Two-dimensional Geometry and the Golden section or
Fascinating Flat Facts about Phi
On this page we meet some of the marvellous flat (that is,
two dimensional) geometry facts related to the golden section
number Phi.
A following page
turns our attention to the solid world of 3 dimensions.
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Contents of this page
The icon means there is a
You do the maths... section of questions to start your own investigations.
Let's start by showing how to construct the golden section points on any line:
first a line phi (0·618..) times as long as the original and then
a line Phi (1·618..) times as long. John Turner has nicknamed the two points
that divide a line at the golden ratio (0.618 of the way from either end) as
gold points.
Constructing the internal golden section points: phi
If we have a line with end-points A and B, how can we find the point which divides
it at the golden section point?
We can do this using compasses for drawing circles
and a set-square for drawing lines at right-angles to other lines,
and we don't need a ruler at all for measuring lengths!
(In fact we can do it with just the compasses, but how to
do it without the set-square is left as an exercise for you.)
We want to find a point G between A and B so that AG:AB = phi (0·61803...)
by which we mean that G is phi of the way along the line. This will also mean
that the smaller segment GB is 0·61803.. times the size of the longer segment AG
too.
AG
=
GB
= phi = 0·618033.. =
√5 – 1
AB
AG
2
Here's how to construct point G using set-square and compasses only:
First we find the mid point of AB. To do this without a ruler,
put your compasses on one end, open them out to be somewhere near the other end
of the line and draw a semicircle over the line AB. Repeat this at the other
end of the line without altering the compass size.
The two points where the
semicircles cross can then be joined and this new line will cross AB at its mid point.
Now we are going to draw a line half the length of AB at point B,
but at right-angles to the original line. This is where you use the set-square
(but you CAN do this just using your compasses too - how?).
So first draw a line at right angles to AB at end B.
Put your compasses on B, open them to the mid-point of AB and draw an arc to
find the point on your new line which is half as long as AB.
Now you have a new line at B at right angles to the AB and BC
is half as long as the original line AB.
Join the point just found to the other end of the original line (A)
to make a triangle. Putting the compass point at the top point of the triangle
and opening it out to point B (so it has a radius along the right-angle line)
mark out a point on the diagonal which will also be half as long
as the original line.
Finally, put the compass point at point A, open it out to the new point
just found on the diagonal and mark a point the same distance along the original
line.
This point is now divides the original line AB into two parts, where the longer
part AG is phi (0·61803..) times as long as the original line AB.
It works because, if we call the top point of the triangle T, then BT is half as long
as AB. So suppose we say AB has length 1. Then BT will have length 1/2. We can find
the length of the other side of the triangle, the diagonal AT,
by using Pythagoras' Theorem:
AT2 = AB2 + BT2
i.e.
AT2 = 12 + (1/2)2
AT2 = 1 + 1/4 = 5/4
Now, taking the square-root of each side gives:
AT = (√5)/2
Point V was drawn so that TV is the same length as TB = AB/2 = 1/2.
So AV is just AT - TV = (√5)/2 – 1/2 = phi.
The final construction is to mark a point G which is
same distance (AV) along the original line (AB) which we do using the compasses.
Constructing the external golden section points: Phi
This time we find a point outside of our line segment AB so that the new point
defines a line which is Phi (1·618..) times as long as the original one.
Here's how to find the new line Phi times as long as the original:
First repeat the steps 1 and 2 above so that we have
found the mid-point of AB and also have a line at
right angles at point B.
Now place the compass point on B and open them out to touch A so that
you can mark a point T on the vertical line
which is as long as the original line.
Placing the compass point on the mid-point M of AB, open them out
to the new point T on the vertical line and draw an arc on the
original line extended past point B to a new point G.
The line AG is now Phi times as long as the original line AB.
If you followed the reasoning for why the first construction (for phi) worked,
you should find it quite easy to prove that AG is Phi times the length of AB, that is,
that AG = (√5/2 + 1/2) times AB.
Hint:
Let AB have length 1 again and so AM=MB=1/2.
Since BT is now also 1, how long is MT? This is the same length
as MG so you can now find out how long AG is since AG=AM+MG.
Hofstetter's 3 Constructions of Gold Points
Using only circles
Kurt Hofstetter has found a beautifully simple construction for a line and its
golden section point
using only compasses to draw 4 circles:
On any straight line S, pick two points X and Y.
With each as centre draw a circle (green) through the other point labelling
their points of intersection G (top) and B (bottom) and the points where they meet line S as P and Q;
With centre X, draw a circle through Q (blue);
With centre Y, draw a circle through P (blue);
labelling the top point of intersection of the blue circles A;
G is a gold point of AB (G is at the golden section of AB).
Kurt Hofstetter has discovered a very simple constructions of the
gold point on a line AB just using circles and one line to find the gold point(s)
on a given line segment AB:
With centre A, draw a circle through B;
With centre B, draw a circle through A;
Extend BA to meet the first circle (centre A) at C;
Label the lower point on both circles as D;
With centre C draw another circle through B;
Where the larger circle meets the circle centred on B, label the upper point E;
Line DE crosses line AB at G. G is a gold point of AB
The other gold point on AB can be constructed similarly.
