Re: Trisecting an arbitrary angle

[bassam king karzeddin]

>
>
>
> Le 19/07/05 12:37, dans
> 4dd7d$42dcd769$d52f93dc$25...@news.chello.at,
> « Jutta Gut » <gut.jutt...@chello.at> a écrit :
>
> >
> > "bassam king karzeddin" <bas...@ahu.edu.jo> schrieb
> im Newsbeitrag
> >
> news:29354512.1121766911405.JavaMail.jakarta@nitrogen.
> mathforum.org...
> >> That is grate,
> >>
> >> This,might open doors to constructible polygons
> >>
> >> In fact,I have deduced & proved the same thing,I
> have mentioned that here:
> >>
> >>
> http://mathforum.org/kb/message.jspa?messageID=3802920
> &tstart=0
> >>
> >> I will provide examples soon.
> >
> > If I understand correctly, you have shown that in
> an triangle with
> > the sides a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) one
> angle is three times
> > another one.
> >
> > The more interesting question would be: given an
> angle, how to
> > construct a triangle with the sides a^3 ,
> a*(b^2-a^2) , b*(b^2-2*a^2)
> > and the given angle?
>
> It tried for some simple angles: pi/8, pi/6, pi/5,
> sides can be computed
> with square roots. For pi/9, you have the 3rd degree
> equation x^3=3*x+1.
> Not surprising, since pi/9 is not constructible. But
> it's still interesting
> to know which triangles have two angles A,B such that
> A=3*B.
>

That is because there is not any EXACT real root for the polynomial (x^3 - 3x -1 = 0), for sure, but nearly an approximated root, and that is why your in mind angle (Pi/9) is never an exact existing angle

So, understanding that fiction unreal and nonexistent number as 2^{1/3} would immediately remove the complete puzzle about trisecting of the arbitrary angle

And let us see what works that had been added by secretive researchers or Wikipedia writers? after that old date reference and what are the rules of this science forum or the rules of professional mathematicians to promote that old interesting issue raised by a mature only to the science community

Regards
Bassam King Karzeddin
3ed, May 2017

[bassam king karzeddin]

Here is the full old thread (in 2005) with 37 messages for interested people, and I simply thought that would be completely visible here

Kindly, confirm the visibility of this old topic with all 37 replies, thanks

http://mathforum.org/kb/thread.jspa?forumID=13&threadID=1175275&messageID=3846213#3846213

BK

[burs...@gmail.com]

Dont get tree-ed in the forest Stevie Wonder, 3ed,
May 2017. Even when it was the 3rd, May 2017.

Am Mittwoch, 3. Mai 2017 20:05:00 UTC+2 schrieb bassam king karzeddin:
> 3ed, May 2017
> BK

[Vinicius Claudino Ferraz]

Let alpha in [0, 2 pi).

Stretch the sphere until it becomes a straight line segment [0, 2 pi ).

Divide your "angle" or line segment into n parts. Let n = 3.

Twist your line segment back to a sphere.

Do the lines OA, OB, OC. Such that AÔC = n AÔB.

[Vinicius Claudino Ferraz]

I know, I know.
Nobody stretched π ever.
Nobody twisted π/n no ever.

[bassam king karzeddin]

But the fools had already stretched it to be visible from a visibly stretched radius before eyes, not only that but they also claimed many exact constructions of (pi), and more foolishly they had constructed its square root also, and soon they would certainly claim construction of its cube root too, and the stupidity would go farther in the future to construct its nth root also, wonder!

but they never can understand that given three non-negative integers
(x, y, z), then it is impossible to find such integers under (pi) power, such that

x^{pi} + y^{pi} = z^{pi},

BK

[bassam king karzeddin]

a link to the above-mentioned equation, with around 75 thousand views and 16 answers (10 collapsed), but none of those answers were even right for sure
https://www.quora.com/Can-we-find-three-positive-integers-x-y-and-z-such-that-x-pi-+-y-pi-z-pi-How-about-three-integers-such-that-x-e-+-y-e-z-e

All of that nonsense was due to so many well-established fictions theorems stories in our current modern mathematics for sure

And one really wonders from where to start exposing them, from (pi) as the mother ghost number, or from the invented arithmetical cube root of two denoted by (\sqrt[3]{2}), or simply 2^{1/3}, to the misuse of negative numbers, to the invented imaginary units ""i"" the basic of baseless complex numbers, or to the polynomial roots that are mainly symptoms of non-existing roots, or from the non-euclidean geometry, or ....etc

But human minds love to stay astray in fictions and so deeply involved in manufacturing more endless fictions from those same basic fictions

But soon there would be no more room sufficient to contain any useless fictions for sure

And the dumps would remain so going blindly in the same wrong directions unless affected by another bigger force that would change and correct their directions forever and for sure

BK

[Vinicius Claudino Ferraz]

Some angles I know how-to trisect. And bisect.

We construct pi/2, then we have already trisected 3 pi/2.
We construct pi/4, then we have already trisected 3 pi/4.
We construct pi/2^n, then we have already trisected 3 pi/2^n.

We construct pi/3 = 60º then we have already trisected pi.
We construct pi/(3 * 2^n), then we have already trisected pi/2^n.

Their sum, too. We construct 60º + 45º.
We construct pi/3 + pi/4 then we have already trisected pi + 3 pi/4.

We construct pi/2^i + pi/(3 * 2^j),
then we have already trisected 3 pi/2^i + pi/(2^j).

But the sum goes from 1 to n.

We construct Sum_{k = 1}^n [ pi/2^i_k + pi/(3 * 2^j_k) ],
then we have already trisected Sum_{k = 1}^n [ 3 pi/2^i_k + pi/(2^j_k).

But the differences too.

Don't forget if alfa can be trissected then alfa + alfa + alfa too can be trissected.

What are the constructible angles? 90, 60, 30, 45. "Esquadros"?

https://pbs.twimg.com/media/C_yxHoQWsAECoXF.jpg:large

All of this comes from euclid's consctruction of an equilateral triangle.
New ideas? Please send. Don't keep the secret. lol

[Vinicius Claudino Ferraz]

I need someone to prove or disprove this procedure.

Draw the main circle C(O, r).

Draw the diameter AB.

Draw the circles C(A, 2r) and C(B, 2r), which intersect at D, E.

Consider the half planes ABD and ABE.

Divide AB into n equal lengths AA_1, A_1 A_2, A_2 A_3, ..., A_{n-1} B.

Fix n = 10.

DA_1 intersects the main circle in the half plane ABE at point E_1.
DA_3 intersects the main circle in the half plane ABE at point E_3.
DA_5 intersects the main circle in the half plane ABE at point E_5.
DA_7 intersects the main circle in the half plane ABE at point E_7.
DA_9 intersects the main circle in the half plane ABE at point E_9.

EA_1 intersects the main circle in the half plane ABD at point D_1.
EA_3 intersects the main circle in the half plane ABD at point D_3.
EA_5 intersects the main circle in the half plane ABD at point D_5.
EA_7 intersects the main circle in the half plane ABD at point D_7.
EA_9 intersects the main circle in the half plane ABD at point D_9.

I wanna know if E_1 E_3 E_5 E_7 E_9 D_9 D_7 D_5 D_3 D_1 is a regular DECAGON.

If my teacher was right [1997] we can build
the central angles 2 pi / n.
And it's internal angles.
And it's external angles.

