Well, read 25-gon were I wrote "pentagon"

Thankyou

sfwc
<><

Regarding your second question on Galois theory, I didn't talk about Galois theory as a matter of authority, to put a theorem with name, but to indicate where you could find some ideas that (still) help for your existence problem. The theorem and the way it is proved, among other things, tells us what you can do without the "marking" operation.

Let me explain how it helps:

Constructing points in the plane is equivalent to finding segments whose lengths are the x-component and y-component of the point, and gluing them ortogonally together (constructin the point is equivalent to constructing segments whose length are its coordinates). Your problem is that of finding segments of the appropriate length. If you have an angle a, fixing an unit, its two components allow you to recover (cos a) and (sin a). You want the angle a/5, or more precisely, the lengths (cos a/5) and (sin a/5).

What Galois Theory, among many other things, tells us for compass-ruler constructions is that if you have a system of points in the complex plane and consider the field K of points generated (with +,·) by them (which can be constructed with ruler and compass), any set of points obtained by ruler and compass from these are contained in an extension of K of order 2^n. Essentially, that you can add, multiply and find square roots for any given complex numbers, and construct points whose real and imaginary parts are obtained in these ways.

The only question remaining is what kind of extensions does the use of a marked straightedge allow. I don't know much about the whole set of constructions allowed by the marked straightedge, but at least it'serves to trisect an angle and find cubic roots.

If this is the case, having the angle is equivalent to having the real numbers cos(a), sin(a). We want to find cos(a/5) and sin(a/5).

You may compute cos 5x in terms of cos x, so:

cos a=16 (cos a/5)^5 - 20 (cos a/5)^3 + 5 (cos a/5)

So your problem is equivalent to solving a quintic equation:

b= 16 x^5 - 20 x^3 + 5 x

for a given b.

Here, taking x/2 instead of x and b/2 instead of b, it is equivalent to solving:
b = x^5 - 5 x^3 + 5 x

b = x (x^4 -5 x^2 +5)
for a given b.

I'm not completely sure of this one, but general quintics are not solvable, and this equation doesn't seem to be solvable by radicals, in the general case (arbitrary b).

If you find out that this equation can be solved by means of square and cubic roots in Q(b) (there must be ways to determine if it is so in terms of b), then you can quinsect the angle with your ruler and marked straightedge.

It is known that the general quintic cannot be solved by radicals (using n-roots), and if it can be solved by radicals, then only square and cubic roots are involved. You already know how to find square and cubic roots.

So this far:

1) If the value of b (which in your question is arbitrary) yields a solvable quintic x^5-5x^3+5x-b, you can quinsect your angle.

2) If the value of b yields a non-solvable quintic, I must ask you if the ruled straightedge can construct numbers other than those generated by addition, multiplication and taking of n-roots, for if it was not the case, you couldn't construct the root and couldn't quinsect your angle.

So, as we knew, the answer depends on the operations you can make with the marked straightedge. If these operations can be reduced to trisecting angles, the answer is No, we cannot 5-sect angles. In any case, I am afraid that the marked straightedge allows for more than what you get by the trisecting operation.

I know this is not an answer to your question, but simplifies the problem from the algebraic point of view. Your question is equivalent to analyze the posibility of finding the roots of all the quintic equations given above with ruler and marked straightedge.

I have the feeling the marked straightedge does serve to solve the general quintic, but the feeling is not enough. Perhaps someone may help you using some of these ideas.

A note: any quintic can be transformed, up to solving a quartic equation (this only involves ruler and compass), in a equation of the form x^5+px+q=0. You may try to solve this equation with a marked straightedge.

And sorry for the lack of geometry in these ideas. I also would have liked to get a geometric solution.

x^5 -5t*x^3 + 5t^2*x -2q = 0 is always solvable(!)

(thus for any value of b )

as a sum of two quintic roots. ( b = 2q ) ( t = 1 )

x = u + t/u will do.

x = (q + (q^2 -t^5 )^(1/2) )^(1/5) + (q - (q^2 -t^5 )^(1/2) )^(1/5)

Gives a perfect solution.

I think it is not possible to constuct with ruler and compass.

Greetings Ruben Hogenhout