% 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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\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 Final Examination}}
\newcommand{\examdate}{June 11 2026}
\newcommand{\timelimit}{2 Hours}

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

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\begin{document} 

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\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}
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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

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% 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[18] ({\bf Euclidean $\R^2$}) Consider the two points $P=(1,2),Q=(-1,-5)\in\R^2$ in the Euclidean plane. Solve the following parts:
\noaddpoints
\begin{parts}
\part[5] Find the Euclidean distance between $P$ and $Q$.\\
\textcolor{blue}{
$d(P,Q)=\sqrt{(-1-1)^2+(-5-2)^2}=\sqrt{4+49}=\sqrt{53}$.
}
\vfill
\part[5] Let $M=\{(x,y)\in\R^2: x=y\}$  and consider the unique line $L\sse\R^2$ equidistant to $P$ and $Q$. Determine the image of $Q$ under the isometry $r_M\circ r_L$.\\
\textcolor{blue}{
By a corollary from class, we know $r_L$ swaps $P$ and $Q$, so $r_L(Q)=P$. Since the reflection across the line $y=x$ precisely swaps the $x$ and $y$-coordinates (this is a simple enough reflection where I do not think it's necessary to prove that it swaps the $x$ and $y$-coordinates), we have that $r_M\circ r_L(Q)=r_M(P)=r_M(1,2)=(2,1)$.
}
\vfill
\part[5] Find all the fixed points of the isometry $r_M\circ r_L$.\\
\textcolor{blue}{
The line $L$ is the perpendicular bisector of $\overline{PQ}$, and $\overline{PQ}$ has slope $\frac{7}{2}$, hence $L$ has slope $-\frac27$. So $M,L$ are not parallel, and by a Theorem from class, $r_M\circ r_L$ is a rotation with exactly one fixed point at the point of intersection of $M,L$. $L$ passes through the midpoint of $P,Q$, which is $(\frac{1-1}{2},\frac{2-5}{2})=(0,-\frac32)$, so $L=\{(x,y)\in\R^2:y+\frac32=-\frac27x\}$. $M,L$ intersect when $x+\frac32=-\frac27x\implies \frac{9}{7}x=-\frac32\implies x=-\frac{7}{6}=y$. So the fixed point of $r_M\circ r_L$ is $(-\frac76,-\frac76)$.
}
\vfill
\part[3] Show that there is no line $N\sse\R^2$ such that $r_N\circ r_M\circ r_L=\mbox{id}$.\\
\textcolor{blue}{
We've deduced from part (c) that $r_M\circ r_L$ is a rotation, which is orientation-preserving. However, for any line $N$, $r_N$ is orientation-reversing, and hence $r_N\circ r_M\circ r_L$ is orientation-reversing. But id is orientation-preserving, and it follows that $r_N\circ r_M\circ r_L\neq$ id.
}
\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[18] ({\bf $\Gamma$-Geometry for the cylinder}) Let $C=\R^2/\Gamma$ be the Euclidean cylinder, where $\Gamma=\langle t_{(1,0)}\rangle\sse\mbox{Iso}(\R^2)$ is the group generated by the translation
$$t_{(1,0)}:\R^2\lr\R^2.$$

\noaddpoints % If you remove this line, the grading table will show 20 points for this problem.
\begin{parts}
	\part[5] Draw the $\Gamma$-orbits $\Gamma P$ and $\Gamma Q$ in $\R^2$ of the two points $$P=(-1,4),Q=(3.25,-7.75)\in\R^2.$$
    \textcolor{blue}{
    \begin{figure}[htp]
        \centering
        \includegraphics[width=0.3\linewidth]{Midterm/gammaP,GammaQ.png}
    \end{figure}
    }
	\vfill
	\part[5] Compute the distance in $C$ from $\Gamma P$ to $\Gamma Q$.\\
    \textcolor{blue}{
    We take the points $(0,4)\in\Gamma P, (0.25,-7.75)\in\Gamma Q$, so $d_C(P,Q)=\sqrt{0.25^2+11.75^2}.$
    }
	\vfill
	