[Barry Schwarz]

There is nothing special about 10. Does it work for n = 6 to build a
regular hexagon.

Start with the unit circle at the origin. Let A = (-1,0) and B =
(1,0) (the horizontal diagonal) be the diagonal of choice. So
A = (-1,0)
A_1 = (-2/3,0)
A_2 = (-1/3,0)
A_3 = (0,0)
A_4 = (1/3,0)
A_5 = (2/3,0)
B = (1,0)

Circle (A,2r) is (x+1)^2 + y^2 = 4
Circle (B,2r) is (x-1)^2 + y^2 = 4
These two curves intersect at D = (0,sqrt(3)) and E = (0,-sqrt(3)).

Note that given the horizontal and vertical symmetry, if D_i = (x,y)
then E_i must be (-x,y), Thus we know the hexagon is "standing on one
of its vertices" and the far left and far right sides are vertical. We
also know that the central angle of a hexagon is 60 degrees. Thus
angle AOD_1 is 30 degrees. sin(30) = 1/2 and cos(30) = sqrt(3)/2. D_1
must be (-sqrt(3)/2,1/2). Is this point on the line EA_1?

The equation of a line passing through (x1,y1) and (x2,y2) is
(y-y1)/(y2-y1) = (x0x1)/(x2-x1)
Substituting E and A_1 yields
(y+sqrt(3))/(0+sqrt(3)) = (x-0)/(-2/3-0)
(y+sqrt(3))/sqrt(3) = x/(-2/3)
(y+sqrt(3)) = -3x*sqrt(3)/2
Substituting the presumed values for D_1
(1/2+sqrt(3)) = -3(-sqrt(3)/2)*sqrt(3)/2
(1/2+sqrt(3)) =/= 3sqrt(3)*sqrt(3)/2/2 = 9/4

The procedure does not appear to generate regular polygons.

--
Remove del for email

[burs...@gmail.com]

Can we derive some of Ramanujan's infinite root
series, by impossibility of some n-gon construction

but then symbolically do some iterative approximation,
an iteration that has square root in it,

Ramanujan's infinite root and its crazy cousins
https://www.youtube.com/watch?v=leFep9yt3JY

BTW: there is something fishy:

3 = sqrt(1+2*sqrt(...)) = 4

(Hey JG and BKK why pick not on this, showing
the big bad wolf BIGGUS STUPIDUS failure?)

[bassam king karzeddin]

If you mean integer degree angle where (Pi = 180 degrees), then we had provided the solution quite many times, but maybe you did not notice, wonder!

And in short, the constructible integer degree angles are of the form (3n), where (n) is integer number, and if (n) is divisible by (3), then the angle (3n) is trisectible, otherwise not at all

And when the integer degree angle is (m) integer, where (m) is not divisible by (3), then the integer angle (m) does not exist for sure

Otherwise, show it in any triangle with exactly known three sides, where you cannot for sure

So, there is no hidden secret in this regard for sure LOL

BKK

[Leon Aigret]

On Sun, 14 May 2017 08:33:50 -0700 (PDT), Vinicius Claudino Ferraz
<borala...@gmail.com> wrote:

>I need someone to prove or disprove this procedure.
>
>Draw the main circle C(O, r).
>
>Draw the diameter AB.
>
>Draw the circles C(A, 2r) and C(B, 2r), which intersect at D, E.
>
>Consider the half planes ABD and ABE.
>
>Divide AB into n equal lengths AA_1, A_1 A_2, A_2 A_3, ..., A_{n-1} B.
>
>Fix n = 10.
>
>DA_1 intersects the main circle in the half plane ABE at point E_1.
>DA_3 intersects the main circle in the half plane ABE at point E_3.
>DA_5 intersects the main circle in the half plane ABE at point E_5.
>DA_7 intersects the main circle in the half plane ABE at point E_7.
>DA_9 intersects the main circle in the half plane ABE at point E_9.
>
>EA_1 intersects the main circle in the half plane ABD at point D_1.
>EA_3 intersects the main circle in the half plane ABD at point D_3.
>EA_5 intersects the main circle in the half plane ABD at point D_5.
>EA_7 intersects the main circle in the half plane ABD at point D_7.
>EA_9 intersects the main circle in the half plane ABD at point D_9.
>
>I wanna know if E_1 E_3 E_5 E_7 E_9 D_9 D_7 D_5 D_3 D_1 is a regular DECAGON.

No, it is just a good approximation. Arc E_1 - E_3 spans an angle of
about 0.1968 pi radians and arc E_3 - E_5 about 0.1997 pi radians.

>If my teacher was right [1997] we can build
>the central angles 2 pi / n.
>And it's internal angles.
>And it's external angles.

Starting with a blank sheet of paper, apart from a single line segment
with nominal length 1, constructions with ruler and compass can only
provide line segments with lenghts that are derived from 1 by
additions, substractions, multiplications, divisions and square root
takings.

Galois theory shows that the only constructible regular n-gons are
those with n = 2^k p_1 p_2 ... p_j where the p_i are different Fermat
primes.

Leon

[bassam king karzeddin]

How many times you had been informed and learned that infinity is the first step towards that Fake Paradise of those BIG FOOLS (whether they had been classified illegally as genius people or not yet),

Get rid of it moron, infinity gives you more factions than you can imagine for sure

BKK

[Vinicius Claudino Ferraz]

180 bissected
90 bissected
45 trissected
15 trissected
5
and 5º = pi/36 ? how-to construct?

[Vinicius Claudino Ferraz]

i think the real question is how to five-sectioning pi.
build pi/5.

[Vinicius Claudino Ferraz]

I meant
a retired mathematician could take pi and bissect 1234 times.
Then multiply by the prime 11.

==> 11 pi/(2^1234)

I also meant
a retired mathematician could take pi/3 and bissect 2345 times.
Then multiply by the prime 7.

==> 11 pi/(2^1234) ± 7 pi/(3 * 2^2345)

I could also mean how-to find cos(x/3) and sin(x/3)
expanding
cos 3x = cos 2x cos x - sin 2x sin x
sin 3x = sin 2x cos x + cos 2x sin x

Then solving a cubic equation by radicals ax^3 + bx^2 + cx ^ d = 0.

I don't know if the roots of the polynomial are ruler and compass constructible.

Then it could be easy do the triangle (cos x/3, sin x/3, 1)

If this were possible (all we need is time, isn't it?)

Then I could also mean
a retired mathematician could take pi and trissect 3456 times.
Then bissect 4567 times.
Then multiply by the prime 13.

==> 11 pi/(2^1234) ± 7 pi/(3 * 2^2345) ± 13 pi/(3^3456 * 2^4567)

We would see a good radical, not a collection of digits at the calculator.
Maybe it would be a long long tEx.

[Peter Percival]

Vinicius Claudino Ferraz wrote:

> Then solving a cubic equation by radicals ax^3 + bx^2 + cx ^ d = 0.
>
> I don't know if the roots of the polynomial are ruler and compass constructible.

Generally, no. This was proved by Wantzel even before Galois' work was
known.

--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan

[bassam king karzeddin]

Vinicius Claudino Ferraz wrote:

> and 5º = pi/36 ? how-to construct?

Your question is exactly equivalent to my following question

How to construct an angle that is really a fiction and unreal angle that doesn't exist for sure, wonder!

