Constructions for the Golden Ratio
Here's an easy method to show the golden section by making a Knotty
Pentagram;
it doesn't need a ruler and it doesn't involve any maths either!
Take a length of paper from a roll - for instance the type that
supermarkets use to
print out your bill - or cut off a strip of paper a couple of
centimetres wide from the
long side of a piece of paper.
If you tie a knot in the strip and put a strong light behind it,
you will see a pentagram with all lines
divided in golden ratios.
This is my favourite method since it
involves a Knot(t)!
Here are 5 pictures to help (well it is a pentagram so I had to make 5
pictures!)
- although it really is easy
once you practice tying the knot!
- As you would tie a knot in a piece of string ...
- ... gently make an over-and-under knot, rolling the paper round as in the diagram.
- (This is the slightly tricky bit!)
Gently pull the paper so that it tightens and you can crease the folds as shown to make it lie perfectly flat. - Now if you hold it up to a bright light, you'll notice you almost have the pentagram shape - one more fold reveals it ...
- Fold the end you pushed through the knot back (creasing it
along the edge of
the pentagon) so that the two ends of the paper
almost meet. The knot will then hang like a medal at the end of a
ribbon.
Looking through the knot held up to the light will show a perfect pentagram, as in the diagram above.
Flags of the World and Pentagram stars
Its Colouring Book link has small pictures of the flags useful for answering the questions in this Quiz.
Things to do
- How many five-pointed stars are there on the USA flag? Has this always been the case? What is the reason for that number?
- Many countries have a flag which contains the 5-pointed star above. Find at least four more.
- Which North African country has a pentagram on its flag?
- Some countries have a flag with a star which does not have 5 points: Which country has a six-pointed star in its flag?
- Find all those countries with a flag which has a star of more than 6 points.
- Project Make a collection of postage stamps containing flags
or specialise
in those with a
five-pointed star or pentagram. You might also include mathematicians
that have
appeared on stamps too. Here's a Swiss stamp to start off your (virtual) collection.
Prof Robin Wilson has a Stamp Corner section in the Mathematical Intelligencer.
He has produced a book Stamping Through Mathematics Springer Verlag, 136 pages, (2001)
which has got some good reviews, eg this and this.
There is also a comprehensive web site called
Sci-Philately with several sections on maths and also
Jeff Miller's Mathematicians on Stamps page is a large catalogue of stamps
with pictures.
Jim Kuzmanovich also has a page on
mathematical stamp collecting.
Phi and Triangles
Phi and the Equilateral Triangle
Chris
and Penny at Regina University's Math Central (Canada) show how
we can use any circle to construct on it a hexagon and
an equilateral triangle. Joining three pairs of points then reveals
a line and its golden section point as follows:
- On any circle (centre O), construct the 6 equally spaced points A, B, C, D, E and F on its circumference without altering your compasses, so they are the same distance apart as the radius of the circle. ABCDEF forms a regular hexagon.
- Choose every other point to make an equilateral triangle ACE.
- On two of the sides of that triangle (AE and AC), mark their mid-points P and Q by joining the centre O to two of the unused points of the hexagon (F and B).
- The line PQ is then extended
to
meet the circle at point R.
Q is the golden section point of the line PR.
Q is a gold point of PR
The proof of this is left to you because it is a nice exercise either
using coordinate geometry and the
equation of the circle and the line PQ to find their point of
intersection or
else using plane geometry to find the lengths PR and QR.
The diagram on the left has many golden sections and yet contains only equilateral triangles. Can you make your own design based on this principle?
Chris
and Penny's page shows how to continue using your compasses
to make a pentagon with QR as one side.
Equilateral Triangles and the
Golden ratio J F Rigby, Mathematical Gazette vol 72 (1988),
pages 27-30.
Phi and the Pentagon Triangle
Earlier we saw that the 36°-72°-72°
triangle shown here as ABC occurs in both the pentagram and the decagon.
Its sides are in the golden ratio (here P is actually Phi)
and therefore we have lots of true golden ratios in the
pentagram star on the left.
But in the diagram of the pentagram-in-a-pentagon on the left,
we not only have the tall 36-72-72 triangles, there is a flatter on
too. What about its sides and angles?
Phi and another Isosceles triangle
If we copy the BCD triangle from the red diagram above
(the 36°-72°-72° triangle), and put another triangle on the side
as we see in this green diagram, we are again using P=Phi as above
and get a similar shape -
another isosceles triangle - but a "flat" triangle.
The red triangle of the pentagon
has angles 72°, 72° and 36°,
this green one has 36°, 36°, and 72°.
Again the
ratio of the shorter to longer sides is Phi, but the two equal sides
here are the shorter
ones (they were the longer ones in the "sharp" triangle).
These two triangles are the basic building shapes of Penrose tilings (see the section mentioned previously for more references). They are a 2-dimensional analogue of the golden section and make a very interesting study in their own right. They have many relationships with both the Fibonacci numbers and Phi.
Phi and Right-angled Triangles
Pythagorean triangles are right-angled triangles that have sides which are whole numbers in size. Since the golden section, Phi, is not a pure fraction (it is irrational), we will not be able to find a Pythagorean triangle with two sides in golden section ratio.
However, there is as a right-angled triangle that does have sides in the golden ratio. It arises if we ask the question:
the ratio of two of its sides is also the ratio of two different sides in the same triangle?
If there is such a triangle, let its shortest side be of length a and let's
use r as the common ratio in the geometric progression so the
sides of the
triangle will be a, ar, ar2.
Since it is right-angled, we can use Pythagoras' Theorem to get:
a2r4 = a2r2 + a2
We can divide through by a2
:
and if we use R to stand for r2 we get a quadratic equation:
R2 – R – 1 = 0
which we can solve to find that
| R = |
|
= Phi or –phi |
Since R is r2 we cannot have R as a negative number, so
r = √Phi
The sides of the triangle as therefore
and any right-angled triangle with sides in Geometric Progression has two pairs of sides in the same ratio √Phi and one pair of sides in the Golden Ratio!
We will meet this triangle and its angles later on this page in the section on Trigonometry and Phi.
A trigonometric intersection D Quadling, Math. Gaz. (2005), Note 89.70
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