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\title{log(m) meeting 8}
%\date{\today}
\lhead{ Log(M): Origami on lattices \\}
\rhead{ Meeting 8 \\ 15 March 2019}
\begin{document}


\section{Logistics}

\ul{Next meeting}:  Friday March 22, 9:30 - 11am
\medskip

    
\bigskip
{Expectations} for next meeting: 
\begin{itemize}

    
    
    \item mountain / valley labellings
    for hexagon crease pattern
    
    \textbf{Problem 9}: 
    In a diagram with $6$ creases coming together at a vertex at equal angles,
the unit null cone near the flat space is two disjoint spheres.
We consider cutting up  these spheres into regions according to mountain/valley labellings.
\begin{enumerate}[(a)]
    \item How many zero-dimensional, one-dimensional, and two-dimensional regions are there on each sphere?
    
    \item What shapes are the two-dimensional regions? 
    How do they fit together? (i.e. draw out a diagram)
\end{enumerate}
    
    \item  area formula for spherical $n$-gons
    
    \textbf{Problem 10}: For a spherical $n$-gon with internal angles $\alpha_1,\ldots,\alpha_n$, show that
    \[ \text{Area} = \sum_{i=1}^n \alpha_i - (n-2)\pi \]
    where area is measured on the surface of a unit sphere.
   Equivalently, for an $n$-gon
    \[ \text{Area} = 2\pi - \sum_{i=1}^n (\pi - \alpha_i) .\]
     You may assume the triangle area formula $\text{Area} = (\alpha_1 + \alpha_2 + \alpha_3) - \pi$.
    
    
    \item angle constraints on hexagon crease patter
    
    \textbf{Problem 11}:
    Suppose we want to fix three consecutive creases on a hexagon at fold angles
    $\theta_1$, $\theta_2$, and $\theta_3$. 
    (By ``fold angle'' we mean flat $\Leftrightarrow \, \theta_i=0$.)
    \begin{enumerate}[(a)]
    \item When is this possible, and when is this impossible?
    \item When this is possible, what are the values of angles $\theta_4,\, \theta_5,\, \theta_6$?
    Note: there may be more than one answer. How many answers are possible?
    \end{enumerate}
    For example, 
    \begin{itemize}
    \item if $\theta_1 = \theta_2 = \theta_3=0$,
    then this is possible and we must have $\theta_4 = \theta_5 = \theta_6=0$.
    
    \item if $\theta_1 = \theta_2 = 0$ and $\theta_3 = \pi$,
    then this is possible and we must have
    $\theta_4 = \theta_5 = 0$ and $\theta_6 = \pi$.
    \end{itemize}
    
    \item \textbf{[Writing]} 
    Write up notes for this meeting, and
    continue writing up relevant discussion from this week in draft of final report 
    

\end{itemize}




\end{document}