Date: May 11, 2001 2:40 PM
Author: Kirby Urner
Subject: [math-learn] A short talk on geometry

A Short Talk on Geometry

A key thing to understand about any discipline is how
people have used it over time. Geometry has everything
to do with navigation, getting from here to there safely,
especially by sea. But it also have everything to do with

Since ancient times, people have been fascinated by
various dimensional relationships or ratios. You all
know what it means for a line segment to be doubled in
size. Do you know that if you have a triangle, square
or any shape, and double all of it's edges, what happens
to its area? How about if you have a pyramid, and blow
it up (in the photography sense -- to make bigger, not
to explode). What happens to its volume? <pause>

Yes, area goes up as a 2nd power of the linear change,
and volume as a 3rd power. So if a pyramid has volume
blah, and you double all its edges, it now has volume
8*blah, where 8 is 2 times 2 times 2 or 2 to the third
power. That's important to keep in mind, and will help
you at many points in the future.

Anyway, one of the key ratios that people have found
fascinating is called the divine proportion or the golden
ratio or phi. You can tell just from the name that it's
important. So what is it? First, it's important to keep
in mind that it's a relationship. It doesn't matter how
long segment A is, only how long it is in relationship
to segment B. So A might be a mile, an inch, a millimeter
or a furlong. Doesn't matter. The important thing is
that A:B as B:(A+B).

In other words, start with a single segment and make a
mark dividing it into A and B. A is the shorter part,
and it relates to B in the same way that B relates to
the whole segment:



A:B = B:(A+B)

This progression goes on in a self-similar manner in both
directions: bigger or more macro, and smaller, or more micro.
In other words, relative to A, there's some shorter segment that
is in divine proportion to it, in the same way as A relates to
B. And likewise (A+B) will relate to a yet longer segment by
the same scale factor. And what is that scale factor? We can
use algebra:

A/B = B/(A+B)

cross-multiply: A(A+B) = B^2

Let's just have A=1 for simplicity, and find B (remember, it makes
no difference what A is by itself -- it's the relationship to B
that matters):

1(1+B) = B^2 or B^2-B-1=0

What are the solutions? You can use completing the square or the
quadratic formula or maybe you don't remember these or haven't learned
them. We can attack the problem empirically at first: We know from
the sketch that B is bigger than A, but not _that_ much bigger. So
let's assume B is double the length of A: 2^2-2-1 = 4-3=1. Too big
-- we need the result to be 0. We know B=1 is too small: 1-1-1=-1.
So the answer we're looking for is somewhere between 2 and 1. Let's
play around at the command line for a sec.

Booting Python here (command window projected on screen):

Python 2.1 (#15, Apr 16 2001, 18:25:49) [MSC 32 bit (Intel)]
on win32
Type "copyright", "credits" or "license" for more information.
IDLE 0.8 -- press F1 for help

>>> from mathobjects import *
>>> p = Poly([1,-1,-1])

We're building a polynomial. Remember from your ancient greek (what
little we teach) that 'poly' means many. That's one of those roots
it pays to remember. Shows up so often: polyhedron, polymorphic,
polygamous, polygon, polymath, polywannacracker (bad joke)...

Polynomials, you may remember, are made up of monomials, which are
coefficents multiplied by your variable (often x or t, by convention)
to some power. The highest power in play is the degree of your
polynomial. B^2 - B - 1 is an example (variable is B). Here we
switch to x -- doesn't matter.

>>> p
x**2 - x - 1

Now we pass in values for our x. Behind the scenes, Python is
evaluating the string 'x**2 - x - 1' and using whatever argument
we pass for x. ** means "to the power of" i.e. x**2 = x to the
2nd power or x*x.

>>> p(2)
>>> p(1)

...the examples we used.

>>> p(1.5)
>>> p(1.6)
>>> p(1.7)
>>> p(1.65)
>>> p(1.63)
>>> p(1.61)

We're narrowing the search for B, the solution to the polynomial,
simply by going higher and lower, getting better and better
approximations. We now know that B is in the neighborhood of 1.61.
That's what you need to scale A by to get B, or scale B by to get
(A+B) -- approximately.

