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:

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:

We can divide through by a²

:

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

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

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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Phi and Rectangles

Phi and the Trapezium (Trapezoid)

Scott Beach has invented the trisoceles trapezoid (or trapezium as we say in the UK). It is a tall isosceles triangle with its top part cut off to form a quadrilateral with the following properties:

1. the top and bottom edges are parallel
2. the angles at A and B are equal as are those at C and D
3. all the sides except the top are of equal length (hence trisoceles in Scott's name for this shape)
4. the top and bottom edges are in the proportion of the golden section: the top edge is phi = φ = 0.618... times as long as the base or, equivalently, the base is Phi = Φ = 1.618... times as long as the top edge.
   So we could call this a traphizium or, in the USA, a traphisoid!

It has the interesting property that the diagonals bisect the top two angles - the yellow angles in the diagram are all equal.

The diagram shows the whole isosceles triangle AEB with the missing part of the sides DE and CE given the length x.
Since the triangles AEB and DEC are similar (the angles are the same in each triangle) then their sides are in the same ratio (proportion) to each other.

In the large triangle AEB the base is 1 and the sides are 1+x.
In the smaller triangle DEC the base is φ and the sides are x.
Therefore the ratio of 1 to (1+x) is the same as the ratio of φ to x.
or 1/(1+x) = φ/x
i.e. x = (1+x)φ
Collecting the x's on one side we have (1 - φ)x = φ
so that x = φ/(1 – φ)
If we divide top and bottom by φ and using Φ = 1/φ we see this is the same as 1/(Φ – 1) = 1/φ = Φ
So x is the larger golden section number and the cut-off point on the side of the isosceles triangle is a golden section point!

If we split the triangles in half from E to the base, we can see that the sine of the green angle is 1/(2 (1+Φ)) = 1/( 2Φ²) = φ²/2 = (1 – φ)/2.
This makes it about 78.98984..° = 1.37863... radians.

Things to do

1. Here is another trapezium PQRS that is constructed using the other gold point on a the equal sides of an isosceles triangle RQT.
   Also, the gold point makes the three equal-length sides SP=PQ=QR in the resulting trapezium so it is trisoceles (to use Scott's phrase) and so is a special isosceles triangle.
   PS is Phi times as long as ST and QR is Phi times as long as RT.
   Note that PQT is not the same shape as ABE above!

   What is the length of the top edge, x, in this new trapezium?

2. Can you find any other properties of the angles or lines in this trapezium?

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Phi and the Root-5 Rectangle

If we draw a rectangle which is 1 unit high and √5 = 2·236 long, we can draw a square in it, which, if we place it centrally, will leave two rectangles left over. Each of these will be phi=0·618.. units wide and, of course, 1 unit high.

Since we already know that the ratio of 1 : phi is the same as Phi : 1, then the two rectangles are Golden Rectangles (one side is Phi or phi times the other).

This is nicely illustrated on Ironheart Armoury's Root Rectangles page where he shows how to construct all the rectangles with width any square root, starting from a square.

This rectangle is supposed to have been used by some artists as it is another pleasing rectangular shape, like the golden rectangle itself.

The shape of a piece of paper

Modern paper sizes have sides that are in the ratio √2 : 1. This means that they can be folded in half and the two halves are still exactly the same shape. Here is an explanation of why this is so:

"A" series Paper

Take a sheet of A4 paper.
Fold it in half from top to bottom.
Turn it round and you have a smaller sheet of paper of exactly the same shape as the original, but half the area, called A5.
Since its area is exactly half the original, its sides are √(1/2) of the originals, or, an A4 sheet has sides √2 times bigger than a sheet of A5.
Do this on a large A3 sheet and you get a sheet of size A4.
The sides must be in the ratio of 1:√2 since if the original sheet has the shorter side of length 1 and the longer side of length s, then when folded in half the short-to-longer-side ratio is now s/2:1.
By the two sheets being of the same shape, we mean that the ratio of the short-to-long side is the same. So we have:
1/s = s/2 /1 which means that s² = 2 and so s = √2

