Two-dimensional Geometry and the Golden section

or

Fascinating Flat Facts about Phi

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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. Theicon means there is asection of questions to start your own investigations.

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Constructions for the Golden Ratio

gold points

Constructing the internal golden section points: phi

(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

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. 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:i.e.Now, taking the square-root of each side gives:Point V was drawn so that TV is the same length asSo AV is justThe final construction is to mark a point G which is same distance (AV) along the original line (AB) which we do using the compasses. So AG is phi times as long as AB!

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Constructing the external golden section points: Phi

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

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;

A Simple Construction of the Golden Section, Kurt Hofstetter in Forum Geometricorum Vol 2 (2002) pages 65-66 which has the proof too.

An even simpler method!

gold point

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;

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

so A is the gold point of CG

Kurt Hofstetter's proof was published as Another 5-step Division of a Segment in the Golden Section in Forum Geometricorum Vol 4 (2004) pages 21-22.

Lemoine's Construction of the Golden Section

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;

A 5-step Division of a Segment in the Golden Section K Hofstetter in Forum Geometricorum Vol 3 (2003) 205-206.

K Hofstetter in Vol 3 (2003) 205-206. 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

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.

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

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

angle Repeat moving it along the sides and turning through the rest of the orange angles until it lies back on the top edge

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

angles So the sum of the 5 orange angles is one turn or 360°

angles is one turn or 360° Each orange angle is therefore 360/5=72°

green

orange

yellow

72°

72°

36°

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°.

The 36°-72°-72° triangle

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

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:

This is my favourite method since it involves a Knot(t)!

This is my favourite method since it involves a Knot(t)!

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.

Robin Wilson 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

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.

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.

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?



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

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



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?

a, ar, ar2

(ar2)2 = (ar)2 + (a)2

a2r4 = a2r2 + a2

a2

r4 = r2 + 1

R

r2

R2 = R + 1 or

R2 – R – 1 = 0

R = 1 ± √5 2 = Phi or –phi

R

r2

R

R = r2 = Phi so

r = √Phi

a, a √ Phi , a Phi

√Phi

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

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Phi and Rectangles

Phi and the Trapezium (Trapezoid)

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!

So what was the shape of the isosceles triangle which had its top part removed to make the traphizium?

x

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?

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Phi and the Root-5 Rectangle

1

√5 = 2·236

phi=0·618..

1

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.

√2 : 1

√(1/2)

√2

1:√2

1

s

s/2:1

1/s = s/2 /1

s2 = 2

s = √2

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Phi and other Polygons

Decagons

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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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Recently, 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. 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 isor. When they appear in nature in crystals, they are called. 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 ofor, 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. Dissecting the Sharp and Flat Triangles:

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

The dart is Phi times the area of the kite

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.

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.

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

.. ..

Φ

√ Φ = Φ1/2 = φ –1/2 = 1.27201964951406

Φ

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...

√φ = φ 1/2 = 1/ √Φ = Φ –1/2 = 0.78615137...

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!

and I cannot recommend them highly enough! Aperiodic Tiles, R.Ammann, B. Grünbaum, G.C. Shephard, Discrete and Computational Geometry (1992), pages 1-25.

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More Geometrical Gems

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.

x(w+z)/2

yw/2

(x+y)z/2

x(w+z) = yw and x(w+z) = z(x+y) .

x = yw/(w+z)

zx

xw = zy

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 .

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.

in , 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



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

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.

What ratio of circle sizes (radii) makes the ellipse equal in area to the ring between the two circles?

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

a = b

a

b

Area of ring = π (b2 – a2)

π (b2 – a2) = π a b

b2 - a2 - a b = 0

b/a

K

a2

K2 - K - 1=0

K

Phi

–phi

K

b = Phi a

a = phi b

The equation of an ellipse is

(x/b)2 + (y/a)2 = 1

a = b

a(=b)

(x/a)2 + (y/a)2 = 1 or

x2 + y2 = a2 as it is more usually written.

(x,y)

a

Note 79.13 A Note on the Golden Ratio , A D Rawlings, Mathematical Gazette vol 79 (1995) page 104.

, A D Rawlings, 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.

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The Fibonacci Squares Spiral

This section answers the question:

What is the equation of the Fibonacci spiral?

