Challenge: splitting an angle into three equal parts

[MartinV]

I recently decided to take a whack at this problem. Came up with an interesting approach, thought it would make a good conversation topic.

Anyone else tried to do this? What were your results?

 

[fresh_42]

What do you use for splitting? Computers?
With compass and ruler it's impossible (proven) unless you have some certain angles for which it can be done.
You might always get some close approximations, but that's what you get with a computer, too.

 

[micromass]

Curiously, it can be done with origami. It's a very neat question which problems can be solved with origami as opposed to simple ruler and compass.

 

[fresh_42]

Curiously, it can be done with origami. It's a very neat question which problems can be solved with origami as opposed to simple ruler and compass.

Wow! Never heard about it. The classical three induced a lot of mathematics. Do you know whether anyone has explored origami methods in greater detail, will say which objects allowed transformations lead to?

 

[micromass]

Wow! Never heard about it. The classical three induced a lot of mathematics. Do you know whether anyone has explored origami methods in greater detail, will say which objects allowed transformations lead to?

Yes, it has been explored in a lot of details. Origami allows the doubling of the cube, the trisection of an angle and solving cubic and quartic polynomial equations. http://www.cs.mcgill.ca/~jking/papers/origami.pdf

 

[fresh_42]

I'm going to love $\mathbb{O}$. What would have happened to geometry etc., if Euclid and Archimedes had been Japanese scholars? This would be an interesting topic for a science fiction novel.

 

[micromass]

Squaring the circle would still be impossible though

 

[fresh_42]

Squaring the circle would still be impossible though

Yes, but this is cool: A folding with a center where all folds meet by angles $α_1, \dots , α_{2n}$ can be flattened if and only if
$${\displaystyle \alpha _{1}+\alpha _{3}+\cdots +\alpha _{2n-1}=\pi }$$
(Kawasaki's theorem - a version of)