Fibonacci paper

If we take a sheet of paper and fold a corner over to make a square at the top and then cut off that square, then we have a new smaller piece of paper.
If we want the smaller piece to have the same shape as the original one, then, if the longer side is length f and the short side length 1 in the original shape, the smaller one will have shorter side of length f-1 and longer side of length 1.
So the ratio of the sides must be the same in each if they have the same shape: we have 1/f = (f-1)/1 or, f²-f=1 which is exactly the equation from which we derived Phi.
Thus if the sheets are to have the same shape, their sides must be in the ratio of 1 to Phi, or, the sides are approximately two successive Fibonacci numbers in length!

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Phi and other Polygons

Decagons

The smallest angle in the "pentagram triangle" was 36° so we can pack exactly ten of these together round a single point and make a decagon as shown here.

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Here is a decagon - a 10-sided regular polygon with all its angles equal and all its sides the same length - which has been divided into 10 triangles.
Because of its symmetry, all the triangles have two sides that are the same length and so the two other angles in each triangle are also equal.
In each triangle, what is the size of the angle at the centre of the decagon?
We now know enough to identify the triangle since we know one angle and that the two sides surrounding it are equal. Which triangle on this page is it?

From what we have already found out about this triangle earlier, we can now say that

The radius of a circle through the points of a decagon is Phi times as long as the side of the decagon.

This follows directly from Euclid's Elements Book 13, Proposition 9.

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Penrose tilings

Recently, Prof Roger Penrose has come up with some tilings that exhibit five-fold symmetry yet which do not repeat themselves for which the technical term is aperiodic or quasiperiodic. When they appear in nature in crystals, they are called quasicrystals. They were thought to be impossible until fairly recently. There is a lot in common between Penrose's tilings and the Fibonacci numbers.
The picture above is made up of two shapes of rhombus or rhombs - that is, "pushed over squares" where each shape has all sides of the same length. The two rhombs are made from gluing two of the flat pentagon triangles together along their long sides and the other from gluing two of the sharp pentagon triangles together along their short sides.

Dissecting the Sharp and Flat Triangles:

Here the sharp triangle is dissected into two smaller sharp triangles and one flat triangle, the flat triangle into one smaller flat and one sharp triangle. At each stage all the triangles are dissected according to this pattern.
Repeating gives rise to one version of a Penrose Tiling.

Note that the top "flat" diagram shows the sharp and flat triangles have the same height and that their bases are in the ration Phi:1 (or 1: Phi-1 which is 1:phi):

Since the Kite and Dart are made of two identical triangles, then

The diagram on the right shows the relationship between the kite and dart and a pentagon and pentagram.
You can make similar tiling pictures with Quasitiler 3.0, a web-based tool and its link mentions more references to Penrose tilings.
A floor has been tiled with Penrose Rhombs at Wadham College at Oxford University.

Here are some interesting links to the Penrose tilings at other sites.

Here are some ready-to-photocopy Penrose tiles for you to photocopy and cut-out and experiment with making tiling patterns.
The Geometry Junkyard has a great page of Penrose links
Ivars Peterson's ScienceNewsOnline has an interesting page about quasicrystals showing how Penrose tilings are found in nature.
Pentaplex sell puzzle tiles based on a Penrose tiling.
Eric Weisstein's Penrose Tilings entry in his World of Mathematics online encyclopaedia.
Penrose Tiles to Trapdoor Ciphers, 1997, chapters 1 and 2 are on Penrose Tilings and, as with all of Martin Gardner's mathematical writings they are a joy to read and accessible to everyone.
A Near Golden Cuboid by Graham Hoare in Mathematics Today Vol 41, April 2005, page 53 gives the relationship between the pentagon/pentagram and Penrose's kite and dart.