	\part[5] Show that there are infinitely many lines in $C$ through $\Gamma P$ and $\Gamma Q$.\\
    \textcolor{blue}{
    Consider the point $P\in\Gamma P$. For each $Q'\in\Gamma Q$, we get a line from $P$ to $Q'$, each with a different slope, hence they are all distinct when we quotient by $\Gamma$. Since there are infinitely many such $Q'$, we conclude that there are infinitely many lines in $C$ through $\Gamma P$ and $\Gamma Q$.
    }
	\vfill
	
	\part[3] Consider the projections $\pi(L_1),\pi(L_2)$ to $C$ of the two lines
    $$L_1=\{x=0\}\sse \R^2,\quad L_2=\{47x+y=4\}\sse\R^2.$$ Explicitly find all the intersections points of $\pi(L_1)$ and $\pi(L_2)$ in $C$.\\
    \textcolor{blue}{
    Lifting to $\R^2$, we have that $\pi^{-1}(L_1)$ is identified with all lines $\{x=n\}$ for $n\in\Z$, so for $\pi(L_1),\pi(L_2)$ to intersect, we look for the points $(n,y)\in L_2$, which get mapped to $(0,y)\in\pi(L_1)\cap \pi(L_2)$. Rewriting, we have $L_2=\{y=-47x+4\}$, and thus the intersection points $(0,y)$ of $\pi(L_1)\cap \pi(L_2)$ are precisely $(0,-47n+4)$ for all $n\in\Z$.
    }
	\vfill
\end{parts}

\newpage
\addpoints
\question[18] ({\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] Consider the points $P=(1,0,0)\in S^2$ and $Q=\frac{1}{\sqrt{3}}(1,-1,1)$. Compute the spherical distance $d_{S^2}(P,Q)$ from $P$ to $Q$.\\
\textcolor{blue}{
We compute the angle between $\langle 1,0,0\rangle$ and $\langle \frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\rangle$ using the formula $\mathbf{u}\cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}|\cos \theta\implies \theta=\cos^{-1}(\mathbf{u}\cdot \mathbf{v})$, so $d_{S^2}(P,Q)=\cos^{-1}(\frac{1}{\sqrt{3}})$.
}

