11
$\begingroup$

I am working on a project about Mathematics and Origami. I am working on a section about how origami can be used to solve cubic equations. This is the source I am looking at:
http://origami.ousaan.com/library/conste.html
I understand the proof they go on to explain about how the slope of the crease line is the solution to the general cubic equation. I am also interested in understanding how this is used to double the cube and trisect the angle which they go on to mention at the end. However, I am having difficulty filling in the missing gaps and understanding how to use this origami method to solve those equations given and how this tells us we can trisect the angle, and double the cube. Any help is much appreciated

$\endgroup$
1
  • 2
    $\begingroup$ I posted some references in this 17 December 2008 sci.math post at Math Forum. However, this particular post consists of items I mostly collected back around 2002, so there are probably a lot of more recently published items you can find by searching the internet using some of these older titles in your search (in order to pick up more recent papers and other items that cite papers I give in that 2008 post). $\endgroup$ Jul 16, 2014 at 13:32

2 Answers 2

3
$\begingroup$

For the trisection of an angle note that $\cos3\alpha=4\cos^3\alpha-3\cos\alpha$ with $3\alpha$ constructible (this is equivalent to say that $\cos 3 \alpha$ is constructible). It is an equation of degree 3. In order to obtain the desired $\alpha$ angle, take a squared paper and do the crease $L_1$ corresponding to $3\alpha$ (the angle between $L_1$ and the $x-$axe). Make now the crease $L_2$ parallel to the $x-$axe and that divide in two stripes of equal heigh the entire paper. Call $P_1$ the intersection point of $L_2$ and the $y-$axe. Do the same with the first stripe, constructing a new crease $L_3$. Now using one of the Huzita-Hatori axioms you can do the crease that sends $P_1$ over $L_1$ and the origin over $L_3$. Call $P_1'$ and $O'$ the points obtained by this reflection. It turns that the axis (call it $L_4$) of the segment $\overline{P_1'O'}$ passes through the origin. The reflection of the line $L_1$ (call it $L_5$) with respect $L_4$ also passes through $O$, and in particular it trisect the angle $3\alpha$.

$\endgroup$
2
$\begingroup$

Thre's a good video showing the trisection bit here:

https://www.youtube.com/watch?v=SL2lYcggGpc&feature=em-subs_digest

$\endgroup$
1
  • 3
    $\begingroup$ Welcome to MSE. Link-only answers are discouraged here because, in this case, the video might get taken down later. Usually we ask answerers to summarize the information at the link, though it's difficult to summarize a video. Perhaps this would be better as a comment. $\endgroup$
    – Null
    Dec 17, 2014 at 18:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .