Post by: f.point on 02/09/2015 18:30:11
AB divider, circuit 1 (c pictured), AG (Analytical Geometry)
Ruler, the line referred to in Section A and B (
pictured) , AG 
provided the point D (intersection circle 1 and line )Divider AB from point D cuts the line
and afforded the point EBisection angle DAB, a point C
Ruler, the line referred to in Section B and C (b pictured), AG [latex]-x-y=-\sqrt{d}[/latex]
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Post by: evan_au on 03/09/2015 03:38:55
The reason is that these instruments cannot construct a cube root (on the other hand, today we can easily calculate cube roots on a calculator, but that is not permitted by the rules defined by the ancient Greek mathematicians).
See: https://en.wikipedia.org/wiki/Angle_trisection
I think I'll stick with Pierre....
Post by: f.point on 03/09/2015 18:45:24
Pierre Wantzel
a man he did not know well mathematics,
SECOND PART
Divider AB, at point A (arm angle rotates around point C), from point B to create a circle 2 (g pictured), AG
Divider BC, at point B, cut the circle 2, we get the point F
Ruler, line through the points A and F, AG
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Post by: Bored chemist on 03/09/2015 19:30:19
It looks like some man doesn't understand mathematics.
a man he did not know well mathematics,
You seem to be trying to draw a sphere with a compass.
Post by: wolfekeeper on 04/09/2015 20:48:09
Post by: f.point on 06/09/2015 18:31:33
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
ladies and gentlemen looking for a mistake ...
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Post by: chiralSPO on 06/09/2015 19:29:21
Post by: f.point on 16/09/2015 19:04:04
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
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Next - my character
- Solution of the construction of a regular n (n> 2) of the polygon
Post by: f.point on 17/09/2015 18:59:22
should - Divider DH, from the point I, intersects a circular arc FG, obtained point J
Divider DH, referred to in Clause, cuts a circular arc FG, obtained point J
Post by: Bored chemist on 19/09/2015 21:55:26
In any even, your construction is wrong because it is known that no correct construction exists.
I would suggest making it clearer- but there's no point. The fact that I'm not sure what you have done doesn't matter because you have not trisected the angle.
Do you understand that this has been proven to be impossible?