Uli Eggermann of Germany points out that we also have A as the gold point of CG!
It is quite simple to prove too using the properties of Phi and phi that phi=1/Phi and
1+phi=Phi:
If AB is 1 then so is CA: they are both radii of the same circle
Above we constructed G, the gold point of AB (=1), so that AG is phi
CG = CA + AG = 1 + phi = Phi
Since CA is 1 and CG is Phi then CA/CG = 1/Phi = phi
In this construction to find the gold point of a given line segment AB
rediscovered by K Hofstetter but going back to 1902,
we again need to construct only one line:
With A as centre, draw the first circle through B;
With B as centre, draw a second circle through A
using labels C (top) and D (bottom) for the two points of intersection with the first
circle;
Through the top point, C, draw a third circle through A and B and use label
E for its other intersection with the first circle;
Draw the line CD and where it meets the third circle, label this point F;
With centre E draw a circle though F; where it meets the line AB label this point G;
G is a gold point of AB.
Also, if BA is extended to meet the larger circle with centre E at G', then A is a gold point of G'B.
The earliest reference to this construction appears to be in
Géométrographie ou Art des Constructions Géométriques by E. Lemoine,
published by C. Naud, Paris, 1902, page 51.
Phi and Pentagons
There is an intimate connection between the golden section and the regular 5-sided shape called a pentagon and its
variation - the pentagram - that we explore first.
Pentagons and Pentagrams
The pentagram is a symmetrical 5-pointed star that fits inside a pentagon.
Starting from a pentagon, by joining each vertex to the next-but-one
you can draw a pentagram
without taking your pen off the paper.
The pentagram has 5 triangles on the edges of another pentagon at its centre.
Let's focus on one of the triangles and the central pentagon as shown here.
All the orange angles at the vertices of the
pentagon are equal. They are called the external angles of the polygon. What size are they?
This practical demonstration will give us the answer:
Take a pen and lay it along the top edge pointing right
Turn it about the top right vertex through the orange
angle so that
it points down to the lower right
Move the pen down that side of the pentagon to the next vertex and turn it
through the next orange angle
Repeat moving it along the sides and turning through the rest of the
orange angles until it lies back on the top edge
The pen is now back in its starting position, pointing to the right so
it has turned through one compete turn.
It has also turned through each of the 5 orange
angles
So the sum of the 5 orange angles is one turn or 360°
Each orange angle is therefore 360/5=72°
The green angle is the same size as
the orange angle so that
the two "base"
angles of the blue triangle are both 72°.
Since the
angles in a triangle sum to 180° the yellow angle is 36°
so that 72° +
72° +
36° = 180°.
The basic geometrical facts we have used here are:
The external angles in any polygon sum to 360°.
The angles on a straight line sum to 180°.
The angles in a triangle sum to 180°.
So the pentagram triangle has angles of 36 °, 72° and 72°.
Now let's find out how long its sides are.
The 36°-72°-72° triangle
In this diagram, the triangle ABC is isosceles,
since the two sides, AB and AC, are equal as are the two angles at B and C.
[Also, angles ABC and ACB are twice angle CAB.]
If we bisect the base angle at B by a line from B to point D on AC
then we have the angles as shown and also angle BDC is also 72°.
BCD now has two angles equal and is therefore an isosceles triangle;
and also we have BC=BD.
Since ABD also has two equal angles of 36°, it too is isosceles
and so BD=AD. So in the diagram the three sides BC, BD and DA are all the same
length.
We also note that the little triangle BCD and the whole
triangle ABC are similar since they are both 36°-72°-72° triangles.
Let's call the smallest segment here, CD, length 1 and find the lengths of the others
in relation to it. We will therefore let the ratio of the smaller to longer sides in
triangle BCD be r so that if CD is 1 then BC is r.
In the larger triangle ABC, the base is now r and as it is the same shape as BCD, then
its sides are in the same ratio so Ab is r times BC, e.d. AB is r2.
Also, we have shown BC=BD=AD so AD is r (and CD is 1).
From the diagram we can see that AC=AD+DB. But AC=1+r
and in isosceles triangle ABC, AB (which is r2) is the same length as AC (which is 1+r), so r2 = 1 + r and this is the equation which defined the golden ratio.
r is Phi or -phi and since lengths are positive, we therefore have that r is Phi!
So the triangle with angles 36°, 72°, 72° has sides that are proportional to
Phi, Phi and 1 (which is the same as 1,1,phi).
Pentagrams and the 36°-72°-72° triangle
If we look at the way a pentagram is constructed,
we can see there are lots of lines divided in the golden ratio:
Since the points can be joined to make a pentagon,
the golden ratio appears in the pentagon also and
the relationship between its sides and the diagonals
(joining two non-adjacent points).
The reason is that Phi has the value 2 cos (π/5) where the angle is
described in radians, or, in degrees, Phi=2 cos (36°).
