trisection of angle

Given an angle of measure $\alpha$ such that $0<\alpha \le \frac{\pi}{2}$ , one can construct an angle of measure $\frac{\alpha}{3}$ using a compass and a ruler with one mark on it as follows:

1. Construct a circle $c$ with the vertex $O$ 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.
   
   
   
2. Draw the ray $\overrightarrow{AO}$ .
   
   
   
3. 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.)
   
   
   

Let $m$ denote the measure of an angle. Then this construction is justified by the following:

• Since $\angle AOB$ is an exterior angle of $\triangle BOD$ , we have that $m(\angle AOB)=m(\angle OBD)+m(\angle ODB)$ ;
• Since $OC=OB=CD$ , we have that $\triangle BOC$ and $\triangle OCD$ are isosceles triangles;
• Since the angles of an isosceles triangle are congruent, $m(\angle OBC)=m(\angle OCB)$ and $m(\angle COD)=m(\angle CDO)$ ;
• Since $\angle OCB$ is an exterior angle of $\triangle OCD$ , we have that $m(\angle OCB)=m(\angle COD)+m(\angle CDO)$ ;
• Note that $\angle OBC=\angle OBD$ and $\angle ODB=\angle CDO$ ;
• Thus,
  

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 $0<\beta\le 2\pi$ :

• If $0<\beta\le\frac{\pi}{2}$ , then use the construction given above.
• 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.
• 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.
• 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.

This construction is attributed to Archimedes.

Bibliography

1

Rotman, Joseph J. A First Course in Abstract Algebra. Upper Saddle River, NJ: Prentice-Hall, 1996.