% 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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% two packages if you're not making pictures with tikz.
\documentclass[12pt]{exam}
\RequirePackage{amssymb, amsfonts, amsmath, latexsym, verbatim, xspace, setspace}
\RequirePackage{tikz, pgflibraryplotmarks}

% By default LaTeX uses large margins.  This doesn't work well on exams; problems
% end up in the "middle" of the page, reducing the amount of space for students
% to work on them.
\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}}
\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}

% For an exam, single spacing is most appropriate
\singlespacing
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% For an exam, we generally want to turn off paragraph indentation
\parindent 0ex

\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}
\runningheadrule

\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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\gradetablestretch{2}
\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}

% Question with parts
\addpoints
\question[20] ({\bf Rotations in $\R^2$}) Consider the two points $P=(0,0),Q=(1,0)\in\R^2$ in the Euclidean plane. Solve the following parts:
\noaddpoints
\begin{parts}
\part[5] Let $R_{P,\pi/2}$ be a rotation of angle $\pi/2$ centered at $P$. Compute the image $R_{P,\pi/2}(3,3)$ of the point $(3,3)\in\R^2$ under the isometry $R_{P,\pi/2}$.\\
\textcolor{blue}{
We have the formula $\displaystyle R_\theta=\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix},$ so $\displaystyle R_{P,\frac{\pi}{2}}=\begin{bmatrix}
    0 & -1 \\
    1 & 0
\end{bmatrix},$ thus $\displaystyle R_{P,\frac{\pi}{2}}(3,3)=\begin{bmatrix}
    0 & -1 \\
    1 & 0
\end{bmatrix}\begin{bmatrix}
    3\\
    3
\end{bmatrix}=\begin{bmatrix}
    -3\\
    3
\end{bmatrix}=(-3,3).$
}
\vfill
\part[5] Let $R_{Q,-\pi/2}$ be a rotation of angle $-\pi/2$ centered at $Q$. Compute the image $R_{Q,-\pi/2}(4,5)$ of the point $(4,5)\in\R^2$ under the isometry $R_{Q,-\pi/2}$.\\
\textcolor{blue}{
By conjugation, we have $\displaystyle R_{Q,-\frac{\pi}{2}}=t_{(1,0)}R_{P,-\frac{\pi}{2}}t_{(-1,0)}=t_{(1,0)}\begin{bmatrix}
    0 & 1\\
    -1 & 0
\end{bmatrix}t_{(-1,0)},$ so \\
$\displaystyle R_{Q,-\frac{\pi}{2}}(4,5)=t_{(1,0)}\begin{bmatrix}
    0 & 1\\
    -1 & 0
\end{bmatrix}\begin{bmatrix}
    3\\
    5
\end{bmatrix}=t_{(1,0)}\begin{bmatrix}
    5\\
    -3
\end{bmatrix}=(6,-3).$
}
\vfill
\part[5] Let $(x,y)\in\R^2$ be any point. Where does $(x,y)\in\R^2$ get send under the composition $R_{Q,-\pi/2}\circ R_{P,\pi/2}$ ?
\textcolor{blue}{
\begin{align*}
    R_{Q,-\frac{\pi}{2}}\circ R_{P,\frac{\pi}{2}}(x,y)&=t_{(1,0)}\begin{bmatrix}
        0 & 1\\
        -1 & 0
    \end{bmatrix}t_{(-1,0)}\begin{bmatrix}
        0 & -1\\
        1 & 0
    \end{bmatrix}\begin{bmatrix}
        x\\
        y
    \end{bmatrix}\\
    &=t_{(1,0)}\begin{bmatrix}
        0 & 1\\
        -1 & 0
    \end{bmatrix}t_{(-1,0)}\begin{bmatrix}
        -y\\
        x
    \end{bmatrix}\\
    &=t_{(1,0)}\begin{bmatrix}
        0 & 1\\
        -1 & 0
    \end{bmatrix}\begin{bmatrix}
        -y-1\\
        x
    \end{bmatrix}\\
    &=t_{(1,0)}\begin{bmatrix}
        x\\
        y+1
    \end{bmatrix}=(x+1,y+1).
\end{align*}
}
\vfill
\part[5] Show that $R_{Q,-\pi/2}\circ R_{P,\pi/2}=t_{(1,1)}$.\\
\textcolor{blue}{It is shown in part (c) that $R_{Q,-\frac{\pi}{2}}\circ R_{P,\frac{\pi}{2}}=t_{(1,1)}$.}
\vfill
\end{parts}