But this is an absolute fact that the whole mathematics stands behind for sure

However, so much written about this SILLY issue by me, and so unfortunately nither the ancient mathematicians or scientists (including of course the Greeks) nor the modern professional mathematicians or scientists could ever understand such a very simple impossibility, nor they want to understand for so silly reasons regarding their so little and too negligible egoistic incurable and inherited diseases for more than sure

And do not tell me please any kind of any approximations, since this is certainly a very exposed game for kids mainly

And the angle case is much stronger than the real number case, wherein the second they used to play an epsilon-delta game to create infinitely many fiction unreal and non-existing numbers, but here they cannot for sure

And to save the situation they have no other choice but to listen carefully to the KING and drop all those invented fiction real numbers with endless digits that are not constructible numbers

And like the vast majorities of mathematicians, you most likely would not accept this simplest shining fact, but it doesn't at all any matter for sure.

BKK

[Vinicius Claudino Ferraz]

to build pi/36 = 2 pi/72 is the central angle of a 72-sided polygon.

to build pi/5 = 2 pi/10 is the central angle of a decagon.

to build 1 degree = pi/180 = 2 pi / 360 is the central angle of a 360-sided polygon.

i thought you said integer degrees!

[bassam king karzeddin]

On Tuesday, May 16, 2017 at 7:51:01 PM UTC+3, Vinicius Claudino Ferraz wrote:
> to build pi/36 = 2 pi/72 is the central angle of a 72-sided polygon.

72 - Sided polygon doesn't exist, and impossible to construct, for sure

>
> to build pi/5 = 2 pi/10 is the central angle of a decagon.
>
> to build 1 degree = pi/180 = 2 pi / 360 is the central angle of a 360-sided polygon.

Also, it is impossible to build one-degree angle, one-degree angle doesn't exist basically, for sure

>
> i thought you said integer degrees!

Yes here we are describing the integer degree angle that can exist and thus can be constructed exactly, those angles are simply of this form (3n), where (n) is integer number

And if (n) is divisible by (3), then the angle (3n) is trisectible for sure

But the integer degree angle (m), where (m) is integer not divisible by (3), then the integer angle (m) is a fiction, unreal and nonexisting angle for sure

However there are smilar analogy when using (Pi), I had explained it in my posts, and generally most of our known angles are actualy fiction angles, exactly similar to the case of our real numbers that are mostly fiction, unreal and non existing numbers for 100% sure

BKK

[Leon Aigret]

By first constructing 2 pi/5 and then halving it?

https://en.wikipedia.org/wiki/Pentagon#Construction_of_a_regular_pentagon

Leon

[bassam king karzeddin]

So, nothing new in the reference, they simply show some arts of constructed polygons by Euclid and before thousands of years back, wonder!

BKK

[Vinicius Claudino Ferraz]

Em quarta-feira, 17 de maio de 2017 04:49:46 UTC-3, bassam king karzeddin escreveu:

> > https://en.wikipedia.org/wiki/Pentagon#Construction_of_a_regular_pentagon
> >
> > Leon
>
> So, nothing new in the reference, they simply show some arts of constructed polygons by Euclid and before thousands of years back, wonder!
>
> BKK

So there are guys who think we can not take n = 360 icecream sticks
and exhibit a a regular polygon of n = 360 sides?

just because 360/n ≠ 3k
n = 72 sticks idem
just because 360/72 ≠ 3k

There aren't some radius
There aren't some angles
There are many holes in (ρ, θ) polar cohordinates.

Has anybody seen a "deaf number"?
I searched it at wiki and there are lots of deaf people. :-)

In a book, after proving irrationality and transcendence of e and pi,
there was a theorem
every deaf \circ deaf \circ deaf \circ ... \circ deaf number is algebrical.
So there is a polynomium.
Which is it?

"Todo número surdo é algébrico."

As I remember, a/b + sqrt(c/d)
e/f + sqrt{a/b + sqrt(c/d)}
and so on.

twitter.com/mathspiritual

[Vinicius Claudino Ferraz]

I'm still imagining, because it's too weird

There are four classes of angles. 3ℤ, 3ℤ + 1, 3ℤ + 2 AND ℝ - ℤ degrees.

A = ℤ pi/60
B = ℤ pi/60 + pi/180
C = ℤ pi/60 + 2 pi/180
D = ℝ - A - B - C

"From each 3 integers, 1 is constructible, 2 do not exist."

What about D? How many times we can bissect pi?

"f(n) = pi/(2^n) exists"

[Vinicius Claudino Ferraz]

Everything which is imaginable (and consistent) does exist.

[bassam king karzeddin]

On Thursday, May 18, 2017 at 12:27:01 AM UTC+3, Vinicius Claudino Ferraz wrote:
> Everything which is imaginable (and consistent) does exist.

Yes, being imaginable and also consistent does exist indeed in mind only, but never in the existing reality for sure
exactly like chess game mainly for entertainment, but never should be related to any real science, since it well ruins it completely for sure
exactly the case with the damaged physics mainly due to the consistent imagination in mathematics, for sure!

BKK

[bassam king karzeddin]

On Wednesday, May 3, 2017 at 8:50:05 PM UTC+3, bassam king karzeddin wrote:
> >
> >
> >
> > Le 19/07/05 12:37, dans
> > 4dd7d$42dcd769$d52f93dc$25...@news.chello.at,
> > « Jutta Gut » <gut.jutt...@chello.at> a écrit :
> >
> > >
> > > "bassam king karzeddin" <bas...@ahu.edu.jo> schrieb
> > im Newsbeitrag
> > >
> > news:29354512.1121766911405.JavaMail.jakarta@nitrogen.
> > mathforum.org...
> > >> That is grate,
> > >>
> > >> This,might open doors to constructible polygons
> > >>
> > >> In fact,I have deduced & proved the same thing,I
> > have mentioned that here:
> > >>
> > >>
> > http://mathforum.org/kb/message.jspa?messageID=3802920
> > &tstart=0
> > >>
> > >> I will provide examples soon.
> > >
> > > If I understand correctly, you have shown that in
> > an triangle with
> > > the sides a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) one
> > angle is three times
> > > another one.
> > >
> > > The more interesting question would be: given an
> > angle, how to
> > > construct a triangle with the sides a^3 ,
> > a*(b^2-a^2) , b*(b^2-2*a^2)
> > > and the given angle?
> >
> > It tried for some simple angles: pi/8, pi/6, pi/5,
> > sides can be computed
> > with square roots. For pi/9, you have the 3rd degree
> > equation x^3=3*x+1.
> > Not surprising, since pi/9 is not constructible. But
> > it's still interesting
> > to know which triangles have two angles A,B such that
> > A=3*B.
> >
>
> That is because there is not any EXACT real root for the polynomial (x^3 - 3x -1 = 0), for sure, but nearly an approximated root, and that is why your in mind angle (Pi/9) is never an exact existing angle
>
> So, understanding that fiction unreal and nonexistent number as 2^{1/3} would immediately remove the complete puzzle about trisecting of the arbitrary angle
>
> And let us see what works that had been added by secretive researchers or Wikipedia writers? after that old date reference and what are the rules of this science forum or the rules of professional mathematicians to promote that old interesting issue raised by a mature only to the science community
>
> Regards
> Bassam King Karzeddin
> 3ed, May 2017

You have to understand well first the true meaning of the word arbitrary in maths, though mathematics doesn't define it correctly

An arbitrarily chosen distance or length or angle is an existing distance or length or angle (just before your big eyes), thus is constructible for sure (Did not you just construct it arbitrarily, also by the rules?) wonder!

