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.

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.

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+(a^{2}+b^{2})(x_{0}^{2}+y_{0}^{2})-c^{2}=0, where A=(a^{2}+b^{2})x_{0}+ac and B=(a^{2}+b^{2})y_{0}+bc. Let the coordinate of P_{2} be (h,0), and p_{2} has the equation y^{2}-2hx+h^{2}=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}-C^{2}=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}-C^{2}, 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.)

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 x^{3}+ax^{2}+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 x^{3}+ax^{2}+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=-t^{2}-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, t^{3}+at^{2}+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 t^{2}+at+b=0. Therefore t^{3}+at^{2}+bt+c=0.

We can double a cube by solving the equation x^{3}-2=0.

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

Hatori Koshiro (Koshiro is my first name.)