% 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 $\Gamma$-Geometry for the 2-Torus}) Let $T^2=\R^2/\Gamma$ be the Euclidean Torus, where $\Gamma=\langle t_{(0,1)},t_{(1,0)}\rangle\sse\mbox{Iso}(\R^2)$ is the group generated by the two translations
$$t_{(0,1)},t_{(1,0)}:\R^2\lr\R^2.$$

\noaddpoints % If you remove this line, the grading table will show 20 points for this problem.
\begin{parts}
	\part[5] Draw the $\Gamma$-orbits of the two points $P=(2,3),Q=(0.5,-7.5)\in\R^2$.\\
    \begin{figure}[htp]
        \centering
        \includegraphics[width=0.3\linewidth]{Midterm/points on torus.png}
    \end{figure}
    \textcolor{blue}{The black points are $\Gamma(P)$ and the red points are $\Gamma(Q)$. \\
    In particular $\Gamma(P)=\{(n,m):n,m\in\Z\}$ and $\Gamma(Q)=\{(n+0.5,m+0.5):n,m\in\Z\}$. }
	\vfill
	\part[5] Find a fundamental domain $D_\Gamma\sse\R^2$ which contains $P\in\R^2$.\\
	\textcolor{blue}{A valid fundamental domain is the square $1\leq x\leq 2, 2\leq y\leq 3$.}
    \vfill
	
	\part[5] Consider $P=(2,3),Q=(0.5,-7.5)\in\R^2/\Gamma$ as points in the 2-torus. Show that the line $\{(x,y)\in T^2:x=y\}\sse T^2$ contains both $P$ and $Q$.\\
    \textcolor{blue}{
    The line $y=x$ goes through the points $(0,0),(0,5,0,5)$, and we know that on $T^2$, $(0,0)\sim (2,3)=P$ and $(0.5,0.5)\sim (0.5,-7.5)=Q$, hence this line contains both $P$ and $Q$.
    }
	\vfill
	
	\part[5] Find {\it all} lines $L\sse T^2$ such that $P,Q\in L$.\\
    \textcolor{blue}{
    We claim that the collection $\mathcal{L}:=\{(x,y)\in T^2:y=\frac{m+0.5}{n+0.5}x, \hspace{2mm}m,n\in\Z\}$ form all the lines in $T^2$ containing $P,Q$. By construction, these lines all go through $(0,0)\sim P,(n+0.5,m+0.5)\sim Q$. So it remains to show that any line going through $P,Q$ must be of this form. If a line $\widehat{L}$ goes through $P,Q$, then it must go through $(a,b),(c+0.5,d+0.5)$ for some $a,b,c,d\in\Z$, and we may apply the translation $t_{(-a,-b)}$ on everything, thus yielding the points $(0,0),(n+0.5,m+0.5)$, where $n=d-b,m=c-a$, and the line through these two points is $\widehat{L}=\{y=\frac{m+0.5}{n+0.5}x\}$, which is in $\mathcal{L}$.
    }
	\vfill
\end{parts}
\newpage
\addpoints
\question[20] ({\bf Geometry in the Twisted Cylinder}) In this problem, {\it all} points and lines are considered in the twisted cylinder $M=\R^2/\Gamma$, where $\Gamma=\langle t_{(1,0)}\circ \overline{r}\rangle\sse\mbox{Iso}(\R^2)$. Solve the following parts:\\
\noaddpoints

