APPENDIX I

THE GOLDEN SECTION
or
THE EXTREME AND MEAN RATIO

Among the many relationships that exist between numbers, there are some that may be expressed as ratios and proportions.  Thus when we say that the ratio of 3 to 12 is the same as the ratio of 12 to 48, we are recognizing the fact that there are 4 threes in 12, and that there are also 4 twelve's in 48.  This may be written in the form of a proportion, thus:

The numbers 3 and 48 are known as the extremes, and the number 12 is known as the mean.  This may also be written in a fractional form:

The relationship becomes obvious when we reduce the fractions to ¼.

This is the simplest type of proportion because only three quantities appear: 3, 12 and 48, and it will serve our purposes adequately.

It is a property of all such proportions that the product of the extremes is equal to the square of the mean.  Thus, we would find that the area of a square that is 12 inches on a side is equal to the area of a rectangle that is 3 inches wide and 48 inches long.  Each would have an area of 144 square inches.

A proportion of this kind may be written algebraically by the use of the letters a, b, and c, thus:

a  :  b = b : c     and       a x c = b²

We may now develop a sequence of numbers based upon this pattern: 3, 12, 48, 192, 768, and so on, by simply multiplying each term by 4.  Such a sequence of numbers is known as a geometrical progression, with a constant ratio between any two adjacent terms.  In this instance the ratio is 1 : 4.

Another kind of numerical series is generated by a process of addition, wherein the series increases by a fixed increment.  This is known as an arithmetical series.  A good example of this is seen in the column of a calendar.  Take the column which reads 4, 11, 18, 25, wherein each term is merely the previous term increased by 7.  Here there is no fixed ratio between any two adjacent terms.  Thus, we cannot say that 4 : 11 as 11 : 18, nor as 18 : 25.

Now we are in search of a series of numbers which will combine the properties of the arithmetical series with the geometrical progression.  We shall reach a halfway point in our search if we consider the so-called Fibonacci Series, named after its discoverer, an Italian mathematician who was born about 1180 AD.  This series starts with 0, and the second term is 1.  Thereafter, each term is the sum of the two preceding terms, thus:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, etc., etc.

This series of numbers has a remarkable property, for while there is not a fixed ratio between the terms, nevertheless if we take them two at a time, starting them as fractions, and find their decimal values, we shall discover that these values come closer and closer to a certain number, already familiar to the readers of this book.  Carried to six significant figures, this number is .618034…  Actually, it is an indeterminate decimal, and for most purposes it is sufficient to carry it to three significant places only, as has been done throughout this book.

Let us now make a table of fractions taken from the Fibonacci Series.  The (+) and the (-) signs indicate that the value is either above or below the decimal .618034; but notice that these values converge more and more closely to that figure as the process is carried on.  Actually, the final value is reached only when this series is carried to infinity:

Figure 22 shows a graph of these values.  Note the bold line indicating .618034… and the manner in which the values approach it, until the eye cannot distinguish them from the line.  However, if we wish to be very precise in our study, we must recognize that it is only when this series is carried out to infinity, as said before, that this ratio between the consecutive terms becomes .618034…

We discover the series for which we are looking if we will take .618034… as one of its terms, and subtract it from 1.000, giving us .381966… By adding any two consecutive terms in order to give us the following one, we shall find that the series grows as follows:

.382,   .618,   1.00,   1.618,    2.618,   4.236,   6,854,  and so on.

This series of numbers is an arithmetical series for the reason that it increases to the right by a process of addition.  Thus, each term is the sum of the two preceding terms.

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Fig. 22 — The Golden Section and the Fibonacci Series.
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It is also a geometrical progression for the reason that it is generated by a process of multiplication.  The factor .618 is a constant.  Thus, each term is .618 times the term to the right.

This is a series of numbers which extends indefinitely in both directions, so it does not really start with .382.  We may extend it indefinitely to the left.  So it is now presented with a value 1.00 as occupying a central position:

The number under each term of this series indicates its place to the left or the right of 1.00.  This is done in order to bring out a most curious property of this series.  To multiply two or more terms of this series, take the algebraic sum of their subscripts and locate the product above the new subscript.  Two examples will make this clear: Multiply .236 by .6854.  Their subscripts are —3 and 4, the algebraic sum of which is 1.  Locate 1.618 over subscript 1.