\part[5] Let $R_{P,\pi/2},R_{Q,\pi/2}\in\mbox{Isom}(S^2)$ be the rotations of angle $\pi/2$ centered at $P$ and $Q$. Show that $R_{P,\pi/2}\circ R_{Q,\pi/2}$ is a rotation and find its center and angle.\\
\textcolor{blue}{
By a theorem from class, we know products of rotations on $S^2$ are also rotations, so $R_{P,\pi/2}\circ R_{Q,\pi/2}$ is a rotation. Now, the hard part, using only things that we've learned: We know from class that rotations on $S^2$ centered at a point with angle $\theta$ are products of reflections across planes in $\R^2$ intersecting at the point and its antipodal, with angle of intersection $\theta/2$. Define $G$ to be the unique plane through $O=(0,0,0),P,Q$. By choice of reflecting planes, we may have $R_{Q,\pi/2}, R_{P,\pi/2}$ be written as products of reflections, where one of the reflections is across $G$. So we write $R_{P,\pi/2}\circ R_{Q,\pi/2}=r_A\circ r_G\circ r_G\circ r_B=r_A\circ r_B$, where $A,B$ are planes to be determined. The idea is that the intersection point $A\cap B\cap S^2$ will be the center of rotation, and the angle of rotation is twice the angle between $A,B$. The plane $A$ is easier to find, by looking at the sphere from the $x$-axis (like a bird's eye view):
\begin{figure}[htp]
    \centering
    \includegraphics[width=0.3\linewidth]{Midterm/r_A.png}
\end{figure}
We see that the line \textcolor{red}{$z=0$} forms an angle of $\frac{\pi}{4}$ with $G$, so $A$ is the $xy$-plane. $B$ is a little more tricky. We know already that $G,A$ form an angle of $\frac{\pi}{4}$, and we want $B,G$ to form an angle of $\frac{\pi}{4}$ as well (be careful with the sign of the angle; we had the angle $\frac{\pi}{4}$ from $G$ to $A$, and now we want the angle $\frac{\pi}{4}$ from $B$ to $G$). Consider the following picture, viewing from the axis $y=-x,z=0 $:
\begin{figure}[htp]
    \centering
    \includegraphics[width=0.3\linewidth]{Midterm/plane B.png}
\end{figure}\\
By symmetry, if we were to have $B,G$ form an angle of $\frac{\pi}{4}$, then we must have $d_{S^2}(P,(x,y,0))=d_{S^2}(Q,(x,y,0))$, and $(x,y,0)\in S^2$. But we may also just compare the Euclidean distances between them, hence we have the equation 
\begin{align*}
    &\sqrt{(x-1)^2+y^2}=\sqrt{\left(x-\frac{1}{\sqrt{3}}\right)^2+\left(y+\frac{1}{\sqrt{3}}\right)^2+\frac13}\\
    &\iff (x-1)^2+y^2=\left(x-\frac{1}{\sqrt{3}}\right)^2+\left(y+\frac{1}{\sqrt{3}}\right)^2+\frac13\\
    &\iff x^2-2x+1+y^2=x^2-\frac{2}{\sqrt{3}}x+\frac13+y^2+\frac{2}{\sqrt{3}}y+\frac13+\frac13\\
    &\iff \left(1-\frac{1}{\sqrt{3}}\right)x+\frac{1}{\sqrt{3}}y=0\\
    &\implies y=-\left(1-\frac{1}{\sqrt{3}}\right)x\cdot \sqrt{3}=(1-\sqrt{3})x.
\end{align*}
Then, we substitute this into the equation $x^2+y^2=1$ to get 
$$x^2+(1-\sqrt{3})^2x^2=1\implies x^2=\frac{1}{1+(1-\sqrt{3})^2}\implies x=\frac{1}{\sqrt{1+(1-\sqrt{3})^2}}$$
and $$y=\frac{1-\sqrt{3}}{\sqrt{1+(1-\sqrt{3})^2}}.$$
So, $B$ is the plane through $O,Q,\left(\frac{1}{\sqrt{1+(1-\sqrt{3})^2}},\frac{1-\sqrt{3}}{\sqrt{1+(1-\sqrt{3})^2}},0\right)$. Using 21C methods, we can construct two vectors from these three points, compute their cross product to get a normal vector, and then using the 21C definition (or theorem?) that the equation of a plane through $(0,0,0)$ with normal vector $\langle a,b,c\rangle$ is $ax+by+cz=0$, we ultimately arrive at $B$ being the plane $(\sqrt{3}-1)x+y+(2-\sqrt{3})z=0$, or some scaling of this. Here's what the planes look like:
\begin{figure}[htp]
    \centering
    \includegraphics[width=0.4\linewidth]{Midterm/S2reflections.png}
\end{figure}\\
By the construction above, $\left(\frac{1}{\sqrt{1+(1-\sqrt{3})^2}},\frac{1-\sqrt{3}}{\sqrt{1+(1-\sqrt{3})^2}},0\right)$ lies on both $A$ and $B$, and therefore is fixed by $r_A\circ r_B$. That is, the center of rotation of $R_{P,\pi/2}\circ R_{Q,\pi/2}$ is $\left(\frac{1}{\sqrt{1+(1-\sqrt{3})^2}},\frac{1-\sqrt{3}}{\sqrt{1+(1-\sqrt{3})^2}},0\right)$. Finally, we compute the angle of rotation, which is twice the angle between $A,B$. From 21C, we know the angle between two planes is equal to the angle between their normal vectors. $A$ has normal vector $\mathbf{n}_1=\langle 0,0,1\rangle$ and $B$ has normal vector $\mathbf{n}_2=\langle \sqrt{3}-1,1,2-\sqrt{3}\rangle$. The angle between them is 
$$
\cos^{-1} \frac{\mathbf{n}_1\cdot \mathbf{n}_2}{|\mathbf{n}_1||\mathbf{n}_2|}=\cos^{-1} \frac{2-\sqrt{3}}{\sqrt{(\sqrt{3}-1)^2+1+(2-\sqrt{3})^2}}=\cos^{-1}\frac{2-\sqrt{3}}{\sqrt{12-6\sqrt{3}}}\approx 77.8^\circ
$$
and thus the angle of rotation is $2\cos^{-1}\frac{2-\sqrt{3}}{\sqrt{12-6\sqrt{3}}}\approx 155.6^\circ$. 
}

