Quintisection of an Angle

One of the most famous geometrical constructions in Euclid's
"Elements" is to divide a line segment in "extreme and mean 
ratio" (Book VI, Prop 30) or, equivalently, to divide a line 
segment into two parts a,b such that a rectangle with sides
a+b and b has the same area as a square of size a^2 (Book II,
Prop 11).  Such a figure is often called a "golden rectangle.

Obviously by equating (a+b)b to a^2 we have the equation
ab + b^2 = a^2, and if we divide both sides by b^2 this gives
x + 1 = x^2  where x = a/b.  The roots of this equation are
r = (1+SQR(5))/2  and r' = (1-SQR(5))/2.  One of the many
interesting properties of r is that it is the ratio of a 
diagonal to a side of a regular pentgon.  Thus it allows us 
to construct a regular pentagon and quintisect a circle.

We can also construct a right triangle with hypotenuse 1 and 
sides of length (1/r) and SQR(1/r).  Letting "beta" denote the 
angle opposite the leg of length (1/r) we have sin(beta)=1/r.
We might wonder if the angle beta can be quintisected by 
elementary Euclidean methods (ruler and compass).  To prove 
that it can't, observe that if sin(beta) = 1/r then we have 
tan(beta) = SQR(1/r), which is constructible.  Now we can 
express  tan(beta/5)  in terms of tan(beta) by making use 
of the trigonometric relation
 
                          tan(x) + tan(y)
            tan(x+y) =    ----------------
                          1 - tan(x)tan(y)

Letting T denote tan(beta/5), we have the identity

                            5 - 10T^2 + T^4
            tan(beta) =  T -----------------
                            1 - 10T^2 + 5T^4

We've defined tan(beta) as the quantity SQR((sqrt(5)-1)/2), so
if we square both sides of the above equation, multiply by 2,
add 1, and square again, and make the substitution x = T^2,
we arrive at

 x^10 - 15x^9 - 605x^8 + 5260x^7 - 15070x^6 + 23126x^5

            - 16370x^4 + 4460x^3 - 595x^2 + 65x - 1  =  0

Now we only need to prove that this polynomial neither factors nor 
decomposes into quadratics.  In other words, from elementary Galois
theory

  If the magnitude w is constructible by Euclidean methods, then 
  the degree of its minimum polynomial with rational coefficients 
  is a power of 2.

Thus, if the minimal polynomial with root w is NOT a power of 2, 
it follows that w is not constructible by Euclidean means.  (Note 
that the converse of this theorem is not true, because it's 
actually possible to have a magnitude w whose minimal polynomial 
is of degree 2^n but which cannot be constructed.)  

If beta/5 could be constructed, then tan(beta/5)^2 could also be 
constructed, and this magnitude is a root of the 10th degree 
polynomial given above.  To prove that it is the minimum polynomial
for that magnitude, make the substitution x = y + 2, which gives

 y^10 + 5y^9 - 695y^8 - 5620y^7 - 15910y^6 - 9090y^5

     + 58850y^4 + 183500y^3 + 259525y^2 + 191525y + 58805 = 0

By Eisenstein's criterion this is irreducible, so the minimum 
polynomial over the rationals with root tan(beta/5)^2 has degree 
10, which is not a power of 2.  Therefore, beta/5 cannot be 
constructed by Euclidean methods.

This is exactly similar to the method of proving that an arbitrary
angle cannot be trisected by Euclidean means.  The ancient Greeks 
sought a way of trisecting any angle, but since pi/3 is constructible
this would imply that we could construct the magnitude w = 2cos(pi/9), 
which is a root of p(x) = x^3 - 3x - 1.   We know this is the minimum 
polynomial because if we make the substitution x = y + 1 we have 
p(y) = y^3 + 3y^2 - 3, which is irreducible by Eisenstein's criterion.  
Thus the degree of the minimum field extension of 2cos(pi/9) over the 
rationals is of degree 3, which is not a power of 2, so it cannot be
constructed by Euclidean methods.

Notice that in both examples we had to perform a substitution of
variables into the polynomial to bring a it into a form where 
Eisenstein's criterion applies.  I don't know of any systematic 
way of finding the right substitution, other than trial and error.  
This raises an interesting question:  If a polynomial p(x) is 
irreducible over the rationals, does there necessarily exist a
bi-linear change of variables, i.e., x -> (ay+b)/(cy+d)  such 
that the resulting polynomial in y satisfies the conditions of 
Eisenstein's criterion?  Also, assuming p(x) is irreducible, 
how many substitutions would we expect to try (in some natural 
order) before finding one that works?  Also, what is the analog 
of Eisenstein's criterion for testing primality of integers?