% method: 
% mpost example ; latex conwayfig ; make conwayfig.ps ; gv conwayfig.ps
\documentclass{article}
\usepackage{epsf}
\begin{document}
%\input{psfig.tex}
\newcommand{\dofigure}[3]{\begin{figure}[htbp]\[\epsfbox{#2}\]\caption[a]{#3}\label{fig#1}\end{figure}}
\section{Squares}
\dofigure{2}{example.2}{}

\dofigure{3}{example.3}{}

\newpage
\section*{Morley's theorem}
%\subsection*{Proof by John H Conway and }
%\subsubsection*{Statement of the theorem }
 Take a triangle, and trisect each of its angles.
\dofigure{5}{example.5}{}

 Extend the trisections until they meet at three
 points like this.
\dofigure{6}{example.6}{}

 These three points form an equilateral triangle. 
\dofigure{7}{example.7}{}

 And this is true for any starting 
triangle -- here's an obtuse triangle, for example.
\dofigure{8}{example.8}{}

%\subsubsection{Proof}
\subsection*{Proof by John H Conway and }
 We can prove Morley's theorem simply as follows.
 First, please tell me the three angles $A$, $B$, $C$,
 of your original triangle. Thank you.
 Here's the plan.
 I'm going to start from an equilateral triangle of some size
 and build up a sequence of other triangles
 around it, and glue them together to create a triangle 
 that has angles $A$, $B$, and $C$, just like yours; 
 so for some choice of the size of the equilateral 
 triangle, my construction would  exactly
 reproduce your original triangle; and the method of construction
 will prove that if you trisect your triangle's angles,
 you'll find my equilateral triangle in the middle.

 Here are the six triangular  pieces that we will build around
 the equilateral triangle. 
\dofigure{4}{example.4}{}

 The above picture  looks like a shattered version of figure
 \ref{fig3}, and indeed we'll in due course glue the pieces together
 to create figure \ref{fig3}, but to understand the proof correctly,
 you must think of the six new triangles as pieces that we are going
 to {\em define\/}, starting from my equilateral triangle, with the
 help of the supplied values of $A$, $B$, and $C$. Figure \ref{fig3}
 is now our destination, not our starting point.

 We define $\alpha = A/3$,
	$\beta=B/3$, and $\gamma=C/3$. Since $A+B+C=180$, 
 $\alpha+\beta+\gamma= 60$.
 We introduce a piece of notation for angles,  defining
 $\theta^{+}$ to denote $\theta+60$ and  $\theta^{++}$ to denote $\theta+120$.
% \degrees$.
 So, for example, the three interior angles in the equilateral  triangle
 may be marked $0^{+}$.  
 We construct the six new triangles as follows.
 First, each  
% acute 
 triangle that abutts onto the 
 equilateral triangle is defined by fixing the length of one 
 side to that of the equilateral,  and fixing the three angles to 
 be, as shown below, $\{  \alpha,\beta^+, \gamma^+ \}$, $\{  \alpha^+,\beta, \gamma^+ \}$,
 and  $\{  \alpha^+,\beta^+, \gamma \}$. 
\dofigure{9}{example.9}{}

 [Each of these triples  sums to 180, as it 
 must to define a triangle.]  
%The third angle in each of these triangles can be computed;
% the answers are respectively $\alpha$, $\beta$, and $\gamma$,
% as shown in the next figure. 
 Next we make each obtuse triangle by fixing two of its sides to equal the 
 sides in the adjacent  triangles, and by fixing its obtuse angle to be 
 $\alpha^{++}$, 
 $\beta^{++}$, or
 $\gamma^{++}$, respectively.
\dofigure{10}{example.10}{}

 Two tasks now remain. We must confirm that the six pieces thus defined 
 do indeed snugly fit around my equilateral triangle; 
 and we must confirm that the six angles of the vertices meeting at the points 
 $A$, $B$, and $C$ -- currently marked `?' in 
 figure \ref{fig10} -- are all equal respectively to $\alpha$, $\beta$, and $\gamma$,
 as asserted below.
\dofigure{11}{example.11}{}
 We confirm that the fit is snug by checking that all 
 adjacent edges have matching lengths, and checking that  the angles 
 around any one internal vertex  sum to 360 degrees.
 The sum around a typical internal vertex is $\alpha^+ + \beta^{++} + \gamma^+ + 0^+
 = \alpha + \beta + \gamma + 300 = 360$.
%; 
% there are five pluses there, and $\alpha + \beta + \gamma = 60$, so the total 
% is indeed 360.
% `$+$'

 We pin down the values of the unknown angles in the obtuse triangles by adding a couple 
 more lines to our diagram.
\dofigure{12}{example.12}{}

\newpage
\section{Knots}
\dofigure{20}{example.20}{}
\dofigure{21}{example.21}{}

\end{document}