# Huzita–Hatori axioms

The **Huzita–Hatori axioms** or **Huzita–Justin axioms** are a set of rules related to the mathematical principles of paper folding, describing the operations that can be made when folding a piece of paper. The axioms assume that the operations are completed on a plane (i.e. a perfect piece of paper), and that all folds are linear. These are not a minimal set of axioms but rather the complete set of possible single folds.

The axioms were first discovered by Jacques Justin in 1989.^{[1]} Axioms 1 through 6 were rediscovered by Italian-Japanese mathematician Humiaki Huzita and reported at *the First International Conference on Origami in Education and Therapy* in 1991. Axioms 1 though 5 were rediscovered by Auckly and Cleveland in 1995. Axiom 7 was rediscovered by Koshiro Hatori in 2001; Robert J. Lang also found axiom 7.

## The seven axioms

The first 6 axioms are known as Huzita's axioms. Axiom 7 was discovered by Koshiro Hatori. Jacques Justin and Robert J. Lang also found axiom 7. The axioms are as follows:

- Given two points
*p*_{1}and*p*_{2}, there is a unique fold that passes through both of them. - Given two points
*p*_{1}and*p*_{2}, there is a unique fold that places*p*_{1}onto*p*_{2}. - Given two lines
*l*_{1}and*l*_{2}, there is a fold that places*l*_{1}onto*l*_{2}. - Given a point
*p*_{1}and a line*l*_{1}, there is a unique fold perpendicular to*l*_{1}that passes through point*p*_{1}. - Given two points
*p*_{1}and*p*_{2}and a line*l*_{1}, there is a fold that places*p*_{1}onto*l*_{1}and passes through*p*_{2}. - Given two points
*p*_{1}and*p*_{2}and two lines*l*_{1}and*l*_{2}, there is a fold that places*p*_{1}onto*l*_{1}and*p*_{2}onto*l*_{2}. - Given one point
*p*and two lines*l*_{1}and*l*_{2}, there is a fold that places*p*onto*l*_{1}and is perpendicular to*l*_{2}.

Axiom 5 may have 0, 1, or 2 solutions, while Axiom 6 may have 0, 1, 2, or 3 solutions. In this way, the resulting geometries of origami are stronger than the geometries of compass and straightedge, where the maximum number of solutions an axiom has is 2. Thus compass and straightedge geometry solves second-degree equations, while origami geometry, or origametry, can solve third-degree equations, and solve problems such as angle trisection and doubling of the cube. However, in practice the construction of the fold guaranteed by Axiom 6 requires "sliding" the paper, or neusis, which is not allowed in classical compass and straightedge constructions. Use of neusis together with a compass and straightedge does allow trisection of an arbitrary angle.

## Details

### Axiom 1

Given two points *p*_{1} and *p*_{2}, there is a unique fold that passes through both of them.

In parametric form, the equation for the line that passes through the two points is :

### Axiom 2

Given two points *p*_{1} and *p*_{2}, there is a unique fold that places *p*_{1} onto *p*_{2}.

This is equivalent to finding the perpendicular bisector of the line segment *p*_{1}*p*_{2}. This can be done in four steps:

- Use
**Axiom 1**to find the line through*p*_{1}and*p*_{2}, given by - Find the midpoint of
*p*_{mid}of*P*(*s*) - Find the vector
**v**^{perp}perpendicular to*P*(*s*) - The parametric equation of the fold is then:

### Axiom 3

Given two lines *l*_{1} and *l*_{2}, there is a fold that places *l*_{1} onto *l*_{2}.

This is equivalent to finding a bisector of the angle between *l*_{1} and *l*_{2}. Let *p*_{1} and *p*_{2} be any two points on *l*_{1}, and let *q*_{1} and *q*_{2} be any two points on *l*_{2}. Also, let **u** and **v** be the unit direction vectors of *l*_{1} and *l*_{2}, respectively; that is:

If the two lines are not parallel, their point of intersection is:

where

The direction of one of the bisectors is then:

And the parametric equation of the fold is:

A second bisector also exists, perpendicular to the first and passing through *p*_{int}. Folding along this second bisector will also achieve the desired result of placing *l*_{1} onto *l*_{2}. It may not be possible to perform one or the other of these folds, depending on the location of the intersection point.

If the two lines are parallel, they have no point of intersection. The fold must be the line midway between *l*_{1} and *l*_{2} and parallel to them.

### Axiom 4

Given a point *p*_{1} and a line *l*_{1}, there is a unique fold perpendicular to *l*_{1} that passes through point *p*_{1}.

