GOLDEN SECTION

The golden section is the division of a segment into two parts such that the greatest part of the segment is the mean proportional between the entire segment and the smallest part.

In practice, starting from a segment AB, if one of its internal points C satisied the proportion:

AB : AC = AC : CB

or

AC² = AB x CB

then it is said that the part AC of the segment AB is the mean proportional between the entire segment AB and its remaining part CB. Thus:

AC is the golden section of AB, also defined “mean ratio”.

The ratio between the whole segment AB and its golden section is called golden ratio.

The golden ratio is numerically expressed by an irrational number, called "golden number" and it is commonly indicated with the letter φ, which is the initial of the name of the Greek sculptor Phidias who used this ratio to design the Parthenon in Athens:

φ = AB/AC = (1 + √5)/2 = 1.6180339887...

DIVISION OF A GENERIC SEGMENT IN GOLDEN RATIO

The following figure illustrates, through a sequence of drawings, the division of a generic segment AB in the golden ratio:

Image source: https://online.scuola.zanichelli.it/fava-geometriaedisegno/files/2015/02/AppB2-5_Sezione-aurea_Mondrian.pdf

CALCULATION OF THE GOLDEN RATIO

In the figure, the length of the segment AB has been indicated with “a” and the length of the longest part AC of the segment AB, previously defined as “golden section”, has been indicated with “x”.

The golden ratio is calculated starting from the proportion:

a : x = x : (a – x)

namely:

a/x = x/(a – x)

x² = a(a – x)

Simple steps allow to obtain the quadratic equation:

x² + ax – a² = 0

The solutions of the quadratic equation are:

The positive solution of the quadratic equation gives the length of the segment "x", which is the "golden section" of the segment which has length "a":

The ratio between the length "a" of the segment AB and its "golden section" is the "golden ratio":

The numerical value of the golden ratio is the "golden number φ":

CONSTRUCTION OF THE GOLDEN SECTION OF A UNIT SEGMENT

• Consider the segment AB which has length equal to 1;

• starting from A, draw the segment AP which has length equal to 1/2 and perpendicular to the segment AB;

• connect the point P with the point B;

• the Pythagorean theorem, applied to the right-angled triangle BAP, gives the length of the segment PB (hypotenuse of the right-angled triangle), whose value is √5/2;

• using a compass with the tip on the point P and opening PA, report the length of AP on the hypotenuse PB, as indicated in the figure:

• the length of the remaining part of the hypotenuse will be: (√5/2 – 1/2);

• report the length (√5/2 – 1/2) on the segment AB.

The segment CB which has a length (√5 – 1)/2 = 0.618033... is the golden section of AB.

The golden number is the ratio between the unit segment AB and its golden section:

φ = 1/[(√5 – 1)/2] ~ 1.618...

THE GOLDEN RECTANGLE

The golden rectangle is a rectangle that is built with the golden proportions.

The below figure shows how to do it graphically:

• initially consider the square which has vertices AEFD;

• draw the segment A'H which divides the side DF and the upper opposite side AE in half;

• place the tip of the compass on the point A' with opening A'E. Draw a right-handed arc, starting from E;

• extend the segment DF until the point C is determined;

• draw the segment CB perpendicular to DC and the segment BE perpendicular to EF.

The rectangle ABCD is a golden rectangle in which the side AB is exactly divided in the golden section by the point E, that is:

AB : AE = AE : EB

In a golden rectangle the ratio between the longest side DC and the shortest side DA is the golden number φ.

THE GOLDEN SECTION AND THE REGULAR PENTAGON

A regular polygon has congruent sides and angles.

The sum of the internal angles of a regular polygon is given by the relation:

S_int = (n – 2) x 180⁰

where "n" is the number of the sides of the polygon,

then:

the sum of the interior angles of a regular pentagon is 540⁰

therefore

each interior angle of a regular pentagon measures 108⁰ = 540⁰/5

The diagonals of a regular pentagon, obtained by connecting all the opposite vertices of the polygon, form a pentagram, that is a five-pointed star, emblem of the Pythagoreans:

Pentagon with relative pentagram.

There is a connection between the geometry of the regular pentagon and the golden section. It can be proved by examining the triangles that are obtained when the diagonals of the pentagon are traced.

Initially, consider the triangle EDC in the figure 1:

Fig.1

This triangle is isosceles since its oblique sides are congruent (ED = DC) and the angles at the base are also congruent, in fact, both of them measure 36⁰ = (180⁰ – 108⁰)/2. The same thing can be said for the triangles EAB, ABC, BCD, AED, identifiable within the regular pentagon, which are also triangles isosceles.

Now, consider the triangle ECA in the figure 2:

Fig.2

This triangle is isosceles since the diagonals of a regular pentagon are congruent, therefore CE = CA. Furthermore, in this triangle both of the angles at the base measure 72⁰ = 108⁰ – 36⁰, while the value of the vertex angle is 36⁰.

Within a regular pentagon, each side forms, with two diagonals, an isosceles triangle; the base angles measure 72⁰ and the vertex angle measures 36⁰.