[See below for more angles whose sines and cosines involve Phi!]
Since the ratio of a pair of consecutive Fibonacci numbers is roughly
equal to the golden section,
we can get an approximate
pentagon and pentagram using the Fibonacci numbers as lengths of lines:
There is another flatter triangle inside the pentagon here. Has this any
golden sections in it? Yes! We see where further down this page, but first,
a quick and easy way to make a pentagram without measuring angles or using
compasses:
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.
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.
You do the maths...
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.
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.
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?
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 108° and the new combined triangle, BDE also has angles
36°, 36°, and 108°.
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, the "sharp" 36-72-72 and the "flat" 36-36-108,
are the basic building shapes of Penrose tilings
(see the Penrose Tilings section
mentioned later 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 - the Kepler Triangle
The German astronomer and mathematician
Johannes Kepler (1571-1630)
had a great interest in both Pythagorean triangles and the golden ratio which was known then mainly by the term used by Euclid
"the division of a line into extreme and mean ratio":
Geometry has two great treasures;
one is the Theorem of Pythagoras; the other,
the division of a line into extreme and mean ratio.
The first we may compare to a measure of gold;
the second we may name a precious jewel. Mysterium Cosmographicum 1596
Here we have both together in a single unique triangle.
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 R2 – R – 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 are 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 some interesting properties of its angles later on this page in the section
on Trigonometry and Phi.
How to construct Kepler's Triangle
It is very easy to construct Kepler's Triangle if we start from a golden rectnagle
as described above....
Make the golden rectangle...
and draw a circle centred on one corner and having the longer side of the rectangle as radius...
The point where the circle crosses the other longer side marks one vertex of Kepler's triangle,
the centre of the last circle is another and the right angle of the rectangle is the third.
A trigonometric intersection
D Quadling, Math. Gaz. (2005), Note 89.70
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:
the top and bottom edges are parallel
the angles at A and B are equal as are those at C and D
all the sides except the top are of equal length (hence trisoceles in Scott's name
for this shape)
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.
So what was the shape of the isosceles triangle which had its top
part removed to make the traphizium?
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.
You do the maths...
Here is another trapezium PQRS that is constructed using the other gold point
on a the equal sides of an isosceles triangle PQT.
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?
Can you find any other properties of the angles or lines in this trapezium?
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.
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:
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
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!
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.
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.
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 here is made up of two kinds of rhombus or rhombs, that is, 4-sided shapes with all sides of equal 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 as shown in the diagams below.
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 ratio Phi:1 (or 1: Phi-1 which is 1:phi):
The sharp triangle is Phi times the area of the flat triangle
if the shortest sharp triangle's side = the longest flat triangle's side
Since the Kite and Dart are made of two identical triangles, then
The dart is Phi times the area of the kite
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.
Ivars Peterson's ScienceNewsOnline has
an interesting page
about quasicrystals showing
how Penrose tilings are found in nature.
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.
Another Irregular tiling using Phi: the Ammann Chair
The Ammann Chair Tiling
..
..
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):
√ Φ = Φ1/2 = φ –1/2 = 1.27201964951406
which means it can
be split into two smaller tiles of exactly the same shape. They all depend on the property of Φ that
Φ n+2 = Φ n+1 + Φ n
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
φ = 1/Φ = Φ –1 = 0.618039...
times the size and one that is
√φ = φ 1/2 = 1/√Φ = Φ –1/2 = 0.78615137...
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.
You do the maths...
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 Grünbaum, 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. Grünbaum, G.C. Shephard, Discrete and Computational Geometry
(1992), pages 1-25.
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.
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 top-right triangle is yw/2.
The area of the bottom triangle is (x+y)z/2.
Making these equal means:
x(w+z) = yw
and x(w+z) = z(x+y).
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
Both sides of the rectangle are divided in the same proportion.
Returning to xw = zy, we put
x = yw/(w+z) into it giving
yw2/(w+z)=zy.
We can cancel y from each side and rearrange it to give
w2 = z2 + zw.
If we divide by z2 we have a quadratic equation in w/z:
(w/z)2 = 1 + w/z
Let X=w/z then X2 = 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:
Each side of the rectangle is divided in the same ratio
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 follow-up 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 right-angled!
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 573-4.
Balancing a Crescent
Here is a similar problem but this time using circles: to remove the largest circle and still have a
pivotal point:
The full circle balances at its centre point
Removing a small circle at the top moves the balance point down a little
Removing a large circle moves the balance point outside the crescent shape
The limit point is when the radii of the two circles are in the golden ratio and
the centre of gravity is just on the edge of the inner circle
Ellipse and Ring
The third little geometrical gem is about a ring and an ellipse. An ellipse is the shape that a plate or
anything circular appears when viewed at an angle.
Here are two circles, one inside the other. The yellow and red areas define an outer ring.
Also the orange and red parts
form an ellipse (an oval).
If the ring is very narrow, the two circles are similar in size and the ellipse has a much bigger
area than the ring.
