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.

The axioms were first discovered by Jacques Justin in 1989. [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.] Axioms 1 through 6 were rediscovered by Italian-Japanese mathematician Humiaki Huzita in 1992, and Axiom 7 was rediscovered by Koshiro Hatori in 2002.

The seven axioms

The first 6 axioms are known as Huzita's axioms. Axiom 7 was discovered by Koshiro Hatori. 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.

It should be noted that 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 :

:$F(s)=p\_1\; +s(p\_2\; -\; p\_1)$

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 $P(s)=p\_1+s(p\_2-p\_1)$
* 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::$F(s)=p\_mathrm\{mid\}\; +\; scdotmathbf\{v\}^\{mathrm\{perp$

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:

:$mathbf\{u\}\; =\; (p\_2-p\_1)\; /\; left|(p\_2-p\_1)\; ight|$:$mathbf\{v\}\; =\; (q\_2-q\_1)\; /\; left|(q\_2-q\_1)\; ight|$

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

:$p\_mathrm\{int\}\; =\; p\_1+s\_mathrm\{int\}cdotmathbf\{u\}$

where

:$s\_\{int\}\; =\; -frac\{mathbf\{v\}^\{perp\}\; cdot\; (p\_1\; -\; q\_1)\}\; \{mathbf\{v\}^\{perp\}\; cdot\; mathbf\{u$

The direction of one of the bisectors is then:

:$mathbf\{w\}\; =\; frac\{left|mathbf\{u\}\; ight|\; mathbf\{v\}\; +left|mathbf\{v\}\; ight|\; mathbf\{u\{left|mathbf\{u\}\; ight|\; +left|mathbf\{v\}\; ight$

And the parametric equation of the fold is:

:$F(s)\; =\; p\_mathrm\{int\}\; +\; scdotmathbf\{w\}$

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:

:$F(s)\; =\; p\_1\; +\; scdotmathbf\{v\}$

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:

:$x\; =\; x\_1\; +\; s(x\_2\; -\; x\_1)$:$y\; =\; y\_1\; +\; s(y\_2\; -\; y\_1)$

Let the circle be defined by its center at "p"₂=("x_c", "y_c"), with radius "r" equal to the distance from "p"₁ to "p"₂. Then the circle can be expressed as:

:$(x-x\_c)^2\; +\; (y-y\_c)^2\; =\; r^2$

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:

:$(x\_1\; +\; s(x\_2-x\_1)\; -\; x\_c)^2\; +\; (y\_1\; +\; s(y\_2\; -\; y\_1)\; -\; y\_c)^2\; =\; r^2$

Or, simplified:

:$as^2\; +\; bs\; +\; c\; =\; 0$

Where:

:$a\; =\; (x\_2\; -\; x\_1)^2\; +\; (y\_2\; -\; y\_1)^2$:$b\; =\; 2(x\_2\; -\; x\_1)(x\_1\; -\; x\_c)\; +\; 2(y\_2\; -\; y\_1)(y\_1\; -\; y\_c)$:$c\; =\; x\_c^2\; +\; y\_c^2\; +\; x\_1^2\; +\; y\_1^2\; -\; 2(x\_c\; x\_1\; +\; y\_c\; y\_1)-r^2$

Then we simply solve the quadratic equation:

:$frac\{-bpmsqrt\{b^2-4ac\{2a\}$

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:

:$l\_1\; =\; overline\{p\_1\; d\_1\}$:$l\_2\; =\; overline\{p\_1\; d\_2\}$

A fold "F"₁("s") perpendicular to "l"₁ through its midpoint will place "p"₁ on the line at location "d"₁. Similarly, a fold "F"₂("s") perpendicular to "l"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:

:$F\_1(s)\; =\; p\_1\; +frac\{1\}\{2\}(d\_1-p\_1)+s(d\_1-p\_1)^\{perp\}$:$F\_2(s)\; =\; p\_1\; +frac\{1\}\{2\}(d\_2-p\_1)+s(d\_2-p\_1)^\{perp\}$

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. The two parabolas have foci at "p"₁ and "p"₂, respectively, with directrices defined by "l"₁ and "l"₂, respectively.

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.

Koshiro Hatori has discovered this axiom, and Robert J. Lang has proven that this list of axioms completes the axioms of origami.

References

* [http://www.merrimack.edu/~thull/omfiles/geoconst.html Origami Geometric Constructions] by Thomas Hull
* [http://nyjm.albany.edu:8000/j/2000/6-8.html A Mathematical Theory of Origami Constructions and Numbers] by Roger C. Alperin
* cite paper
author = Lang, Robert J.
title = Origami and Geometric Constructions
publisher = Robert J. Lang
date = 2003
url = http://www.langorigami.com/science/hha/origami_constructions.pdf
format = PDF
accessdate = 2007-04-12