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\begin{document}


\title{Equilateral Triangles}

\maketitle



\begin{figure}
\includegraphics{napol1.gif} 


\caption{\label{ET}}
\end{figure}

\textbf{Phil} Today we prove the following theorem.  If we
construct equilateral triangles on the sides of any triangle (all
outward or all inward), the centers of those equilateral
triangles themselves form an equilateral triangle, as illustrated
above.  

\textbf{Stephan} If the original triangle is an
equilateral triangle, I think I can easily prove it.

\textbf{Phil}  Of course you can. The figure in that case is
completely symmetric.  So no one will doubt the theorem is true.  


\textbf{Jessica}  Since a triangle is equilateral if and only if
it has the equal three angles, I want to try first if I can prove
that the traiangle $LMN$ has three equal angles. For that
purpose, I draw lines from $A$ to $L$ and $N$, lines from $B$ to
$L$ and $M$, and
lines from $C$ to $M$ and $N$ as below. 

[formulates some equalities among angles.]

\begin{figure}
\includegraphics{napol2.gif} 


\caption{\label{ET}}
\end{figure}

\textbf{Jessica}  No I can't.  Too many angles are involved for
the angle L, M, and N, and I cannot get the desired equalities.  

\textbf{Stephan}  Probably we must calculate the lengths of  LM
MN, and NL from the data about the original triangle ABC and
prove these lengths are equal.

\textbf{Phil}  Yes, that is a safe approach.

\textbf{Stephan}  Since the angle $\angle MAC$ is $30^{\circ }$,
I can claculate the length of $AM$ by drawing a perpendicular 
$MQ$ to $AC$.  Then clearly
$$
|AM| = 1/\sqrt{3}\times |AC|.       
$$
Similarly
$$
|AN| = 1/\sqrt{3}\times |AB|.
$$

\begin{figure}
\includegraphics{napol3.gif} 

\caption{\label{ET}}
\end{figure}

\textbf{Stephan}  Now, the triangle MAN is determined, since we
also have
$$
     \angle MAN = \angle A + 60^{\circ }.
$$

\textbf{Stephan}  To calculate the length $MN$ I must apply the
law of cosines. First, let
$$
     a = |BC|, \quad b = |CA|, \quad c = |AB|.
$$
Then 
$$
     3|MN|^{2} = b^{2} + c^{2} -
2bc\,\mathrm{cos}(\angle A + 60^{\circ }).
$$    
Similarly,
$$
     3|NL|^{2} = c^{2} + a^{2} -
2ca\,\mathrm{cos}(\angle B + 60^{\circ }).
$$
and
$$
     3|LM|^{2} = a^{2} + b^{2} -
2ab\,\mathrm{cos}(\angle C + 60^{\circ }).
$$
We must prove all these three values are equal for the triangle
$ABC$. But how? Can any exert on trigonometry help me?. 

\textbf{Jessica} I'm the last one you can ask for help on that
matter.  I must find some purely geometric solution. [Thinks for
a couple of minutes.]

\textbf{Jessica}  I think I've got it.  If we 
extend both $AM$ and $AN$ to their $\sqrt{3}$ times to create a
new triangle $M'AN'$ and rotate it counterclockwise by $30^{\circ
}$ around $A$, we get the triangle $TAB$ in the following figure. 
Similarly, if we extend both $CM$ and $CL$ to their $\sqrt{3}$
times to create a new triangle $M'CL'$ and rotate it clockwise by
$30^{\circ }$ around $C$, we get the triangle $TCB$. That means
$|MN|$ is equal to $|LM|$.  Similarly we can prove $|NN| = |NL|$. 
That's it.

\textbf{Phil}  That is great.  In this problem, we have learned
we'd sometmes better not know trigonometry to get an elegant
geometric proof.

\vspace{0.3cm}
{\begin{figure}
{\centering \includegraphics{napol4.gif} \par}


\caption{\label{Napol4}}
\end{figure}\par}
\vspace{0.3cm}



\end{document}