\vfill
\part[5] Determine whether $R_{P,\pi/2}\circ R_{Q,\pi/2}$ is equal to $R_{Q,\pi/2}\circ R_{P,\pi/2}$.\\
\textcolor{blue}{
We claim that $R_{P,\pi/2}\circ R_{Q,\pi/2}\neq R_{Q,\pi/2}\circ R_{P,\pi/2}$, and it suffices to show that $R_{Q,\pi/2}\circ R_{P,\pi/2}$ is a rotation with a different center. As with part (b), we write $R_{Q,\pi/2}\circ R_{P,\pi/2}=r_C\circ r_G\circ r_G\circ r_D=r_C\circ r_D$, where $C,D$ are planes to be determined. From the same picture above,
\begin{figure}[htp]
    \centering
    \includegraphics[width=0.3\linewidth]{Midterm/r_A.png}
\end{figure}
we see that the line $y=0$ forms an angle of $\frac{\pi}{4}$ with $G$, so $D$ is the $xz$-plane. By symmetry, if we were to have $G,C$ form an angle of $\frac{\pi}{4}$, then we must have $d_{S^2}(P,(x,0,z))=d_{S^2}(Q,(x,0,z))$, and $(x,0,z)\in S^2$, and a similar computation as above yields $z=(\sqrt{3}-1)x$, and substituting into $x^2+z^2=1$ gives $$x^2+(\sqrt{3}-1)^2x^2=1\implies x=\frac{1}{\sqrt{1+(\sqrt{3}-1)^2}},z=\frac{\sqrt{3}-1}{\sqrt{1+(\sqrt{3}-1)^2}}.$$
With the same reasoning as above, we deduce that $R_{Q,\pi/2}\circ R_{P,\pi/2}$ is a rotation with center $\left(\frac{1}{\sqrt{1+(\sqrt{3}-1)^2}},0,\frac{\sqrt{3}-1}{\sqrt{1+(\sqrt{3}-1)^2}}\right)$, and it follows that $R_{P,\pi/2}\circ R_{Q,\pi/2}\neq R_{Q,\pi/2}\circ R_{P,\pi/2}$.
}

\vfill
\part[3] Is there a spherical isometry $f\in\mbox{Isom}(S^2)$ such that $f\circ(R_{P,\pi/2},R_{Q,\pi/2})$ has no fixed points?\\
\textcolor{blue}{
Yes (I'm assuming the professor meant $f\circ(R_{P,\pi/2}\circ R_{Q,\pi/2})$). By above, we know that $R_{P,\pi/2}\circ R_{Q,\pi/2}$ is a rotation, which fixes only the center and its antipodal point. So we take $f$ to be the reflection across the set of equidistant points to the center and its antipodal. Then clearly, $f\circ(R_{P,\pi/2}\circ R_{Q,\pi/2})$ does not fix the center and its antipodal. Points on the line of reflection remain on the line of reflection, but are translated by $R_{P,\pi/2}\circ R_{Q,\pi/2}$, then don't move by $f$. The remaining points get reflected to the opposite side of the line of reflection, hence we conclude that $f\circ(R_{P,\pi/2}\circ R_{Q,\pi/2})$ has no fixed points.
}

