I will give some info about polyhedra in this section which I find interesting. I'll try to give references when I can.

The importance of the following proposition, I think, can not be over stated. So I will give some introductary remarks.

First of all I need to point out that any polyhedron can be divided into tetrahedra. That is, all polyhedra are made up of face bond tetrahedra.

So, if we can prove that all the tetrahedra making up a polyhedron have integer volumes, with respect to the defining unit of measure regular tetrahedron volume, then the orginal polyhedron will also have an integer volume (it being just the sum of all the constituent, integer-volumed tetrahedra.)

This is what the following proposition and proof shows for the case of all the polyhedron's vertices are part of an IVM.

Proposition: | The volume of *any* tetrahedron |

formed by any 4 points in the IVM | |

(not all in the same plane) in terms of | |

the IVM's unit tetrahedron's volume will | |

be an *integer* number. |

We now calculate a normal vector to this plane by the vector cross product as follows:

n = (y2*z3 - y3*z2)i + (x3*z2 - x2*z3)j + (x2*y3 - x3*y2)k

which we write as

n = Ai + Bj + Ck

where i, j, k are unit vectors along the x, y, z positive axis.

Note that A, B, C are integers and that they are *all* even or *all* odd when we require that x2+y2+z2=even and x3+y3+z3=even. (See below for proof of this.)

Remember that the vector cross product also gives the *area* of the parallelogram defined by the 2 vectors. That is mag(n) = area of parallelogram defined by V2 and V3.

So the *area of the base triangle* is (1/2) mag(n).

area = (1/2) sqrt(A*A + B*B + C*C) = (1/2) mag(n)

Now we look at the 4th point forming the tetrahedron in the IVM V4 = (x4, y4, z4) with x4, y4, z4 all integers.

The distance from this point (V4) to the plane formed by V2 and V3 (which is also the altitude of the tetrahedron) is given by

L = (A*x4 + B*y4 + C*z4) / sqrt(A*A + B*B + C*C)

L = (A*x4 + B*y4 + C*z4) / mag(n)

Now, the volume of the tetrahedron is given by

Vol = (1/3)*[base triangle area]*[tetra altitude]

= (1/3)*[(1/2)mag(n)]*[(A*x4 + B*y4 + C*z4) / mag(n)]

= (1/3)(1/2)(A*x4 + B*y4 + C*z4)

And since A, B, C, x4, y4, z4 are integers, the volume of the tetrahedron will be a rational number.

To show that the volume will be an integer in terms of the IVM's unit tetrahedron's volume:

The "unit" tetrahedron has a volume of 1/3 in the IVM. So the formula for the volume of any tetrahedron in terms of the "unit" tetrahedron volume is given by

Vol-t = (1/2)(A*x4 + B*y4 + C*z4)

We show that (A*x4 + B*y4 + C*z4) is always an even number.

With the condition that x2+y2+z2 = even and x3+y3+z3 = even it can be shown that A, B, C are either *all* even or *all* odd.

For example (vector cross product):

(e,e,e)x(e,e,e) = (e,e,e)

(e,e,e)x(e,o,o) = (e,e,e)

(e,o,o)x(o,e,o) = (o-e)i+(o-e)j+(e-o)k=(o,o,o)

and so on.

Now, x4+y4+z4 = even can only happen in 4 ways:

(e,e,e), (e,o,o), (o,e,o), (o,o,e)

Then we have for (A*x4 + B*y4 + C*z4):

CASE: A, B, C, all even:

(e*e + e*e + e*e) = even

(e*e + e*o + e*o) = even

(e*o + e*e + e*o) = even

(e*o + e*o + e*e) = even

CASE: A, B, C, all odd:

(o*e + o*e + o*e) = even

(o*e + o*o + o*o) = even

(o*o + o*e + o*o) = even

(o*o + o*o + o*e) = even

Therefore, in terms of the IVM's unit tetrahedron's volume, *any* tetrahedron formed by using the IVM vertices will have an *integer* volume.

For a more general case, we have the following:

Proposition: | The volume of *any* tetrahedron |

formed by any 4 points in which the 4 (x,y,z) | |

coordinates are all integers will have | |

either an integer or half integer volume | |

with respect ot the unit tetrahedron. |

We now calculate a normal vector to this plane by the vector cross product as follows:

n = (y2*z3 - y3*z2)i + (x3*z2 - x2*z3)j + (x2*y3 - x3*y2)k

which we write as

n = Ai + Bj + Ck

where i, j, k are unit vectors along the x, y, z positive axis.

Note that A, B, C are integers.

Remember that the vector cross product also gives the *area*
of the parallelogram defined by the 2 vectors. That is

mag(n) = area of parallelogram defined by V2 and V3.

So the *area of the base triangle* is (1/2) mag(n).

area = (1/2) sqrt(A*A + B*B + C*C) = (1/2) mag(n)

Now we look at the 4th point forming the tetrahedron.

V4 = (x4, y4, z4) with x4, y4, z4 all integers.

The distance from this point (V4) to the plane formed by V2 and V3 (which is also the altitude of the tetrahedron) is given by

L = (A*x4 + B*y4 + C*z4) / sqrt(A*A + B*B + C*C)

L = (A*x4 + B*y4 + C*z4) / mag(n)

Now, the volume of the tetrahedron is given by

Vol = (1/3)*[base triangle area]*[tetra altitude]

Vol = (1/3)*[(1/2)mag(n)]*[(A*x4 + B*y4 + C*z4) / mag(n)]

Vol = (1/3)(1/2)(A*x4 + B*y4 + C*z4)

A, B, C, x4, y4, z4 are integers.

The "unit" tetrahedron has a volume of 1/3. So the formula for the volume of any tetrahedron in terms of the "unit" tetrahedron volume is given by

Vol-t = (1/2)(A*x4 + B*y4 + C*z4)

Which proves the assertion that the volume will be either an integer or half-integer with respect to the unit tetrahedron.

(RWG) Theorem 1: Given *any* tetrahedron, regular or
asymmetrical, anywhere in space, if the 4 vertex coordinates
are all rational then so will the 4 vertex coordinates of the
complementary polyhedron when face bond to the original polyhedron.

The "complementary" polyhedron is the "inside-out" mirror image of the original polyhedron. These complementary polyhedra will be indicated by an "*" appended to their name when it becomes necassary to distinguish them from the original polyhedra.

Comments

This is a very useful theorem. It tells us, for example, that all the coordinates to the vertices of the tetrahelix of any fixed length can be given integer numbers (with the appropiate orientation and scaling.) This is quite remarkable since the spiral rotation of the tetrahelix is irrational.

(RWG) Theorem 2: The resulting vertex coordinates for the A, A*, B or B* quantum module when face bond to a given A, A*, B or B* quantum module will all be rational numbers if the original quantum module has rational vertex coordinates.

Comments

Theorem 1 does not cover the case of face bonding *different*
kinds of polyhedra together. It only deals with a signal polyhedron and its
complement mirror image polyhedron. Theorem 2 tells us that
*any* polyhedron, not just nice symmetrical convex polyhedra,
composed of face bond A and B quantum modules, can be assigned
integer vertex coordinates when appropiately orientated and scaled.

Go To: Table of Contents
or Home Page

Usage Note: My work is copyrighted. You may use my work but
you may *not* include my work, or parts of it, in *any* for-profit project
without my consent.