If the inner circle is very small,
the ellipse will be very narrow and the outer ring will be much bigger in area than the ellipse.
So the question is
What ratio of circle sizes (radii) makes the ellipse equal in
area to the ring between the two circles?
The answer is again when the inner radius is 0.618 of the larger one, the golden ratio.
This is quite easy to prove using these two formulae:
The are of a circle of radius r is π r2
The area of an ellipse with "radii" a and b (as shown above)
is πab
(Note how that, when a = b in the ellipse, it becomes a circle and the two formulae are the same.)
So the outer circle has radius a, the inner circle radius b and the area of the ring between them
is therefore:
Area of ring = π (b2 – a2)
This is equal to the area of the ellipse when
π (b2 – a2) = π a b
b2 - a2 - a b = 0
If we let the ratio of the two circles radii = b/a, be K, say, then dividing the equation
by a2 we have
K2 - K - 1=0
which means K is either Phi or –phi.
The positive value for K means that b = Phi a
or a = phi b.
The equation of an ellipse is
(x/b)2 + (y/a)2 = 1
When a = b, we have the equation of a circle of radius a(=b):
(x/a)2 + (y/a)2 = 1 or
x2 + y2 = a2as it is more usually written.
You might have spotted that this equation is merely Pythagoras' Theorem that all
the points (x,y) on the circle are the same distance from the origin, that distance being a.
Note 79.13 A Note on the Golden Ratio, A D Rawlings,
Mathematical
Gazette vol 79 (1995) page 104.
The
Changing Shape of Geometry C Pritchard (2003) Cambridge University Press paperback
and hardback,
is a collection of popular, interesting and enjoyable articles selected from the
Mathematical
Gazette . It will be of particular interest to teachers and students in school
or indeed anyone interested in Geometry.
The three gems above are given in more
detail in the section on The Golden Ratio.
The Golden section squares are shown in red here, the axes in blue and all the points of the
squares lie on the green lines, which pass through the origin (0,0).
Also, the blue (axes) lines and the green lines are each separated from the next
by 45° exactly.
The large rectangle ABDF is the exactly the same shape as CDFH,
but is (exactly) phi times as large. Also
it has been rotated by a quarter turn. The same applies to CDFH and HJEF
and to all the golden rectangles in the diagram.
So to transform OE (on the x axis) to OC (on the y axis), and indeed any point on the spiral
to another point on the spiral, we expand lengths by phi times for every rotation of 90°:
that is, we change (r,theta) to (r Phi,theta + π/2) (where, as usual,
we express angles in radian measure, not degrees).
So if we say E is at (1,0),
then C is at (Phi,π/2),
A is at (Phi2, π), and so on.
Similarly, G is at (phi,–π/2),
and I is at (phi2, –π) and so on
because phi = 1/Phi.
The points on the spiral are therefore summarised by:
r = Phinand theta= n π/2
If we eliminate the n in the two equations, we get a single equation
that all the points on the spiral satisfy:
r = Phi2 theta / π
or
r = Mthetawhere M = Phi2/π = 1.35845627...
For ordinary (cartesian) coordinates, the x values are y values are generated from the polar coordinates as follows:
x = r cos(theta) y = r sin(theta)
which we can then use in a Spreadsheet to generate a chart as shown here.
Such spirals, where the distance from the origin is a constant to the power of the angle,
are called equiangular spirals. They also have the property that
a line from the origin to any point
on the curve always finds (the tangent to) the curve meeting it at the same angle.
Another name is a logarithmic spiral because the angle of any point
from the x axis through the origin
is proportional to the logarithm of the point's distance from the origin.
To see that the Fibonacci Spiral here is only an approximation
to the (true) Golden Spiral above note that:
at its start there are two squares making the first rectangle
but the true golden spiral above has no "start"
those two squares make a rectangle 1x2 but all rectangles in the true spiral are
true Golden Rectangles 1xPhi.
All the other rectangles on the right are ratios of two neighbouring Fibonacci numbers
and are therefore only approximations to Golden Rectangles
The Golden or Phi Spiral
In the spiral above, based on Fibonacci squares spiralling out from an initial two 1x1 squares,
we noted that one quarter turn produces an expansion by Phi in the distance of a point on the curve
from the "origin". So in one full turn we have an expansion of Phi4.
In sea-shells, we notice an expansion of Phi in one turn, so that not only has the
shell grown to Phi times as far from its origin (now buried deep inside the shell).
Also, because of the properties of the golden section, we can see that the distances measured
on the outside of the shell also have increased by Phi and it is often easier to measure this distance
on the outside of a shell, as we see in the picture here on the left.
In this case, the equation of the curve is
r = Phi theta / (2 π) when theta is measured in radians or r = Phi theta / 360 if theta is in degrees.
Notice that an increase in the angle theta of 2 π radians (360° or
one full turn)
makes r increase by a factor of Phi
because the power of Phi has increased by 1.
I will call this spiral, that increases by Phi per turn, the Golden Spiral
or the Phi Spiral because of this property and also because it is the one we find in
nature (shells, etc.).