\vfill


\end{parts}

\newpage
\addpoints

\addpoints
\question[18] ({\bf Hyperbolic distances and lines in $\mathbb{H}^2$}) Let $P=i,Q=3+4i\in\H^2$ be points in the hyperbolic upper-half plane $\H^2$. Solve the following parts:
\noaddpoints
\begin{parts}
\part[5] Show that $L=\{z\in\H^2:|z-4|^2=17\}\sse\H^2$ is the unique hyperbolic line through the points $P$ and $Q$.\\
\textcolor{blue}{
We construct the geodesic through $P,Q$. $\overline{PQ}$ has slope $1$, so its perpendicular bisector has slope $-1$, and passes through $\frac{P+Q}{2}=\frac32+\frac52i$. So the perpendicular bisector has equation $y-\frac52=-(x-\frac32)=-x+\frac32$. It intersects the $x$-axis at $y=0$, so $-\frac52=-x+\frac32\implies x=4$. Therefore, the center of the geodesic is $(4,0)$. The radius is $d_{\mathbb{E}^2}((4,0),(0,1))=\sqrt{17}$, and thus the geodesic through $P,Q$ is $\{(x,y)\in\H^2:(x-4)^2+y^2=17\}= \{z\in\H^2:|z-4|^2=17\}=L$.
}
\part[5] Compute the hyperbolic distance $d_{\H^2}(P,Q)$.\\
\textcolor{blue}{
We have $y=\sqrt{17-(x-4)^2},\frac{dy}{dx}=\frac{-(x-4)}{\sqrt{17-(x-4)^2}}$, and from $P$ to $Q$, we have $0\leq x\leq 3$, so
\begin{align*}
    d_{\H^2}(P,Q)&=\int_0^3 \frac{\sqrt{1+\frac{(x-4)^2}{17-(x-4)^2}}}{\sqrt{17-(x-4)^2}}dx=\int_0^3 \frac{\sqrt{\frac{17}{17-(x-4)^2}}}{\sqrt{17-(x-4)^2}}dx\\
    &=\int_0^3 \frac{\sqrt{17}}{17-(x-4)^2}dx
\end{align*}
}
\part[5] Find the unique hyperbolic line $M$ equidistant to $P$ and $Q$. \\
\textcolor{blue}{
First, we know from a theorem from class that reflection across $M$ exchanges $P,Q$. In particular, $M$ is perpendicular to the Euclidean line segment $\overline{PQ}$. By the construction of a circle, the radius from the center to any point on the circle forms an angle of $\pi/2$, and thus we may extend $\overline{PQ}$ to intersect the $x$-axis, from which we get that the center of $M$ is $(-1,0)$. Next, we find the radius. As $M$ has center $(-1,0)$, and must pass through some point on the segment $\overline{PQ}$, it follows that $M$ must pass through the $y$-axis as well. So, consider a point $y_0 i\in\H^2$ equidistant to $P,Q$. We've seen from class that $d_{\H^2}(P,y_0 i)=\ln(y_0)$, and we set $d_{\H^2}(Q,y_0 i)=\ln(y_0)$. Let's derive $d_{\H^2}(Q,y_0 i)$: The perpendicular bisector has slope $-\frac{3}{4-y_0}$, and passes through the midpoint $\frac{3+(4+y_0)i}{2}$, from which we get the equation of the perpendicular bisector is $y-\frac{4+y_0}{2}=-\frac{3}{4-y_0}(x-\frac32)$. When $y=0$, we get the center of this geodesic to be $x=\frac{(4+y_0)(4-y_0)}{6}+\frac32=\frac{16-y_0^2+9}{2}=\frac{25-y_0^2}{6}$. The radius is the Euclidean distance $d_{\mathbb{E}}(\frac{25-y_0^2}{6},3+4i)=\sqrt{\left(\frac{25-y_0^2-18}{6}\right)^2+16}=\sqrt{\left(\frac{7-y_0^2}{6}\right)^2+16}$. Thus, this geodesic has equation $\left(x-\frac{25-y_0^2}{6}\right)^2+y^2=\left(\frac{7-y_0^2}{6}\right)^2+16$. We have the following simplification of the distance formula for a geodesic with center $(\alpha,0)$ and radius $r$: \\
\begin{align*}
\int\frac{\sqrt{1+(dy/dx)^2}}{y}dx&=\int\frac{\sqrt{1+\frac{(-(x-\alpha))^2}{r^2-(x-\alpha)^2}}}{\sqrt{r^2-(x-\alpha)^2}}dx=\int\frac{\sqrt{\frac{r^2}{r^2-(x-\alpha)^2}}}{\sqrt{r^2-(x-\alpha)^2}}dx\\