Multiply .618 by .146.  Their subscripts are — 1 and — 4, the algebraic sum of which is —5.  Locate .090 above subscript —5.

To find the square of any term, multiply its subscript by 2, and locate the answer.  Thus the square of .382 is .146.  The square of 1.618 is 2.618.  Since each term is the sum of the two terms preceding it, the formula

a : b = b : c       becomes       a : b = b: (a + b)

We are ready to move into a geometrical consideration of the Golden Section; this should be introduced by stating that a rectangle constructed such that the height and width are any two consecutive terms of this series, will be proportioned as .618 : 1.00.  This is held to be the most beautifully proportioned of all rectangles, and the Greeks recognized this.  They used it widely in their architecture, their ornaments, statuary, vases and paintings.  We shall henceforth refer to it as a Golden Rectangle.

In order that we may construct this rectangle and other figures, we must learn how to divide a line at the point of the Golden Section.  See upper part of Fig. 22.

Upon a given line AB, erect a perpendicular AC at point A, equal to one half of AB.

Join BC.

With C as center, and with radius CA, describe an arc cutting CB at K.

With center at B, and with radius BK, describe an arc cutting AB at point P.  P is the point of the Golden Section, and AP : PB as PB : AB.  Expressed in words, the shorter line segment is to the longer line segment as the longer is to the whole line.  This satisfies the equation a : b = b : ( a + b).

To present the geometrical proof of this might be confusing to some readers, as a knowledge of several geometrical principles involving the relationship between tangents and secants to a circle would be required.  This material is readily available in any standard textbook on Plane Geometry, and it is better to confine ourselves to a consideration of the numerical relationships which apply to the Golden Section of the line AB.

Let us consider that the perpendicular AC equals 1.00, and the line AB equals 2.00.  BC is the hypotenuse of the right triangle ABC and is therefore the square root of the sum of the squares of the sides AC and AB, 1² + 2² = 5.  Thus, BC equals the square root of 5, or 2.236.  BK equals BC — KC, or 2.236 - 1.00.  This means that BK equals 1.26.  Marking off this length on AB at the point P, gives us a ratio of 1.26 : 2.  This reduces to .618 : 1.00, or the Golden Section of the line.   

Now there is a second point, P', which balances the point P.  It may be found by erecting the perpendicular at B; but for simplicity of explanation this point P' is shown without the construction which would only duplicate what has already been done.  The point P' not only divides the line AB at the point of the Golden Section, but it also divides PB in the same manner.  Thus we have four terms in the Golden Series, since PP' is .236.

If we wish to construct a Golden Rectangle, we have only to construct a square as in Fig. 23.  M is the midpoint of the base of the square; with M as a center, and the length MC as a radius, describe an arc cutting AB produced, at the point E.  Erect a perpendicular at point E, to meet DC produced, at the point F.  Rectangle AEFD is a Golden Rectangle, and the width (or height) EF is to the length (or base) AE as .618 : 1.00.

The diagonal of the Golden Rectangle, AF, is interesting.  If we were to construct a circle with radius AE we would find that the diagonal AF is the length of the side of an inscribed regular pentagon.  Also, the height of the rectangle, EF, is the side of an inscribed regular decagon.  The smaller rectangle at the side of the square (standing on its shorter side) is also a Golden Rectangle, and therefore similar to the rectangle AEFD, and its diagonal, EC, is perpendicular to the longer diagonal AF.

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Fig. 23 — The “Golden” Rectangle.
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Fig. 24 — The Rectangle of the Whirling Squares
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Their intersection is the “eye” of an intersecting series of squares which appear to rotate about (Figure 24 shows this as a “rectangle of the whirling squares”).

Figure 25 shows one of the triangle segments of a regular Decagon, generated by the moving circle as in Chapter II.  Its sides would be two of the radii of the circle, and its base is one side of the polygon.  We demonstrated at that time that numerically its length is .618 of the radius, but the geometrical proof that it actually is a regular decagon was left to this later time.  thus we must prove that the angle at the apex of the triangle is 36º.  Having done this, we shall not only find that the proposition of the moving circle is geometrically correct, but that we have here a triangle of truly remarkable properties.

