Challenge: splitting an angle into three equal parts

Click For Summary

Discussion Overview

The discussion centers around the challenge of splitting an angle into three equal parts, exploring various methods including traditional geometric approaches and origami techniques. Participants share their experiences and insights on the feasibility of this problem, as well as its implications in mathematics and geometry.

Discussion Character

  • Exploratory
  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant mentions their interest in the problem and invites others to share their results.
  • Another participant asserts that splitting an angle with a compass and ruler is impossible in general, except for specific angles, and suggests that computers can provide approximations.
  • Some participants note that origami provides a method for angle trisection, contrasting it with traditional ruler and compass methods.
  • A participant inquires about the exploration of origami methods in greater detail, particularly regarding the transformations allowed by folding.
  • Another participant confirms that origami techniques have been extensively studied, allowing for the trisection of angles and other geometric constructions.
  • One participant speculates on the historical implications of geometry had ancient scholars like Euclid and Archimedes been from Japan.
  • There is a repeated assertion that squaring the circle remains impossible, with a reference to Kawasaki's theorem regarding the conditions for folding angles.

Areas of Agreement / Disagreement

Participants express differing views on the methods for angle trisection, with some supporting the impossibility of traditional methods while others advocate for origami as a viable solution. The discussion remains unresolved regarding the implications of these methods and their historical context.

Contextual Notes

Some claims depend on specific definitions and assumptions about geometric constructions, and the discussion includes references to unresolved mathematical concepts.

MartinV
Messages
68
Reaction score
0
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?
 
Mathematics news on Phys.org
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.
 
  • Like
Likes   Reactions: mfb
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.
 
micromass said:
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?
 
fresh_42 said:
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
 
  • Like
Likes   Reactions: 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.
 
Squaring the circle would still be impossible though
 
micromass said:
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)
 

Similar threads

Replies
1
Views
2K
  • · Replies 125 ·
5
Replies
125
Views
20K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 86 ·
3
Replies
86
Views
24K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
7
Views
3K
  • · Replies 26 ·
Replies
26
Views
6K
  • · Replies 67 ·
3
Replies
67
Views
16K