&=\int\frac{r}{r^2-(x-\alpha)^2}dx=\int \frac{A}{r-(x-\alpha)}+\frac{B}{r+(x-\alpha)}dx\\
&\implies r=Ar+A(x-\alpha)+Br-B(x-\alpha)\\
&\implies Ar+Br=r, A(x-\alpha)-B(x-\alpha)=0\\
&\implies A+B=1\implies A=B=\frac12\\
&=\int \frac{\frac12}{r-(x-\alpha)}+\frac{\frac12}{r+(x-\alpha)}dx\\
&=\frac12\left(-\ln|r-(x-\alpha)|+\ln|r+(x-\alpha)| \right)=\frac12\ln \left|\frac{r+(x-\alpha)}{r-(x-\alpha)}\right|,
\end{align*}
So we get:
\begin{align*}
    d_{\H^2}(Q,y_0 i)&=\int_0^3 \frac{\sqrt{\left(\frac{7-y_0^2}{6}\right)^2+16}}{\left(\frac{7-y_0^2}{6}\right)^2+16-\left(x-\frac{25-y_0^2}{6}\right)^2}dx=\frac12 \ln\left|\frac{\sqrt{\left(\frac{7-y_0^2}{6}\right)^2+16}+\left(x-\frac{25-y_0^2}{6}\right)}{\sqrt{\left(\frac{7-y_0^2}{6}\right)^2+16}-\left(x-\frac{25-y_0^2}{6}\right)}\right|_0^3\\
    &=\frac12 \ln\left|\frac{\sqrt{\left(\frac{7-y_0^2}{6}\right)^2+16}+\left(\frac{y_0^2-7}{6}\right)}{\sqrt{\left(\frac{7-y_0^2}{6}\right)^2+16}-\left(\frac{y_0^2-7}{6}\right)}\right|-\frac12 \ln\left|\frac{\sqrt{\left(\frac{7-y_0^2}{6}\right)^2+16}-\frac{25-y_0^2}{6}}{\sqrt{\left(\frac{7-y_0^2}{6}\right)^2+16}+\frac{25-y_0^2}{6}}\right|.
\end{align*}
Setting this equal to $\ln(y_0)$, we can simplify this to
$$y_0^2=\frac{\sqrt{\left(\frac{7-y_0^2}{6}\right)^2+16}+\left(\frac{y_0^2-7}{6}\right)}{\sqrt{\left(\frac{7-y_0^2}{6}\right)^2+16}-\left(\frac{y_0^2-7}{6}\right)}\cdot\frac{\sqrt{\left(\frac{7-y_0^2}{6}\right)^2+16}+\frac{25-y_0^2}{6}}{\sqrt{\left(\frac{7-y_0^2}{6}\right)^2+16}-\frac{25-y_0^2}{6}}.$$
As this gets a bit too messy for me to TeX it, we shall skip the details. But one may solve this and get $y_0^2=7$, hence $y_0=\sqrt{7}$. Finally, we compute the radius of $M$ to be the Euclidean distance from the center $(-1,0)$ to the point $(0,\sqrt{7})$, which is $\sqrt{1+7}=\sqrt{8}$. Hence, $M=\{(x,y)\in\H^2:(x+1)^2+y^2=8\}$
.
%We know that $M$ passes through the midpoint $\frac{P+Q}{2}$. By a theorem from class, the reflection across $M$ swaps $P$ and $Q$, so $M$ must be perpendicular to $\overline{PQ}$. Now, by the construction of a circle, we know that any tangent line to a point on a circle is perpendicular to the radius to that point, and since the perpendicular bisector of $\overline{PQ}$ is tangent to $M$ at $\frac{P+Q}{2}$, we may extend $\overline{PQ}$ to intersect the $x$-axis to find the center $C=(-1,0)$ of $M$. Then, the radius is the Euclidean distance between $C$ and $\frac{P+Q}{2}$, which is $\frac52\cdot \sqrt{2}$ (we have a $45^\circ-45^\circ-90^\circ$ triangle with base $\frac52)$. Hence, $M=\{(x,y)\in\H^2:(x+1)^2+y^2=\frac{25}{2}\}$.
%\begin{figure}[htp]
%    \centering
%    \includegraphics[width=0.5\linewidth]{Midterm/set of equidistant points.png}
%\end{figure}
}
\part[3] Let $r_L,r_M\in\mbox{Isom}(\H^2)$ be the hyperbolic inversions along $L$ and $M$. Compute the image of $P$ and $Q$ under the composition $r_M\circ r_L$.\\
\textcolor{blue}{
By part (a), we know $P,Q\in L$, and therefore $r_L(P)=P,r_L(Q)=Q$. We also know by a theorem from class that $r_M$ swaps $P$ and $Q$, and it follows that $r_M\circ r_L(P)=r_M(P)=Q, r_M\circ r_L(Q)=r_M(Q)=P$.
}
\vfill
\end{parts}

