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Thread: final solution

  1. #1
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    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
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  2. #2
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    Re: final solution

    Easier to agree than read/understand all that
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  3. #3
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    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.
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    Re: final solution

    Quote Originally Posted by HallsofIvy View Post
    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)
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    Forum Admin topsquark's Avatar
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    Re: final solution

    Quote Originally Posted by msbiljanica View Post
    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
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  6. #6
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    Re: final solution

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  7. #7
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    Re: final solution

    Quote Originally Posted by msbiljanica View Post
    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..

    .
    Last edited by zzephod; Jun 6th 2015 at 12:59 AM.
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  8. #8
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    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}
    final solution-slika-1.png
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  9. #9
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    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.
    Last edited by johnsomeone; Sep 2nd 2015 at 11:09 AM.
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  10. #10
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    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.
    Last edited by HallsofIvy; Sep 2nd 2015 at 11:18 AM.
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  11. #11
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    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
    final solution-112.png
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  12. #12
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    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 ...
    final solution-3.png
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  13. #13
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    Re: final solution

    Quote Originally Posted by msbiljanica View Post
    ladies and gentlemen looking for a mistake ...
    Why not "looking for something correct"?
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    Re: final solution

    Quote Originally Posted by DenisB View Post
    Why not "looking for something correct"?
    question - whether it's my procedure is correct, if I followed the rules
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  15. #15
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    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}

    final solution-333.png

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