Do you know the quadratic formula? We can solve any polynomial of
the 2nd degree simply by substituting as follows:

Given: a*x**2 + b*x + c = 0

x1 = (a**2 + (b**2 - 4*a*c)**0.5)/2*a
x2 = (a**2 - (b**2 - 4*a*c)**0.5)/2*a

A 2nd degree polynomial has 2 solutions, although they might not
be real numbers, and certainly they don't have to be positive
numbers. But we're looking for a positive, real number, i.e.
something that makes sense as "the length of B" (remember A is 1,
so A*scalefactor = B, and A=1, so B=scalefactor, numerically

Let's write this as a function and pass the coefficients as

>>> def quadequa(a,b,c):
x1 = (a**2 + (b**2 - 4*a*c)**0.5)/2*a
x2 = (a**2 - (b**2 - 4*a*c)**0.5)/2*a
return [x1,x2]

I'm entering this function directly at the command line, without
going into edit mode. Guido's IDLE lets me do this. quadequa is
now a part of my working namespace...

>>> quadequa(1,-1,-1)
[1.6180339887498949, -0.6180339887498949]

I've entered my coefficients, and here we have a positive value for
the variable. By the way, don't confuse b the coefficient of the
2nd monomial with B, what we used as a variable in reference to
segment B. B is our x, our unknown, our variable, in the polynomial
B^2 - B - 1 = 0. b was our coefficient in the 2nd term, i.e. -1.

We derive the quadratic equation using algebra and I've included that
in a hand-out. We can go over that later. But we don't want to lose
the thread here: the importance of certain ratios in architecture,
in design science, and the fact that geometry has everything to do
with these fields, as a discipline with a long history.

We also want to simplify the solution for the scalefactor
algebraically. (a**2 + (b**2 - 4*a*c)**0.5)/2*a in more traditional
math notation would be (1 + SQRT(1 + 4))/2 once you do the
substitution. And that simplifies further to (1 + SQRT(5))/2.
We can check that:

>>> import math
>>> (1+math.sqrt(5))/2


Yep, same answer. So our scale factor is (1+sqrt(5))/2. That's the
golden ratio, the divine proportion, or phi. Given how it goes up
and down the size spectrum, ever smaller or larger, it's kind of like
a fractal. The self-similarity of A:B shows up at every level.
You might even say "Phi is the Phirst Phractal" (see the handout for
how I do the spelling -- haha, pretty silly, but memorable).

The golden rectangle is a rectangle of dimensions A X B, i.e. B is
1.6180339887498949 times longer than A, making for a taller or wider
rectangle that you find all over the place in architecture. Of course
our measurements, when working in inches or meters or whatever, don't
need that kind of precision, so the architect will typically just use
1.618 -- close enough for folk music. This has been going on for
centuries. People even find phi in the proportions of the human body
itself -- evidence, to some, of a divine creator (the so-called
sacred geometers imbued their math with a lot of philosophy and
metaphysics -- didn't separate all these aspects of thought into
walled-off-from-each-other compartments, the way we do today).

Note that if we say the area of this rectangle is 1.618, and then
scale all the edges by 3, the area goes up by a factor of 9, i.e.
(3**2)*1.618 = 14.562.

The golden cuboid is like a brick, and has edges of 1/phi, 1 and
phi. Or .618, 1 and 1.618. 1/phi has the pleasing property of
being phi-1. Indeed (phi-1)=1/phi is restatement of the basic
algebra, i.e. may be transformed into our polynomial: B^2-B-1=0.
B is phi here, since A was 1.

If you halve the edges of the golden cuboid, all the angles stay
the same. The *shape* is unchanged because *shape* relates to
angles, and these are constant through a scaling operation. But
the volume is affected. If it started at 1*phi*(1/phi)=1 cube
(same as a cube with edges 1), it will now be (1/2)**3 or 1/8 of
the same unit cube -- because volume changes as a 3rd power of
the change in linear dimensions (this is part of the 7th grade
standard in California in 2001, so you better know this if you
want to pass out of the 7th grade in California).

Note also that sqrt(5) is irrational, i.e. the algorithm behind it
will keep giving you decimal digits until you stop it. So we're
dealing with an infinite loop, a loop with no defined break point.
That's what an irrational number is -- an algorithm with no terminus.
So yes, phi is one of those. It's not transcendental though, because
it shows up as a solution to a polynomial with rational coefficients,
as we've just seen -- transcendentals don't do that.

In our talk tomorrow, we'll go into how polyhedra, not just line
segments, may be ratioed. This has a lot to do with how they can
be nested, or fit inside one another, in a pleasing, memorable way.
This kind or ratio-ing, or relationship formation, has also been very
important to geometers through the ages, and is a good place to pick
up as we continue to explore this discipline.


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