[bob boben]

The solution of mathematical tasks in the ancient Greek
Trisection of angles
angle=0° - no solution
180°>angle>0° - general solution (consists of 4 parts)

the first part



1.ruler AB
2.ruler AC
3.caliper A-AD
4.ruler DE



5.caliper D-DE
6.caliper E-DE
7.ruler FG intersects DE the point H ,DH=HE



8.caliper H-HE

[John Cuthber]

The proof of the trisection of an angle is mathematically equivalent to proving that 3 is an even number.
Do you really want your first post here to say

"I think three is an even number!"?

[bob boben]

John Cuthber, on 4 September 2011 - 01:17 PM, said:

The proof of the trisection of an angle is mathematically equivalent to proving that 3 is an even number.
Do you really want your first post here to say

"I think three is an even number!"?

when you look at the other three parts , then you will make surethat it is possible, see you for 20-30h,

This post has been edited by bob boben: 4 September 2011 - 01:28 PM

[doG]

Wantzel proved this impossibility in 1836. See http://www.jimloy.co...try/trisect.htm

[John Cuthber]

bob boben, on 4 September 2011 - 01:27 PM, said:

when you look at the other three parts , then you will make surethat it is possible, see you for 20-30h,

Should we sell tickets?
The point is that since it has been proved impossible, all you will do is post up an error of some sort. There's every chance that someone here will spot that mistake.

[DrRocket]

John Cuthber, on 4 September 2011 - 05:53 PM, said:

Should we sell tickets?
The point is that since it has been proved impossible, all you will do is post up an error of some sort. There's every chance that someone here will spot that mistake.

There is a better chance that no one who understands the issue will bother to read it. Every math department regularly encounters someone who insists that he has solved the general trisection problem.

It is one of the things that one encounters in an abstract algebra class on Galois theory. The trisection problem is impossible. This does not mean that no one has found the answer. It means that it has been proved rigorously that no classical straightedge and compass construction can exist that would trisect an arbitrary angle.

[bob boben]

second part


9.caliper D-DH , gets the point D1
10.straightedge (ruler) HD1
11.caliper D-DH
12.caliper D1-D1D
13.straightedge ( ruler ) HI1 , gets the point D2

14.caliper D2-D2D
15.caliper D-DD2
16.straightedge (ruler) HI2 , gets the point D3
17.caliper D3-D3D
18.caliper D-DD3
19.straightedge (ruler) HI3 , gets the point D4
20.caliper D4-D4D

21.caliper D-DD4
22.straightedge (ruler) HI4 , gets the point D5
23.this procedure with a compass and a ruler is actually a series of numbers (finite, infinite)

[michel123456]

Where is trisection? I see bisection in your drawings.

[imatfaal]

michel123456, on 5 September 2011 - 01:26 PM, said:

Where is trisection? I see bisection in your drawings.

From what I gather this explanation comes in three parts - and yes so far I can only see bisection. I presume the OP is going to make many many very small divisions at the right hand side of the diagram and then claim that two of lines formed cross the arc or line between AB at the trisection points - this will be an estimation and not generalisable (as has been pointed out above)

[bob boben]

third part
computer program-coreldraw 13
http://www.fileserve.com/file/sZpPSyf/T ... -DOKAZ.cdr
png image
http://www.fileserve.com/file/EpEKHxx/p ... ection.png
the picture is
-circle
-diameter circle AB
-points on the diameter of the circle A(0°),B(180°), C (AD=DC=CB) , D , F(center of the circle ,FB-radius circle)
-tendon EF

I went from the assumptions that there are lines in a circle which is the first point (180 °> point> 0 °), the last point is (360 °> point> 180 °) to the intersection with the circle diameter (AB) be the point C (D).
I shared a circle with corners (second part , 9 to 23).
I found the string (EF, 3.75 ° -187.5 °) which passes through the point C ,which means that we have the angle trisection.
fourth part

24.caliper D-DD5-(22.)
25.caliper E-DD4-(19.)
26.straightedge (ruler) D4D5 , gets the point H1
27.straightedge (ruler) AH1
28.caliper H1-H1D , gets the point H2
29.straightedge (ruler) AH2

[imatfaal]

From point 8 onwards your construction seems independent of the position of A with regards to H - and thus independent to the angle you claim to be trisecting

[doG]

bob boben, on 6 September 2011 - 10:51 AM, said:

I shared a circle with corners (second part , 9 to 23).

Are you aware of the fact that circles don't have corners?

[John Cuthber]

DrRocket, on 4 September 2011 - 07:45 PM, said:

There is a better chance that no one who understands the issue will bother to read it. Every math department regularly encounters someone who insists that he has solved the general trisection problem.

It is one of the things that one encounters in an abstract algebra class on Galois theory. The trisection problem is impossible. This does not mean that no one has found the answer. It means that it has been proved rigorously that no classical straightedge and compass construction can exist that would trisect an arbitrary angle.

I know that, and you know that but I suspect that there's someone else here who doesn't know it (yet).
However I'm sure someone will find it in amusing challenge to find the flaws in the argument (they have already started)

I wasn't kidding about this being the equivalent of proving that 3 is an even number.
The crassly oversimplified version is that you can find square roots with a straight edge and compasses, you can find the square roots of square roots too. In fact you can find any 2^nth root by repeating the process.
But 3 isn't of the form 2^n
(not least, because all such numbers are even apart from the trivial case of 2^0).
You need to be able to solve a cubic equation to trisect an angle and you can't solve cubic equations with only even powers.
So, if he proves that he can trisect an angle, he has shown that 3 is an even number.

However notwithstanding reality, I confidently predict that the OP will triumphantly bring forth the 4th part of his series: which will be tosh.

(Incidentally, I know that's a dumbed down version but it does make it clear that you are probably wasting your time trying to double a cube or trisect an angle. The full proof that it's really impossible is on the web (isn't everything?) if you want to look for it.)
The hand waving argument proof has the advantage that the maths isn't difficult. Many people are not that familiar with Galois and his work.

[bob boben]

angle=180°

30.straightedge (ruler) AB
31.caliper C-CD
32.caliper D-DC ,gets the point E
33.straightedge (ruler) CE
34.caliper E-ED , gets the point F
35.straightedge (ruler) CF

[michel123456]

It is half the construction of an hexagon. Bob trisected correctly an angle of 180°.

[John Cuthber]

OK, and now once more with feeling!
Try 179 degrees
( this might be a long thread)

[michel123456]

90° is feasable too.

Curiously, I was never interested in the question.
Yesterday I made a simple thought: it is possible to make a circle, draw 3 smaller circles (of random radius) upon its circumference, and create an arc, divided in 3 equal parts, forming a random angle.
I wondered why the reverse construction was impossible (beginning from the angle and dividing the arc). As if geometry had a direction that cannot always be reversed, like entropy and time...

[Hal.]

bob boben ,

I like to see people attempting to develop these types of things , good luck in your supposed impossible task . By reliable accounts it can't be done , that doesn't stop you trying and if you have the conclusion some time in the future that it can't be done we may then all be gone .

[John Cuthber]

This isn't a development because it can't go anywhere or do anything.
It's not "supposed" impossible; its proven impossible.
It would be better if he spent his time helping little old ladies across the road or, at least, trying to do something where he might succeed.

[Pantaz]

Reported as spam -- He's been trying to promote this idea all over the place...
http://www.google.co...ngles-similarly