Problem 14.4 Trisecting Angles

It is often stated in popular literature that it is impossible to trisect angles with straightedge and compass. However, it has been known since ancient Greek times that any angle can be can be trisected using only straightedge and compass. For example, Archimedes (287211b.c.) showed how to trisect any angle (less than 145°) using a marked straightedge (a straightedge with two points marked on it). See [Tx: Martin], page 49.

OK, so then maybe it is impossible to trisect angles with compass and unmarked straightedge?

a. Construct the figure on the left in Figure 14.13a with compass and unmarked straightedge starting with any two points AB.

Figure 14.13 "Shoemaker's knife" or "tomahawk" trisector

b. Construct the figure in 14.13a on a transparent sheet and lay it on the angle a (see Figure 14.13b) so the vertex, V, of the angle lies on DB, one side contains C, and the other is tangent to the semicircle with center A. Prove that ÐBVC trisects angle a.

In order to give a correct statement of what is impossible, it is necessary first to define a compass and unmarked straightedge sequence to be a finite sequence of points, lines, and circles that starts with two distinct given points and such that: (1) Each of the other points in the sequence is the intersection of lines or circles that occur before it in the sequence. (2) Each circle has its center and one point on the boundary occurring before it in the sequence. (3) Each line contains two distinct points which occur before it in the sequence.

For a proof of this theorem and related discussions about various compass and straightedge constructions, see [Tx: Martin].