Given isosceles triangle ABC such that AC and BC are equal, and the base AB is .618 of AC.  to prove that the angle at C contains 36º.

With A as center, and with radius AB, describe an arc cutting BC at K.  Join AK.

Triangle ABK is isosceles because AB equals AK.  Therefore Ð AKB = ÐABK.  But ÐCAB = ÐABC, as they are the base angles of the isosceles triangle ABAC.

Therefore ÐCAB = ÐAKB, and the two triangles are similar.

Therefore ÐKAB = ÐACB.

Since AB : AC = .618 : 1.00, KB : AB = .618:1.00.

Corresponding parts of similar triangles are proportional.

The Golden Series of terms that we have just studied shows us that KB = .382.

CK = CB - KB, or  1.00 - .382.  Therefore  CK = .618.

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Fig. 25 — A “Golden” Triangle
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But AK, being equal to AB, is also equal to .618; therefore the triangle ACK is isosceles with the sides AK and CK equal: the base angles ACK and CAK are also equal.

Therefore CAK = ACK, which also equals KAB, and CAB is 2 KAB.

Therefore each of the base angles of triangle ABC is equal to 2 KAB.

But KAB = ACB, so that each of the base angles is equal to 2 ACB.

Therefore the sum of all the angles of triangle ABC = 5 ACB.

The sum of the angles of any triangle is 180°.  Therefore 180° is 5 ACB.

Therefore ACB = 36°, and the polygon within the circle is proven to be a regular decagon.

The base angles of triangle ABC are each equal to 72°.  Much importance has already been attached to the number 72.

It is worth noting that in the original triangle, the base is to the side as .618 : 1.00, and the triangle ACK, the side is to the base as .618 : 1.00.  These triangles were held by the Greeks to be so important that they were named after the two stars in the sky associated with the constellation Gemini, the Twins: Castor and Pollux.  These triangles are repeated many times in the five-pointed star diagrams that we have studied.

By repeating the process in the formation of these triangles from the original triangle ABC, we can subdivide endlessly, making a series of  “whirling triangles,” and we find that they form the basis of an interesting logarithmic spiral which is similar in shape to the shells of certain sea-creatures.  Numerous instances can be found in nature, wherein the Golden Section plays its part in natural design.

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Fig. 26 — Constructing a Pentagon Within a Circle
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Fig. 26 shows how to construct an inscribed regular Pentagon within a circle.  Given circle with center O, construct the diameter AB and CD mutually perpendicular.

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Fig. 27 — Constructing a Pentagon, having given one Side.
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M is the midpoint of OB, and with M as a center, and with radius MC, describe an arc cutting AO at the point P.  With C as center, and with radius CP, describe an arc cutting the circumference at P'.  CP' is the side of the inscribed pentagon.

Figure 27 shows how to construct a regular Pentagon having one of its sides given.  Given side AB, on A construct a perpendicular AC equal to ½ AB.

Join BC and extend.

With C as center, and radius CA, describe an arc cutting BC extended at point P.

With B as center, and radius BP, describe an arc; and then with A as center, and with the same radius, describe another arc.

Point X is the intersection of these arcs.

With A as center, and with radius AB, describe an arc cutting arc PX at P'.  With B as center, and with radius AB, describe an arc cutting arc XY at Q.  Join points A, P;, X, Q and B to complete the pentagon.

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Fig. 28 — Constructing a Pentagram (Five-pointed Star), having given one Side.
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Figure 28 shows how to construct a five-pointed star, having given one of its lines, AB.

Upon AB locate the two points P and P' of the Golden Section by the method already explained.

With P as center, and with radius PA, describe an arc upwards.

With P' as center, and with radius P'B, describe an arc upwards.  Locate Point C where the arcs intersect.  Join CP and extend.  Join CP' and extend.

Set the compass to radius AB.  With C as center, describe an arc cutting CP extended at point D, and CP' extended at point E.  Join AE and BD to complete the star.