** Theorem 4.** The trisection of the angle by an unmarked ruler and
compass alone is in general not possible.

This problem, together with * Doubling the Cube*, * Constructing the
regular Heptagon* and * Squaring the Circle* were posed by the Greeks in
antiquity, and remained open until modern times.

The solution to all of them is rather inelegant from a geometric perspective. No geometric proof has been offered [check?], however, a very clever solution was found using fairly basic results from extension fields and modern algebra.

It turns out that trisecting the angle is equivalent to solving
a cubic equation. Constructions with ruler and compass may
only compute the solution of a limited set of such equations,
even when restricted to integer coefficients. In particular,
the equation for *theta = 60* degrees cannot be solved by
ruler and compass and thus the trisection of the angle is not
possible.

It is possible to trisect an angle using a compass and a ruler marked in 2 places.

Suppose *X* is a point on the unit circle such that *angle XOE* is
the angle we would like to ``trisect''. Draw a line *AX* through a
point *A* on the *x*-axis such that *|AB| = 1* (which is the same as
the radius of the circle), where *B* is the intersection-point of the
line *AX* with the circle.

**Figure 7.1:** Trisection of the Angle with a marked ruler

Let *theta * be *angle BAO*.
Then *angle BOA = theta *, and *angle XBO = angle BXO = 2 theta *

Since the sum of the internal angles of a triangle equals *pi *
radians (*180* degrees) we have *angle XBO + angle BXO + angle BOX = pi *, implying *4 theta + angle BOX = pi *. Also, we have that
*angle AOB + angle BOX + angle XOE = pi *, implying *theta + angle BOX + angle XOE = pi *. Since both quantities are equal to *pi *
we obtain

*4 theta + angle BOX = theta + angle BOX + angle XOE *

From which

*3 theta = angle XOE *

follows. QED.

Fri Feb 20 21:45:30 EST 1998