\begin{parts}
	\part[5] Consider the points $P=(0,0),Q=(0.9,0.2),R=(5.9,-0.2)\in M$. Find the three distances $d(P,Q),d(P,R),d(Q,R)\in M$.\\
    \textcolor{blue}{
    $d(P,Q)=d(P,(-0.1,-0.2))=\sqrt{0.1^2+0.2^2}$\\
    $d(P,R)=d(P,(-0.1,-0.2))=\sqrt{0.1^2+0.2^2}$\\
    $d(Q,R)=0$, since we find that $Q\sim R$ in $M$.
    }
\vfill

	\part[5] Find the intersection points between the line $\{(x,y)\in M:x=0.5\}\sse M$ and the line $\{(x,y)\in M:x=-y\}\sse M$.\\
    \textcolor{blue}{
    Define $L_1=\{(0.5,y):y\in\R\},L_2=\{(x,-x):x\in\R\}\sse\R^2$, and let $\pi:\R^2\rightarrow M=\R^2/\Gamma$ be the natural projection. In $M$, we have the relation $(x,-x)\sim (x+n,(-1)^n(-x))$ for all $n\in\Z$. So for $\pi(L_1),\pi(L_2)$ to intersect in $M$, we must have $x=k+0.5$ for any $k\in\Z$. If $k$ is even, say $k=2j$ for some $j\in\Z$, then we have $(x,-x)=(2j+0.5,-2j-0.5)\sim (0.5,(-1)^{2j}(-2j-0.5))=(0.5,-2j-0.5)$. If $k$ is odd, say $k=2i+1$ for some $i\in \Z$, then we have $(x,-x)=(2i+1.5,-2i-1.5)\sim (0.5,2i+1.5)$. So $\pi(L_1)\cap\pi(L_2)=\{(0.5,-2j-0.5):j\in\Z\}\cup \{(0.5,2i+1.5):i\in\Z\}$. But we also observe that for any $2i+1.5$, we may take $j=-(i+1)$, so that $2i+1.5=2(i+1)-0.5=-2j-0.5$, and thus the latter set is contained in the first, hence $\pi(L_1)\cap\pi(L_2)=\{(0.5,-2j-0.5):j\in\Z\}$.
	}
	\vfill
	\part[5] Find two lines $K,L\sse M$ such that $|L\cap K|=2$.\\
    \textcolor{blue}{
    Consider $L=\{x=0.5\},K=\{y=0.5\}$. Note that the horizontal line $K$ traverses the segment $0\leq x\leq 1, y=0.5$, crossing $L$ once. Then, since $(1,0.5)\sim (0,-0.5)$, $K$ traverses the segment $0\leq x\leq 1,y=-0.5$, crossing $L$ a second time. Since $(1,-0.5)\sim (0,0.5)$, $L_2$ closes back up to its initial starting point. Thus, $|L\cap K|=2$.
    }
	
	\vfill
	
	\part[5] Show that given two points $S,T\in M$ in the complement of the line $H=\{(x,y)\in M:y=0\}\sse M$, there exists a continuous path $\gamma\sse M$ from $S$ to $T$ such that $|H\cap\gamma|=0$.\\
    \textcolor{blue}{
    We may take $S,T$ to be in the fundamental domain $0\leq x\leq 1$. Write $S=(x_1,y_1),T=(x_2,y_2),$ with $y_1,y_2\neq 0.$ If $y_1y_2>0$ (i.e. either $S,T$ are both above $H$, or both below $H$), then we may take $\gamma$ to be the straight line through $S,T$ (drawing such a line in the fundamental domain may not look continuous, but keep in mind that with the identification given by the quotient, these lines are indeed continuous in $M$). If $y_1y_2<0$, then note that $T=(x_2,y_2)\sim (x_2+1,-y_2)$, and $(x_2+1,-y_2),S$ are either both above $H$ or both below $H$, so we may take $\gamma$ to be the straight line through $S$ and $(x_2+1,-y_2)$.
    }
	
	\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.

