% Exam Template for UMTYMP and Math Department courses % % Using Philip Hirschhorn's exam.cls: http://www-math.mit.edu/~psh/#ExamCls % % run pdflatex on a finished exam at least three times to do the grading table on front page. % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % These lines can probably stay unchanged, although you can remove the last % 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 II}} \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 % \onehalfspacing % \doublespacing % 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} %\cellwidth{3em} \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. % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %http://ksuweb.kennesaw.edu/~plaval/math4381/seqlimthm.pdf \begin{questions} % Question with parts \addpoints \question[20] ({\bf Isometries in $\R^2$}) Consider the three points $P=(0,0),Q=(1,0),R=(0,1)\in\R^2$ in the Euclidean plane. Let $f:\R^2\lr\R^2$ be an isometry such that $f(P)=(2,2)$, $f(Q)=(2,3)$ and $f(R)=(3,2)$. \noaddpoints \begin{parts} \part[5] Find the images $f(-1,0)$ and $f(8,2)$ of the points $(-1,0)$ and $(8,2)$ under the isometry $f$.\\ \textcolor{blue}{ First, we claim that $f$ is the glide reflection $t_{(2,2)}\overline{r}_{y=x}$ (the reflection across the line $y=x$ followed by $t_{(2,2)}$). Note that $\overline{r}_{y=x}(x,y)=(y,x)$ for any $(x,y)\in\R^2$, so $t_{(2,2)}\overline{r}_{y=x}$ does indeed send $P\mapsto(2,2),Q\mapsto (2,3),R\mapsto (3,2)$, and by uniqueness of isometries on three noncollinear points, it follows that $f=t_{(2,2)}\overline{r}_{y=x}$. Then, we compute $f(-1,0)=t_{(2,2)}(0,-1)=(2,1)$, and $f(8,2)=t_{(2,2)}(2,8)=(4,10)$. } \vfill \part[5] Prove that the isometry $f$ is not a translation, i.e. there exists no vector $(\alpha,\beta)\in\R^2$ such that $f=t_{(\alpha,\beta)}$.\\ \textcolor{blue}{ We showed in part (a) that $f$ is a glide reflection, which is orientation-reversing. Since translations are orientation-preserving, $f$ cannot be a translation. } \vfill \part[5] Show that there exists no point $S\in\R^2$ such that $f(S)=S$.\\ \textcolor{blue}{ Assume for a contradiction that $f(x,y)=(x,y)$. We have $(x,y)=f(x,y)=t_{(2,2)}(y,x)=(y+2,x+2)$, which implies $y+2=x,x+2=y$, and it follows that $y+4=y$ and thus $4=0$, which is absurd. So there is no point $S\in\R^2$ such that $f(S)=S$. } \vfill \part[5] Find a set of {\it at most} three reflection $\{\overline{r}_{L_1},\overline{r}_{L_2},\overline{r}_{L_3}\}\in\mbox{Iso}(\R^2)$ such that $f$ is a composition of these reflections.\\ \textcolor{blue}{ Let $L_1=\{y=2-x\}, L_2=\{y=-x\}, L_3=\{y=x\}$. We have a Theorem from class that any translation is the product of two reflections, in particular a translation by $d$ units in a given direction is the product of two reflections across lines perpendicular to the direction of $d$ that are distance $\frac{d}{2}$ apart. So we note that $t_{(2,2)}$ is a translation by $2\sqrt{2}$ units along the line $y=x$, and $L_1,L_2$ are both perpendicular to $\{y=x\}$ and are distance $\sqrt{2}/2$ apart, so $t_{(2,2)}=\overline{r}_{L_1}\circ \overline{r}_{L_2}$, and it follows that $f=\overline{r}_{L_1}\circ \overline{r}_{L_2}\circ\overline{r}_{L_3}$. (One