\documentclass[12pt]{article}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\usepackage{geometry}

%TCIDATA{OutputFilter=LATEX.DLL}
%TCIDATA{Version=5.00.0.2570}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{Created=Wednesday, September 21, 2005 19:23:38}
%TCIDATA{LastRevised=Sunday, January 01, 2006 15:56:10}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{<META NAME="DocumentShell" CONTENT="Standard LaTeX\Blank - Standard LaTeX Article">}
%TCIDATA{CSTFile=40 LaTeX article.cst}
%TCIDATA{ComputeDefs=
%$f\left( -10\right) +f\left( 15\right) =\left( 12\left( -10\right)
%-42\right) =\allowbreak -162$
%}


\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\geometry{left=0.6in,right=0.6in,top=0.6in,bottom=0.6in}
\input{tcilatex}

\begin{document}


\begin{center}
\bigskip {\Large Puzzle 12 - SOLUTION}
\end{center}

\bigskip

There is a lot of different proofs of the Pythagorean Theorem. This is one I
really like. Consider the right triangle shown on the picture below. \FRAME{%
dtbpF}{1.9545in}{1.0957in}{0pt}{}{}{insert1.bmp}{\special{language
"Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display
"USEDEF";valid_file "F";width 1.9545in;height 1.0957in;depth
0pt;original-width 1.9268in;original-height 1.0672in;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename 'insert1.bmp';file-properties
"XNPEU";}}

We used four identical triangles to create the figure shown on the picture
below.\FRAME{dtbpF}{1.9882in}{2.0349in}{0pt}{}{}{insert2.bmp}{\special%
{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "USEDEF";valid_file "F";width 1.9882in;height 2.0349in;depth
0pt;original-width 1.9597in;original-height 2.0072in;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename 'insert2.bmp';file-properties
"XNPEU";}}

\begin{enumerate}
\item Find the area of the big square in terms of $c$.

Solution: $c^{2}$

\item Find the area of the small rectangle in the middle in terms of $a$ and 
$b$.

Solution: The rectangle in the middle is a square with sides $b-a$ units
long. Thus the area is $\left( b-a\right) ^{2}$

\item Find the area of the big square as the sum of 5 areas: four triangles
and the small rectangle in the middle.

Solution: 
\[
4\cdot \dfrac{ab}{2}+\left( b-a\right) ^{2}=2ab+b^{2}-2ab+a^{2}=a^{2}+b^{2}
\]

So the area is:%
\begin{eqnarray*}
A &=&c^{2}\text{ \ and} \\
A &=&a^{2}+b^{2}\text{ \ \ which means} \\
c^{2} &=&a^{2}+b^{2}
\end{eqnarray*}

This completes the proof of the Pythagorean Theorem.
\end{enumerate}

\bigskip

\end{document}