So, to trisect any existing angle or any arbitrary angle just consider the following circle with arbitrary radius and center at point (C)

then chose any arbitrary two points at the circumference as (A & B)

So, you have a triangle (ABC), say CLOCKWISE

Bisect the angle (ABC), and let the bisector intersects the radius (CA) at point (D)

And let your assumed arbitrary angle be simply (CDB) clockwise, then the angle (DBC) is the exact trisection angle of your assumed arbitrary angle for sure

But, you will get completely confused whenever you assume (but cannot construct) in mind a fiction and nonexisting angle for angle CAB, that explained well enough in my posts

Even the Greeks who posed this OLDEST problem did not understand it the way I expose it here for you, and for sure

(c)

Regards
Bassam King Karzeddin
May 20, 2017

[Vinicius Claudino Ferraz]

I liked that.
∠ABD = ∠BDC = x
∠ABC = 2x
|CA| = |CB|
ACB is isósceles
∠ABC = ∠CAB = 2x
∠CDB = 2x + x

Do you have more?

Now we can build
19 pi/(3^i_1 * 2^j_1) ± 17 pi/(5 * 3^i_2 2^j_2)

bissetriz
trissetriz
do you have a "pentassetriz"?

[burs...@gmail.com]

I guess you have problems with the notion "construction", and
especially whats the input and output of trisceting. Lets recap:

1) Input an angle x

2) Output an angle x/3

So if you have an angle x, at some point D, how do you choose
C, and then A,B, so that you can read off the angle x/3?

Your construction starts with C, and unless you have a kind of
neusis, you will hardly hit D. You could construct a new form

of neusis, a bisector neusis, and use some intial C and A, and
then slide B until it stops at D. The neusis would look like

an expandable trellis, which can do the bisection mechanical.
But this is not the machanic of rule and compas.

https://en.wikipedia.org/wiki/Neusis_construction

[bassam king karzeddin]

On Saturday, May 20, 2017 at 10:31:38 PM UTC+3, burs...@gmail.com wrote:
> I guess you have problems with the notion "construction", and
> especially whats the input and output of trisceting. Lets recap:
>
> 1) Input an angle x
>
> 2) Output an angle x/3
>
> So if you have an angle x, at some point D, how do you choose
> C, and then A,B, so that you can read off the angle x/3?

How easy I made it for you but still not working with you, wonder!

Did not you notice that others had got it from the first sight, wonder!

Still asking about C trying hopelessly to confuse the innocent students
so deliberately

Again, C is the center of any arbitrary circle, where (A, B) are two points on the circumference of that circle, is that so difficult for you to understand? really wonder!

>
> Your construction starts with C, and unless you have a kind of
> neusis, you will hardly hit D. You could construct a new form

Again trying to cheat and show that nonsense neusis is the origin, just because it is documented in that Wikipedia, so what?

My solution is also PUBLISHED in a SCIENCE site supposedly for MATHEMATICS, wonder! and I don't know what are the duties of those site owners or moderators here really

>
> of neusis, a bisector neusis, and use some intial C and A, and
> then slide B until it stops at D. The neusis would look like
>

MY solution is only a few steps by the sacred rules of construction, then representation, whereas that neusis is not certainly by the rules and also tedious, expandable, trellises, ....etc

> an expandable trellis, which can do the bisection mechanical.
> But this is not the machanic of rule and compas.
>
> https://en.wikipedia.org/wiki/Neusis_construction

Why do not you hurry up before anyone else write it in Wikipedia so professionally, since the idea is so silly,

However, the same idea included in that symbolic triangle I provided in 2004

And naturally, there are too many subsequences or more generalizations to this one, (read my old PUBLISHED posts in this regard), they are certainly much more advanced than wiki mini pages

And not that whenever you choose arbitrarily your points (A, B), SO for your arbitrary angle (CDB) is changing arbitrarily and always have a trimester exact angle (CBD)

But the whole issue is that whenever you assume a fiction nonexisting
and fake angle then obviously you would get sucked since no arbitrary angle can be fiction angle but differently a constructible angle, for sure

Had the Greeks understood the constructible numbers the way it is explained completely for you, then they would never ask such a silly questions, for sure

But, certainly, the Greeks never understood what is the deep meaning of a constructible number, since rational numbers (they only used to understand) are certainly SO HELPLESS in this regards, for sure

Bassam King Karzeddin
May 21, 2017

[burs...@gmail.com]

Dream on Unicorn sodomizer, similar to JG and AP,
you only produced irrelevant nonsense so far.

[burs...@gmail.com]

Hey AP, this is not meant as a weak-up call and to disturb
your napping. If you pile more shit over shit, your posts

don't get any better. You have a track record of 30-years
posting zero math, spanning idiotic topics like Boole-T-Shirts

and the 10^604 infinity border. Neither Martin Gardner nor
Henry Dudeney would call you a disciple, the recretional

value is too low. Its just struggle, and struggle, and
struggle, of some imbecile...

[burs...@gmail.com]

Well as a freak show, its quite amusing...

[bassam king karzeddin]

On Sunday, May 21, 2017 at 5:00:50 PM UTC+3, burs...@gmail.com wrote:
> Dream on Unicorn sodomizer, similar to JG and AP,
> you only produced irrelevant nonsense so far.

OK, Let that be your opinion or anyone else opinion (it doesn't matter) for sure

Anyway, Facts are written, by me, by anyone else no matter, and if people do not get them, certainly a future artificial intelligence will, for sure

Who are you after all to decide the truth? wonder!

BKK

[Vinicius Claudino Ferraz]

Let me try with Descartes

D = (0, 0)
C = (c, 0)
A = (a, 0)
c < 0 < a
where is B = (u, v)?
|CA| = a - c
|CB| = a - c ==> (u - c)² + v² = (a - c)² (1)
x = ∠CDB = 3 ∠DBC (2)
∠DAB = 2 ∠DBC (3)
∠DBA = 1 ∠DBC (4)

With four equalities we have to find c, a, u, v.

cos ∠DBC = <D - B, C - B> / (|DB| |CB|)
= <(-u, -v), (c - u, -v)> / ( sqrt{u² + v²} sqrt{(c - u)² + v²} )
cos ∠CDB = ...
cos ∠DAB = ...
cos ∠DBA = ...

Patience.

[burs...@gmail.com]

Its very easy, use sum of angles in a triangle:

Triangle C,A,B: alpha + alpha + beta = pi
Triangle C,D,B: gamma + alpha/2 + beta = pi
Triangle D,A,B: alpha + alpha/2 + (pi-gamma) = pi

Now the last equation does already the job:

alpha + alpha/2 + (pi-gamma) = pi

alpha + alpha/2 - gamma = 0

3 * alpha/2 = gamma

So the last equation shows, that gamma is indeed
trice of alpha/2.

But this says nothing that we can construct alpha/2
from gamma.

BKK is notorious for this game:

https://hsm.stackexchange.com/q/3291/4900

https://math.stackexchange.com/q/1419425/4414

But I am currently not yet sure, whether Descarte wise
the triangle from this post here, is related to the
triangle in the MSE and HSM posts.