should check that this composition does indeed send $P,Q,R$ to $f(P),f(Q),f(R)$, whether by explicit computations, or by drawing a picture of $\R^2$ and depicting how the points $P,Q,R$ behave under the three reflections) } \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. \addpoints \newpage \question[20] ({\bf $\Gamma$-Geometry for the Klein Bottle}) Let $K=\R^2/\Gamma$ be the Euclidean Klein Bottle, where $\Gamma=\langle t_{(0,1)},\overline{r}\circ t_{(1,0)}\rangle\sse\mbox{Iso}(\R^2)$. \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 following points: $$P=(0,0),Q=(0.5,2),R=(1,-5),S=(3,-232)\in\R^2.$$ \begin{figure}[htp] \centering \includegraphics[width=0.3\linewidth]{Midterm/points on klein bottle.png} \end{figure}\\ \textcolor{blue}{ Note that $P\sim R\sim S$ since we can first apply a vertical translation by integer units to get $R,S$ to lie on the $x$-axis, then apply the glide reflection to get to $P$. The black points are $\Gamma(P)=\Gamma(R)=\Gamma(S)$, and the red points are $\Gamma(Q)$. } \vfill \part[5] Find a fundamental domain $D_\Gamma\sse\R^2$ which is {\it not} a square.\\ \begin{figure}[htp] \centering \includegraphics[width=0.1\linewidth]{Midterm/fund dom klein.png} \end{figure}\\ \textcolor{blue}{ Starting from a unit square, we have that the two horizontal edges are identified, so we may ``cut" the square into two pieces (in this picture, we're cutting along the line $y=1-x$), and glue the top piece to the bottom. Note that the resulting top and bottom edges of the parallelogram are identified via the translation $t_{(0,1)}$. } \vfill \newpage \part[5] Consider the lines $$L=\{(x,y)\in K:x=2y\},\quad M=\{(x,y)\in K:x=0\}.$$ Find {\it all} the intersection points $L\cap M$.\\ \textcolor{blue}{ Lifting to $\R^2$, we have $L=\{(x,\frac{x}{2}):x\in\R\},M=\{(0,y):y\in\R\}$. In $K$, we have the two identifications: $(x,\frac{x}{2})\sim (x,\frac{x}{2}+n), (x,\frac{x}{2})\sim (x+n,(-1)^n(\frac{x}{2}))$ for any $n\in\Z$. In order for $L,M$ to intersect, we must have $x+n=0$, so $x\in\Z$. If $x$ is even, then $\frac{x}{2}\in\Z$, so $(x,\frac{x}{2})\sim (x,0)\sim (0,0)$, and if $x$ is odd, then $\frac{x+1}{2}\in\Z$, so $(x,\frac{x}{2})\sim (x-x,(-1)^x(\frac{x}{2}))= (0,-\frac{x}{2})\sim(0,-\frac{x}{2}+\frac{x+1}{2})=(0,\frac12)$. Thus, $L\cap M=\{(0,0),(0,\frac12)\}$. } \vfill \part[5] Consider the line $N=\{(x,y)\in K:x=\pi\cdot y\}$. Is the number of intersection points $M\cap N$ finite or infinite ?\\ \textcolor{blue}{ We claim that $|M\cap N|=\infty$. Lifting to $\R^2$, we have that points in $N$ are of the form $(x,\frac{x}{\pi}),x\in\R$. If $M,N$ are to intersect, then we must have $x\in\Z$. In order for two intersection points $(0,\frac{x_1}{\pi}),(0,\frac{x_2}{\pi})$ (assuming $x_1\neq x_2$) to be equivalent, we must have $\frac{x_1}{\pi}=\frac{x_2}{\pi}+n$ for some $n\in\Z$. Multiplying throughout by $\pi$, we have $x_1=x_2+n\pi\implies n\pi=x_1-x_2\in\Z\setminus\{0\}$, which is a contradiction because $\pi$ is irrational. Since there are infinitely many distinct nonzero integers $x_i$, the number of intersection points $M\cap N$ must be infinite. } \vfill \end{parts} \newpage \addpoints \question[20] ({\bf The Cylinder}) In this problem, {\it all} points and lines are considered in the cylinder $C=\R^2/\Gamma$, where $\Gamma=\langle t_{(1,0)}\rangle\sse\mbox{Iso}(\R^2)$. Solve the following parts:\\ \noaddpoints \begin{parts} \part[5] Consider the points $P=(0.5,0),Q=(0.3,0.2),R=(5.9,-0.2)\in M$. Find an isometry $f:C\lr C$ such that $$f(P)=(0.7,0),\quad f(Q)=(0.5,-0.2),\quad f(R)=(6.1,0.2).