Trisection of an arbitrary angle using only a straightedge and compass is known as one of the "three construction problems of antiquity." As the ancient Greeks were able to bisect an angle and divide a segment into any number of congruent segments, it would seem that a method of trisecting an angle should also be possible. They were unable to perform this construction and suspected that it was not possible, but it was not until 1837 that Pierre Wantzel (1814–1848) used algebra to prove it to be impossible. However, several mathematicians from ancient Greece discovered methods to trisect an arbitrary angle when mechanical means are used. Mechanical means, such as marking a straightedge for measurement, are techniques that do not follow normal construction rules for a straightedge and compass. This article will discuss four mechanical solutions to the angle trisection problem, and GeoGebra applets will be used to illustrate the processes.

One solution involves the quadratrix, a curve discovered by Hippias (c. 420 BCE). Hippias was one of the first sophists; that is, he was one of the first persons to be paid for his teaching. Dinostratus (c. 350 BCE) is often credited as the first person to use the quadratrix to square a circle, the property for which the quadratrix was named.

The only mathematical achievement attributed to Hippias is the quadratrix, a locus of points formed by the intersection of a horizontal segment and a ray as each moves at a constant rate from its initial position to its final position. The movements are coordinated to begin and end at the same time. The formation of the quadratrix is illustrated in the Quadratrix applet:**Quadratrix applet:** Line segment \(GH\) moves from initial position \(\overline{EF}\) to final position \(\overline{CD}\) as point \(I\) moves from initial position \(E\) to final position \(D\). Notice the curve traced by point \(J\), the intersection of \(\overline{GH}\) and \(\overline{CI}\). This is the quadratrix. To perform these actions, click the Play (arrow) button in the lower left corner of the applet window; to stop the animation, click this button (now a Pause button) again.

To trisect \(\angle{ICD}\), line segment \({NO}\) has been constructed so that \(NC = \frac{1}{3} GC\). Drag point \(P\) until \(\overline{CP}\) passes through the point of intersection of \(\overline{NO}\) and the quadratrix. The measure of \(\angle{PCD}\) should be one-third the measure of \(\angle{ICD}\). Note that the angle values shown may not be exact since we are estimating the point of intersection of \(\overline{CP}\) and \(\overline{NO}\) instead of constructing it; however, theoretically the angle is trisected exactly because \[NC:GC = mDP:mDI=1:3.\]

A second method for trisecting an angle was discovered by Archimedes (287–212 BCE). His method involves the mechanical processes of marking and sliding a straightedge until conditions are satisfied so that an angle is trisected. The method is illustrated in the Archimedes applet:

**Archimedes applet:** To trisect \(\angle{BOA}\), a line parallel to \(\overline{OA}\) is constructed through point \(B\), and a perpendicular to \(\overline{OA}\) is constructed through point \(B\). A segment \(QM\) is marked on a straightedge so that it is congruent to \(\overline{OB}\), and a circle of radius \(OB\) is constructed with center \(B\). To trisect \(\angle{BOA}\), slide point \(Q\) along \(\overline{BD}\) until point \(M\) lies on the circle. The measure of \(\angle{QOA}\) should be one-third the measure of \(\angle{BOA}\). To prove this method, notice that \(\angle{QBM}\) is congruent to \(\angle{BQM}\) because of isosceles triangle \(BQM\). Because triangle \(BOM\) is isosceles, \(m\angle{BMO} = m\angle{BOM}\), and since \(\angle{BMO}\) is an exterior angle of triangle \(BQM\), \(m\angle{BMO} = 2(m\angle{BQM})\). Therefore, \(m\angle{QOA}=\frac{1}{3}m\angle{BOA}\). This method also works for right and obtuse triangles. Slide point \(B\)* *until a right or obtuse angle \(BOA\) is formed and repeat the process above. The proof for these solutions is similar to the one for acute angles.

A second solution developed by Archimedes involves the spiral with equation \(r = \theta\), where \(\theta\) is measured in radians. This solution is illustrated by the Archimedes Spiral applet: **Archimedes Spiral applet:** Segment \(OC\) is trisected to give point \(I\). Point \(J\) is the intersection of the spiral and the circle with radius \(OI\) and center \(O\). As you drag point \(C\) along the spiral, notice that \(m\angle{JOB} = \frac{1}{3}(m\angle{COB})\). This can be proven by noting that the distance from \(O\) to \(J\) is equal to the radian measure of \(\angle{JOB}\). Since this distance is one-third the length \(OC\), the angle measure is one-third of \(m\angle{COB}\).

Nicomedes (c. 240 BCE) is generally credited with discovering a method of trisecting an angle similar to Archimedes’ marked straightedge solution. A parallel to \(\overline{OA}\) is constructed through point \(B\), and a perpendicular is constructed to \(\overline{OA}\) through point \(B\) (see the Nicomedes applet, below). Points \(Q\), \(M\), and \(P\) are marked on a straightedge so that \(PM = MQ = OB\). The straightedge is aligned with point \(O\) while point \(Q\) slides freely along \(\overline{DB}\), as illustrated in the Nicomedes applet:

**Nicomedes applet:** Drag point \(Q\) until point \(P\) lies on \(\overline{BC}\). The measure of \(\angle{QOA}\) should be one-third the measure of \(\angle{BOA}\). Note again that since we are estimating the location of point \(P\) on \(\overline{BC}\), the angle measures may not be exact. To prove this method, notice that point \(M\) is the midpoint of the hypotenuse of right triangle \(QBP\), resulting in congruent segments \(\overline{BM}\) and \(\overline{QM}\). Therefore, \(\angle{MBQ}\) is congruent to \(\angle{MQB}\). Exterior angle \(BMO\) has a measure of \(2(m\angle{MQB})\). Segment \(OB\) is congruent to \(\overline{BM}\), so \(\angle{BOQ}\) also has a measure of \(2(m\angle{MQB})\). By the property of alternate interior angles, \(\angle{BQM}\) is congruent to \(\angle{QOA}\). Angle \(BOA\) has a measure of \(3(m\angle{QOA})\). Nicomedes’ method works for obtuse and right angles as well. Slide point \(B\) to the left to form an obtuse or right angle, then slide point \(Q\) until point \(P\) lies on \(\overline{BC}\).

The applets illustrate procedures of trisecting an angle when mechanical means are incorporated with construction methods. I use the applets in two classes of mathematics education majors, History of Mathematics and Modern Geometry, to illustrate the procedures and their proofs. The dynamic nature of the applets allows me to show my students several examples of a method and to provide them with a clear illustration to be used in proving the method. Then I have the students perform these procedures on paper with marked straightedge and compass to experience the actual methods used in ancient Greece. I feel that the students gain a deeper understanding of how the processes actually work when performing them with paper, straightedge, and compass.

### References

Burton, David M. *The History of Mathematics: An Introduction* (6th ed.). New York: McGraw-Hill, 2007.

Smart, James R. *Modern **Geometries* (5th ed.). Pacific Grove, CA: McGraw-Hill, 1998.