ANGLE TRISECTION

[bassam king karzeddin]

Dear Mathematicians

I have posted in the geometry research the following problem about angle trisection,but did not get a clear opinion ,and, since, here is a larger groub.

I will be glad to know if I wrote nonsense mathematics or something useful.here is the problem.

An arbitrary angle and its exact trisection angle fits exactly in the following symbolic triangle with the following sides:

a^3 , a*(b^2-a^2) , b*(b^2-2*a^2)

Where : 2 >= b/a >= sqrt(2)

(a,b):are positive real numbers

Of course, I have a hand written proofs for this fact.

Thanking you.

Bassam Karzeddin
Al Hussein Bin Talal University
JORDAN
********************************

[Richard Fateman]

bassam king karzeddin wrote:

>
>
> An arbitrary angle and its exact trisection angle fits exactly in the following symbolic triangle with the following sides:
>
> a^3 , a*(b^2-a^2) , b*(b^2-2*a^2)
>
> Where : 2 >= b/a >= sqrt(2)
>
> (a,b):are positive real numbers
>

Solvetriangle in Macsyma tells us that there are
triangles with your specified sides and computes the
angles.
It is easy to confirm numerically that the second angle
is 3 times the first.
This appears to have no bearing on the impossibility of
trisecting an angle with ruler and compass, if that
is why you are interested in this.

See http://mathworld.wolfram.com/AngleTrisection.html
for related info.

[SIDES = [A^3,A*B^2-A^3,B^3-2*A^2*B],
ANGLES = [ACOS(B/(2*A)),
ACOS((B^3-3*A^2*B)/(2*A^3)),
%PI-ACOS((B^4-4*A^2*B^2+2*A^4)/(2*A^4))]]

[Rouben Rostamian]

In article <3607653.11216984251...@nitrogen.mathforum.org>,

bassam king karzeddin <bas...@ahu.edu.jo> wrote:
>
>I have posted in the geometry research the following problem about
>angle trisection,but did not get a clear opinion ,and, since, here is a
>larger groub.
>
>I will be glad to know if I wrote nonsense mathematics or something
>useful.here is the problem.
>
>An arbitrary angle and its exact trisection angle fits exactly in the
>following symbolic triangle with the following sides:
>
> a^3 , a*(b^2-a^2) , b*(b^2-2*a^2)
>
>Where : 2 >= b/a >= sqrt(2)
>
> (a,b):are positive real numbers
>
>Of course, I have a hand written proofs for this fact.

Let s1, s2, s3 be the sides of the triangle, as you have defined them:

s1 := a^3
s2 := a*(b^2-a^2)
s3 := b*(b^2-2*a^2)

Let v1, v2, v3 be the angles opposite sides s1, s2, s3, respectively.
The from the law of cosines, we get:

s1^2 = s2^2 + s3^2 - 2*s2*s3*cos(v1)

whence:

cos(v1) = b/(2*a);

Similarly:

cos(v2) = b*(b^2 - 3*a^2)/(2*a^3)

cos(v3) = (4*a^2*b^2 - 2*a^4 - b^4)/(2*a^4)

Now apply the trigonometric identity:

cos(3*x) = 3*cos^3(x) - 3*cos(x)

to angle v1 and simplify:

cos(3*v1) = 3*cos^3(v1) - 3*cos(v1)
= 3*[ b/(2*a)]^3 - 3*[ b/(2*a)]
= b*(b^2 - 3*a^2)/(2*a^3).

Thus cos(3*v1) = cos(v2) which is cute. However I don't see a
relation between this and the classic angle trisection problem.

--
Rouben Rostamian

[bassam karzeddin]

[bassam karzeddin]

On Monday, July 18, 2005 at 5:53:15 PM UTC+3, bassam king karzeddin wrote:

> Dear Mathematicians
> I have posted in the geometry research the following problem about angle trisection,but did not get a clear opinion ,and, since, here is a larger groub.
> I will be glad to know if I wrote nonsense mathematics or something useful.here is the problem.
> An arbitrary angle and its exact trisection angle fits exactly in the following symbolic triangle with the following sides:
> a^3 , a*(b^2-a^2) , b*(b^2-2*a^2)
> Where : 2 >= b/a >= sqrt(2)

Correction: (2 > b/a > Sqrt(2))

> (a,b):are positive real numbers
> Of course, I have a hand written proofs for this fact.
> Thanking you.
> Bassam Karzeddin
> Al Hussein Bin Talal University
> JORDAN
> ********************************

And I later discovered the secret of non-existing angles

Bkk

[Jeff Barnett]

I suggest you beg, borrow, steal, or (simply) buy a copy of the
following book:

The Trisectors (Spectrum) Paperback – September 5, 1996
by Underwood Dudley (Author)

Amazon USA carries it and I'm sure other book sellers do too. It's a fun
read.
--
Jeff Barnett

[Jeff Barnett]

PS It's available and less expensive in E-book form at
https://bookstore.ams.org/view?ProductCode=SPEC/16

--
Jeff Barnett

[bassam karzeddin]

On Saturday, September 30, 2023 at 7:04:54 PM UTC+3, Jeff Barnett wrote:
> On 9/30/2023 9:52 AM, Jeff Barnett wrote:
> > On 9/30/2023 5:15 AM, bassam karzeddin wrote:
> >> On Monday, July 18, 2005 at 5:53:15 PM UTC+3, bassam king karzeddin
> >> wrote:
> >>> Dear Mathematicians
> >>> I have posted in the geometry research the following problem about
> >>> angle trisection,but did not get a clear opinion ,and, since, here is
> >>> a larger groub.
> >>> I will be glad to know if I wrote nonsense mathematics or something
> >>> useful.here is the problem.
> >>> An arbitrary angle and its exact trisection angle fits exactly in the
> >>> following symbolic triangle with the following sides:
> >>> a^3 , a*(b^2-a^2) , b*(b^2-2*a^2)
> >>> Where : 2 >= b/a >= sqrt(2)
> >>
> >> Correction: (2 > b/a > Sqrt(2))
> >>> (a,b):are positive real numbers
> >>> Of course, I have a hand written proofs for this fact.
> >>> Thanking you.
> >>> Bassam Karzeddin
> >>> Al Hussein Bin Talal University
> >>> JORDAN
> >>> ********************************
> >>
> >> And I later discovered the secret of non-existing angle

> > I suggest you beg, borrow, steal, or (simply) buy a copy of the
> > following book:
> >
> > The Trisectors (Spectrum) Paperback – September 5, 1996
> > by Underwood Dudley (Author)
> >
> > Amazon USA carries it and I'm sure other book sellers do too. It's a fun
> > read.
> PS It's available and less expensive in E-book form at
> https://bookstore.ams.org/view?ProductCode=SPEC/16
>
> --
> Jeff Barnett

Did your holly books mention anything about the non-existing angles as the most famous angle of (Pi/9 = 20) Degrees Angle?

Most likely they will arrange it somehow from a foreged historical sources in the near future FOR SURE

BKK

[bassam karzeddin]

The simple theme that humans generally were incapable to understand it correctly as I only did is that "something doesn't exist then it is absolutely impossible to construct by any tools or any means as well

However, proofs of Wantzel in 1836 about the impossibility of constructing the angle Pi/9 is not a true mathematical rigorous proof but only a true conclusion that was associated to tools of unmarked straigt edge & a compass

However, tools in mathematics are for skilled carpenters & never for any true genius mathematicians FOR SURE

BKK