\documentclass[12pt]{article}
\usepackage{amssymb,amsmath,amsthm}

\title{Tiling of Sphere by Congruent Pentagons}
\author{Min Yan}
\date{}


\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}

\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}


\theoremstyle{remark}
\newtheorem{exercise}{Exercise}
\newtheorem{problem}{Problem}
\newtheorem{anx}{Answer}

\numberwithin{equation}{section}


\begin{document}

\maketitle

\begin{anx}
Consider $x+y=z$.
\end{anx}

\begin{theorem}
We have $1+2=3$.
\end{theorem}

\begin{definition}
$x$ is positive if $x>0$.
\end{definition}


We consider tilings of the sphere by congruent pentagons. 
The basic example is dodecahedron, which 
has $12$ pentagonal tiles.

To simplify the problem, we assume the tiling is 
edge-to-edge and all vertices have 
degree $\ge 3$.

\section{Write Mathematics}

Mathematics (inline) should inside dollar sign $(\sin x)'=\cos x$. $f'(x)=e^x\tan 2x$. not in dollar sign: f'(x)=e x tan 2x

\bigskip

\bigskip

More inline math: Suppose $x>0$ and $y>0$. Then $xy>0$.

\medskip

Small gap: $x      y$, $x\, y$, $x\; y$, $x\;\; y$, $x\quad y$.

\vspace{2cm}

Display mathematics 
\[
(\sin x)'=\cos x
\]

$$ (\sin x)'=\cos x $$

$$ 
(\sin x)'=\cos x 
$$

\begin{equation}
(\sin x)'=\cos x
\end{equation}

\begin{equation*}
(\sin x)'=\cos x
\end{equation*}

Display several maths

\[
\frac{d}{dx}\sin x=\cos x,\quad
\frac{d}{dx}\cos x=-\sin x,\quad\quad
\dfrac{de^x}{dx}=e^x.
\]

\begin{align}
\frac{d}{dx}\sin x=\cos x, \\ % <- \\ for going to next line
\frac{d}{dx}\cos x=-\sin x, \\
\frac{de^x}{dx}=e^x.
\end{align}

\begin{align*}
\frac{d}{dx}\sin x &=\cos x, \\ % parts in one line separated by &
\frac{d}{dx}\cos x &=-\sin x, \\
&\vdots \\
\frac{de^x}{dx} &=e^x.
\end{align*}


More math codes
\[
\frac{1}{n+\frac{2}{m}},\quad
\int_0^af(x^2)dx,\quad  % <- *_* for subscript, *^* for superscript
\sqrt{2+i},\quad
\sqrt[3]{2+i},\quad
\log_2(xy)=\log_2x+\log_2y
\]

\[
2^{\log_2x}=x,\quad
\log_2 2^x=x.
\]


\subsection{Numerical}


List with number

\begin{enumerate}
\item The first item
\begin{enumerate}
\item Nested item 1
\item Nested item 2
\end{enumerate}
\item The second item
\item The third etc \ldots
\end{enumerate}

List with bullet

\begin{itemize}
\item item 1
\item item 2
\end{itemize}


\subsubsection{Funny}

\subsection*{No Number Section}

\subsection{Number Back Section}

\section{Bad Mathematics}

\section{Forget Mathematics}

Let $v,e,f$ be the numbers of vertices, edges and faces in a spherical pentagonal tiling. Then we have 
\[
v-e+f=2,\quad
5f=2e.
\]
Let $v_i$ be the number of vertices of degree $i$. Then
\begin{align*}
v &=v_3+v_4+v_5+v_6+\cdots, \\
2e &=3v_3+4v_4+5v_5+6v_6+\cdots.
\end{align*}

\begin{exercise}
Derive similar equalities for tiling of sphere by quadrilaterals. What about triangles? What about tiling of torus?
\end{exercise}


We conclude that vast majority of vertices have degree $3$, and $f$ must be an even number $\ge 12$. We call vertices of degree $\ge 4$ {\em high degree} vertices.

\begin{theorem}\label{th1}
A pentagonal tiling of the sphere cannot have only one high degree vertex. If the tiling has exactly two high degree vertices, then the tiling is one of five families of {\em earth map tilings}.
\end{theorem}



\section{Edge and Angle}

As suggested by Theorem \ref{th1}, we may consider putting edges and angles into three possible neighborhood tilings, such that all pentagons are congruent. If we ignore the angles and try to achieve {\em egde congruence}, then we get the following.

\begin{proposition}\label{th3}
In a tiling of the sphere by (edge) congruent pentagons, the edges of the pentagon must be one of the five kinds: $a^5,a^4b,a^3b^2,a^3bc,a^2b^2c$, where $a,b,c$ are distinct. Moreover, the edges are arranged in one of six ways.
\end{proposition}


The basic fact about angles is that the sum of angles ({\em angle sum}) at a vertex is $2\pi$. Moreover, we have the angle sum equation for the pentagon ($4\pi$ is the area of the sphere)
\begin{equation}\label{asum}
\sum(\text{five angles in pentagon})-3\pi
=\text{Area}(\text{pentagon})
=\frac{4\pi}{f}.
\end{equation}
It is also possible to count the distribution of angles in an {\em angle congruent} tiling.


\begin{thebibliography}{1}

\bibitem{ay}
Y.~Akama, M.~Yan.
\newblock On deformed dodecahedron tiling.
\newblock {\em preprint}, arXiv:1403.6907, 2014.



\end{thebibliography}



\end{document}