Take into consideration the isosceles triangle ADB, in the figure 3, which has the characteristics mentioned above:

Fig.3

Draw the bisector AE of the angle at A and consider the triangle EAB. The bisector divides the angle at A (whose value is 72⁰) into two angles which measure 36⁰; knowing the value of the angle at B (72⁰), the angle at E measures 72⁰ = 180⁰ – 36⁰ – 72⁰. The triangle EAB, having two congruent angles, is isosceles, therefore the segment EA is congruent to the segment BA (EA = BA). Moreover, the angles of this triangle are orderly congruent to those of the triangle ADB, thus:

the triangles ADB and EAB are similar.

Also the triangle DEA, whose vertex angle measures 108⁰ = 180⁰ - 36⁰ - 36⁰, is isosceles since the two base angles are congruent, therefore EA = ED.

Ultimately it was obtained that:

EA = BA

and

EA = ED

therefore

BA = ED

Referring again to the similar triangles ADB and EAB, the following proportion can be written:

BA : BD = BE : BA

since:

the side BD of the triangle ADB is corresponding to the side BA of the triangle EAB; in fact, the sides BD and BA are opposite sides at equal angles which measure 72⁰;

the base BA of the triangle ADB is corresponding to the base BE of the triangle EAB; in fact, BA and BE are opposite sides at equal angles which measure 36⁰.

If, in the previous proportion, the means and the extremes are inverted, it can be written:

BD : BA = BA : BE

but BA = ED, thus:

BD : ED = ED : BE

It can be deduced, from this proportion, that the part ED of the segment BD, is mean proportional between the entire segment BD and the remaining part BE. Therefore, for the definition of golden section of a segment, it can be said that:

ED is the golden section of the segment BD

but BD is congruent to AD, therefore

ED is the golden section of the segment AD

Furthermore, ED is congruent to EA, which is congruent to BA, thus:

BA is the golden section of the segment BD

namely

the side of a regular pentagon is the golden section of one of its diagonals

therefore

the ratio between the diagonal BD of the regular pentagon and its side BA is the golden number φ:

φ = BD/BA

A further examination of the isosceles triangle DEA (figure 3), whose base angles measure 36⁰ and the vertex angle measures 108⁰, allows to state that each shorter sides (ED side congruent to EA side) is "golden section" of longer side AD, i.e. of the base. Therefore, the ratio between the base AD of this isosceles triangle and one of the two congruent oblique sides is still the golden number.

Thus

the isosceles triangle DEA and the isosceles triangle ADB are golden triangles since, for each of them, the ratio between the longest side and the shortest side is equal to the golden number φ.

It should be underlined also that, if all the diagonals of a regular pentagon are traced, another regular pentagon is formed in the center, with as many golden triangles. This procedure can be iterated until an infinite series of pentagons, pentagrams and gloden triangles is obtained, as indicated in the figure:

Fig.4

THE GOLDEN SPIRAL

The golden spiral is a logarithmic spiral which has the constant ratio between consecutive radii equal to the golden number φ.

The golden spiral is an open polycentric curve which develops on the partitions of a golden rectangle. The geometric construction can be obtained in the following way:

• Inside a golden rectangle draw a square with the side equal to the smallest side of the rectangle. The difference between the two geometric figures will be a golden rectangle, whose sides are in golden ratio.

• Repeat this procedure at least five times, in order to obtain an acceptable visual effect.

• Place the tip of the compass on the vertex B of the square which lies on the longest side of the rectangle and trace the arc AB which joins the extremes of the two sides of the square, which form the chosen angle. Repeat this operation for each drawn square in order to create a continuous line.

• The subsequent partitions and the tracing of the relative arcs will compose the golden spiral, whose development is theoretically infinite.

THE GOLDEN SPIRAL AND THE FIBONACCI SEQUENCE

The golden spiral can be obtained, with good approximation, using the numbers belonging to the Fibonacci sequence.

The construction process is described below:

1. Initially consider the square ABCD which has the sides are equal to 1. On the side AB construct the square ABEF which has the sides still equal to 1 and, placing the tip of the compass on the point A, trace the circumference arc BF.

2. On the side DF = DA + AF = 1 + 1 = 2 construct the square FGHD and, placing the tip of the compass on the point D, trace the circumference arc FH.

3. On the side CH = CD + DH = 1 + 2 = 3 construct the square CHIL and, placing the tip of the compass on the point C, trace the circumference arc HL.

4. On the side EL = EC + CL = 2 + 3 = 5 construct the square LMNE and, placing the tip of the compass on the point E, trace the circumference arc LN.

5. On the side GN = GE + EN = 3 + 5 = 8 construct the square NOPG and, placing the tip of the compass on the point G, trace the circumference arc NP.

6. On the side PI = PG + GI = 8 + 5 = 13 construct the square PQRI and, placing the tip of the compass on the point I, trace the circumference arc PR.

The iteration of this procedure allows to obtain the golden spiral.