In the MSE and HSM posts its only one triangle if
I remember well, but here its a whole family. So
if you have a Decarte solution this would

be quite interesting, whether and how the MSE and HSM
case would drop out from it.

[burs...@gmail.com]

Corr.:

So the last equation shows, that gamma is indeed

thrice of alpha/2.

https://en.wikipedia.org/wiki/English_numerals#Multiplicative_adverbs

[Vinicius Claudino Ferraz]

From x, is easy to build 3x.

But from an angle theta = 3x = XDY

We need to extend X to some determined C
We need to extend Y to some determined B
We need to extend D to some determined A

and the most dificult is ∠CBD = ∠DBA

It's too dificult the system that proves these points are well determined.

Take a look: https://pbs.twimg.com/media/DAdWeJIXkAE_M4D.jpg:large

Em domingo, 21 de maio de 2017 14:22:42 UTC-3, burs...@gmail.com escreveu:
> alpha/2.

[Vinicius Claudino Ferraz]

Em segunda-feira, 22 de maio de 2017 17:51:22 UTC-3, Vinicius Claudino Ferraz escreveu:
From x, is easy to build 5x.

But from an angle theta = 5x = XDY

We need to extend XD to some determined line segment CXD
We need to extend DY to some determined line segment DYB
We need to extend CD to some determined CA

and the most dificult is ∠CBD = ∠DBA

The first new construction is that E ∈ CD, such that ∠CBE = x
Therefore ∠ECB = 3x + x

The second new construction is that F ∈ EC, such that ∠FBE = x
Therefore ∠FEB = 4x + x

Pentassetriz!

The third new construction is that G ∈ FE, such that ∠GBF = x
Therefore ∠GFB = 5x + x

Hexassetriz!

The fourth new construction is that H ∈ GF, such that ∠HBG = x
Therefore ∠HGB = 6x + x

Heptassetriz!

And so on.

Is the system is well-determined?

If yes, then we have the construction of x/n, ∀ n > 2.

I think this is only a construction of 2x, 3x, 4x, 5x, 6x, 7x, ..., nx for dummies. lol

Vinicius
twitter.com/mathspiritual

[bassam king karzeddin]

In my provided reference, and from old posts nearly (2004)

I had already defined those triangles that have an angle and it's multisection angle in the same triangle

and before sinking too deep into Cardano solutions for cubic polynomials with complex roots...etc, (which unfortunately had been refuted in general)

And for a trisection of an arbitrary angle, provided here, tell me please which is that angle that you can construct by proper placing of points
(A & B) such that it is not trisectiable angle, I confirm no angle

BKK

[Vinicius Claudino Ferraz]

Bassam,

I wanna see that. Send me, please.
You went to Cardano's work and concluded we cannot tri.sect 3º/3 = 1º ???
20º/3
40º/3
80º/3
88º/3

[Tim Golden BandTech.com]

Hi Bassam. I have a mechanical solution that does not involve iteration.

Firstly, we know that we can draw an equilateral triangle and its center using a compass and ruler on flat paper with pencil. draw a baseline, mark a position on the line, compass distance to another point on the line, and then the mutual distance from those points to one side of the line establishes an equilateral triangle. repeating this operation again a couple of times allows the center of the first equilateral triangle to be established. Mark critical lines from the center of the equilateral out to its vertices. Call this the trisection.

Now on another piece of paper we take the arbitrary angle ABC and mark out a constant radius from B. Call this aBc. Now we cut out aBc using the straight aB and aC and the radially curved ac. Now we join a and c so that we have a right cone.

Lay the cone so that its center is directly over the center of the trisection. Trace onto the cone the trisectional intersections d,e,f. Cut open the cone along d to the vertex of the cone. e and f are the trisections of the initial angle ABC. I suppose we could have forced the ac joint to lay upon one of the triradials and saved a cut. Sorry to require a knife in the kit, and some tape as well. At least the square root of three was taken care of without them.

> Here is the full old thread (in 2005) with 37 messages for interested people, and I simply thought that would be completely visible here
>
> Kindly, confirm the visibility of this old topic with all 37 replies, thanks
>
> http://mathforum.org/kb/thread.jspa?forumID=13&threadID=1175275&messageID=3846213#3846213
>
> BK

[bassam king karzeddin]

Hellow Tim G

Most likely it is not so clear what a mechanical method by knife or so would resolve this oldest historical puzzle (especially for the expert specialists) where they still claim so many false constructions for sure

We do understand that this type of problems had become as a rich source of business making for the TOP professional's mathematicians by finding other rules instead of stated original tools which are mind blowing tools, as if the tools were really the problem, not at all for sure, the main problem is because of the impossibility to understand the impossibility of constructing and also trisecting something that really doesn't exist in reality but only on those incompetent little minds for the sake of falling into so many fiction stories that were devilishly or so foolishly cooked up and were made illegally as real numbers

And, It is never any puzzle (at least for me) since this was proved quite many times and beyond any little doubt for sure

BKK

[Tim Golden BandTech.com]

Well, there may be a way to take my method and refine it down to the more classical geometry. It is very close. It happens that the compass does trace out a cone, but the sector which forms the critical angle has to be scaled down by a factor of one over two pi. Can we extract the straight line circumference from a compass? If so then we can trisect. The conversion from a radial curve to a straight segment then means that the trisection of an angle is no different than the trisection of a segment. In some regards the usage of pi as a value implies this ability, even in its symbolic form. Under this logic then trisection has been established. Still, we will see that the means of arriving at a value for pi will involve slightly more mechanics than the traditional ruler and compass and pencil and paper.

It's a good puzzle and it does cause some analysis into what spaces with pi variance would look like. I have a generalization of the cone at my website that covers this from an angular point of view, which is how I arrived at this solution.
http://bandtechnology.com/ConicalStudy/conic.html

[bassam king karzeddin]

On Wednesday, May 31, 2017 at 4:38:54 PM UTC+3, Vinicius Claudino Ferraz wrote:
> Bassam,
>
> I wanna see that. Send me, please.
> You went to Cardano's work and concluded we cannot tri.sect 3º/3 = 1º ???
> 20º/3
> 40º/3
> 80º/3
> 88º/3

Yes, you cann't further trisect (say in integer degrees) any integer degree angle (n), where (n) is not divisible by (9)

The trisect able integer degrees angle are of the form (9m), where (m) is integer, and below is the set of positive integer trisect able angles

{9, 18, 27, 36, 45, 54, 63, 72, 81, 90, ...}

BKK

[Simon Roberts]

I don't want to be arbitrary, but I can tri-sect a 180 degree angle.

[burs...@gmail.com]

Thats even not mean.

[bassam king karzeddin]

You can always trisect any existing trisect able arbitrary angle, sure

BKK

[Vinicius Claudino Ferraz]

I tried to conclude the same as you, but I couldn't.

Let ABC be a triangle whose sides are a, b, c.
angle B = 3 angle A
angle C = 180° - 4 Â > 0
180° > 4 Â
 < 45°
angle B < 3 * 45° = 135°

So could not we trissect anything <= 3/4 pi ???