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% 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[18] ({\bf Hyperbolic isometries in $\mathbb{H}^2$}) Consider the map $f,g:\H^2\lr\H^2$
$$f(z)= \frac{\bar{z} + 9}{3\bar{z} - 5}.$$

\noaddpoints
\begin{parts}
\part[5] Show that $f$ is a hyperbolic isometry and it has no fixed points.\\
\textcolor{blue}{
By a theorem from class, we know that orientation-reversing hyperbolic isometries are of the form $\overline{f}(z)=\frac{-a\overline{z}+b}{-c\overline{z}+d}$, with $ad-bc=1$. So it suffices to write $f$ in this form. We normalize $f$ by dividing by $\sqrt{(-1)\cdot(-5)-(-3)\cdot 9}=\sqrt{5+27}=\sqrt{32}$. So $\displaystyle f(z)=\frac{\overline{z}+9}{3\overline{z}-5}=\frac{\frac{1}{\sqrt{32}}(\overline{z}+9)}{\frac{1}{\sqrt{32}}(3\overline{z}-5)}=\frac{-\frac{-1}{\sqrt{32}}\overline{z}+\frac{9}{\sqrt{32}}}{-\frac{-3}{\sqrt{32}}\overline{z}-\frac{5}{\sqrt{32}}}$, and we check: $\frac{-1}{\sqrt{32}}\cdot \frac{-5}{\sqrt{32}}-\frac{-3}{\sqrt{32}}\cdot \frac{9}{\sqrt{32}}=\frac{5+27}{32}=1$, and thus $f$ is a hyperbolic isometry. Next, we assume that $z_0=\frac{\overline{z_0}+9}{3\overline{z_0}-5}$ for some $z_0=x+yi\in\H^2$. Then, we have 
\begin{align*}
    x+9-yi&=(x+yi)(3x-5-3yi)\\
    &=3x^2-5x-3xyi+3xyi-5yi+3y^2\\
    \implies & 3x^2-6x-9+3y^2=0, -4y=0.
\end{align*}
The latter equation implies $y=0$, in which case $z_0\not\in\H^2$. Therefore, $f$ has no fixed points.
}
\vfill
\part[5] Let $g:\H^2\lr\H^2$ be a hyperbolic isometry such that
$$g\left(-\frac{21}{17} + \frac{16}{17}i\right)=i,\quad g\left(-\frac{33}{61} + \frac{64}{61}i\right)=2i,\quad g\left(-\frac{17}{13} + \frac{32}{13}i\right)=1+i.$$
Determine the image of the point $2+3i$ under the composition $f\circ g$.\\
\textcolor{blue}{
We first do a cheeky observation:
\begin{align*}
    &f(i)=\frac{-i+9}{-3i-5}=\frac{-i+9}{-5-3i}\cdot \frac{-5+3i}{-5+3i}=\frac{5i+3-45+27i}{34}=-\frac{21}{17}+\frac{16}{17}i\\
    &f(2i)=\frac{-2i+9}{-6i-5}=\frac{-2i+9}{-5-6i}\cdot \frac{-5+6i}{-5+6i}=\frac{10i+12-45+54i}{25+36}=-\frac{33}{61}+\frac{64}{61}i\\
    &f(1+i)=\frac{1-i+9}{3-3i-5}=\frac{10-i}{-2-3i}\cdot \frac{-2+3i}{-2+3i}=\frac{-20+30i+2i+3}{4+9}=-\frac{17}{13}+\frac{32}{13}i.
\end{align*}
By a theorem from class, hyperbolic isometries are uniquely determined by the images of three noncollinear points, thus it follows that $g=f^{-1}$. Therefore, $f\circ g(2+3i)=2+3i$.
}
\vfill
\part[5] Find a hyperbolic line $L\sse\H^2$ such that $f(L)=L$.\\
\textcolor{blue}{
We look for the two fixed points on $\partial \H^2$, continuing from the computations in part (a). We have $0=3x^2-6x-9\implies 0=x^2-2x-3=(x-3)(x+1)\implies x=-1,3$. So the hyperbolic line invariant under $f$ has endpoints at $-1,3$, and thus has center $\frac{-1+3}{2}=1$, radius $2$. Therefore, $L=\{(x,y)\in\H^2:(x-1)^2+y^2=4\}$.
}
\vfill
\part[3] Show that there exists no $n\in\N$ such that $f^n=\mbox{id}$.\\
\textcolor{blue}{
Since $f$ is orientation-reversing, we know that any product of odd numbers of $f$ will also be orientation-reversing, hence could not be id. So to show that there exists no $n\in\N$ such that $f^n=$ id, it suffices to check for even powers of $f$. We compute $f^2$:
\begin{align*}
    f^2(z)&=\frac{\overline{\frac{\overline{z}+9}{3\overline{z}-5}}+9}{3\overline{\frac{\overline{z}+9}{3\overline{z}-5}}-5}=\frac{\frac{z+9}{3z-5}+\frac{27z-45}{3z-5}}{\frac{3z+27}{3z-5}+\frac{-15z+25}{3z-5}}=\frac{28z-36}{-12z+52}=\frac{7z-9}{-3z+13}\\
    &\longleftrightarrow \begin{bmatrix}
        7 & -9\\
        -3 & 13
    \end{bmatrix}\sim \begin{bmatrix}
        \frac{7}{8} & -\frac98\\
        -\frac38 & \frac{13}{8}
    \end{bmatrix}.
\end{align*}
Observe that given any two matrices $\begin{bmatrix}
    a&-b\\-c&d
\end{bmatrix},\begin{bmatrix}
    e&-f\\-g&h
\end{bmatrix}$ with $a,b,c,d,e,f,g>0$, we have $\begin{bmatrix}
    a&-b\\-c&d
\end{bmatrix}\begin{bmatrix}
    e&-f\\-g&h
\end{bmatrix}=\begin{bmatrix}
    ae+bg&-af-bh\\-ce-dg&cf+dh
\end{bmatrix}$, with entries $ae+bg,cf+dh>0;-af-bh,-ce-dg<0$. In particular, the top right and bottom left entry will never be $0$, so the product of such matrices is never the identity. Since $f^2$ is of this form, it follows that $f^{2k}=(f^2)^k\neq \begin{bmatrix}
    1&0\\0&1
\end{bmatrix}$ for any $k\in\N$. We conclude that there exists no $n\in\N$ such that $f^n=$ id.
}
\vfill
\end{parts}

