1. ## final solution

Solution trisecting (n-sections) and n-proper construction of the polygon using divider and straightedge.

The ball shall be awarded in two equal parts, obtained half contains two areas, the circle (represents a planar geometry) and a semi-sphere (represents spherical geometry), the circle is the boundary between the circuit and semi-sphere
View photo (below)
CIRCLE
within a given arbitrary angle BAC,straightedge (straightedge is flexible, it can draw on a sphere)
straight line the BA to extend the circle to give the point D
SPHERE
straightedge connect points B and D, you get the curve BD
straightedge and divider - a procedure divisions curve into two equal parts is the same as the process of division shall exceed the level in two equal parts, we get the point E
straightedge - connect points C and E and get the curve CE
https://2bl3tq.bn1302.livefilestore..../q1.png?psid=1

Proportion longer exists in plane geometry, I found that the process can be applied to the sphere
choose point G
divider EC, from point G we get a point Hdivider EC, from point H we get a point I
divider EC, from point E twe get a point Jdivider EC, from point J get the point K
divider EC, from point K get the point L
straightedge point L and point I and connect, we get curve LI
EG divider, from the point L we get a point P
EG divider, from the point P and get the point Ostraightedge join the dots E and P and proceed to the circle, we get the point Q
straightedge merge points E and O and proceed to the circle, we get the point R
CIRCLE
straightedge connect point A and point Q, we get straight line AQ
straightedge connect point A and point R, we get straight line AR
https://dc4f8a.bn1302.livefilestore..../q2.png?psid=1
these have carried out the production of a given trisection angle, is obtained from the rest of the (n-section, a regular n-polygon) ...

Now proclaim everywhere that I decided 2- millennium math problems

2. ## Re: final solution

Easier to agree than read/understand all that

3. ## Re: final solution

The moment you said "the straightedge is flexible" you were no longer doing the classical "trisect a general angle using only a straightedge and compasses". The straightedge cannot be flexible.

4. ## Re: final solution

Originally Posted by HallsofIvy
The moment you said "the straightedge is flexible" you were no longer doing the classical "trisect a general angle using only a straightedge and compasses". The straightedge cannot be flexible.
in terms of writing to be used unmarked straightedge (straightedge for no additional conditions) and divider, I complied with the conditions and find a solution to the semi-sphere, so why should I get recognition for these two solutions (I think it's clear from this how construct a regular polygon, that does not explain the procedure)

5. ## Re: final solution

Originally Posted by msbiljanica
in terms of writing to be used unmarked straightedge (straightedge for no additional conditions) and divider, I complied with the conditions and find a solution to the semi-sphere, so why should I get recognition for these two solutions (I think it's clear from this how construct a regular polygon, that does not explain the procedure)
You are missing the point. There are certain rules to be followed and your method, which I will assume works, does not follow them. There have been many techniques proposed to trisect angles and none of them have done it using "traditional" Euclidean geometry methods. But they did find a way to trisect the angle.

-Dan

7. ## Re: final solution

Originally Posted by msbiljanica
Now proclaim everywhere that I decided 2- millennium math problems
You did not decide a 2-Millennium maths problem, the trisection of an arbitrary angle with straight edge and compass has been proven to be impossible under the rules that apply to the problem. Your use of a "straight edge" that can draw parts of great circles on a sphere is not allowed for the trisection of the angle. Methods of trisecting arbitrary angles using other means and instruments than straight edge and compass are almost as old as the problem itself.

It is considered bad manners and gauche to proclaim a proof of a theorem/result that you evidently have not done any but the most superficial background reading/research on. The first two search results of "angle trisection" would have warned you that you probably did/do not understand what you think you are doing..

.

8. ## Re: final solution

FIRST PART
AB divider, circuit 1 (c pictured), AG (Analytical Geometry) $x^2+y^2=d$
Ruler, the line referred to in Section A and B ( $a$ pictured) , AG $y=0$ provided the point D (intersection circle 1 and line $a$)
Divider AB from point D cuts the line $a$ and afforded the point E
Bisection angle DAB, a point C
Ruler, the line referred to in Section B and C (b pictured), AG $-x-y=-\sqrt{d}$

9. ## Re: final solution

Here's a thought:

The rules of the geometric-construction game involve a theoretically ideal/perfect straight-edge and compass, and take place on a plane.

OK. Fine. You STILL might be able to trisect the angle - in the spirit of msbiljanica's attempt here - while still abiding by those restrictions. How?

Because when we say ideal & perfect for a straight edge and plane and circle and so forth, we assume that puts us in the realm of the mathematically "flat" $\mathbb{R}^2$.

But when one refers to a theoretically ideal & perfect straight-edge/compass/plane, that could be fairly interpreted to mean a theoretically ideal & perfect as determined by the rules within our actual universe.

And by General Relativity, theoretically and ideally straight lines are anything but "straight" in certain regions of space!

So perhaps it IS possible to trisect the angle, using a theoretically ideal & perfect straight-edge/compass/plane... so long as the construction is done near a black hole or something.

Heh.

10. ## Re: final solution

I have no idea what you are saying here or what you are posting it for. What does this circle and line have to do with you previous posts?

You say "Divider AB from point D cuts the line a and afforded the point E". AB is the line segment from A(0, 0) to B(1, 0). D is the point (-2, 0). What does "divider AB from point D" mean? I don't see any point "G" on your picture so I don't know what "AG" means.

What is this a "solution" to??

(By the way, the "millennium problems" were a list of open mathematics problems posted at the beginning of the second millennium. Neither of the problems you mention was one of them.

11. ## Re: final solution

SECOND PART
Divider AB, at point A (arm angle rotates around point C), from point B to create a circle 2 (g pictured), AG $x ^ 2 + z ^ 2 = d$
Divider BC, at point B, cut the circle 2, we get the point F
Ruler, line through the points A and F, AG $y = 0, x = 0$

12. ## Re: final solution

PART THREE
Divider BF from point B, cuts line e is afforded point J
point G on circuit 1 (free choice),
ruler connect points A and G, we get along AG, we get the angle BAG
Divider GB, from the point B, we cut a circle 1, we get the point I
Divider GB, from the point I, we cut a circle1, we get the point H
Ruler join the dots G and J, we get along GJ
Ruler join the dots H and J, we get along JH, we get the angle GJH
angle GAB = angle GJH
Ruler merge point B and J, JB get along, we get the angle GJB
Ruler merge points I and J, we get along IJ, BJI get the angle, we get the angle IJH
$\angle GJB = \angle BJI = \angle IJH = {\angle GJH \over 3}$
ladies and gentlemen looking for a mistake ...

13. ## Re: final solution

Originally Posted by msbiljanica
ladies and gentlemen looking for a mistake ...
Why not "looking for something correct"?

14. ## Re: final solution

Originally Posted by DenisB
Why not "looking for something correct"?
question - whether it's my procedure is correct, if I followed the rules

15. ## Re: final solution

I found how to determine the proportion of angles, and thus solve the trisection angles

Given the angle CAB
Divider AD (Point D is the free choice of the branch AB), from point A, creates a circular arc ED
Bisection circular arc ED obtained item H
Divider AD, from point D, obtained point L
Divider AD, from the point of L, we get the point F
Divider AF, from point A, creates a circular arc FG
Divider DH, from the point F, intersects a circular arc FG, obtained point I
Divider DH, referred to in Clause, cuts a circular arc FG, obtained point J
Divider FJ, from point J, cuts a circular arc FG, obtained point K

$\angle GAK = \angle KAJ = \angle JAF = {\angle GAF \over 3}$

Next - my character
- Solution of the construction of a regular n (n> 2) of the polygon

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