[Tim Golden BandTech.com]

On Friday, June 2, 2017 at 8:11:34 PM UTC-4, Tim Golden BandTech.com wrote:
> Hi Bassam. I have a mechanical solution that does not involve iteration.
>
> Firstly, we know that we can draw an equilateral triangle and its center using a compass and ruler on flat paper with pencil. draw a baseline, mark a position on the line, compass distance to another point on the line, and then the mutual distance from those points to one side of the line establishes an equilateral triangle. repeating this operation again a couple of times allows the center of the first equilateral triangle to be established. Mark critical lines from the center of the equilateral out to its vertices. Call this the trisection.
>
> Now on another piece of paper we take the arbitrary angle ABC and mark out a constant radius from B. Call this aBc. Now we cut out aBc using the straight aB and aC and the radially curved ac. Now we join a and c so that we have a right cone.

Oops. typo. that should read
straight aB and Bc and the radially curved ac. This forms the cone which then gets fitted to the trisected plane for tracing.

[Vinicius Claudino Ferraz]

Today we're going to trissect 3º.

Let  = 3º, B = 1º, C = 176º.

By the sins' rule,

a / sin 3º = b / sin 1º = c / sin 176º

Fix b = 1.

a = sin 3º / sin 1º
c = sin 176º / sin 1º

Do you claim it's impossible to construct these numbers?
(1, sin 3º / sin 1º, sin 176º / sin 1º)

I guess you have understood.
But I like this illustration:

https://pbs.twimg.com/media/DB6hyXIWsAQhb2_.jpg:large

Att,

Vinícius Claudino Ferraz
twitter.com/mathspiritual

Em quarta-feira, 3 de maio de 2017 14:50:05 UTC-3, bassam king karzeddin escreveu:

[Vinicius Claudino Ferraz]

b = 1
a = 2,9987816540381914600124868800879...
c = 3,9969544598219302258830072345067...

[bassam king karzeddin]

On Thursday, June 8, 2017 at 5:45:18 PM UTC+3, Vinicius Claudino Ferraz wrote:
> I tried to conclude the same as you, but I couldn't.
>
> Let ABC be a triangle whose sides are a, b, c.
> angle B = 3 angle A
> angle C = 180° - 4 Â > 0
> 180° > 4 Â
> Â < 45°
> angle B < 3 * 45° = 135°
>
> So could not we trissect anything <= 3/4 pi ???

You must consider the adjacent angles, it is more than easy if the sum of two angles on a straight line (180) degrees, then once you trisect one angle you can easily trisect the neighbouring angle for sure

BKK

[bassam king karzeddin]

The whole secret issue here is the common wrong be leaf among all mathematicians about the true existence of absolute (Pi) as any other real existing length, where then everything becomes easily attainable

But the so sad fact that (Pi) was the first ghost number that never existed except as convenient approximation of a constructible number

BKK

[bassam king karzeddin]

It is impossible for any integer degree angle (n) to exist if (n) is not divisible by (3), thus no one degree angle exist, except in the wrong immaginations of mathematicians

And I know this seems ridiculous to all mathematicians (especially those who had passed away not knowing this simplest fact in their life's

I am also not saying this arbitrary, but because I had proved it rigorously in my posts, where no one was capable of previewing it correctly

Or maybe the so obvious fact that common sense among professional mathematicians are not at all working

BKK

[bassam king karzeddin]

And the one-degree angle is a fiction angle that is impossible to exist

But the shameless current mathematics had already and historically documented those many angles with their many trig functions as reference angles for sure

Now, when shall such a big shame upon the foreheads of mathematicians must be removed to rubbish? wonder!

BKK

[burs...@gmail.com]

If you could draw x^(2/3) + y^(2/3) = 1, you can trisect,
didn't verify it (whats the proof?), but it is shown here:

"given an astroid (= quadricuspid hypocycloid (x^(2/3) + y^(2/3) = 1) = red thing). Draw segment (blue) from center at angle A (= pi/3 in illustration) w.r.t. horizontal, of length 1/4. At its end, draw a circle (green) of radius 3/4. This will cut the astroid in eight places. The cyan, magenta, and yellow radii are at angles -A/3, (2 pi - A)/3, and (4 pi - A)/3."
http://www.tweedledum.com/rwg/trisect.htm

[burs...@gmail.com]

You could roll a smaller circle inside a bigger circle.

Picture: The hypocycloid construction of the astroid.
https://en.wikipedia.org/wiki/Astroid

[Chris M. Thomasson]

On 6/23/2017 11:26 AM, burs...@gmail.com wrote:
> You could roll a smaller circle inside a bigger circle.
>
> Picture: The hypocycloid construction of the astroid.
> https://en.wikipedia.org/wiki/Astroid

Also, a Cardioid comes to mind... ;^)

https://en.wikipedia.org/wiki/Cardioid

[burs...@gmail.com]

Etienne Pascal (1588-1640), the father of Blaise ( 1 623-1 662) the mathematician, philosopher, and writer, trisected with a cardioid
https://www.amazon.com/Trisectors-Spectrum-Underwood-Dudley/dp/0883855143

but searching for trisection and astroid, doesnt give me much

[bassam king karzeddin]

My question is so simple and very naive too, I would like to repeat it again and again till it becomes completely understood by all kid students on earth, but most likely not by any alleged top professional mathematician, Wonder!

"Provide only one triangle with exactly known sides length (constructible), such that (at least one of its angles in integer degrees say (n), where (n) is not divisible by (3)"

And this question may be considered as a real mirror to the world professional mathematicians, where they simply avoid to look into for very obvious psychological reasons for sure

And this question is mainly designed to start healing and enlightening all the alleged top mathematicians on earth from the huge inherited fictions they had already acquired mainly from their current business education that considers many big historical figures in mathematics as real knowledgeable humans

And the reasons to prefer fictions to mere facts by humans are generally so silly and also understandable

So, please don't understand me correctly, since I generally direct my inquiries not to so innocent people like you here, but to the responsible people for sure

Regards
Bassam King Karzeddin
June 24, 2017

[bassam king karzeddin]

Nobody except a very big cheater would trisect the (3) degree angle for sure

and the reason is so simple and beyond the so wrong beliefs of the top professional mathematicians for sure

The reason is that "one-degree angle doesn't exist" for sure

And please do not jump into so many meaningless imaginations or tell me any APPROXIMATIONS, since we know it is ENDLESS, for sure

And the word endless was never correctly comprehended in mathematics, wonder!

BKK

[burs...@gmail.com]

If you play cards, with or without friends, by some rules, or
if you play board game, with or without friends, by some rules,
nonthing is fictious, the game exists.

Same for some numbers, they are just governed by some rules.
Take the cyclotomic numbers, we definition:

ε_n = e^(2*i*pi/n)

They are simply governed by the rule that they satisfy the
polynomial equation:

x^n - 1 = 0

In an old post you said numbers are constructible, by compass
and ruler, if they satisfy the form:

r1^(1/2) + r2^(1/4) + ...

If we take "constructible" as playing the game of math, for
example with polynomial equations, then you have much more
constructible numbers.

I am currently research software that can do cyclotomic numbers.
This looks already promissing:
https://www.gap-system.org/Manuals/doc/ref/chap18.html

Can you play the game and verify, or are you too stubborn
and too mind limited any such math games? Not able to
play the game of math?

ε_9 = -ε_9^4-ε_9^7

ε_9^3 = ε_3

ε_6 = -ε_3^2

ε_12 / 3 = -1/3*ε_12^7

BTW: I am also a little slow here, only the second equation
is immediate for me, but the rest I am still chewing on.

