Constructions for the Golden Ratio
.. ..
A single tile which produces an "irregular" tiling was found by Robert
Ammann in 1977.
Its dimensions involve powers of the square root of Phi (Φ):
which means it can be split into two smaller tiles of exactly the same shape. They all depend on the property of Φ that
By splitting the larger tile, and any others of identical size, in the
same way,
we can produce more and more tiles all of exactly the same shape, but
the whole tiling will never
have any
periodic pattern.
Each time we split a tile, we make one that is
times the size and one that is
times the size. There are only two sizes of tile in each dissection.
The buttons under the tile show successive stages as the dissections get more numerous.
This tiling is irregular or aperiodic which means that no
part of it will appear as an indefinitely recurring pattern
as in the regular tilings.
Things to do
 There are just two sizes of tile in each stage of
the dissections
in the diagram, a larger and a smaller tile.
 How many smaller tiles are there at each stage?
 How many larger tiles are there at each stage?
 How many larger and smaller tiles are there at stage 15?
 Use the button by the diagram to
show the dimensions. Pick one side of the Amman Chair tile.
 What is its length in the original tile?
 In the first split, what is the length of your chosen side on the larger of the two tiles?
 In the first split, what is the length of your chosen side on the smaller of the two tiles?
 What is its length on the subsequent two sizes of tile at each subsequent stage?
 What is the length of your chosen side on each of the two sizes of tile at stage 15?
 Use your answer to the first question to find how many tiles get split at each stage.
 There are three new sides added to the
diagram at the first
split.
 How many new sides are added at the second split (stage)? Use you answer to the previous question.
 How many new sides are added at the third split (stage)?
 Find a formula for the number of edges at each stage.
 Following on from the previous question, how many sides are there in total at each stage?
 Rotations and reflections of the
original tile:
 Does the tile appear rotated 180°?
 Does the tile appear rotated 90° and –90°?
 Does the tile appear reflected in a vertical mirror?
 Does the tile appear reflected in a horizontal mirror?
Enrique Zeleny's Ammann Chair Mathematica demonstration shows these dissections
either animated on the page or
using the free Mathematica Player, and in colour too.
Tilings
and Patterns B Gr�nbaum, G C Shephard (a Dover paperback of the original 1987 edition
is due out in 2009)
ISBN 0486469816 is encyclopaedic in its depth and range of tilings and
patterns. It mentions this Ammann
tiling on page 553.
Penrose Tiles to Trapdoor Ciphers: And the Return of Dr Matrix, M
Gardner,
(The Mathematical Association of America; Revised edition 1997), the
second chapter
gives more on Ammann's work but omits this simplest of aperiodic
tilings given above. As with
all mathematical books by Martin Gardner, they are excellent and I
cannot recommend them highly enough!
Aperiodic Tiles, R.Ammann, B. Gr�nbaum, G.C. Shephard, Discrete
and Computational Geometry (1992), pages 125.
1·61803 39887 49894 84820 45868 34365 63811 77203 09179 80576 ..More..
More Geometrical Gems
Here are some ore little gems of geometry showing Phi in unexpected
places.
The first is about cutting three triangles from a rectangle.
The next two are about balancing a flat shape on a pin
head or pencil point. The final one is about areas of flat circular
shapes.
First the two balancing puzzles.
The centre of gravity of a flat shape is the point at which the
shape balances horizontally
on a point such as the tip of a pencil. If you are careful, you can
get the shape to spin on that point, also
called the pivot(al) point.
A RectangleTriangle dissection Problem
The problem is, given a rectangle, to cut off three triangles from
the corners of
the rectangle
so that all three triangles have the same area.
Or, expressed another way, to find a triangle inside a given rectangle
(any rectangle)
which when it is removed
from the rectangle leaves three triangles of the same area.
As shown here, the area of the leftmost triangle is x(w+z)/2.
The area of the topright triangle is yw/2.
The area of the bottom triangle is (x+y)z/2.
Making these equal means:
The first equation tells us that x
= yw/(w+z).
The second equation, when we multiply out the brackets and cancel the zx terms on
each side, tells us that xw =
zy.
This means that y/x = w/z.
Or in other words, we have our first deduction that
Returning to xw = zy,
we put
x = yw/(w+z) into
it giving
yw^{2}/(w+z)=zy.
We can cancel y from each side and rearrange
it to give
w^{2} = z^{2} + zw.
If we divide by z^{2} we have a
quadratic equation in w/z:
(w/z)^{2} = 1 + w/z
Let X=w/z then X^{2} = 1 + X
The positive solution of this is X
= Phi, that is,
w/z = Phi or w = z Phi.
Since we have already seen that y/x = w/z then:
This ratio is Phi = 1·6180339... i.e. 1:1·618 or 0·618:1.
The Golden Section strikes again!
This puzzle appeared in J A H Hunter's
Triangle Inscribed In a Rectangle in The Fibonacci Quarterly, Vol 1, 1963, Issue 3, page 66.
A followup article
by H E Huntley entitled Fibonacci Geometry in volume 2 (1964)
of the
Fibonacci Quarterly on page 104
shows that, if the rectangle is itself a golden rectangle (the ratio
of the longer
side to the shorter one is Phi) then the
triangle is both isosceles and rightangled!
Balancing an "L" shape
In this first problem, suppose we take a square piece of card. Where
will the pivot point be? That should be easy to guess  at the centre of the square. 

Now suppose we remove a small square from one corner to make an "L"
shape.
Where will the pencil point be now if the "L" is to balance and turn
freely? The centre of gravity this time will be close to the centre but down a little on one of the square's diagonals. 

If we took off a very large square to make the "L" shape quite thin, then the centre of gravity would lie outside the L shape and we could not balance it at all!  
So we are left with the question
What is the largest size of square that we can
remove so that we can still just balance
the L shape?
Nick Lord found it was when the ratio of the big square's side and the
removed square's side are
in the golden ratio phi! 
Note 79.59: Balancing and Golden Rectangles N Lord, Mathematical Gazette vol 79 (1995) pages 5734.