$$ \textcolor{blue}{ Let $f=t_{(0.2,0)}\circ \overline{r}$. We compute that $t_{(0.2,0)}\circ \overline{r}(P)=t_{(0.2,0)}(0.5,0)=(0.7,0), t_{(0.2,0)}\circ \overline{r}(Q)=t_{(0.2,0)}(0.3,-0.2)=(0.5,-0.2), t_{(0.2,0)}\circ \overline{r}(R)=t_{(0.2,0)}(5.9,0.2)=(6.1,0.2)$. } \vfill \part[5] Find infinitely many distinct lines $\{L_i\}\sse C$, $i\in\N$, which contain $P,Q$, i.e. $P,Q\in L_i$, for all $i\in\N$.\\ \textcolor{blue}{ Fixing the point $P$ and defining $Q_i:=(0.3+i,0.2)\in\Gamma(Q)$, we see that the the lines through $P,Q_i$ have slope $\frac{0.2}{i-0.2}$. Such a line passes through $P=(0.5,0)$, so we have the collection $\{L_i\}=\{y=\frac{0.2}{i-0.2}x-\frac{0.5\cdot 0.2}{i-0.2}\}$ of infinitely many lines containing $P,Q$ with different slopes, hence the $L_i$'s are all distinct. } \vfill \part[5] Let $t_{(0,\pi)}:C\lr C$ be a vertical translation, and $H=\langle t_{(0,\pi)}\rangle$ the group of isometries of $C$ it generates. Does the $H$-orbit of the point $R\in C$ have limit points in the cylinder $C$ ? (Justify your answer.)\\ \textcolor{blue}{ No. The $H$-orbit of any point in the cylinder consists of points whose nearest points are distance $\pi$ away, so any $\epsilon$-neighborhood for $\epsilon<\frac{\pi}{2}$ of a point in the $H$-orbit will not contain another point in the $H$-orbit. } \vfill \part[5] Consider the group $A=\langle t_{(0,\sqrt{2})},t_{(0,1)}\rangle$ as a subgroup of the group of isometries of $C$. Prove that the $A$-orbit of $P$ inside the cylinder $C$ has limit points.\\ \textcolor{blue}{ We prove that $P$ is a limit point in $A(P)$. i.e. we want to construct a sequence of points $P_n\in A(P)$ so that $\displaystyle \lim_{n\to\infty}P_n=P$. Note that $A(P)$ consists of points of the form $(0.5,n\sqrt{2}+m)$ for $n,m\in\Z$ (in particular, the $y$-coordinate is a linear combination of $\sqrt{2}$ and $1$). Observe that $|\sqrt{2}-1|<1$, so $\displaystyle\lim_{n\to\infty}(\sqrt{2}-1)^n=0$. By the binomial theorem, we have $\displaystyle (\sqrt{2}-1)^n=\sum_{k=0}^n\binom{n}{k}\sqrt{2}^{n-k}$. We know $n-k\in\Z$, and if $n-k$ is even, then $\sqrt{2}^{n-k}\in \Z$, i.e., an integer multiple of $1$. If $n-k$ is odd, then $\sqrt{2}^{n-k}=\sqrt{2}^{n-k-1}\sqrt{2}$ is an integer multiple of $\sqrt{2}$. By definition, $\binom{n}{k}\in\Z$, thus $(\sqrt{2}-1)^n$ is a linear combination of $\sqrt{2}$ and $1$. Now, define $P_n:=(0.5,(\sqrt{2}-1)^n)\in A(P)$. It follows that $\displaystyle \lim_{n\to\infty}P_n=P$. } \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] Compute the distance between $(1,0,0)$ and $(0,0,1)$.\\ \textcolor{blue}{ Recall that the distance between two points on $S^2$ is the (shorter) arc length between the two points, which is equal to the angle between them. Note that the angle between $(1,0,0)$ and $(0,0,1)$ is $\frac{\pi}{2}$, hence $d_{S^2}((1,0,0),(0,0,1))=\frac{\pi}{2}$. } \vfill \part[5] Determine the set of points in $S^2$ whose distance to $(0,1,0)$ equals their distance to $(0,0,1)$.