Origami Construction

Origami versus Straight-Edge-and-Compass

We usually use a straight edge and a compass for mathematical constructions. The straight-edge-and-compass construction is a finite sequence of the following three procedures on a plane with some points placed arbitrarily.

1. Given two points P 1 and P 2 , to draw a line passing through both P 1 and P 2 .

2. Given one point O and one segment P 1 P 2 , to draw a circle whose center is O and whose radius is the length of P 1 P 2 .

3. Given some lines and circles, to locate points of intersection.

We can also construct some lines and points with origami. We use folds in the origami construction, of course. But what kind of folds? We have many kinds of them, and not all are appropriate for constructions.

I will exclude curved folds, because we do not know about them very much. I will also avoid folding along plural straight lines at a time. Otherwise, it gets too complicated. Moreover, let's avoid using any tool such as a ruler, another sheet of paper, or even a pencil.

Under these conditions, we cannot place a point on the plane directly. If we make a fold arbitrarily, we get a line, not a point. To locate a point, we need at least two creases. In other words, any point in the origami construction has at least two lines passing through it.

Then, the origami construction would be a finite sequence of either "to fold a line at a time without any tool" or "given some creases, to locate points of intersection" on a plane with some lines placed arbitrarily.

Six Folds Plus One

Humiaki Huzita listed six folds which we can use in the origami construction.

A. Given two lines L 1 and L 2 , to fold a line placing L 1 onto L 2 .

B. Given two points P 1 and P 2 , to fold a line placing P 1 onto P 2 .

C. Given two points P 1 and P 2 , to fold a line passing through both P 1 and P 2 .

D. Given one point P and one line L, to fold a line passing through P and perpendicular to L.

E. Given two point P 1 and P 2 and one line L, to fold a line placing P 1 onto L and passing through P 2 .

F. Given two point P 1 and P 2 and two lines L 1 and L 2 , to fold a line placing P 1 onto L 1 and placing P 2 onto L 2 .

I found another.

G. Given one point P and two lines L 1 and L 2 , to fold a line placing P onto L 1 and perpendicular to L 2 .

Robert Lang proved that these seven folds are complete. That is, there is no other fold in origami construction. We do not need more than seven folds to execute all possible construction. Then, do we need all seven of them? I will argue that we need only one fold, instead of seven.

Formulation of Origami Construction

First, we can consider the fold A to be the fold F with P 1 being on L 2 and P 2 being on L 1 . When there are two lines L 1 and L 2 , we can construct, using an arbitrary line, two points P 2 and P 1 on L 1 and L 2 , respectively.

Second, we can also consider the fold B to be the fold F with P 1 being on L 2 and P 2 being on L 1 . In the origami construction, there are always two lines L 1 which passes through P 2 but does not pass through P 1 and L 2 which passes through P 1 but does not pass through P 2 when there are two points P 1 and P 2 .

Now let's look at the fold F. It is known that the crease is a common tangent of two parabolas p 1 whose focus is P 1 and whose directrix is L 1 and p 2 whose focus is P 2 and whose directrix is L 2 . What happens if P 2 comes close to L 2 ?

Set a Cartesian coordinate system so that P 2 is on the x-axis and L 2 equals the y-axis. Let the equation of L 1 be ax+by+c=0 and the coordinate of P 1 be (x 0 ,y 0 ). Then p 1 has the equation (bx-ay)2-2Ax-2By+(a2+b2)(x 0 2+y 0 2)-c2=0, where A=(a2+b2)x 0 +ac and B=(a2+b2)y 0 +bc. Let the coordinate of P 2 be (h,0), and p 2 has the equation y2-2hx+h2=0, since the equation of L 2 is x=0.

Let the crease be tangent to p 1 at (x 1 ,y 1 ), and the crease has the equation (bC-A)x-(aC+B)y+Ax 1 +By 1 -C2=0, where C=bx 1 -ay 1 . Let the crease be tangent to p 2 at (x 2 ,y 2 ), and the crease has the equation hx-y 2 y+h(x 2 -h)=0. Since these equations express the same line, we get three equations: kh=bC-A, ky 2 =aC+B, and kh(x 2 -h)=Ax 1 +By 1 -C2, where k is a nonzero constant.

When P 2 comes close to L 2 , h approximates to 0. So do kh and kh(x 2 -h). At the limit, P 2 gets onto L 2 . If kh=0, the equation of the crease is (aC+B)(y-y 1 )=0. That means the crease is a tangent of p 1 perpendicular to the y-axis, namely L 2 . If kh(x 2 -h)=0, the equation of the crease is (bC-A)x-(aC+B)y=0. That means the crease is a tangent of p 1 passing through the origin, namely P 2 .

Therefore, we can define a line placing point P onto line L when P is on L as either a line perpendicular to L or a line passing through P. Then, we can consider the fold E and G as the fold F with P 2 being on L 2 . We can also consider the fold C and D as the fold F with P 1 being on L 1 and P 2 being on L 2 at the same time.

All things considered, we need only one fold in the origami construction. Therefore, we can define the origami construction as a finite sequence of the following two procedures on a plane with some lines placed arbitrarily.

1. Given some lines, to locate points of intersection.

2. Given two point P 1 and P 2 and two lines L 1 and L 2 , to fold a line placing P 1 onto L 1 and placing P 2 onto L 2 . (A line placing point P onto line L when P is on L is either a line perpendicular to L or a line passing through P.)

Cubic Equation

It is known that the straight-edge-and-compass construction is equivalent to solving quadratic equations. On the other hand, the origami construction is equivalent to solving cubic equations. So, any line or point can be constructed with origami if it can be constructed with a straight edge and a compass. In addition, we can solve with origami some problems which cannot be solved with a straight edge and a compass, such as doubling a cube or trisecting an angle.

Now, let's solve the cubic equation x3+ax2+bx+c=0 with origami. Let two points P 1 and P 2 have the coordinates (a,1) and (c,b), respectively. Also let two lines L 1 and L 2 have the equations y+1=0 and x+c=0, respectively. Fold a line placing P 1 onto L 1 and placing P 2 onto L 2 , and the slope of the crease is the solution of x3+ax2+bx+c=0.

I will explain why. Let p 1 be a parabola having the focus P 1 and the directrix L 1 . Since the crease is not parallel to the y-axis, we can let the crease have the equation y=tx+u. Let the crease be tangent to p 1 at (x 1 ,y 1 ), and (x 1 -a)2=4y 1 . Because the crease has the equation (x 1 -a)(x-x 1 )=2(y-y 1 ), we get t=(x 1 -a)/2 and u=y 1 -x 1 (x 1 -a)/2. From these equations, we get u=-t2-at.

When c is not 0, let p 2 be a parabola having the focus P 2 and the directrix L 2 . Let the crease be tangent to p 2 at (x 2 ,y 2 ), and (y 2 -b)2=4cx 2 . Because the crease has the equation (y 2 -b)(y-y 2 )=2c(x-x 2 ), we get t=2c/(y 2 -b) and u=y 2 -2cx 2 /(y 2 -b). From these equations, we get u=b+c/t. (t is not 0 because the crease is not parallel to the x-axis.) Therefore, t3+at2+bt+c=0.

When c is 0, P 2 is on L 2 . So, either the crease is perpendicular to L 2 or the crease passes through P 2 . In the former case, t=0. In the later case, u=b, and t2+at+b=0. Therefore t3+at2+bt+c=0.

We can double a cube by solving the equation x3-2=0.

We can also trisect an angle by solving the equation x3+3tx2-3x-t=0, where t=1/tanθ and x=tan(θ/3-π/2).