# trisection of angle

Given an angle of measure (http://planetmath.org/AngleMeasure) $\alpha$ such that $0<\alpha\leq\frac{\pi}{2}$, one can construct an angle of measure $\frac{\alpha}{3}$ using a compass and a ruler (http://planetmath.org/MarkedRuler) with one mark on it as follows:

1. 1.

Construct a circle $c$ with the vertex (http://planetmath.org/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.

2. 2.

Draw the ray $\overrightarrow{AO}$.

3. 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,

$\begin{array}[]{rl}\alpha&=m(\angle AOB)\\ &=m(\angle OBD)+m(\angle ODB)\\ &=m(\angle OBC)+m(\angle CDO)\\ &=m(\angle OCB)+m(\angle CDO)\\ &=m(\angle COD)+m(\angle CDO)+m(\angle CDO)\\ &=3m(\angle CDO).\end{array}$

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

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

• If $\frac{\pi}{2}<\beta\leq\pi$, then trisect an angle of measure $\beta-\frac{\pi}{2}$ and add on an angle of measure $\frac{\pi}{6}$ to the result.

• If $\pi<\beta\leq\frac{3\pi}{2}$, then trisect an angle of measure $\beta-\pi$ and add on an angle of measure $\frac{\pi}{3}$ to the result.

• If $\frac{3\pi}{2}<\beta\leq 2\pi$, 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.

## References

• 1 Rotman, Joseph J. A First Course in Abstract Algebra. Upper Saddle River, NJ: Prentice-Hall, 1996.
Title trisection of angle TrisectionOfAngle 2013-03-22 17:16:35 2013-03-22 17:16:35 Wkbj79 (1863) Wkbj79 (1863) 11 Wkbj79 (1863) Algorithm msc 01A20 msc 51M15 VariantsOnCompassAndStraightedgeConstructions