\\ \textcolor{blue}{ The set of points in $\R^3$ equidistant to two points is a plane bisecting the segment between them. In particular, the segment between the two points is a normal vector to the plane, and the plane passes through the midpoint of the two points. The normal vector is $\langle 0,1,-1\rangle$, and the midpoint is $(0,0.5,0.5)$. Then from 21C, we know that the equation of the plane through $(0,0.5,0.5)$ with normal vector $\langle 0,1,-1\rangle$ is $(y-0.5)-(z-0.5)=0 \iff z=y$. So the set of points in $S^2$ whose distance to $(0,1,0)$ equals their distance to $(0,0,1)$ is $$\{z=y\}\cap S^2=\{(x,y,y)\in\R^3:x^2+2y^2=1\}.$$ } \vfill \part[10] Let $E=\{z=0\}\sse S^2$ be the equator. Find an orientation-reversing isometry $f:S^2\lr S^2$ such that $f(E)=E$ but $f$ has no fixed points on the equator.\\ \textcolor{blue}{ Let $R_{z,\theta}$ be the rotation about the $z$-axis with angle $\theta$, and let $\overline{r}_{z=0}$ be the reflection across the $xy$-plane. Then define $f:=R_{z,\theta}\circ \overline{r}_{z=0}$. By construction, $f$ is a composition of isometries on $S^2$, hence $f$ itself is an isometry on $S^2$. $f$ is the product of an orientation-reversing isometry and an orientation-preserving isometry, hence $f$ is orientation-reversing. Since $R_{z,\theta},\overline{r}_{z=0}$ both map $E$ to itself, we have that $f(E)=R_{z,\theta}(E)=E$. But $\overline{r}_{z=0}$ fixes all points on $E$ while $R_{z,\theta}$ fixes no points on $E$ whenever $\theta\not\equiv 0 \pmod {2\pi}$, it follows that $f$ has no fixed points on $E$. } \vfill \end{parts} \newpage \addpoints \addpoints \question[20] For each of the five sentences below, circle the {\bf unique} correct answer. 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] Let $(0,0),(0.5,0)\in C$ be two points in the cylinder. The set of points equidistant to $(0,0)$ and $(0.5,0)$ consists of exactly:\\ (1) A line,\qquad \qquad(2) Empty\qquad \qquad \textcolor{blue}{(3) Two lines}\qquad \qquad(4) Infinite Lines \vfill \part[2] Two lines $L,M\sse T^2$ in the two torus must have:\\ (1) Finitely Many Intersection Points \qquad (2) Infinitely Many Intersection Points\\ \\ (3) No Intersection Points. \qquad\qquad \textcolor{blue}{(4) None of the other answers.} \vfill \part[2] A non-trivial subgroup $\Gamma\sse\mbox{Iso}(\R^2)$ must:\\ (1) Contain a non-trivial translation, \qquad (2) Be generated by at most two elements,\\ \\ (3) Be fixed point free, \qquad \textcolor{blue}{(4) Contain a product of reflections.} \vfill \part[2] There exists a unique isometry which fixes\\ (1) Three collinear points \qquad \textcolor{blue}{(2) Three non-collinear points}\\ (3) Four collinear points \qquad (4) The origin. \vfill \part[2] Let $f:S^2\lr S^2$ be an isometry. Then\\ (1) $f$ cannot have fixed points,\\ \textcolor{blue}{(2) $f$ is a product of at most three reflections},\\ (3) $f$ is a product of at most two reflections,\\ (4) the fixed point set of $f$ must be two antipodal points. \vfill \end{parts} \end{questions} \end{document}