## Frequently Asked Questions

#### How do you prove that trisecting an angle is impossible?

**Theorem**

Trisecting an angle using compass and straight edge is impossible.

**Proof**

Please note that this proof assumes the knowledge that any construction that can be shown to be algebraically equivalent to a cubic containing rational coefficients and having irrational roots is impossible, proved in the Impossible Constructions document.

In order to demonstrate that a general angle cannot be trisected we need only show that one particular angle cannot be trisected.

Clearly a 60 degree angle can be constructed, so we will show that 20 degrees cannot be done. By using the identity cos3θ = 4cos^{3}θ – 3cosθ, we get cos60 = 4cos^{3}20 – 3cos20. We know that cos60 = ½ and by substituting `c` = cos20, we arrive at the cubic equation 8`c`³ – 6`c` – 1 = 0.

As this can be written (2`c`)^{3} – 3(2`c`) – 1 = 0, let `q` = 2`c`, so `q`^{3} – 3`q` – 1 = 0.

If we can show that this cubic equation has no rational roots, we demonstrate that the construction of a 20 degree angle is impossible, hence the general construction of trisecting an angle cannot be done.

Let us assume that a rational root, `q` = `a/b` exists, where `a` and `b` have no common factors.

Therefore `a`^{3}/`b`^{3} – 3`a`/`b` – 1 = 0, giving `a`^{3} – 3`a``b`^{2} – `b`^{3} = 0.

Rearranging we get `a`(`a`^{2} – 3`b`^{2}) = `b`^{3}. From this we deduce that `b`^{3} is a multiple of `a`, that is `b`^{3} = `ka` and so `a` and `b` must have common factors. However we wrote `q` = `a/b`, such that `a` and `b` had no common factors, therefore `a` = ±1. Similarly, by writing `a`³ = `b`(`b`² + 3`ab`) we deduce that `a`³ must be a multiple of `b`, and by the same reasoning `b` = ±1. However `a/b` = ±1 does not satisfy the cubic equation, so we deduce that no rational roots must exist. Hence the trisection of an angle is not possible by compass and straight edge.

**Note**

Despite the Greeks being unable to trisect an angle using compass and straightedge (which was proved, in 1847, by the method above), Archimedes knew of another method for trisecting the angle. Consider the diagram.

It can be demonstrated that if length `AE` is equal to the radius of the semicircle, then angle `EAB` is one-third angle `DOC`.

Let angle `EAO = x` = angle `EOA` (isosceles triangle). So angle `AEO` = 180 – 2`x` and so angle `DEF` = 2`x`. By using the result that the angle at the centre, `DOF`, is twice the angle at the circumference, `DEF`, we deduce that angle `DOF` = 4`x`. By the X angle property, angle `EOB` = angle `COF` = `x`. Hence angle `DOC` = 3`x`; which is precisely three times angle `EAO`.