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

[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.

Folding a line through two points

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.

Folding a line putting one point on another

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 pmid of P(s)
  • Find the vector vperp perpendicular to P(s)
  • The parametric equation of the fold is then:
F(s)=p_\mathrm{mid} + s\cdot\mathbf{v}^{\mathrm{perp}}

[edit] Axiom 3

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

Folding a line putting one line on another

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:

\mathbf{u} = (p_2-p_1) / \left|(p_2-p_1)\right|
\mathbf{v} = (q_2-q_1) / \left|(q_2-q_1)\right|

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

p_\mathrm{int} = p_1+s_\mathrm{int}\cdot\mathbf{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}\right| \mathbf{v} + \left|\mathbf{v}\right| \mathbf{u}} {\left|\mathbf{u}\right| + \left|\mathbf{v}\right|}

And the parametric equation of the fold is:

F(s) = p_\mathrm{int} + s\cdot\mathbf{w}

A second bisector also exists, perpendicular to the first and passing through pint. 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.

Folding through a point perpendicular to a line

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:

F(s) = p_1 + s\cdot\mathbf{v}

[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.

Folding a point onto a line through another point

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=(xc, yc), with radius r equal to the distance from p1 to p2. Then the circle can be expressed as:

(x − xc)2 + (y − yc)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) − xc)2 + (y1 + s(y2 − y1) − yc)2 = r2

Or, simplified:

as2 + bs + c = 0

Where:

a = (x2 − x1)2 + (y2 − y1)2
b = 2(x2 − x1)(x1 − xc) + 2(y2 − y1)(y1 − yc)
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{-b\pm\sqrt{b^2-4ac}}{2a}

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:

l_1 = \overline{p_1 d_1}
l_2 = \overline{p_1 d_2}

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:

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}

[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.

Image:Huzita axiom 6.png

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.

Image:Huzita-Hatori axiom 7.png

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

Origami Portal

[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.