[burs...@gmail.com]

They are also doing stuff like: (Whats a quick
verification of this, i.e. a proof?)

sqrt(5) = ε_5-ε_5^2-ε_5^3+ε_5^4

I dont think I want this expansion, i.e. left to right,
on the other hand right to left would be neat.

i.e. computing radical expressions for de Moivre numbers.

[bassam king karzeddin]

So, it is the attitude of a person that defines maths for her/his own sake, wonder!

And if the player is more skilled, then the outcome is better

But my attitude is completely very different since I look at mathematics as

an existing constructible physical reality that must reflect original independent and perpetual perfection, and not simply temporarily perfect games that must create unclear gaps too

And if the case is such, then must be declared so openly as games but not science for only interested people, exactly like chess game fans, not simply imposed on every person in schools

BKK

[burs...@gmail.com]

These games, for example x^n-1 = 0, are ethernally perfect,
and do have of course also physical reality applications.

I am digging right now where the phyiscs exactly comes in, what
the examples are, something can be possibly found in the

vincinity of the following Kronecker Weber theorem,
we find sqrt(5) = ε_5-ε_5^2-ε_5^3+ε_5^4 again (sic!):
https://en.wikipedia.org/wiki/Kronecker%E2%80%93Weber_theorem

You know what you ask relates to Hilberts sixth problem,
something I also learnt just now, the internet is a wonderful

thing, why don't you use it? BTW: This PDF has some nice
pictures: Geometry in Physics
http://www.thp.uni-koeln.de/alexal/pdf/geometry.pdf

[bassam king karzeddin]

Your Wikipedia reference was just playing and ploughing with constructible numbers just to show the validity of (e) as a real number

Note that the basic angle is 72 degrees, which is a constructible angle, where simply (cos [2*(pi/5)] = (sqrt(5) - 1)/4), so the number sqrt(5) is already there included in the square root operation

The source is really cheating and wasting people time and was making things that sound like magic for very poor minds and beginners, exactly like this way:

[X = (2/5)X - 7X^2 + 7(X - 1/35)^2 - 1/1225], which is neither an equation or any meaningful statement, but full nonsense for sure

Where as the other huge source is complete nonsense like any modern standard mathematics

And more important that they can never trisect exactly the (3) degree angle because I say it is impossible with proofs I published

So, why are you like your great ancestors mathematickers who loves to cheat people forever? wonder!

Aren't stealing people time and efforts thefts and crimes too, wonder!

BKK

[bassam king karzeddin]

Didn't you realize the obvious cheating at that unnamed Wikipedia reference? wonder!

Then why do people still publish more nonsense that is more cheating?

Why don't keen people clean the rubbish immediately? wonder!

BKK

[bassam king karzeddin]

*

[bassam king karzeddin]

Do you see this full link related to this question?

https://www.quora.com/How-can-we-exactly-construct-a-triangle-with-known-sides-such-that-at-least-one-of-its-angles-is-in-integer-degree-n-where-n-is-not-a-multiple-of-3

Because, professional moderator mathematics are generally so frustrated to hear any facts, hoping that isn't repeated here at this moderated site that seems much better for free opinions
BKK

[bassam king karzeddin]

And still, people refuse to understand this very simple puzzle, wonder!

They still believe that any named angle can so simply exist (at least inside their nutty skulls), wonder again about the unlimited foolishness associated with most top professional and so genius mathematicians

And they would never be able to produce in reality only one angle that I claimed EARLIER to be fake and non-existing angle, but they would keep arguing aimlessly and very foolishly forever and for sure

BKK

[bassam king karzeddin]

And the whole magic trick with mathematics is only APPROXIMATION, where simply people are excited, forgetting that they were competing on a level of a carpenter skill,

But truly speaking, mathematics is perfection and exactness and never was an APPROXIMATION except for the silly matters (which isn't any mathematics) for sure

BKK

[Dan Christensen]

On Saturday, September 30, 2017 at 11:02:04 AM UTC-4, bassam king karzeddin wrote:

> >
> > And still, people refuse to understand this very simple puzzle, wonder!
> >
> > They still believe that any named angle can so simply exist (at least inside their nutty skulls), wonder again about the unlimited foolishness associated with most top professional and so genius mathematicians
> >
> > And they would never be able to produce in reality only one angle that I claimed EARLIER to be fake and non-existing angle, but they would keep arguing aimlessly and very foolishly forever and for sure
> >
> > BKK
>
> And the whole magic trick with mathematics is only APPROXIMATION, where simply people are excited, forgetting that they were competing on a level of a carpenter skill,
>

Crank Boy here actually believes that 40 degree angles don't exist. Really. Needless to say, his goofy system of geometry has gone nowhere over the past decade or so. He blames it on a vast international conspiracy of self-serving mathematicians. He imagines them in secret gatherings all around the world, plotting and scheming against him. Typical crank.


Dan

Download my DC Proof 2.0 software at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

[Pancho ValveJob]

Idiot: shut up.

[Zelos Malum]

We can find the result for that easily, the issue is more you do not understand the rules of the game.

[bassam king karzeddin]

Yes, and true, they made mathematics as a game, but they must confess it clearly as a game and not bothering the whole population about its UNLOGICAL rules for sure

BKK

[Zelos Malum]

First if, you negate "logic" with "il-" prefix, secondly, they are entirely logical you moron. Just cause you are too stupid doens't make them illogical.

[bassam king karzeddin]

On Wednesday, May 3, 2017 at 8:50:05 PM UTC+3, bassam king karzeddin wrote:
> >
> >
> >

And did you truly understand what does the arbitrary angle means in mathematics? I doubt it for sure
BKK

[bassam king karzeddin]

The arbitrary angle means exactly constructible angle! no wonder!

But, you are still suffering mentally a lot to comprehend this very simple fact, for sure
BKK

[Zelos Malum]

No, an arbitrary angle means any possible angle, not just constructible. Moron.

[bassam king karzeddin]

On Thursday, February 1, 2018 at 9:58:24 AM UTC+3, Zelos Malum wrote:
> No, an arbitrary angle means any possible angle, not just constructible. Moron.

You don't understand big issues moron nor the so elementary issues as well and especially in mathematics for sure, so keep silent and learn bad boy

BKK

[bassam king karzeddin]

[bassam king karzeddin]

> Here is the full old thread (in 2005) with 37 messages for interested people, and I simply thought that would be completely visible here
>
> Kindly, confirm the visibility of this old topic with all 37 replies, thanks
>
> http://mathforum.org/kb/thread.jspa?forumID=13&threadID=1175275&messageID=3846213#3846213
>
> BK

let us see now who truly understood the angle trisection problem?

I guess, nobody else, sure

BKK

[Zelos Malum]

It's been understood for ages and it has been solves, both as to why it doesn't work under one set of rules and why it works under another.

Now do you need anything else you moron?

[bassam king karzeddin]

ZM is not only parroting so empty words that he never understands but also parking aimlessly everywhere I go

He still and would never understand why it is truly impossible to trisect the arbitrary angle (I earlier added "by any means") for sure

Who is truly on earth still dumber than him for not understanding that the whole problem was completely solved, and never by trisecting any arbitrary angles but by DISCOVERING that most of the angles (in mathematics) are truly impossible existence, FOR SURE

So to say, why it is impossible to trisect the angle (pi/3 = 60 degrees)
that Wantzel proved in 1837 (but never understood the true reason)

And the answer is funnily so simple to believe for a nut mathematicians

It is strictly because of the trisection angle (pi/9 = 20 degrees) is a "non-existing angle" FOR SURER

Otherwise, how can something truly existing but also impossible to construct? wonder!

