In the period of Greece we are studying (Thales, Pythagoras, Euclid, Archimedes), there were three famous problems which were studied with passion by the greats of the time. They are the following: *Using only a compass and a straightedge (with no markings on it)* can you

- construct the edge of a cube having twice the volume of a given cube (duplication of the cube problem)
- trisect a given angle (angle trisection problem)
- construct a square having an area equal to a given circle (quadrature of the circle problem)

The amazing result is that one can’t solve *any* of these problems. What is even more amazing is that the solution was not obtained until over 2000 years after they were posed.

Let’s focus on the angle trisection problem. In particular, let us focus on exactly what we mean by this problem. First there is the clear restriction that the only tools are a straightedge (with no markings on it) and a compass. Second, we need to do more than just identify the angle measure or else the problem is rather trivial. Indeed, if I were to say, trisect the angle of 60 degrees and you say. Sure thing, the answer is 20 degrees. The problem is more complicated than that. The angle is drawn for you on a piece of paper and you need to draw another line, coming from the vertex of the given angle, which trisects the given angle. Third, let’s understand what exactly we mean by can and can’t solve this problem. There are* some *angles which *can* be trisected. Indeed, as we shall see in a moment, a 90 degree angle can be trisected – as can a 45 degree angle. However, in order to say you solve the angle trisection problem, you need to show that *any* angle can be trisected. We shall see in a little while that a 60 degree angle can *not* be trisected.

To get started on ruler and compass constructions, we will review some constructions you may have done in high school.

**Proposition.*** Any given angle can be bisected.*

**Proposition.** *Given a line L and a point P not on this line, it is possible to construct a line passing through Q and perpendicular to L.*

**Proposition.*** Given a like L and a point P not in this line, it is possible to construct a line passing through P and parallel to L.*

**Proposition**. *It is possible to bisect any given line segment.*

Here is a new construction.

**Proposition**. *It is possible to trisect any given line segment.*

**Proof** (this is from J. B. Conway’s book – *Mathematical Connections*): Let OP be the given line segment. Draw a second line segment OX.

Using any radius r draw a circle with center O and let C be where this circle intersects OX. Then, with the same radius r, make another circle with center C and let B be where this second circle intersects OX. One more time, with the same radius r, but with center B, let A be where this third circle intersects OX.

Now draw a line from A to P. Using of of the earlier propositions, draw lines from B to Q and from C to R which are parallel to AP.

We now claim that |OR| (the length of OR) is 1/3 |OP|. From high school geometry, we see that angle A = angle B = angle C. Since the angle at O is shared by the triangles OAP and OCR we conclude that OAP and ORC are similar triangles. By Thales’ theorem on similar triangles we get |OP|/|OR| = |OA/|OC|. But since, by our construction, |OC| = r and |OA| = 3r, we get that |OP| = 3|OR|. Thus the segment OR trisects OP. **QED.**

**Corollary: **

*Given a line segment and an integer n, it is possible, using only a straightedge and compass, to divide the given line segment into n equal pieces.*

**Proof:** Exercise.

Let’s go back to our original problem: can we trisect any angle? We will see in a while that, in general, the answer is no. In other words, there are some angles which can be trisected and some which can’t. Here is one positive result.

**Proposition:** *A 90 degree angle can be trisected.*

**Proof** (from Conway’s book Mathematical Connections): Let AOB be a right angle.

Choose any radius r of the compass and draw a circle of radius r with center O. This circle will meet the line segment OB at a point P. Now draw a circle with radius r and center P. This second circle will meet the first circle at a point Q.

Notice that the triangle OPQ is an equilateral triangle and this each angle is a 60 degree angle. This means that the angle QOA is a 30 degree angle. **QED**.

**Proposition**: *The 45 degree angle can be trisected.*

**Proof:** Exercise.

**Proposition:** A regular hexagon can be constructed.

**Proof.** Exercise.

It turns out that if, along with our straightedge and compass, we include the use of the *spiral of Archimedes*, we can trisect any angle. The spiral of Archimedes is a spiral starting at the origin O such that at any position on the spiral, the radian angle that position makes with the horizontal axis is equal to the distance from the origin. For those of you who had some calculus, it is the curve r = Θ in polar coordinates.

**Theorem (Archimedes)**: *Given a straightedge, compass, and the spiral of Archimedes, any angle can trisected*.

**Proof:** We start with the spiral of Archimedes which just sits there on our paper along with our given angle at O. The spiral will intersect our given angle line at a point P. Since we can trisect a line segment, we do so and produce the point R on the angle line. Draw a circle with radius |OR| centered at O. This circle will intersect the spiral at a point T. Form the line OT.

Since |OT| is one-third of the angle from the horizontal (by the definition of the spiral of Archimedes), we get that the line OT trisects our original angle. **QED.**

As a historical note, the above can be found in Archimedes’ work titled *On Spirals*.

**Theorem:** *A 60 degree angle can not be trisected.*

We will not prove this theorem here since the proof requires some material in an advanced mathematics course called abstract algebra. Who proved this theorem? Well, there is some controversy here. C. F. Gauss, one of the three great mathematicians of all time (Archimedes and Newton being the other two) claims to have shown this but his proof was never found. A formal proof was finally published by P. Wantzel in 1837. Who gets the credit? Perhaps one can say that Gauss was one of the great mathematicians of all time and he proved plenty of other amazing things in his life. So, if he said he did it, then he probably did. On the other hand, one could argue that the person who actually takes the time to write down a formal proof and put it out for public critique is the one who deserves the credit. In this case the prize goes to Wantzel.