Trisection of angles-similarly

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

Fremen

I don't really understand what you're saying here. Is this meant to be a construction for trisecting an arbitrary angle using a compass and straightedge? It looks more like a bisection from (7). A result from Galois theory says that trisecting an arbitrary angle is impossible.

See here:
http://terrytao.wordpress.com/2011/08/10/a-geometric-proof-of-the-impossibility-of-angle-trisection-by-straightedge-and-compass/

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)

LeixlipRed

bob boben, this is a discussion forum. Contribute to the discussion please and don't just post random stuff without giving an opinion on it.

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

LeixlipRed

Did you read my post? Are you some sort of maths spam bot? Post up declaring what your interest in this is, contribute some discussion or I'm going to lock the thread.

bob boben

LeixlipRed wrote: »

Did you read my post? Are you some sort of maths spam bot? Post up declaring what your interest in this is, contribute some discussion or I'm going to lock the thread.

solving math problems that are not resolved, I want to show that mathematics can be based on a different basis (resolution - trisection of angles-doublin gcube-squaring the circle, as well as proving that there are three errors in the current mathematics). Of courseif you want to familiarize yourself with my math

Eliot Rosewater

Fremen wrote: »

I don't really understand what you're saying here. Is this meant to be a construction for trisecting an arbitrary angle using a compass and straightedge? It looks more like a bisection from (7). A result from Galois theory says that trisecting an arbitrary angle is impossible.

See here:
http://terrytao.wordpress.com/2011/08/10/a-geometric-proof-of-the-impossibility-of-angle-trisection-by-straightedge-and-compass/

Unless you've an infinite amount of time, in which case you can use the fact that

[LATEX]\displaystyle \sum_{n=1}^{\infty}\frac{1}{(-2)^{n+1}} = \frac{1}{3}[/LATEX]

to do an infinite number of bisections to trisect!

I love bastardized use of infinity - it makes all of one's wildest dreams possible!

Fremen

bob boben wrote: »

solving math problems that are not resolved, I want to show that mathematics can be based on a different basis (resolution - trisection of angles-doublin gcube-squaring the circle, as well as proving that there are three errors in the current mathematics). Of courseif you want to familiarize yourself with my math

Those.problems have been solved. The proofs are universally accepted among mathematicians. The problems you're talking about are infamous for attracting amateur mathematicians who claim to.have found counterexamples. See http://en.m.wikipedia.org/wiki/Underwood_Dudley

Note the title.of his book...

Pherekydes

bob boben wrote: »

solving math problems that are not resolved, I want to show that mathematics can be based on a different basis (resolution - trisection of angles-doublin gcube-squaring the circle, as well as proving that there are three errors in the current mathematics). Of courseif you want to familiarize yourself with my math

Pseudomathematics

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
Pseudomathematics -There are theories, just what is true can be proved by experiment in science, everything is sci fi