final solution

msbiljanica

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

msbiljanica

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
$$\displaystyle \angle GJB = \angle BJI = \angle IJH = {\angle GJH \over 3}$$
ladies and gentlemen looking for a mistake ...

DenisB

ladies and gentlemen looking for a mistake ...
Why not "looking for something correct"?

msbiljanica

Why not "looking for something correct"?
question - whether it's my procedure is correct, if I followed the rules (Yes)

msbiljanica

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

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

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

topsquark

Forum Staff
Next - my character
- Solution of the construction of a regular n (n> 2) of the polygon
Very pretty. Also very wrong. Trisecting a general angle was shown to be impossible by Pierre Wantzel in 1837 so I wouldn't go on your victory tour just yet. I'm sure Wikipedia has a copy of the proof somewhere. Maybe you should look it up.

-Dan

1 person

msbiljanica

Very pretty. Also very wrong. Trisecting a general angle was shown to be impossible by Pierre Wantzel in 1837 so I wouldn't go on your victory tour just yet. I'm sure Wikipedia has a copy of the proof somewhere. Maybe you should look it up.

-Dan
man does not know mathematics , the application procedure and check other methods, so make sure whether it is true or not

msbiljanica

Divider DH, referred to in Clause, cuts a circular arc FG, obtained point J
should - Divider DH, from the point I, intersects a circular arc FG, obtained point J

topsquark

Forum Staff
man does not know mathematics , the application procedure and check other methods, so make sure whether it is true or not
But have you looked it up? Can you find the flaw in his proof? Or are you simply so sure of your method that you don't feel you have to check it out for yourself?

-Dan

Archie

man does not know mathematics , the application procedure and check other methods, so make sure whether it is true or not
This is just rubbish. If the world relied only on empirical arguments, we'd never have determined that the earth is not the center of the universe.