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

\begin{parts}
\part[5] Show that there exists a unique line $L\sse S^2$ containing the points $(1,0,0)$ and $(0,1,0)$.\\
\textcolor{blue}{
By definition, a line in $S^2$ is the intersection of a plane through $O$ in $\R^3$ with $S^2$. We know from 21C that there exists a unique plane through $3$ noncollinear points in $\R^3$, and since $(0,0,0),(1,0,0),(0,1,0)$ are noncollinear, this plane is unique, hence there exists a unique line through $(1,0,0),(0,1,0)$ in $S^2$.
}
\vfill
\part[5] Prove that there are infinitely many lines in $S^2$ containing the points $(1,0,0)$ and $(-1,0,0)$.\\
\textcolor{blue}{
Since the three points $(0,0,0),(1,0,0),(-1,0,0)$ are collinear, there exists infinitely many planes containing the three points, hence there are infinitely many lines in $S^2$ containing $(1,0,0)$ and $(-1,0,0)$.
}
\vfill
\part[10] Let $R_1:S^2\lr S^2$ be the rotation with center $(1,0,0)$ and angle $\pi/2$, and $R_2:S^2\lr S^2$ be the rotation with center $(0,0,1)$ and angle $\pi/2$. Determine whether the isometry $R_1\circ R_2$ is equal to $R_2\circ R_1$.\\
\textcolor{blue}{
We show that $R_1\circ R_2\neq R_2\circ R_1$. Consider the point $(0,1,0)$, and observe that $R_2$ is just rotation about the origin in $\R^2$, with the $z$-coordinate fixed, so $(0,1,0)$ gets sent to $(-1,0,0)$. Then $(-1,0,0)$ is on the axis of revolution of $R_1$, so it is fixed by $R_1$, hence $R_1\circ R_2(0,1,0)=R_1(-1,0,0)=(-1,0,0)$. Similarly, observe that $R_1$ is the rotation about the origin in the $yz$-plane, with the $x$-coordinate fixed, so $(0,1,0)$ gets sent to $(0,0,1)$. Then $(0,0,1)$ is on the axis of revolution of $R_2$, so it is fixed by $R_2$, hence $R_2\circ R_1(0,1,0)=R_2(0,0,1)=(0,0,1)$.
}

\vfill


\end{parts}

\newpage
\addpoints

\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] Two lines $K,L\sse T^2$ cannot intersect at more than one point.\\
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill
	
	\part[2] Let $\Gamma\sse\mbox{Iso}(\R^2)$ be generated by translations. Then an isometry $g\in\Gamma$ cannot have fixed points.\\
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill
	
	\part[2] The composition of an even number of reflections cannot be a reflection.\\
	
	\textcolor{blue}{(1) True.} \qquad \qquad \qquad \qquad (2) False.
	\vfill
	
	\part[2] A glide reflection admits infinitely many fixed points.\\
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill
	
	\part[2] For any glide reflection $\overline{r}_1\in\mbox{Iso}(\R^2)$, there exists a glide reflection $\overline{r}_2\in\mbox{Iso}(\R^2)$ such that $\overline{r}_2\circ\overline{r}_1=\mbox{Id}$.\\
	
	\textcolor{blue}{(1) True.} \qquad \qquad \qquad \qquad (2) False.
	\vfill

    \part[2] An isometry $f:S^2\lr S^2$ is uniquely determined by the image of three points.
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill

    \part[2] There are no isometries in the M\"obius band except for the identity.
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill

    \part[2] The product of two rotations in $S^2$ is necessarily a rotation.
	
	\textcolor{blue}{(1) True.} \qquad \qquad \qquad \qquad (2) False.
	\vfill

    \part[2] Any isometry $f:S^2\lr S^2$ can be written as the product of at most three rotations.
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill
	
	
\end{parts}

\end{questions}
\end{document}