This is equivalent to finding a perpendicular to *l*_{1} that passes through *p*_{1}. If we find some vector **v** that is perpendicular to the line *l*_{1}, then the parametric equation of the fold is:

### Axiom 5

Given two points *p*_{1} and *p*_{2} and a line *l*_{1}, there is a fold that places *p*_{1} onto *l*_{1} and passes through *p*_{2}.

This axiom is equivalent to finding the intersection of a line with a circle, so it may have 0, 1, or 2 solutions. The line is defined by *l*_{1}, and the circle has its center at *p*_{2}, and a radius equal to the distance from *p*_{2} to *p*_{1}. If the line does not intersect the circle, there are no solutions. If the line is tangent to the circle, there is one solution, and if the line intersects the circle in two places, there are two solutions.

If we know two points on the line, (*x*_{1}, *y*_{1}) and (*x*_{2}, *y*_{2}), then the line can be expressed parametrically as:

Let the circle be defined by its center at *p*_{2}=(*x _{c}*,

*y*), with radius . Then the circle can be expressed as:

_{c}In order to determine the points of intersection of the line with the circle, we substitute the *x* and *y* components of the equations for the line into the equation for the circle, giving:

Or, simplified:

where:

Then we simply solve the quadratic equation:

If the discriminant *b*^{2} − 4*ac* < 0, there are no solutions. The circle does not intersect or touch the line. If the discriminant is equal to 0, then there is a single solution, where the line is tangent to the circle. And if the discriminant is greater than 0, there are two solutions, representing the two points of intersection. Let us call the solutions *d*_{1} and *d*_{2}, if they exist. We have 0, 1, or 2 line segments:

A fold *F*_{1}(*s*) perpendicular to *m*_{1} through its midpoint will place *p*_{1} on the line at location *d*_{1}. Similarly, a fold *F*_{2}(*s*) perpendicular to *m*_{2} through its midpoint will place *p*_{1} on the line at location *d*_{2}. The application of Axiom 2 easily accomplishes this. The parametric equations of the folds are thus:

### Axiom 6

Given two points *p*_{1} and *p*_{2} and two lines *l*_{1} and *l*_{2}, there is a fold that places *p*_{1} onto *l*_{1} and *p*_{2} onto *l*_{2}.

This axiom is equivalent to finding a line simultaneously tangent to two parabolas, and can be considered equivalent to solving a third-degree equation as there are in general three solutions. The two parabolas have foci at *p*_{1} and *p*_{2}, respectively, with directrices defined by *l*_{1} and *l*_{2}, respectively.

This fold is called the Beloch fold after Margharita P. Beloch who in 1936 showed using it that origami can be used to solve general cubic equations.^{[2]}

### Axiom 7

Given one point *p* and two lines *l*_{1} and *l*_{2}, there is a fold that places *p* onto *l*_{1} and is perpendicular to *l*_{2}.

This axiom was originally discovered by Jacques Justin in 1989 but was overlooked and was rediscovered by Koshiro Hatori in 2002.^{[3]} Robert J. Lang has proven that this list of axioms completes the axioms of origami.

## Constructibility

Subsets of the axioms can be used to construct different sets of numbers. The first three can be used with three given points not on a line to do what Alpern calls Thalian constructions.^{[4]}

The first four axioms with two given points define a system weaker than compass and straightedge constructions: every shape that can be folded with those axioms can be constructed with compass and straightedge, but some things can be constructed by compass and straightedge that cannot be folded with those axioms.^{[5]} The numbers that can be constructed are called the origami or pythagorean numbers, if the distance between the two given points is 1 then the constructible points are all of the form where and are Pythagorean numbers. The Pythagorean numbers are given by the smallest field containing the rational numbers and whenever is such a number.

Adding the fifth axiom gives the Euclidean numbers, that is the points constructible by straightedge and compass constructions.

Adding the neusis axiom 6, the reverse becomes true: all compass-straightedge constructions, and more, can be made. In particular, the constructible regular polygons with these axioms are those with sides, where is a product of distinct Pierpont primes. Compass-straightedge constructions allow only those with sides, where is a product of distinct Fermat primes. (Fermat primes are a subset of Pierpont primes.)

The seventh axiom does not allow construction of further point. The seven axioms give all the single fold constructions that can be done rather than being a minimal set of axioms.

## References

- ↑ Justin, Jacques, "Resolution par le pliage de l'equation du troisieme degre et applications geometriques", reprinted in
*Proceedings of the First International Meeting of Origami Science and Technology*, H. Huzita ed. (1989), pp. 251–261. - ↑ {{#invoke:Citation/CS1|citation |CitationClass=journal }}
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## External links

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- Origami Geometric Constructions by Thomas Hull
- A Mathematical Theory of Origami Constructions and Numbers by Roger C. Alperin
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|CitationClass=journal }}