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\title{Tiling of Sphere by Congruent Pentagons}
\author{Min Yan}
\date{}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}
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\newtheorem{proposition}[theorem]{Proposition}
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\maketitle
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{Numerical}
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}
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