trisection of angleTrisectionOfAngle2007-06-18 11:14:572007-06-23 21:34:11Algorithm\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{pstricks}
\usepackage{psfrag}
\usepackage{graphicx}
\usepackage{amsthm}
\usepackage{xypic}
\PMlinkescapeword{label}
\PMlinkescapeword{measure}
\PMlinkescapeword{ruler}
Given an angle of \PMlinkname{measure}{AngleMeasure} $\alpha$ such that $0<\alpha \le \frac{\pi}{2}$, one can construct an angle of measure $\frac{\alpha}{3}$ using a compass and a \PMlinkname{ruler}{MarkedRuler} with one mark on it as follows:
\begin{enumerate}
\item Construct a circle $c$ with the \PMlinkname{vertex}{Vertex5} $O$ of the angle as its center. Label the intersections of this circle with the rays of the angle as $A$ and $B$. Mark the length $OB$ on the ruler.
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\psline{->}(0,0)(3,0)
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\rput[a](0,-0.3){$O$}
\rput[a](2.1,-0.3){$A$}
\rput[a](1.7,1.4){$B$}
\rput[a](0,-2.2){$c$}
\end{pspicture}
\end{center}
\item Draw the ray $\overrightarrow{AO}$.
\begin{center}
\begin{pspicture}(-5,-3)(3,3)
\rput[l](-5,0){.}
\rput[r](3,2){.}
\psline{->}(0,0)(3,0)
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\psline[linecolor=blue]{->}(2,0)(-5,0)
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\rput[a](0,-0.3){$O$}
\rput[a](2.1,-0.3){$A$}
\rput[a](1.7,1.4){$B$}
\rput[a](0,-2.2){$c$}
\end{pspicture}
\end{center}
\item Use the marked ruler to determine $C\in c$ and $D\in \overrightarrow{AO}$ such that $CD=OB$ and $B$, $C$, and $D$ are collinear. Draw the line segment $\overline{BD}$. Then the angle measure of $\angle CDO$ is $\frac{\alpha}{3}$. (The line segment $\overline{OC}$ is drawn in red. Having this line segment drawn is useful for reference purposes for the justification of the construction.)
\begin{center}
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\rput[l](-5,0){.}
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\psdots(0,0)(2,0)(1.6641,1.1094)(-1.9617,0.3895)(-3.923445,0)
\rput[a](0,-0.3){$O$}
\rput[a](2.1,-0.3){$A$}
\rput[a](1.7,1.4){$B$}
\rput[a](0,-2.2){$c$}
\rput[r](-2,0.6){$C$}
\rput[a](-3.923445,-0.3){$D$}
\end{pspicture}
\end{center}
\end{enumerate}
Let $m$ denote the measure of an angle. Then this construction is justified by the following:
\begin{itemize}
\item Since $\angle AOB$ is an exterior angle of $\triangle BOD$, we have that $m(\angle AOB)=m(\angle OBD)+m(\angle ODB)$;
\item Since $OC=OB=CD$, we have that $\triangle BOC$ and $\triangle OCD$ are isosceles triangles;
\item Since the angles of an isosceles triangle are congruent, $m(\angle OBC)=m(\angle OCB)$ and $m(\angle COD)=m(\angle CDO)$;
\item Since $\angle OCB$ is an exterior angle of $\triangle OCD$, we have that $m(\angle OCB)=m(\angle COD)+m(\angle CDO)$;
\item Note that $\angle OBC=\angle OBD$ and $\angle ODB=\angle CDO$;
\item Thus,
\begin{center}
$\begin{array}{rl}
\alpha & =m(\angle AOB) \\
& =m(\angle OBD)+m(\angle ODB) \\
& =m(\angle OBC)+m(\angle CDO) \\
& =m(\angle OCB)+m(\angle CDO) \\
& =m(\angle COD)+m(\angle CDO)+m(\angle CDO) \\
& =3m(\angle CDO). \end{array}$
\end{center}
\end{itemize}
Note that, since angles of measure $\frac{\pi}{6}$, $\frac{\pi}{3}$, and $\frac{\pi}{2}$ are constructible using compass and straightedge, this procedure can be extended to trisect any angle of measure $\beta$ such that $0<\beta\le 2\pi$:
\begin{itemize}
\item If $0<\beta\le\frac{\pi}{2}$, then use the construction given above.
\item If $\frac{\pi}{2}<\beta\le\pi$, then trisect an angle of measure $\beta-\frac{\pi}{2}$ and add on an angle of measure $\frac{\pi}{6}$ to the result.
\item If $\pi<\beta\le\frac{3\pi}{2}$, then trisect an angle of measure $\beta-\pi$ and add on an angle of measure $\frac{\pi}{3}$ to the result.
\item If $\frac{3\pi}{2}<\beta\le 2\pi$, then trisect an angle of measure $\beta-\frac{3\pi}{2}$ and add on an angle of measure $\frac{\pi}{2}$ to the result.
\end{itemize}
This construction is attributed to Archimedes.
\begin{thebibliography}{9}
\bibitem{unclejoe} Rotman, Joseph J. {\em A First Course in Abstract Algebra}. Upper Saddle River, NJ: Prentice-Hall, 1996.
\end{thebibliography}