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

[Turtle]

well, i think you should have written "straightedge", not ruler for one thing. for another, you have in the second page constructed a vesica piscis and since it is a 120º arc it trisects the angle.

 

this is a special case however and the trisection problem specifies an arbitrary angle and using only a compass & straightedge. it is proven impossible. :(

 

angle trisection

Angle trisection is a classic problem of compass and straightedge constructions of ancient Greek mathematics. It concerns construction of an angle equal to one-third of a given arbitrary angle, using only two tools: an un-marked straightedge, and a compass.

 

With such tools, the task of angle trisection is generally impossible, as shown by Pierre Wantzel (1837). Wantzel's proof relies on ideas from the field of Galois theory—in particular, trisection of an angle corresponds to the solution of a certain cubic equation, which is not possible using the given tools. Note that the fact that there is no way to trisect an angle in general with just a compass and a straightedge does not mean that it is impossible to trisect all angles: for example, it is relatively straightforward to trisect a right angle (that is, to construct an angle of measure 30 degrees). ...

[CraigD]

Hi Bob – welcome to hypography. :)

 

You seem to have omitted the finishing steps of your construction. :(

 

Notwithstanding the problem of construction by straightedge only being able to generate surd-valued (rational numbers, their irrational square roots, and square roots of them) points, while the value of points on lines of the trisection of some angles may require irrational cube roots, the construction in post #1 doesn’t draw 2 lines through point A, evenly or unevenly trisecting angle CAD.

 

To trisect an angle, you’ve gotta ultimately draw these 2 lines.

 

Do you have some more steps you failed to include in your first post?

[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)

[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