Huzita-Hatori axioms

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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.^[1] Axioms 1 through 6 were rediscovered by Italian-Japanese mathematician Humiaki Huzita in 1992, and Axiom 7 was rediscovered by Koshiro Hatori in 2002.

Contents 1 The seven axioms 2 Details 2.1 Axiom 1 2.2 Axiom 2 2.3 Axiom 3 2.4 Axiom 4 2.5 Axiom 5 2.6 Axiom 6 2.7 Axiom 7 3 References

[edit] 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:

1. Given two points p₁ and p₂, there is a unique fold that passes through both of them.
2. Given two points p₁ and p₂, there is a unique fold that places p₁ onto p₂.
3. Given two lines l₁ and l₂, there is a fold that places l₁ onto l₂.
4. Given a point p₁ and a line l₁, there is a unique fold perpendicular to l₁ that passes through point p₁.
5. Given two points p₁ and p₂ and a line l₁, there is a fold that places p₁ onto l₁ and passes through p₂.
6. 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.
7. Given one point p and two lines l₁ and l₂, there is a fold that places p onto l₁ and is perpendicular to l₂.

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.

[edit] Details

[edit] Axiom 1

Given two points p₁ and p₂, 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) = p₁ + s(p₂ − p₁)

[edit] Axiom 2

Given two points p₁ and p₂, there is a unique fold that places p₁ onto p₂.

This is equivalent to finding the perpendicular bisector of the line segment p₁p₂. This can be done in four steps:

• Use Axiom 1 to find the line through p₁ and p₂, given by P(s) = p₁ + s(p₂ − p₁)
• 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:

[edit] Axiom 3

Given two lines l₁ and l₂, there is a fold that places l₁ onto l₂.

This is equivalent to finding a bisector of the angle between l₁ and l₂. Let p₁ and p₂ be any two points on l₁, and let q₁ and q₂ be any two points on l₂. Also, let u and v be the unit direction vectors of l₁ and l₂, 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 l₁ onto l₂. 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 l₁ and l₂ and parallel to them.

[edit] Axiom 4

Given a point p₁ and a line l₁, there is a unique fold perpendicular to l₁ that passes through point p₁.

This is equivalent to finding a perpendicular to l₁ that passes through p₁. If we find some vector v that is perpendicular to the line l₁, then the parametric equation of the fold is:

[edit] Axiom 5

Given two points p₁ and p₂ and a line l₁, there is a fold that places p₁ onto l₁ and passes through p₂.

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 l₁, and the circle has its center at p₂, and a radius equal to the distance from p₂ to p₁. 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, (x₁, y₁) and (x₂, y₂), then the line can be expressed parametrically as:

x = x₁ + s(x₂ − x₁)

y = y₁ + s(y₂ − y₁)

Let the circle be defined by its center at p₂=(x_c, y_c), with radius r equal to the distance from p₁ to p₂. Then the circle can be expressed as:

(x − x_c)² + (y − y_c)² = r²

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:

(x₁ + s(x₂ − x₁) − x_c)² + (y₁ + s(y₂ − y₁) − y_c)² = r²

Or, simplified:

as² + bs + c = 0

Where:

a = (x₂ − x₁)² + (y₂ − y₁)²

b = 2(x₂ − x₁)(x₁ − x_c) + 2(y₂ − y₁)(y₁ − y_c)

Then we simply solve the quadratic equation:

If the discriminant b² − 4ac < 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 d₁ and d₂, if they exist. We have 0, 1, or 2 line segments:

A fold F₁(s) perpendicular to l₁ through its midpoint will place p₁ on the line at location d₁. Similarly, a fold F₂(s) perpendicular to l₂ through its midpoint will place p₁ on the line at location d₂. The application of Axiom 2 easily accomplishes this. The parametric equations of the folds are thus:

[edit] Axiom 6

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.

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 p₁ and p₂, respectively, with directrices defined by l₁ and l₂, respectively.

[edit] Axiom 7

Given one point p and two lines l₁ and l₂, there is a fold that places p onto l₁ and is perpendicular to l₂.

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

[edit] References

1. ^ 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.

• Origami Geometric Constructions by Thomas Hull
• A Mathematical Theory of Origami Constructions and Numbers by Roger C. Alperin
• Lang, Robert J. (2003). "Origami and Geometric Constructions" (PDF). . Robert J. Lang Retrieved on 2007-04-12.