Origami originated in China, although it is now associated to Japan, around 200AD and was then known as Zhezhi. It was brought to Japan in the 6th century by Chinese Buddhist monks, where it obtained the name origami (”ori’ meaning fold and ‘gami’ meaning paper). Paper folding has become of great interest to mathematicians and in today’s post I will discuss the mathematics behind it.
The construction of origami models is often shown by the patterns left by the creases. The main question that mathematicians are interested in is looking at these crease patterns and answering whether they can be folded to a flat model, and if so, how to fold them. There are three rules for making flat-foldable origami crease patterns:
- Maekawa’s Theorem: at any vertex the number of valley (crease goes inwards) and mountain (crease goes outwards) folds always differs by two;
- Kawasaki’s Theorem: if you add up the angle measurements of every other angle around a point, the sum will be 180 degrees;
- A sheet can never penetrate a fold.
Just as Euclid devised axioms for planar geometry, Humiaki Huzita and Koshiro Hatori created a complete set of axioms to describe the geometry of origami.
- Given two points p1 and p2, there is a unique fold that passes through both of them;
- Given two points p1 and p2, there is a unique fold that places p1 into p2;
- Given two lines l1 and l2, there is a fold that places l1 onto l2;
- Given a point p1 and a line l1, there is a unique fold perpendicular to l1 that passes through point p1;
- Given two points p1 and p2 and a line l1, there is a fold that places p1 onto l1 and passes through p2;
- Given two points p1 and p2 and two lines l1 and l2, there is a fold that places p1 onto l1 and p2 onto l2;
- Given one point p and two lines l1 and l2, there is a fold that places p onto l1 and is perpendicular to l2.
[Taken from plus.maths.org]
Origami has also been applied to solve geometric problems, which cannot be solved using a compass and unmarked ruler. I will highlight a few examples.
Doubling the Cube
This problem concerns a cube with a side length s1 and of volume V. The challenge is to find a cube with side length s2, such that its volume is now 2V. This construction is credited to Peter Messer and goes as follows:
Trisecting an angle
This construction is due to Hisashi Abe.
Any suggestions for future posts? M x