How do we prove that a triangle with sides $(one, x, y)$, where $x$ is any constructible length from one to three at the elliptic curve

$$y^2 = x^3 -x^2 -x +1$$then the triangle possess at least one trisectible angle?

I would like to add this note: *if $x$ is a constructible length or (any real number) from zero to one), then you would kindly notice that (3*Alpha - Beta = PI), where (Alpha & Beta) are two angles in the triangle $(one, x, y)$, which implies a trisectible angle*

  • $\begingroup$ Note that, the assumption holds true always for x as any arbitrary length from one to three & the trisection of the angle is exact (without any approximation), also this is not a claim of a solution to the angle trisection problem, but a very little insight to classify the trisectible angle which may be further developed by interested people $\endgroup$ – bassam karzeddin Sep 3 '15 at 8:50
  • $\begingroup$ I dk how,but I see that you want to show that at least one of $cos(A/3),cos(B/3),cos(C/3)$ is constructible ,where $ A,B,C$ are the angles of the triangle. That is,belongs to some field $ F_n $ where $ F_0=Q$ and $F_{j+1}=F_j[x_j]$ with $ x_j^2 \in F_j $. $\endgroup$ – DanielWainfleet Sep 3 '15 at 10:02
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    $\begingroup$ I guess one needs to play with the law of cosines, and show that cubic roots disappear from the solution of (at least one of) the resulting cubic equation with the relevant $\cos(\alpha/3)$ as a root. I didn't see anything right away. This is kinda cool question. Where does it come from? A minor quibble I have is that your equation does not define an elliptic curve, because that cubic polynomial of $x$ has a double zero at $x=1$, and the curve is thus singular. That means that the curve is actually rational with its points parametrized by $t=y/(x-1)$. May be that will come into play? $\endgroup$ – Jyrki Lahtonen Sep 4 '15 at 18:21
  • $\begingroup$ The proof is not so interesting as the result, since this can be done easily, but requires some drawings that is difficult to present here, what I'm trying to find other algebraic methods of proving the alleged statement $\endgroup$ – bassam karzeddin Sep 5 '15 at 8:45
  • $\begingroup$ Many algebraic methods can nowadays quickly be executed with CAS (computer algebra systems). Also there are some freely available for limite use on the internet. You might then tell us yourself what further interesting developments you have in mind? $\endgroup$ – Transfinite Numbers Nov 29 '16 at 12:48

So we then must conclude that $y=(x-1)*\sqrt{x+1}$. And indeed we have $y =< x+1$ for $x$ in the range $1$ to $3$. So the three values define a triangle. As the angles I get the following(*):

$cos(α) = \sqrt{x + 1}/2$

$cos(β) = -\frac{x^2}{2} + x + \frac{1}{2}$

$cos(γ) = \frac{1}{2} (x - 2) \sqrt{x + 1}$

By the formula $s=4*t^3-3*t$ I have also computed the tripple angle for the first angle and I get the third angle. From this it also follows that β = π - 4α.

$cos(3α) = \frac{1}{2} (x - 2) \sqrt{x + 1} = cos(γ)$

But I don't see any classical constructive method behind it, which shouldn't exist. Problem is how do we get from a triangle with $cos(γ)$ to a triangle with $x$?

(*) Was using Sine, Cosine and Tangent rules and Wolfram Alpha.


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