I dont understand why it is proven to be impossible..
> I can trisect any given angle with compass and straightedge
> I can also double the cube
The triple angle product is in terms of a cubic equation. A number that you can construct with straight edge and compass needs to be the root of some polinomial of degree at most 2. There are some angles which when you get your cubic, the polinomial is irreducible so its in terms of numbers you can't construct with a compass or sttaught edge. That doesn't mean all angles can't be trisected by such means, but it's not true in general. I don't know if there's a purely geometric proof though.
The problem is trisecting the ANGLE, not trisecting a line segment. If you make an equilateral triangle "abc" and then trisect BC with the points X and Y that doesn't mean the angles BAX, XAY, and YAC are the same
>>8876751
AB + AC + BC = 3*radius + 3*radius + 3*radius
AB = 3/1r
AC = 3/1r
BC = 3*1/1r = 3/1r
How is BC not exactly 3/3 of angle A?
>>8876745
Every proof I read about trisecting an angle, is based on numbers and lots of stuff I dont understand.
Can't find a pure geometry proof anywhere.
It can be done with a modern locking compass, but not a COLLAPSING compass. Read the whole problem, brainlet.
>>8876819
I think you need to REEEEEEad some more about it youself
https://www.google.nl/url?sa=t&source=web&rct=j&url=http://www.msc.uky.edu/droyster/courses/spring02/classnotes/Chapter03.pdf&ved=0ahUKEwi-l8u-_s_TAhXKPFAKHQXFCVgQFghyMBc&usg=AFQjCNGQA9A-ROrlsbtk_ZmBMObvoX8QhA
>>8876704
Bamp
No one that can explain to me, why it is impossible?
this is called the 3 utilities problem. Basically you want to split 3 utilities into equal angles using only a compass
>>8877345
no. pick a book on galois theory for a proof
>>8877384
If I can prove with geometry that it can be done for any given angle with only a compas and an straightedge, the statement that 'trisecting an angle is impossible" is then FALSE right?
>>8877420
No.
>>8876704
The angles are not exactly right.
Let A be at (0, 0).
Let B be at (1, 3).
Let D be at (1, 1).
Let E be at (1, 0).
Let F be at (1, -1).
Let C be at (1, -3).
Therefore, |segment BD| = |segment DF| = |segment FC| = length 2
The purported trisection produces angles are:
BAD
DAF
FAC
BAD
= BAE - DAE
= arctan(3 / 1) - arctan(1 / 1)
= 26.57 degrees (approx)
DAF
= 2 DAE
= 2 arctan(1/1)
= 90 degrees
The problem becomes obvious as you decrease the length of AE relative to the length of BC. As |BC| >> |AE|, then it becomes about right, in the limit. However, your method is simply wrong.
Also, this is probably a troll OP.
>>8877468
(OP) here,
Thank you for taking the time to prove me wrong.
>already working on a better version right now. I think you will be surprised
>>8877815
https://en.wikipedia.org/wiki/Mathematics_of_paper_folding
Oh look, a field that is still relevant, where your contributions may actually be useful!