Doubling a Cube Using Origami There are three problems in classical geometry that for many centuries were open: trisecting an angle, doubling the cube, and squaring the circle. These are all now known to be impossible using "classical methods". What do I mean by "classical methods"? I mean we start with a Euclidean plane, so flat, and extending infinitely far in all directions. We are given two tools: an unmarked compass and an unmarked straightedge. The compass allows us to draw circles; if we have a point and a length, we can draw a circle whose center is at that point and whose radius is that length. But the compass is unmarked, so we can't actually just pick a length arbitrarily. We have to first construct or be given a line segment of that length. The straightedge allows us to draw line segments. In particular, given two points, we can draw a line segment that joins those two points, and to extend that line segment as far as we want. We can also find and draw lines that are tangent to a circle, so if we have a point outside of the circle we can draw a line that goes through the point and is tangent to the circle, or if we have a point on the circle we can draw the tangent line to the circle through that point. But the straightedge is unmarked, so again we can't pick the lengths of our line segments beforehand; we need points to cordon off any particular lengths we want to use. The problem of doubling a cube is to take a line segment and construct another line segment whose length is the length of the first segment times the cube root of 2. The problem of trisecting an angle is to take two line segments that intersect at a point with some angle, and construct a third line segment that also goes through the point and whose angle with one of the first two line segments is 1/3 of the original angle. The problem of squaring the circle is to take a circle and construct a square whose area is exactly that contained within the circle. All of these can be done if we allow calculus, but the question was how much could be done with just an unmarked compass and an unmarked straightedge? Let's look at the question of doubling a cube. It is really a question of constructing two line segments where the ratio of their lengths is the cube root of 2. So we can ask: can we construct a cube root using an unmarked compass and an unmarked straight edge? Well, actually, first let's ask a simpler question: can we construct a square root using an unmarked compass and an unmarked straightedge? In particular, suppose that we have a line segment of length L; can we construct another line segment of length L√n for any natural number n? The answer turns out to be yes. There's a fairly simple construction, although that doesn't mean it's easy to stumble across. Spoiler: Hint Set up two similar right triangles that share a leg; what does that tell you about the ratios of the leg lengths? So we can take square roots. Indeed, the quadratic formula tells us that since we can take square roots, we can in fact solve general quadratic equations with integer coefficients, at least once you figure out how to formulate them in terms of line segments. Since you can take square roots, you can also take fourth roots, and eight roots, and sixteenth roots, by repeatedly taking square roots. Now consider all the numbers you get by starting with the natural numbers and including everything you can get by repeatedly adding, subtracting, multiplying, dividing, and taking square roots. We end up with a field that we call the "constructible numbers". It turns out that these numbers are precisely the length ratios that you can get via compass and straightedge starting from a single line segment. To briefly give some intuition: compasses and straightedges given you lines and circles. From an algebraic perspective, a line is defined by a degree 1 polynomial: y = mx + b. So you can do linear stuff with lines. A circle is defined by a degree 2 polynomial: x2 + y2 = r2. So you can solve quadratic stuff since you have access to circles. Thus you might expect that anything you can construct with lines and circles are the things you can construct via basic arithmetic and solving quadratic equations. Now the question becomes: is the cube root of 2 a constructible number? I don't remember if I talked about Galois theory in this thread yet, so I'll just say it outright: no, the cube root of 2 is not constructible. Indeed, if an integer is not the cube of another integer, then its cube root is not constructible. So you can't double a cube using just a compass and straightedge. In fact, trisecting an angle is also a form of taking a cube root; the angle addition formulas in trigonometry and/or De Moivre's law will show that being able to trisect an arbitrary angle is equivalent to being able to take the cube root of an arbitrary integer. Just as there are some integers whose cube roots are easy, like 8 or 27, there are some angles that are easy to trisect; but most of the time you're out of luck. However, with origami, you can take the cube root of a length! Firstly, what are the allows moves in origami? You can make a fold along a line that joins two specific points, you can make a fold that takes one specified point and puts it in the same place as another specified point, you can make a fold that takes one specified line and lays it along another specified line, and given two points and two lines, you can make a fold that puts the first point on the first line and the second point on the second line, provided such a fold exists at all. A little work shows that anything you can construct with a compass and straightedge can be achieved via folding, provided you have a big enough piece of paper. But you can also take cube roots! Take two lines that cross at right angles and treat them as coordinate axes; i.e. you specify some length as 1, and you specify one line as the x-axis and the other as the y-axis. Now mark two points, M at (0,-1) and N at (-n, 0). Fold so that M ends up on the line y = 1 and N ends up on the line x = N. The x-intercept of the fold is at 3√n. How does this work? It essentially extends the reasoning behind the construction for square roots. Spoiler: Hint The fold allows you to set up 3 similar right triangles so that the first shares a leg with the second, and the second shares its other leg with the third. Now what can you say about the ratios of leg lengths? So you can double a cube using origami, and you can trisect an angle using origami. Just as you can solve quadratics using a compass and straightedge, you can solve degree 4 polynomials with integer coefficients using origami. However, squaring a circle requires constructing a length ratio that is not the solution to any polynomial with integer coefficients, i.e. a transcendental number, and hence neither compass and straightedge nor origami will be able to square a circle.