## 02. Sectio Aurea – Michel Tombroff

### More about the golden ratio

#### 3.1 Golden triangles

A golden triangle is an isosceles triangle, such that the ratio of the length of the long side to the short side is $$\phi$$. The angle formed by the two equal sides must be equal to $$36^o$$ or $$108^o$$.

The green triangle is an acute or golden triangle, the blue triangle is an obtuse triangle or golden gnomon. They form part of a regular pentagon:

Proof: Angle $$E$$ in triangle $$BEC$$ is equal to $$36^o$$ since it is equal to half of angle $$O$$ in triangle $$BOC$$, which in turn is equal to $$\frac{360^o}{5}=72^o$$ as it is part of a regular pentagon.
Angle $$D$$ in triangle $$CDE$$ is equal to $$108^o$$ since it is twice the size of angle $$D$$ in triangle $$ODC$$; this angle is the same as angle $$C$$; angle $$O$$ is equal to $$72^o$$ and the sum of the angles of a triangle is equal to $$180^o$$.

These are golden triangles: suppose $$x = \frac{L}{\ell} > 1$$; angle $$E$$ in the blue triangle is equal to $$36^o$$, on a $$cos 36^o = \frac{L/2}{\ell}=\frac{x}{2}$$ ; furthermore, by halving the green triangle, we obtain $$sin 18^o = \frac{\ell/2}{L}=\frac{1}{2x}$$. The relation $$cos 36^o=cos^2 18^o-sin^2 18^o$$ devient $$\frac{x}{2}=1-2.\frac{1}{4x^2}$$ thus $$x^3-2x+1=0=(x^2-x-1)(x-1)$$.
The unique solution $$x$$ when $$x > 1$$ is $$x = \phi$$.

These triangles are the base for Roger Penroseās (2020 Nobel prize) aperiodic tiling of a plane. An obtuse triangle or golden gnomon (resp. acute) is made up of two (resp. 3) golden triangles; the length of the long side of each new triangle is equal to $$\frac{1}{\phi}$$ times the length of the long side of the initial triangle (and thus equal to the length of the initial short side):

#### 2. Golden ratio and the Fibonacci sequence

The golden ratio is closely related to the Fibonacci sequence. The Fibonacci sequence $${(F_n)}_{n\ge 0}$$ is a series of integers where each term is the sum of the preceding two terms:

$F_{n+2}=F_n+F_{n+1}$

starting with $$F_0=0$$ and $$F_1=1$$.
The first ten terms in the sequence are $$0, 1, 1, 2, 3, 5, 8, 13, 21$$ et $$34$$.

The Fibonacci sequence, introduced by Leonardo da Pisa (1175- around 1250), aka Leonardo Fibonacci, is considered as perhaps the first mathematical model for population growth. It uses the following (very simplified) hypotheses to describe the growth of a population of rabbits: three months after birth, each pair of rabbits will produce a pair of offspring, and so forth, indefinitely. The first month, one begins with a pair of rabbits that are one month old. The second month, there is still only one pair. The third month, however, the pair gives birth to another pair. There are now two pairs. The following month, the first pair gives birth to an additional pair. The first pair they produced is still only two months old and does not multiply. As a result, there are now three pairs. The fifth month, each of the first two pairs give birth to another pair. The third pair does not multiply. There are now 3 + 2 = 5 rabbit pairs. The Fibonacci sequence $$F_n$$ provides the number of rabbit pairs $$n$$ every month, a description of the population growth.

The terms in the Fibonacci sequence can be directly expressed as a golden ratio:

$F_n=\frac{1}{\sqrt{5}}\left(\phi^n+\left(\frac{-1}{\phi}\right)^n\right)$

Proof: If $$x$$ is a solution to equation $$x^2-x-1=0$$ then $$x^2=x+1=x^1+x^0$$, $$x^3=x^2\times x= x^2+x^1$$ and to recurrence relation $$x^{n+2}=x^{n+1}+x^n$$ then every series $$C_{n}=Kx^{n}$$ satisfies properties $$C_{n+2}=C_n+C_{n+1} \, \forall n\ge 0$$; all the series that satisfy this property is a vectorial space with dimension 2 since this series and its values for $$n=0$$ and $$n=1$$.
The two solutions to the equation $$x^2-x-1=0$$ are $$\frac{1\pm\sqrt{5}}{2}$$, and hence, $$\phi=\frac{1+\sqrt{5}}{2}$$ et $$-\frac{1}{\phi}=-\frac{2}{1+\sqrt{5}}= -\frac{2(1-\sqrt{5})}{1-5}=\frac{1-\sqrt{5}}{2}$$.
Every sequence $${(C_n)}_{n\ge 0}$$ such as $$C_{n+2}=C_n+C_{n+1} \, \forall n\ge 0$$ is defined by
$$C_n=K_1\phi^{n}+K_2\left(\frac{-1}{\phi}\right)^{n}$$ where $$K_1$$ and $$K_2$$ are constant values determined by the values of $$C_0$$ and $$C_1$$.
Notably, for the Fibonacci sequence $$F_n=K_1\phi^n+K_2\left(\frac{-1}{\phi}\right)^n$$ with $$F_0= 0=K_1+K_2$$ and $$F_1=1=K_1 \phi -\frac{K_2}{\phi}$$ which results in $$K_2=-K_1$$ et $$1=K_1 (\phi+\frac{1}{\phi})=K_1(\frac{1+\sqrt{5}}{2}-\frac{1-\sqrt{5}}{2})=K_1\sqrt{5}$$.

The golden ratio can be derived from the Fibonacci sequence. The ratio sequence of two successive integers in this sequence:

$(\frac{F_{n+1}}{F_{n}})_{n\ge 1}$
the first 9 terms are $$1, 2, \frac{3}{2}, \frac{5}{3},\frac{8}{5},\frac{13}{8},\frac{21}{13},\frac{34}{21}$$.

The golden ratio is the limit of this new sequence; it is hence the limit of the ratios of two consecutive terms of the Fibonacci sequence.

Proof: In accordance with the formula above, as $$\frac{1}{\phi}<1$$, for a sufficiently large $$n$$, $$\left(\frac{1}{\phi}\right)^n$$ becomes very small and $$F_n\simeq\frac{1}{\sqrt{5}}\phi^n$$ hence $$\frac{F_{n+1}}{F_n}$$ tends towards $$\frac{\frac{1}{\sqrt{5}}\phi^{n+1}}{\frac{1}{\sqrt{5}}\phi^n}=\phi$$.

#### 3. Golden spirals

The largest possible square within a golden rectangle is, once again, a golden rectangle.

Proof: Start with a golden rectangle whose length is $$L$$ and width is $$\ell$$, or $$\frac{L}{\ell}=\phi$$. Remove a square from side $$\ell$$. What remains is a rectangle whose length is $$\ell$$ and width is $$L-\ell$$.
It is a golden rectangle
$\frac{\ell}{L-\ell}=\frac{1}{\left(\frac{L-\ell}{\ell}\right)}=\frac{1}{\phi-1}=\phi$
since $$\phi(\phi-1)=\phi^2-\phi=1$$.

To obtain a golden spiral, draw quarter circles in each successive square derived from the golden rectangles.

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