Anthony Yuen - 82472010

dmxlite AT gmail DOT com

There are three classical geometric problems that the Greeks are unable to solve using an unmakred straightedge and compass:

- Trisecting an angle
- Squaring a circle
- Doubling a cube

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.

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

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)

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)

From

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

Hull, T., "A Comparison between Straightedge and Compass constructions and Origami", 1997. http://www.merrimack.edu/~thull/omfiles/geoconst.html

Wikipedia, "Ruler-and-compass construction", 2005. http://en.wikipedia.org/wiki/Ruler-and-compass_construction