Click on the Spreadsheet image to open an Excel Spreadsheet to generate the Golden Spiral
in a new window.
What is trigonometry?
We can answer this by looking at the origin of the word trigonometry.
Words ending with -metry are to do with measuring (from the Greek
word metron meaning "measurement").
(What do you think that thermometry measures?
What about geometry? Can you think of any more words ending
with -metry?)
Also, the -gon part comes from the Greek gonia) meaning angle.
It is derived from the Greek word for "knee" which is gony.
The prefix tri- is to do with three
as in tricycle (a three-wheeled cycle), trio (three people), trident
(a three-pronged fork).
Similarly, quad means 4, pent 5 and
hex six as in the following:
a (five-sided and) five-angled shape is a penta-gon meaning literally
five-angles and
a six angled one is called a hexa-gon then we could call
a four-angled shape a quadragon
(but we don't - using the word quadrilateral instead which means
"four-sided") and
a three-angled shape would be a tria-gon
(but we call it a triangle instead)
"Trigon" was indeed the old English word for a triangle.
So trigonon means
"three-angled" or, as we would now say in English, "tri-angular" and hence we have
tri-gonia-metria meaning "the measurement of triangles".
With thanks to proteus of softnet for this information.
Here are the graphs of three familiar trigonometric functions: sin x, cos x
and tan x in the region of x from 0 to π/2 (radians) = 90°:
The graphs meet at
the origin, where tan x = sin x
in the middle, where sin x = cos x i.e. where tan x = 1 or
x = 45° = π/4 radians
at another angle where tan x = cos x
What angle is at the third meeting point?
tan x
= cos x and, since tan x = sin x / cos x
, we have:
sin x
= (cos x)2
=1-(sin x)2because (sin x)2+(cos x)2=1.
or
(sin x)2 + sin x = 1
and solving this as a quadratic in sin x, we find
sin x = (–1 + √5)/2 or sin x = (–1 – √5)/2
The second value is negative and our graph picture is for positive x, so we have our answer:
the third point of intersection is
the angle whose sine is Phi – 1 = 0·6180339... = phi
which is about 0·66623943.. radians or 38·1727076..°
On our graph, we can say that the intersection of the green and blue graphs
(cos(x) and tan(x))
is where the red graph (sin(x)) has the value phi [i.e. at the x value of the point
where the line y = phi meets the sin(x) curve].
Is there any significance in the value of tan(x) where tan(x)=cos(x)?
Yes. It is √phi =
√0·618033988... =
0·786151377757.. .
Here's how we can prove this.
Take a general right-angled triangle and label one side t
and another side 1 so that one angle (call it A) has a tangent of t.
By Pythagoras's Theorem, the hypotenuse is √(1+t2).
So we have:
For all right-angled triangles: tan A = t
cos A =
1
√(1 + t2)
sin A =
t
√(1 + t2)
We now want to find the particular angle A which has tan(A) = cos(A).
From the formulae above we have:
t=1/√(1+t2)
Squaring both sides:
t2 = 1/(1 + t2)
Multiply both sides by 1 + t2:
t2 ( 1 + t2 ) = 1
t2 + t4 = 1
If we let T stand for t2 then we have a quadratic equation in T which we can easily solve:
SinceT is t2 it must be positive, so the value of T
we want in this case is
T = ( √5 − 1 ) / 2 = phi
Since T is t2 then t is
√phi.
Since 1 + Phi = Phi2 then the hypotenuse
√(1+T2) = Phi as shown in the triangle here.
From the 1, √Phi, Phi triangle, we see that
tan(A)=cos(A) tan(B)=1/sin(B) or cot(B)=sin(B) or tan(B)=cosec(B)
What is the angle whose tangent is the same as it cosine?
In the triangle here, it is angle A: A = 38.1727076° = 30° 10' 21.74745" = 0.1060352989.. of a whole turn = 0.666239432.. radians.
The other angle, B has its tangent equal to its cosecant (the reciprocal of the sine): B = 90° – A = 51.8272923..° = 51° 49' 38.2525417.."
= 0.143964701.. of a whole turn = 0.90455689.. radians.
Notice how, when we apply Pythagoras' Theorem to
the triangle shown here with sides 1, Phi and root Phi, we have
Phi2 = 1 + (√Phi)2 = 1 + Phi
which is one of our classic definitions of (the positive number) Phi.....
and we have already met this triangle as Kepler's Triangle earlier on this page.
The next section looks at the other trigonometrical relationships in a triangle and shows that, where they are equal, each involves
the numbers Phi and phi.
Notation for inverse functions
A common mathematical notation for the-angle-whose-sin-is 0.5
is arcsin(0.5) although you will sometimes see this written as
asin(0.5) or sin-1(0.5).
We can prefix arc (or a)
to any trigonometrical function (cos, cot, tan, etc.) to make
it into its "inverse" function which, given the trig's value, returns the angle itself.
