bob boben Posted September 4, 2011 Share Posted September 4, 2011 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 Link to comment Share on other sites More sharing options...
John Cuthber Posted September 4, 2011 Share Posted September 4, 2011 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!"? Link to comment Share on other sites More sharing options...
bob boben Posted September 4, 2011 Author Share Posted September 4, 2011 (edited) 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, Edited September 4, 2011 by bob boben Link to comment Share on other sites More sharing options...
doG Posted September 4, 2011 Share Posted September 4, 2011 Wantzel proved this impossibility in 1836. See http://www.jimloy.com/geometry/trisect.htm Link to comment Share on other sites More sharing options...
John Cuthber Posted September 4, 2011 Share Posted September 4, 2011 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. Link to comment Share on other sites More sharing options...
DrRocket Posted September 4, 2011 Share Posted September 4, 2011 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. Link to comment Share on other sites More sharing options...
bob boben Posted September 5, 2011 Author Share Posted September 5, 2011 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) Link to comment Share on other sites More sharing options...
michel123456 Posted September 5, 2011 Share Posted September 5, 2011 Where is trisection? I see bisection in your drawings. Link to comment Share on other sites More sharing options...
imatfaal Posted September 5, 2011 Share Posted September 5, 2011 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) Link to comment Share on other sites More sharing options...
bob boben Posted September 6, 2011 Author Share Posted September 6, 2011 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 Link to comment Share on other sites More sharing options...
imatfaal Posted September 6, 2011 Share Posted September 6, 2011 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 Link to comment Share on other sites More sharing options...
doG Posted September 6, 2011 Share Posted September 6, 2011 I shared a circle with corners (second part , 9 to 23). Are you aware of the fact that circles don't have corners? 1 Link to comment Share on other sites More sharing options...
John Cuthber Posted September 6, 2011 Share Posted September 6, 2011 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. Link to comment Share on other sites More sharing options...
bob boben Posted September 7, 2011 Author Share Posted September 7, 2011 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 Link to comment Share on other sites More sharing options...
michel123456 Posted September 7, 2011 Share Posted September 7, 2011 It is half the construction of an hexagon. Bob trisected correctly an angle of 180°. Link to comment Share on other sites More sharing options...
John Cuthber Posted September 7, 2011 Share Posted September 7, 2011 OK, and now once more with feeling! Try 179 degrees ( this might be a long thread) Link to comment Share on other sites More sharing options...
michel123456 Posted September 8, 2011 Share Posted September 8, 2011 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... Link to comment Share on other sites More sharing options...
Hal. Posted September 12, 2011 Share Posted September 12, 2011 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 . Link to comment Share on other sites More sharing options...
John Cuthber Posted September 12, 2011 Share Posted September 12, 2011 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. Link to comment Share on other sites More sharing options...
Pantaz Posted September 12, 2011 Share Posted September 12, 2011 Reported as spam -- He's been trying to promote this idea all over the place... http://www.google.com/search?q=Trisection+of+angles-similarly Link to comment Share on other sites More sharing options...
Hal. Posted September 12, 2011 Share Posted September 12, 2011 This isn't a development because it can't go anywhere or do anything. This is an attempt at a development which is what I admire . It's not "supposed" impossible; its proven impossible. I am supposing it is impossible , as I did not see a proof of impossibility , I do know who I am willing to believe if they say it is 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. You could be surprised how a person could receive an A+ for trying to prove something which is impossible to prove as it doesn't really matter sometimes if somebody is right or wrong , it is about the persistence of an argument . Take law as an example , people are committing crime all day long and lawyers are convincing people of their innocence even if only their guilt can be proven . Maybe the original poster could take a lesson from the legal profession in so far as trying not to prove the geometry but trying to convince people of it's proof first which could be much easier . Link to comment Share on other sites More sharing options...
John Cuthber Posted September 13, 2011 Share Posted September 13, 2011 as I did not see a proof of impossibility , it is about the persistence of an argument Then you should have followed the link in the 4th post in this thread. and It is indeed about persistence of an argument. If you persist after it has been proved that you are wrong then, at best, you are a fool. Link to comment Share on other sites More sharing options...
Hal. Posted September 13, 2011 Share Posted September 13, 2011 I supposed correctly , Cuthber , from the posts of two people and you aren't one of them . Link to comment Share on other sites More sharing options...
imatfaal Posted January 19, 2012 Share Posted January 19, 2012 Rockas - I have not had time to read your paper fully, but two things: 1. It can be proven that one cannot in general trisect the angle 2. I think from the brief glance I gave your work that it merely shows that an angle that is constructed is half the size of the original angle. This is not, I believe, something new - what you need to be able to show in geometric steps is how to trisect an angle between two lines and (this time shouting for emphasis) THIS IS IMPOSSIBLE Could you just confirm that you are working from merely two lines call them A and B which intersect? Any points on the lines, tangents, circles around points etc must be uniquely identifiable and given in simple steps. It also seems that your proof that angle D1 is half of D2 is a bit over the top. from your construction DG=DG'=DE - they are points on a circle of which D is centre. G'G=GE - two radii of circle centred at G. thus the two triangles DG'G and DGE are congruent isosceles triangles. the line DZ is easily shown to bisect the angle G'DG. As the angle ZDG is half the angle G'DG it must be half the same angle of the congruent triangle ie GDE. The total angle ZDE is then clearly 3 times larger than ZDG - but that is not because you have trisected an angle, it is because you have mirrored an angle, bisected it, and then taken the sum of the original and the half angles. it is impossible to do this the other way around! Link to comment Share on other sites More sharing options...
doG Posted January 19, 2012 Share Posted January 19, 2012 (edited) Read my work about trisection at http://bestsitetopbuy.com/Trisection-of-an-Angle-English.docx Waste of time...the problem was algebraically proved impossible by Wantzel (1836). Edited January 19, 2012 by doG Link to comment Share on other sites More sharing options...
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