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Huzita's axioms

Huzita's axioms are a set of rules related to the mathematical principles of origami. They were formulated by Italian-Japanese mathematician Humiaki Huzita in 1992, and are the most powerful known set of axioms related to origami. 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 unique 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.

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.

Table of contents
1 Details

Details

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:

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:

Axiom 3

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

This is equivalent to finding the 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 the bisector is then:

And the parametric equation of the fold is:

If the two lines are parallel, on the other hand, they have no point of intersection. The fold must be the line midway between l1 and l2 and parallel to them.

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:

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:

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:

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 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:

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.