Each of these inverse functions
is applied to a number and returns an angle.
e.g. if sin(90°)=0 then 90° = arcsin(0).
What is arccos(0.5)?
The angle whose cosine is 0.5 is 60°.
But cos(120°)=0.5 as does
cos(240°) and cos(300°) and we can add 360°
to any of these angles to find some more values!
The answer is
arccos(0.5) = 60° or 120° or 240° or 300° or ...
With all the inverse trig. functions you must carefully select the answer
or answers that are appropriate to the problem you are solving.
"The angle whose tangent is the same as its cosine" can be written mathematically
in several ways:
tan(A)=cos(A) is the same as
arctan( cos(A) ) = A
Can you see that it can also be written
as arccos( tan(A) )= A?
More trig ratios and Phi (sec, csc, cot)
In a right-angled triangle if we focus on one angle (A),
we can call the two sides round the right-angle the Opposite
and the Adjacent sides and the longest side is the Hypotenuse,
or Adj, Opp and Hyp for short.
You might wonder why we give a name to the ratio Adj/Opp (the tangent) of angle A but not to Opp/Adj.
The same applies to the other two pairs of sides:
we call Opp/Hyp the sine of A but what about Hyp/Opp?
Similarly Adj/Hyp is cosine of A but what about Hyp/Adj?
In fact they do have names:
the inverse ratio to the tangent is the cotangent or cot i.e. Adj/Opp; cot(x)=1/tan(x)
the inverse ratio to the cosine is the secant or sec i.e. Hyp/Adj; sec(x)=1/cos(x)
the inverse ratio to the sine is the cosecant or cosec or sometimes csc i.e. Hyp/Opp;
csc(x) = 1/sin(x)
You'll notice that these six names divide into two groups:
secant, sine, tangent
cosecant, cosine, cotangent
and show another way of choosing one representative for each of the 3 pairs of ratios (x/y and y/x where x and
y are one of the three sides).
Here is a graph of the six functions where the angle is measured in radians:
This extended set of graphs reveals two more intersections
involving Phi, phi and their square roots: e.g.
if A is the angle where cos(A) = tan(A) then
sin(A) = phi and cosec(A) = Phi;
The value of A is A = 38.172..° = 0.666239... radians;
if B is the angle where sin(B) = cot(B) then
cos(B) = phi and sec(B) = Phi;
The value of B is 51.827..° = 0.904556... radians.
Notice that A and B sum to 90°
- as we would expect of any two angles
where the sine of one is the cosine of the other.
Some Results in Trigonometry
L Raphael, Fibonacci Quarterly vol 8 (1970), page 371, continued on page 392.
You do the maths...
Above we solved cos(x)=tan(x) using the 1,t,√(1+t2) triangle. Use the same
triangle and adapt the method
to find the value of sin(x) for which sin(x)=cot(x).
If you use the formulae above then remember that you will find
t, the tangent of the angle for which sin(x)=cot(x).
Since we want the cotangent, just take the reciprocal of t to
solve sin(x)=cot(x)=1/t.
On the previous page we saw two ways to find Phi on your calculator using the 1/x button,
square-root button and just adding 1. Here's how we can do the same thing to find √phi and √Phi.
From tan(x)=cos(x) we found t=phi is a solution to t=1/√(1+t2)
So, to uncover this value using your calculator, follow these steps:
Enter any number to start the process
Square it
Add 1
Take the square-root
Take the reciprocal (the 1/x button)
Write down the number now displayed
Repeat the previous step as often as you like.
Each time, given t, we compute 1/√(1+t2) and keep repeating this transformation.
Eventually, the number we write down does not change.
It is √phi = 0.7861513777... . In fact, no matter how big or small is your starting value, you'll get
√phi to 4 or 5 dps after only a few iterations. Try it!
If you want √Phi, just use the 1/x button on your final answer.
Some Results in Trigonometry, Brother L Raphael, The Fibonacci Quarterly
vol 8 (1970) pages 371 and 392.
Other angles related to Phi
Look again at the sharp and flat
triangles of the pentagon that we saw above. If we divide each in half, we have right angled
triangles with sides 1 and Phi/2 around the 36°
angle in the flat triangle and
sides 1/2 and Phi around the 72° angle in the sharp triangle. So:
cos(72°) = cos
(
2 π 5
)
= sin(18°) = sin
(
π 10
)
=
φ 2
=
1 2 Φ
cos(36°) = cos
(
π 5
)
= sin(54°) = sin
(
3π 10
)
=
Φ 2
=
1 2 φ
We have sin(18°) but what about cos(18°)?
This has a somewhat more awkward expression as:
cos(18°) =
√
Φ √5
2
Now we know the sin and cos of both 30°
and 18° we can find the sin and cos of
the angle differences using:
cos(A – B) = cos(A) cos(B) + sin(A) sin(B)
and get:
cos(12°) =
φ +
√
3 Φ √5
4
AAAAGH! as Snoopy might have said.
Is there a neater (that is, a simpler) expression? Perhaps you can find one.