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% 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[10] 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] \small Two different lines in the twisted cylinder cannot intersect only at two points.\\
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill
	
	\part[2] Any orientation-reversing hyperbolic isometry must have a fixed point.\\
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill
	
	\part[2] A spherical isometry is determined by the image of three distinct points.\\
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill
	
	\part[2] For any two points $P,Q\in S^2$, there exists a unique line between them.\\
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill
	
	\part[2] Any hyperbolic triangle with vertices in $\H^2$ must have area less than $\pi$.\\
	
	\textcolor{blue}{(1) True.} \qquad \qquad \qquad \qquad (2) False.
	\vfill

    \part[2] There exists a regular hyperbolic hexagon whose angles are all $\pi/2$.
	
	\textcolor{blue}{(1) True.} \qquad \qquad \qquad \qquad (2) False.
	\vfill

    \part[2] For any two distinct hyperbolic lines $L,M\sse\H^2$, there exists a hyperbolic isometry $f:\H^2\lr\H^2$ with $f(L)=M$.
	
	\textcolor{blue}{(1) True.} \qquad \qquad \qquad \qquad (2) False.
	\vfill

    \part[2] If $L\sse\R^2$ is the line equidistant to $P,Q\in\R^2$, then any isometry fixing the points of $L$ must send $P$ to $Q$.
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill

    \part[2] If two hyperbolic lines $L,M$ do not intersect, then there exists a unique hyperbolic line that is perpendicular to both $L$ and $M$.
	
	(1) True. \qquad \qquad \qquad \qquad \textcolor{blue}{(2) False.}
	\vfill
	
	
\end{parts}

\end{questions}
\end{document}