(0,0)

45°

phi

x

y

phi

(r,theta)

(r Phi,theta + π/2)

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 = Phin and theta= n π /2

n

r = Phi2 theta / π

r = Mtheta where M = Phi2/ π = 1.35845627...

x = r cos(theta)

y = r sin(theta)

x

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

Phi

Phi4

Phi

Phi

Phi

r = Phi theta / (2 π) when theta is measured in radians or

r = Phi theta / 360 if theta is in degrees.

theta

2 π

r

Phi

Phi

1

Phi

Click on the Spreadsheet image to open an Excel Spreadsheet to generate the Golden Spiral in a new window.





1·61803 39887 49894 84820 45868 34365 63811 77203 09179 80576 ..More..

Trigonometry and Phi

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

meaning literally and a six angled one is called a hexa-gon then we could call



then we could call a four-angled shape a quadragon

(but we don't - using the word quadrilateral instead which means "four- sided ") and

(but we don't - using the word instead which means "four- ") 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.

With thanks to proteus of softnet for this information.

sin x, cos x

tan x

x

π/2

= 90°

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

i.e. where or radians at another angle where tan x = cos x

tan x = cos x and, since tan x = sin x / cos x , we have: sin x = (cos x)2 =1-(sin x)2 because (sin x)2+(cos x)2=1. or (sin x)2 + sin x = 1

sin x

sin x = (–1 + √5)/2

sin x = (–1 – √5)/2

x

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..°

cos(x)

tan(x)

sin(x)

x

y = phi

sin(x)

Is there any significance in the value of tan(x) where tan(x)=cos(x)?

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)

t2 = 1/(1 + t2)

1 + t2

t2 ( 1 + t2 ) = 1

t2 + t4 = 1

T

t2

t4 + t2 – 1 = 0

T2 + T – 1 = 0

T = ( –1 ± √(1+4) ) / 2

T

t2

T

T = ( √5 − 1 ) / 2 = phi

T

t2

t

√phi

1 + Phi = Phi2

√(1+T2) = Phi

1, √Phi, Phi

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?

A

A

B

B

A

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

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

the-angle-whose-sin-is 0.5

arcsin(0.5)

asin(0.5)

sin-1(0.5)

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

More trig ratios and Phi (sec, csc, cot)

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?



the inverse ratio to the tangent is the cotangent or cot i.e. Adj/Opp; cot(x)=1/tan(x)

or 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)

or 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)

secant, sine, tangent

cosecant, cosine, cotangent

Here is a graph of the six functions where the angle is measured in radians:

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 ;

is the angle where then and ; The value of is ; 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 .

A

B

90°

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

1

Phi/2

36°

1/2

Phi

72°



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

sin(18°)

cos(18°)

cos(18°) = √ Φ √5 2

sin

cos

30°

18°

sin

cos

cos(A – B) = cos(A) cos(B) + sin(A) sin(B)

cos(12°) = φ + √ 3 Φ √5 4

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

cos(45°) = sin(45°) = 1 = √2

√2

2

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

Fib(n) = Phin – (–Phi)–n = Phin – (–phi)n

√5

√5

A Simple Trig formula for Fib(n)

cos(π/5) = cos(36°) = Φ/2

cos(2 π/5) = cos(72°) = φ/2

cos(180° – A) = –cos(A)

cos(180° – 72) = – φ/2

Φ = 2 cos( π/5 ), –φ = 2 cos( 3π/5 )

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).

Binet's Formula solely in Trig terms

sin ( π

5 ) sin ( 3 π

5 ) = √5

4 and sin ( 3 π

5 ) sin ( 9 π

5 ) = − √5

4

Fib(n) = 2n+2

5 ( cosn( π /5) sin( π /5) sin(3 π /5) + cosn(3 π /5) sin(3 π /5) sin(9 π /5) )

Fib(n) = 2n+2

5 ( cosn(36°) sin( 36°) sin(108°) + cosn(108°) sin(108°) sin(–36°) )

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

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! + x4 4! – x6 6! + ...

sin( x ) = x − x3

3! + x5 5! – x7 7! + ...

n!

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( π 10 ) = 2 ( π − π 3 + π 5 − π 7 +...) 10 1033! 1055! 1077! = π − π 3 + π 5 − π 7 +... 5 3 000 6 000 000 25 200 000 000

π9

φ

π8

Φ

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:

3. 14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59230 78164 06286 20899 86280 34825 34211 70679 82..

With thanks to John R Goering for suggesting this connection between Phi and π.

1·61803 39887 49894 84820 45868 34365 63811 77203 09179 80576 ..More..

Phi and a Geometric and Calculus problem

Leonardo chases a Rabbit

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

1·61803 39887 49894 84820 45868 34365 63811 77203 09179 80576 ..More..

1·61803 39887 49894 84820 45868 34365 63811 77203 09179 80576 ..More..