Doubling the cube with origami

Math 308 Project - 2005 Winter Semester
Anthony Yuen - 82472010
dmxlite AT gmail DOT com

Geometric problems of antiquity

There are three classical geometric problems that the Greeks are unable to solve using an unmakred straightedge and compass: This project is to show how you can double the cube using origami (folding).


According to legend, there was a plague that was devastating Greece in 430BC. The citizens went to the ocacle of Apollo at Delos which told them that the plague will stop when they've doubled the size of their altar. The citizens doubled the sides of the altar, but the plague did not stop. What went wrong? The citizens did not follow instructions. Doubling of the sides of the altar caused the volume of the altar (which is cubic) to increase eight times (x^3 becomes (2x)^3 = 8x^3). The correct solution is doubling the volume. This problem is also known as the Delian problem.

The problem

Let x be the side of the new altar, and y be the side of the old altar. You want x^3 = 2y^3. Rearranging the terms yield x/y = 2^(1/3). This means to solve the problem, you need to find sides such that the ratio of the new side to the old side is the cube root of 3. How come the greeks did not solve it? At that time, the method to solve geometric problems was using an unmarked straightedge and a compass. You cannot construct the cube root of 2 because this number cannot be computed by a straightedge and compass' operations: addition, subtraction, multiplication, division, and taking square roots.

How origami solves this

Origami can be used to show this ratio because certain origami folds yield answers which are similar to solving a cubic polynomial.


The following axioms are the operations which are allowed in origami folding.

Links will open a postscript file.
Axiom Description Mathematical operation
Axiom I Given two points, p1 and p2, there is a unique fold that passes through both of them This gives a line through p1 and p2
Axiom II Given two points, there is a fold that places p1 onto p2 This is finding the perpendicular bisector of the line through p1 and p2
Axiom III Given two lines, L1 and L2, there is a fold that places one onto the other This is like finding a bisector of the angle between L1 and L2
Axiom IV Given a line, L1, and a point, p1, there is a fold that is perpendicular to L1 and passes through p1 This is like finding a perpendicular line to L1 that passes through p1
Axiom V Given two points, p1 and p2, and a line, L1, there exists a fold that passes through p2 and places p1 onto L1 This is equivalent to finding the intersection of a line with a circle. The line is L1, and the circle is centered at p2 with a radius equal to the distance between p1 and p2
Axiom VI Given two points, p1 and p2, and two lines, L1 and L2, there is a fold that can place p1 and p2 onto L1 and L2 respectively This is like finding a line that is a tangent to two parabolas. The parabolas have foci at p1 and p2, and their directrices are defined by the lines L1 and L2

The above are the 'original' axioms discovered by Humiaki Huzita. The sixth axiom will be important for our cause.
For completeness, there is a seventh axiom, discovered by Koshiro Hatori which completes the axioms of origami.
This axiom says that given a point p1 and two lines, L1 and L2, you can fold a line which is perpendicular to L2 and places p1 onto L1.

Solving the Delian problem

First we need a square piece of paper, which we will divide into thirds. To see how this is done using origami, see Casper Ho's Math 308 project at this link: (Notice how he also shows that the first problem of antiquity can be solved with origami)

To see how to show the cube root of 2, open this postscript file:

Let CB be of a unit length, 1. Let AC be x and BD be y.
The side of the square is s = 1 + x. CD = 1 + y^2
Rewritten in terms of x, y = (x^2 + 2x)/(2x + 2) (1)
AP = 1 /3 * s = (1 + x) / 3
CP = (2x - 1)/3
Triangle PCS is similar to BDC, thus CT/BT => (1 + x) / (2x - 1) = [(1+x)/y] - 1
or y = (2x^2 + x -1)/(3x) (2)
From (1) and (2), 4x^3 + 6x^2 - 2 = 3x^3 + 6x^2
Hence x^3 = 2, thus x = 2 ^ 1/3. The ratio of the length x to the unit length is the cube root of 2


Problem 1054, "Crux Mathematicorum", V12, No10, 1986, pp 284-285

Hull, T., "A Comparison between Straightedge and Compass constructions and Origami", 1997.

Wikipedia, "Ruler-and-compass construction", 2005.