Let me know if you do and it will be added here with your name!
There are several angles whose cosines are similar to this one:
cos(24°) =
Φ +
√
3 φ √5
4
cos(48°) =
−φ +
√
3 Φ √5
4
cos(84°) =
−Φ +
√
3 φ √5
4
What about other angles? From an equilateral triangle cut in half we can easily show
that:
cos(60°)
=
sin(30°) =
1
2
cos(30°)
=
sin(60°) =
√3
2
and from a 45-45-90 degree triangle we can derive:
cos(45°) = sin(45°) =
1
=
√2
√2
2
and not forgetting, of course:
sin(0°) = cos(90°) = 0
sin(90°) = cos(0°) = 1
The form of cos(12°) above is derived from the expression on page 42 of Roots of (H-L)/15 Recurrence Equations in Generalized Pascal Triangles
by C Smith and V E Hoggatt Jr. in The Fibonacci Quarterly
vol 18 (1980) pages 36-42.
Can you find any more angles that have an exact expression (not
necessarily involving Phi or phi)?
Let me know what you find and let's get
a list of them here.
Here is my initial attempt - can you add any more angles?
Binet's Formula for Fib(n) using trigonometry
These two alternative forms of Binet's formula:
Fib(n) =
Phin – (–Phi)–n
=
Phin – (–phi)n
√5
√5
can be written using the cosine function and sine as we see in the next two subsections:
A Simple Trig formula for Fib(n)
Above we saw that cos(π/5) = cos(36°) = Φ/2 and
cos(2 π/5) = cos(72°) = φ/2.
Since cos(180° – A) = –cos(A), we have
cos(180° – 72) = – φ/2.
So
Φ = 2 cos( π/5 ), –φ = 2 cos( 3π/5 )
We can use these in Binet's formula to rewrite it as
Fib(n) =
2n [ cosn(π/5) – cosn(3π/5) ]
√5
The Bee and the Regular Pentagon W Hope-Jones, Mathematical Gazette vol 55, 1971, pg 220 ff
which is a reprint of the original 1921 version (vol 10).
Fibonacci in Trigonometric Form
Problem B-374 proposed and solved by F Stern
in The Fibonacci Quarterly vol 17 (1979) page 93 where another form is also given.
Phi and Powers of Pi
There is a simple (infinite) series for calculating the cosine and the sine
of an angle where the
angle is expressed in radians.
See Radian Measure
(the link opens in a new window - close it to return here) for a fuller explanation.
Basically, instead of
360 degrees in a full turn there are
2π radians. The radian measure makes many
trigonometric equations simpler and so it is the preferred unit of measuring angles
in mathematics.
If angle x is measured in radians then
cos( x ) = 1 −
x2 2!
+
x44!
–
x66!
+ ...
sin( x ) = x −
x3 3!
+
x55!
–
x77!
+ ...
Here, n! means the factorial of n which means
the product of all the whole numbers
from 1 to n.
For example, 4! = 1 × 2 × 3 × 4 = 24.
So, using the particular angles above in sin(π/10) and cos(π/5) we have formulae for
φ and
Φ
in terms of powers of π :-
φ
= 2 sin(
π10
)
= 2 (
π
−
π3
+
π5
−
π7
+...)
10
1033!
1055!
1077!
=
π
−
π3
+
π5
−
π7
+...
5
3 000
6 000 000
25 200 000 000
In the upper formula, going to up to the π9 term only
will give φ to 9 decimal places whereas
stopping at the π8 term in the lower formula
will give Φ to 7 decimal places.
These two formula easily lend themselves as an iterative method for a
computer program (i.e. using a loop) to compute Φ and φ.
To compute the next term from the previous one,
multiply it by (π/5)2 or by (π/10)2 for φ
and divide by two integers to update the factorial
on the bottom, remembering to add the next term if the previous one was subtracted
and vice versa. Finally multiply your number by 2.
You will need and
an accurate value of π. Here is π to 102
decimal places:
This section is for those who have already met calculus in your mathematics courses. In the following problems, look out for Phi!
Leonardo chases a Rabbit
Leonardo spots a rabbit in one corner of his field which just happens to be a square. He starts to chase it at exactly the same moment
as the rabbit bolts for the hole in the hedge at another corner of the field.
If Leonardo always runs always directly towards the rabbit and at a constant speed and if the rabbit runs at a
constant speed in a straight line along the hedge towards the hole then
if both reach the hole at exactly the same time, what is the ratio between their two speeds?
Based on Problem 307 of
536 Puzzles and Curious Problems by H E Dudeney, which is now out of print, but you can often pick up
a cheap second hand version by clicking on this link. Dudeney only considers the farmer moving at twice the speed of
the rabbit (pig in Dudeney's puzzle) but in this variation we are asking for an optimum solution.
