Angle trisection

1. Mar 13, 2007

happyg1

1. The problem statement, all variables and given/known data

Prove that it is possible to trisect 72 degrees.

2. Relevant equations
$$cos(3\theta)=4cos^3(\theta)-3cos(\theta)$$

3. The attempt at a solution
well,72/3=24, So I need to see if cos24 is constructible, Which means that it is a root of a polynomial with a degree that is a power of 2.(?)I think.

I did this:$$cos(72)=\frac{\sqrt5-1}{4}$$
and $$Let \theta=24$$
so then letting $$cos\theta=x$$ in the equation above,
I get:
$$(\sqrt5)=(16x^3+12x+1)$$
Now I get a little confused about what I'm trying to do.
Here's what I need help with:
Firstly, am I on the right track here?
Secondly, Is the "degree of a power of 2" (above) correct?
What is it that I need to do with this polynomial from here?
I've confused myself.
CC

Last edited: Mar 13, 2007
2. Mar 13, 2007

happyg1

3. Mar 14, 2007

HallsofIvy

Staff Emeritus
Yes, that is (almost) correct. A number is "constructible" if and only if it is "algebraic of order a power of 2" which means it is a zero of a polynomial with integer coefficients of degree a power of two. The "with integer coefficients" is why I said "almost".

Not quite. In the first place, the right hand side should be
$$16x^3- 12x+ 1$$
In the second, you can't have $\sqrt{5}$- that's not an integer. What you want to say is that if x= cos(24) , then x satisfies
$$5= (16x^3- 12x+ 1)^2[/itex] which is a 6th degree equation. Now, you need to show that it can be reduced to either a 4th or 2nd degree equation- that it has either rational numbers or square roots as solutions. I'm not at all sure how you are going to do that! 4. Mar 14, 2007 happyg1 Hey, Thanks for looking at this. Now I'm not QUITE so cunfused. I'll fix my algebra and see where I can get. CC 5. Mar 14, 2007 happyg1 OK Please let me know of this is correct. I fixed my sign error(der) and squared the thing to get rid of the square root: [tex]5=256x^6-384x^4+32x^3+144x^2-24x+1$$
$$256x^6-384x^4+32x^3+144x^2-24x-4=0$$
Which factors like this:
$$4(4x^2+2x-1)(16x^4-8x^3-16x^2+8x+1)$$
The quadratic factor has square roots as the zeros and the 4th degree has cos24 as one of the roots. So x=cos24 satisfies that 4th degree polynomial factor with integer coefficients. 4 is a degree of a power of 2, so cos24 is constructible?
Am I any closer?
CC

6. Mar 14, 2007

HallsofIvy

Staff Emeritus
Wow, I am impressed! When I wrote my previous reply, I was tempted to add "Good luck with trying to factor that"!
Yes, if cos(24) satisfies the original equation (and you say it does) then it must satisfy one or the other of the factors equal to 0. You should be able to solve 4x2+ 2x-1= 0 to determine if it is that one but it really doesn't matter- cos(24) is either algebraic of order 4 or algebraic of order 2. In either case it is constructible.

Last edited: Dec 9, 2007
7. Mar 14, 2007

Hurkyl

Staff Emeritus
Some ("most", actually) irreducible 4-th degree polynomials have roots that are not constructible.

8. Mar 14, 2007

HallsofIvy

Staff Emeritus
It ocurred to me after I responded that you do need to show that the 4th degree polynomial is irreducible.

If the second degree polynomial happened to be reducible then it would follow that the root is either algebraic of order 1 or order 2, both of which are algebraic or order a power of 2.

If the 4th degree factor were separable, then it would either be factored into two quadratic factors (in which case x would be algebraic of order 1 or 2, both powers of 2) or factored into a cubic and a first degree factor. In that last case, where there is a first degree factor, then
$$(16x^4-8x^3-16x^2+8x+1)$$
has a rational root. By the "rational root theorem", any rational roots must have denominator that evenly divides 16: i.e. $\pm 1$, $\pm 2$, $\pm 4$, $\pm 8$ or [/itex]\pm 16[/itex] and denominator that evenly divides 1: i.e. $\pm 1$. In other words, the only possible rational roots are $\pm 1$, $\pm \frac{1}{2}$, $\pm \frac{1}{4}$, $\pm\frac{1}{8|}$, or $\pm\frac{1}{16}$.

Do any of those satisfy the equation?

9. Mar 15, 2007

happyg1

I'm a little confused about this. The 4th degree polynomial that I got has no rational roots, so that means that cos24 isn't rational, but it's constructible? Yes?
And also, from the quadratic, can I conclude that cos72 is constructible? That one has$$\frac{\sqrt5-1}{4}=cos72$$ as a root.
And then since cos72 is constructible, so is a regular pentagon?
I thought that this might be true because 360/5=72.Is this the angle of interest? Or is there another method to show the pentagon is constructible?

I have attempted to prove that the hexagon is constructible and I keep getting a cubic. i know I'm wrong. That's another post tho.

BTW, on that factorization, I had gotten stuck because I had the wrong sign the first time. I scrawled about 3 pages trying to figure it out and then I busted out the TI89. Don't be too impressed.
CC

10. Dec 8, 2007

dubi

Some info on trisection

The ever-dubious wikipedia has some info on trisection -- full disclaimer, I wrote it -- at
< http://en.wikipedia.org/wiki/Angle_trisection >. The book by Ian Stewart I own, and is
available in updated editions.

Another reference that trisects 72 degrees (big .pdf file, search for "72") is
< http://www.math.uncc.edu/~droyster/courses/spring02/classnotes/Chapter03.pdf [Broken] >.
The construction is a little messy, but looks valid to me.

dino, not "dubi," -- too fast on the keyboard.

Last edited by a moderator: May 3, 2017
11. Jun 12, 2010

pogovio

Sometimes too much knowledge can distract us in the wrong direction. This problem actually doesn't require anything about the forms of equations for cos 24. Its a much simpler problem.

If you have a 72-degree angle, then construct a 60-degree angle, and subtract from 72, leaving a 12-degree angle. Double it, and you have the 24-degree angle, which is 1/3 of 72.