^2}=\sqrt{2a^2}=2\cdot \sqrt{2}/2$, thus $a=1$. We conclude that $\overline{r}_{M_0}\circ\overline{r}_{L_1}(x,y)=t_{(1,-1)}(x,y)=(x+1,y-1)$. (Note that $L_1$ is above $M_0$, so the composition would be a translation towards the bottom right rather than the top left). You may also compute the formula explicitly by conjugation or some other method.
}

\vfill


\end{parts}


\newpage

\addpoints
\question[20] ({\bf Spherical geometry}) Consider the 2-sphere
$$S^2:=\{(x,y,z)\in\R^3:~x^2+y^2+z^2=1\},$$
endowed with the spherical distance. Solve the following parts:
\noaddpoints % If you remove this line, the grading table will show 20 points for this problem.
\begin{parts}

	\part[5] Consider the point $P=\frac{1}{\sqrt{3}}(1,1,1)$. Compute its spherical distance to each of the points $Q_1=(1,0,0),Q_2=(0,1,0)$ and $Q_3=(0,0,1)$.\\
    \textcolor{blue}{
    We find the angles between $P$ and $Q_i$. By symmetry, these angles will all be the same. 
    \begin{align*}
        d_{S^2}(P,Q_1)&=2\sin^{-1}(\frac12 d_{\R^3}(P,Q_1))\\
        &=2\sin ^{-1}\left(\frac12 \sqrt{(1-\frac{1}{\sqrt{3}})^2+(\frac{1}{\sqrt{3}})^2+(\frac{1}{\sqrt{3}})^2}\right)\\
        &=2\sin ^{-1}\left(\frac12\sqrt{\frac{(\sqrt{3}-1)^2}{3}+2\cdot\frac13}\right)\\
        &=2\sin ^{-1}\left(\frac12 \sqrt{\frac{3-2\sqrt{3}+1+2}{3}}\right),
    \end{align*}
    which is some ugly stuff that I would refuse to simplify much further on an exam.\\
    Alternatively, we may use 21C methods to find the angle between two vectors $\langle \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\rangle,\langle1,0,0\rangle$. Recall $\mathbf{u}\cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}|\cos \theta$, so
    \begin{align*}
        \theta&=\cos ^{-1}\frac{\langle \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\rangle\cdot\langle1,0,0\rangle}{|\langle \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\rangle||\langle1,0,0\rangle|}\\
        &=\cos ^{-1}\frac{1}{\sqrt{3}}.
    \end{align*}
    So $d_{S^2}(P,Q_i)=\cos ^{-1}\frac{1}{\sqrt{3}}$ for $i=1,2,3$.
    }
	
	
	\vfill

    \part[5] Let $L\sse S^2$ be the unique line containing $P$ and $Q_1$, show that the point $\frac{1}{\sqrt{2}}(0,1,1)$ belongs to $L$.\\
    \textcolor{blue}{
    By symmetry, we have that $P,Q_1,(0,0,0)$ are all equidistant from $Q_2,Q_3$. That is, the line $L$ is the intersection of the plane equidistant from $Q_2,Q_3$ with $S^2$. So to show that $\frac{1}{\sqrt{2}}(0,1,1)\in L$, it suffices to show that it is equidistant from $Q_2=(0,1,0),Q_3=(0,0,1)$. But indeed by symmetry, we find that $$d_{\R^3}\left(\frac{1}{\sqrt{2}}(0,1,1),(0,1,0)\right)=d_{\R^3}\left(\frac{1}{\sqrt{2}}(0,1,1),(0,0,1)\right),$$ and it follows that $\frac{1}{\sqrt{2}}(0,1,1)\in L$.
    }
	
	\vfill
	
	\part[5] Let $E$ the unique line containing $Q_1$ and $Q_2$. Find the image of the unique line through $Q_1$ and $Q_3$ under the composition $\overline{r}_{L}\circ \overline{r}_{E}$.\\
    \textcolor{blue}{
    First observe that $E$ is the equator, and the line $\mathcal{L}$ through $Q_1,Q_3$ is perpendicular to $E$, hence $\overline{r}_E(\mathcal{L})=\mathcal{L}$. Then, we note that $L$ is the plane of equidistant points to $Q_3,Q_2$, hence $\overline{r}_L$ maps $Q_3$ to $Q_2$. But $Q_1\in L$, so $\overline{r}_L(Q_1)=Q_1$, and it follows that $\overline{r}_L\overline{r}_E(\mathcal{L})$ is the unique line containing $Q_1,Q_2$, which is $E$.
    }
	
	\vfill
	
	\part[5] Find the fixed points of the isometry $f$ obtained by first applying $\overline{r}_{L}\circ \overline{r}_{E}$ and then applying the reflection $(x,y,z)\lr (-x,y,z)$.\\
    \textcolor{blue}{
    $\overline{r}_L\overline{r}_E$ is a product of reflections across planes intersecting at $Q_1,-Q_1$, so it is a nontrivial rotation about the $x$-axis. In particular, $\overline{r}_L\overline{r}_E(x,y,z)=(x,y',z')$ where $(y,z)\neq (y',z')$ if $(x,y,z)\neq Q_1,-Q_1$. Then applying the reflection $(x,y,z)\lr (-x,y,z)$ sends $(x,y',z')\lr (-x,y',z')$. Now, if $x=0$, then $(x,y,z)\neq Q_1,-Q_1$, and it follows that $(x,y,z)\neq (-x,y',z').$ Otherwise, we have $x\neq 0$, and $x\neq -x$, so $(x,y,z)\neq (-x,y',z')$. Hence, $f$ has no fixed points.
    }
	
	\vfill

	
\end{parts}

\newpage
\addpoints
\question[20] For each of the ten sentences below, circle whether they are {\bf true} or {\bf false}. You do {\it not} need to justify your answer.\\
\noaddpoints % If you remove this line, the grading table will show 20 points for this problem.
\begin{parts}

	\part[2] For any pair of points $P,Q\in K$ in the Klein bottle, there are infinitely many distinct lines $L\sse K$ containing $P,Q\in K$.
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill
	
	\part[2] There exist rotations $R_{P,\theta},R_{Q,\phi}\in\mbox{Iso}(\R^2)$ such that the composition $R_{P,\theta}\circ R_{Q,\phi}$ is {\it not} a rotation.\\
	
	\textcolor{blue}{(1) True.} \qquad \qquad \qquad \qquad (2) False.
	\vfill
	
	\part[2] Two lines $L,K\sse M$ in the twisted cylinder either intersect $0$,$1$ or infinitely many times.\\
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill
	
	\part[2] Let $\Gamma\sse\mbox{Iso}(\R^2)$ be generated by a finite number of translations. Then there exists a fundamental domain $D_\Gamma\sse\R^2$ of finite area.\\
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill

	\part[2] The quotient of the isometry $t_{(0,1)}:\R^2\lr\R^2$ gives a well-defined isometry in the twisted cylinder $\R^2/\langle t_{(1,0)}\circ\bar{r}\rangle$.\\
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill

    \part[2] There are no parallel lines in the cylinder.
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill

    \part[2] Every orientation-preserving isometry of the 2-sphere is a rotation.
	
	\textcolor{blue}{(1) True.} \qquad \qquad \qquad \qquad (2) False.
	\vfill

    \part[2] Every orientation-reserving isometry of the 2-sphere has a fixed point.
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill

    \part[2] There are parallel lines in the 2-sphere.
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill

    \part[2] The spherical distance between two points in the 2-sphere is the same as the Euclidean distance between these two points, as computed in $\R^3$.
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill
	
\end{parts}
\end{questions}
\end{document}