Paul Bruckman gives the solution in H-333 Fibonacci Quarterly 21 (1983) pages 77-78
pdf
New Visual Perspectives on Fibonacci Numbers by K Atanassova, V Atanassova, A Shannon and J Turner,
World Scientific (Oct 2002)
A book at the undergraduate level of mathematics that introduces the idea of
two intertwined Fibonacci-type series (2-Fibonacci series), recurrence trees and
Gray codes, and a new Fibonacci vector as well as John Turner's goldpoint geometry (well known from his
papers and presentations at the International Fibonacci Conferences and printed volumes) and fractals
and tilings. Have a look at the publisher's
description and the chapter titles.
Domingo Gómez shows a construction for the
Isosceles-Fibonacci
partition
to construct points on any line AB which divide it into
AB/2, AB/3, AB/5, AB/8,
and so on, where the denominators are the Fibonacci numbers.
icon means there is a
You do the maths... section of questions to start your own investigations.
If we have a line with end-points A and B, how can we find the point which divides
it at the golden section point?
We want to find a point G between A and B so that AG:AB = phi (0·61803...)
by which we mean that G is phi of the way along the line. This will also mean
that the smaller segment GB is 0·61803.. times the size of the longer segment AG
too. 
First we find the mid point of AB. To do this without a ruler,
put your compasses on one end, open them out to be somewhere near the other end
of the line and draw a semicircle over the line AB. Repeat this at the other
end of the line without altering the compass size.
The two points where the
semicircles cross can then be joined and this new line will cross AB at its mid point.
Now we are going to draw a line half the length of AB at point B,
but at right-angles to the original line. This is where you use the set-square
(but you CAN do this just using your compasses too - how?).
So first draw a line at right angles to AB at end B.
Put your compasses on B, open them to the mid-point of AB and draw an arc to
find the point on your new line which is half as long as AB.
Now you have a new line at B at right angles to the AB and BC
is half as long as the original line AB.
Join the point just found to the other end of the original line (A)
to make a triangle. Putting the compass point at the top point of the triangle
and opening it out to point B (so it has a radius along the right-angle line)
mark out a point on the diagonal which will also be half as long
as the original line.
Finally, put the compass point at point A, open it out to the new point
just found on the diagonal and mark a point the same distance along the original
line.
This point is now divides the original line AB into two parts, where the longer
part AG is phi (0·61803..) times as long as the original line AB.
It works because, if we call the top point of the triangle T, then BT is half as long
as AB. So suppose we say AB has length 1. Then BT will have length 1/2. We can find
the length of the other side of the triangle, the diagonal AT,
by using Pythagoras' Theorem:
Now place the compass point on B and open them out to touch A so that
you can mark a point T on the vertical line
which is as long as the original line.
Placing the compass point on the mid-point M of AB, open them out
to the new point T on the vertical line and draw an arc on the
original line extended past point B to a new point G.
Kurt Hofstetter has found a beautifully simple construction for a line and its
golden section point
using only compasses to draw 4 circles:
Kurt Hofstetter has discovered a very simple constructions of the
gold point on a line AB just using circles and one line to find the gold point(s)
on a given line segment AB:
In this construction to find the gold point of a given line segment AB
rediscovered by K Hofstetter but going back to 1902,
we again need to construct only one line:
All the orange angles at the vertices of the
pentagon are equal. They are called the external angles of the polygon. What size are they?
This practical demonstration will give us the answer:
In this diagram, the triangle ABC is isosceles,
since the two sides, AB and AC, are equal as are the two angles at B and C. 
This is my favourite method since it involves a Knot(t)!
You do the maths...
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:
Q is a gold point of PR
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. 
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 R is r2
we cannot have R as a negative number, so
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:
The diagram shows the whole isosceles triangle AEB with the missing part of the sides
DE and CE given the length x.
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.
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.
Here are two circles, one inside the other. The yellow and red areas define an outer ring.
Also the orange and red parts
form an ellipse (an oval). 
If the ring is very narrow, the two circles are similar in size and the ellipse has a much bigger
area than the ring.
If the inner circle is very small,
the ellipse will be very narrow and the outer ring will be much bigger in area than the ellipse.
The Golden section squares are shown in red here, the axes in blue and all the points of the
squares lie on the green lines, which pass through the origin (0,0).
In the spiral above, based on Fibonacci squares spiralling out from an initial two 1x1 squares,
we noted that one quarter turn produces an expansion by Phi in the distance of a point on the curve
from the "origin". So in one full turn we have an expansion of Phi4.
Here are the graphs of three familiar trigonometric functions: sin x, cos x
and tan x in the region of x from 0 to π/2 (radians) = 90°:
Look again at the sharp and flat
triangles of the pentagon that we saw above. If we divide each in half, we have right angled
triangles with sides 1 and Phi/2 around the 36°
angle in the flat triangle and
sides 1/2 and Phi around the 72° angle in the sharp triangle. So:
Leonardo spots a rabbit in one corner of his field which just happens to be a square. He starts to chase it at exactly the same moment
as the rabbit bolts for the hole in the hedge at another corner of the field. 
The Mathematical Magic of the Fibonacci Numbers
Fibonacci Home Page
The Golden String
© 1996-2016 Dr Ron Knott
