# Constructions for the Golden Ratio

Making a Paper Knot to show the Golden Section in a Pentagon

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!

1. As you would tie a knot in a piece of string ...
2. ... gently make an over-and-under knot, rolling the paper round as in the diagram.
3. (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.
4. Now if you hold it up to a bright light, you'll notice you almost have the pentagram shape - one more fold reveals it ...
5. 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

Here are two flags with just one 5-pointed star: Guinea-Bissau (left) and Puerto Rico (right) from the FOTW Flags Of The World website at http://flagspot.net/flags/ site.

Its Colouring Book link has small pictures of the flags useful for answering the questions in this Quiz.

#### Things to do 1. How many five-pointed stars are there on the USA flag? Has this always been the case? What is the reason for that number?
2. Many countries have a flag which contains the 5-pointed star above. Find at least four more.
3. Which North African country has a pentagram on its flag?
4. Some countries have a flag with a star which does not have 5 points: Which country has a six-pointed star in its flag?
5. Find all those countries with a flag which has a star of more than 6 points.
6. 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:

1. 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.
2. Choose every other point to make an equilateral triangle ACE.
3. 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).
4. 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:

Is there a right-angled triangle with sides in geometric progression, that is
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:

(ar2)2 = (ar)2 + (a)2
a2r4 = a2r2 + a2

We can divide through by a2

:

r4 = r2 + 1

and if we use R to stand for r2 we get a quadratic equation:

R2 = R + 1 or
R2R – 1 = 0

which we can solve to find that

R =
 1 ± √5 2
= Phi or –phi Since R is r2 we cannot have R as a negative number, so

R = r2 = Phi so
r = √Phi

The sides of the triangle as therefore

a, a√Phi, a Phi

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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