Complete and submit Assignment 2.3 according to the directions at the beginning of that assignment.

Comments on Step #7: In this section, we show how the Constructible Root Theorem can be applied to show that it is not possible to trisect an angle of 60 degrees, or duplicate a cube, or square a circle with straightedge and compass alone. In Assignment 2.3 you will also conclude the impossibility of several other constructions by using the tools we have developed. This is the payoff section for the whole module!

Impossibility of trisecting a 60 degree angle with compass and straightedge alone.

We have already shown that if an angle 3m is constructible, then this angle can be trisected to obtain an angle m with compass and straightedge alone if and only if a segment of length equal to the cosine of the angle m can be constructed from a segment length 1. (See bullet item 2) in Part 4 of Unit 1.) If we can find a polynomial expression in powers of x = cos(m) for cos(3m), and if we can show that this polynomial has no constructible roots when m = 20 degrees, then we can conclude that the angle of measure 60 degrees cannot be trisected with straightedge and compass. To this end, we can use standard trigonometric identities to verify that the cosine of 20 degrees satisfies a cubic polynomial equation as follows:

If we let m = 20 degrees, we see that the cosine of 20 degrees is a root of cubic polynomial:

In Problem 2 of Assignment 2.3 for this unit, you are asked to show that P(x) has no rational roots. Consequently, by the Constructible Root Theorem, P(x) has no constructible roots. We conclude that the cosine of 20 degrees is not a constructible number. Therefore, an angle of 60 degrees cannot be trisected with compass and straightedge alone.

Impossibility of duplicating a cube with compass and straightedge alone.

This one is for you to do! Just Do It! 2.3.1

Impossibility of squaring a circle.

In Part 4 of Unit 1, we showed that, given a line segment of length 1 and a circle of radius r, the problem of finding a geometric construction for the side of a square with the same area as the given circle is equivalent to the following problem:

Of course, if a segment of length Pi can be constructed from a segment of length 1, then a segment of length equal to the square root of Pi can also be constructed. Consequently, the problem of squaring the circle is equivalent to that of determining whether or not Pi is a constructible number. However, in Part 2 of Unit 2, we pointed out that every constructible is an algebraic number; that is, it is the root of a polynomial with integer (or rational number) coefficients. However, Pi is well-known to be a transcendental (ie. not an algebraic) number. Consequently, we can conclude that a square of area equal to that of a given circle cannot be constructed with compass and straightedge alone.

Now try giving an impossibility proof of you own by completing the following