Similarly, We had long ago solved the other two impossible construction problems raised strictly by the ancient Greeks for more than two thousand years back

And more funnily and oddly and only in the world of the very stupid world of mathematicians nowadays despite the immediate fast communications, where they had searched the reasons for almost thousands of years but very unsuccessfully for sure

But when it was correctly solved and beyond any little doubt, they simply didn't want even to have the minimum chance to understand it, despite the undeniable fact that they truly fully understood it FOR SURE

It is indeed so easy to understand it even by school students and laypersons for sure

BKK

[Zelos Malum]

Saying things you are too stupid to understand is not parroting, you just parrot the same shit over and over and never adress the issues at hand, that IS parroting.

[bassam king karzeddin]

And the main reason that makes the world mathematicians ignore the true reasons for such impossibilities raised by the ancient Greeks is just to protect all the wrong baseless mathematics that they had dived deeply into for many centuries, they simply understood the unsolvable diaphontine equations, but refuse to understand that most of polynomials od odd degrees are also insolvable by real numbers that are too dense to realize or distinguish like the case with integers, for sure

BKK

[bassam king karzeddin]

Many angles are impossible existence, get it it is too simple FOR SURE
BKK

[bassam king karzeddin]

And when true Old Greek problems solver comes, everyone keeps silent, no wonder!
BKK

[bassam king karzeddin]

You keep juggling these words that you never understand for sure

Have anybody shows you your angle of (2pi/9 = 40) degrees in any triangle with exactly known sides? wonder!

And don't tell me again (1, sin(40), cos(40)), since everybody before you ad tried it unsuccessfully FOR SURE

So, this general MODERN cosine law is too little dwarf imitative of the greatest original theorem of (PT), that was strictly refuted by the PT theorem

Without any little doubt and FOR SURE

So to say, Go to the hill with all your alleged gibberish logic
BKK

[bassam king karzeddin]

On Thursday, May 4, 2017 at 12:54:51 PM UTC+3, Vinicius Claudino Ferraz wrote:
> Let alpha in [0, 2 pi).
>
> Stretch the sphere until it becomes a straight line segment [0, 2 pi ).
>
> Divide your "angle" or line segment into n parts. Let n = 3.
>
> Twist your line segment back to a sphere.
>
> Do the lines OA, OB, OC. Such that AÔC = n AÔB.

Are you sure? wonder!
BKK

[bassam king karzeddin]

On Sunday, May 14, 2017 at 5:21:15 PM UTC+3, Vinicius Claudino Ferraz wrote:
> Some angles I know how-to trisect. And bisect.
>
> We construct pi/2, then we have already trisected 3 pi/2.
> We construct pi/4, then we have already trisected 3 pi/4.
> We construct pi/2^n, then we have already trisected 3 pi/2^n.
>
> We construct pi/3 = 60º then we have already trisected pi.
> We construct pi/(3 * 2^n), then we have already trisected pi/2^n.
>
> Their sum, too. We construct 60º + 45º.
> We construct pi/3 + pi/4 then we have already trisected pi + 3 pi/4.
>
> We construct pi/2^i + pi/(3 * 2^j),
> then we have already trisected 3 pi/2^i + pi/(2^j).
>
> But the sum goes from 1 to n.
>
> We construct Sum_{k = 1}^n [ pi/2^i_k + pi/(3 * 2^j_k) ],
> then we have already trisected Sum_{k = 1}^n [ 3 pi/2^i_k + pi/(2^j_k).
>
> But the differences too.
>
> Don't forget if alfa can be trissected then alfa + alfa + alfa too can be trissected.
>
> What are the constructible angles? 90, 60, 30, 45. "Esquadros"?
>
> https://pbs.twimg.com/media/C_yxHoQWsAECoXF.jpg:large
>
> All of this comes from euclid's consctruction of an equilateral triangle.
> New ideas? Please send. Don't keep the secret. lol

How can you construct all those things, without constructing exactly the pi? wonder!
Don't you learn yet that non-numbers are impossible construction (not because they are too difficult to exactly construct, but because they never existed to be constructed), for sure
BKK

[bassam king karzeddin]

[bassam king karzeddin]

And who is on earth who truly wants to understand "CORRECTLY" those three old Greek impossible construction problems? wonder!
I think firmly that nobody wants to understand and very deliberately for sure because once you do then all very big lies in your mathematics would be uncovered just before your own big eyes for surest
BKK

[bassam king karzeddin]

Why it is impossible to trisect the angle (pi/3 = 60 degrees) and by any means for sure?

Isn't this is truly a very easy exercise for YOU ***NOW***, sure but why are so ashamed to write it? wonder!

It takes only a few minutes to write it, for sure

Do it the way I **ONLY** taught you, and make sure even Wantzel didn't understand the truth the way it is indeed, very sure

Cammon, video makers, Wiki-piki Writers, book authors, ** VERY SECRETIVE RESEARCHERS** hiding here under many fake names, FOR SURE

Didn't this is truly bothering you for many thousands of years but aimlessly for sure
But at least they could recently relatively know that is an indeed impossible task, but funnily, even though they didn't understand, and more oddly they **don't want** even to understand, Wonder!
What a bigger shame than this is truly upon all alleged intellectual people indeed, No wonder!

OR do you like it ***secretly*** your own, NO wonder
BKK

[bassam king karzeddin]

Are you ashamed to rewrite my proof? Sure

Or don't like even to see it again and again? no wonder!

OR is true that you don't understand anything yet, which is so unbelievable?

Maybe you want to make it alone and away from here? no wonder!
But then we would come to know about it, FOR SURE
BKK

[bassam king karzeddin]

On Thursday, May 18, 2017 at 12:27:01 AM UTC+3, Vinicius Claudino Ferraz wrote:
> Everything which is imaginable (and consistent) does exist.

Everything which is naturally existing is consistent and does exist For sure

But not everything imaginable and seems consistent may necessarily exist, for surer

For instance, you may imagine the *EXACT* heptagon or the *EXACT* cube of *EXACT* volume of two units or even the *EXACT* circle were the very *BITTER* fact that none of them *EXACTLY* exists and *IMPOSSIBLE* to exist, FOR SURE

So please to say, leave the imagination away, sure
BKK

[bassam king karzeddin]

[bassam king karzeddin]

See here, how did they keep hiding the above old original references, where an academic old troll-like Robert Israel and otherS one had completely missed all the free chances that had been given to them freely to understand and high light the problem those days,

Most likely they understand it now, but they wouldn't admit it for the very much wrong maths that would reveal about its truth

For instance, that guy Robert Israel claimed that my method is tripling the angle instead of exact trisecting of only the tri-selectable existing angles as if the angle (pi/9 = 20 Degrees) does indeed exist where it's three multiple eventually would give the angle (pi/3 = 60 Degrees), by his too primitive surfical thinking where they suppress the facts that humans were involved searching for it all those thousands of years

And now, they want to hide the discussions and most likely delete all of them in order to delete any older fact to them and most likely steal them, No wonder!

This is how the alleged Nobel professional mathematicians ethic nowadays, No wonder!

My older public profile contains many things and even more important and nobody has any right to hide it or steal it, so is yours also FOR SURE
BKK