Investigation of Golden Ratio in Construction of The Great Pyramid of Giza

Next year we are planning to visit Egypt. I am so excited that I have already started planning out and finding information about the place. The most interesting things there are the pyramids. Reading articles and watching videos about them I came to know a lot of things. One of which is the term golden ratio. It had relations with the Fibonacci series, the one which triggers a bitter memory.

 

I was in class 6 when I started computer programming. I can clearly recall the day my teacher had given a program on Fibonacci series. I did not do it and was punished.  I was made to write the Fibonacci series program 50 times. Later, it so happened that i knew the series very well and always wondered about its applications.

 

Getting back to my research on pyramids, I came to know that the patterns in Pyramid of Giza and the pyramids around it follow Golden Ratio.

 

I was intrigued. I read multiple researches, writings and books to know more about it. I wanted to know to what extent they follow the golden ratio. This question of mine could not be found in any writings and I was very much interested in finding the answer.

 

This is the reason I am doing this IA on the same.

The main motive of this IA is to explore the evidences of Golden Ratio in construction of the Great Pyramid of Giza. In course of that exploration, the concept of Golden Ratio and study about the existence of Golden Ratio is very important to find. Thus, derivation of Golden Ratio is another objective in this IA.

To what extent does the Golden Ratio is followed in the Construction of The Great Pyramid of Giza?

The Great Pyramid of Giza was assumed to be constructed in 2580 – 2560 BC in Cairo, Egypt. Being one of the seven wonders of the world, the great pyramid of Giza is not only one of the most attractive tourist destinations but also a place for tremendous archaeological excavation. Apart from this, the construction of the great pyramid is itself a mystery. Construction at such high scale with such precision needs efficient machineries which were not present at the mentioned period of time. Surprisingly, the golden ratio is observed in several aspects of the pyramid.

 

The dimensions of the pyramid are as follows: Height: 146.7 m, Length of Base: 440 m and the angle of the slope or elevation is 51.85°.

 

Furthermore, the Golden Ratio is observed in several objects in this nature, such as, facial apertures, music, Fibonacci series etc. The Fibonacci series can be used to derive golden ratio expression and the value. The ratio between two successive terms of Fibonacci series is the golden ratio. Fibonacci Series can be expressed using matrix:

 

\(\bigg(\frac{A_n}{A_{n-1}}\bigg)=(1\ 1\ 1\ 0)\bigg(\frac{A_{n-1}}{A_{n-2}}\bigg).........(1)\)

 

where, A_n is the present term of the sequence A_n-1 is the previous term.

 

The generalized form of Eigen Vector and Eigen value is also required in this IA for the derivation of Golden Ratio. The required equation is shown below:

 

\(det\ det (A - λI) = 0……………(2)\)

 

where, A is the matrix, λ is the eigen value and I is the identity matrix.

 

Moreover, eigen vector with its eigen value is also defined as the following equation:

 

[A- λI]v = 0………(3)

I will be deriving the golden ratio using equation (1) and (2) in this section of the IA.

 

\(\bigg(\frac{A_n}{A_{n-1}}\bigg)=(1\ 1\ 1\ 0)\bigg(\frac{A_{n-1}}{A_{n-2}}\bigg)\)

 

For simplicity, let us consider: (1 1 1 0) = P

 

\(∴\bigg(\frac{A_n}{A_{n-1}}\bigg)=P\bigg(\frac{A_{n-1}}{A_{n-2}}\bigg)\)

 

\(=>\bigg(\frac{A_n}{A_{n-1}}\bigg)=P\bigg\{\bigg(\frac{A_{n-2}}{A_{n-3}}\bigg)\bigg\}=P^2\bigg(\frac{A_{n-2}}{A_{n-3}}\bigg)\)

 

\(=>\bigg(\frac{A_n}{A_{n-1}}\bigg)=P^{n-1}\bigg(\frac{A_1}{A_0}\bigg)=P^{n-1}\bigg(\frac{1}{0}\bigg)\)

 

So, we can see that by substituting the values of the A for different values of n with the principle equation (4), we can derive a relationship of n^th term with the first two terms.

 

In order to diagonalize the value of P rather than (P^n-1), let us consider two non-zero matrices N and D such that,

 

P . N = N . D

 

=> D = N . K . N^-1

 

=> D = N^-1. P . N……………………(3)

 

Now, we will find the value of K using the characteristic equation of Eigen Vectors and Eigen Values.

 

det det (B - λI) = 0

 

Here, B = X = (1 1 1 0) and I = (1 0 0 1)

 

∴ det det {1 1 1 0) - λ(1 0 0 1)} = 0

 

=> det det (1 - λ 1 1  -λ) = 0

 

=> (1 - λ )(- λ) - (1 × 1) = 0

 

=> λ² - λ - 1 = 0………………………(4)

 

Using Sreedhar Acharya’s Formula,

 

λ₁ = \(\frac{1+\sqrt{5}}{2}\)

 

λ₂ = \(\frac{1-\sqrt{5}}{2}\)

 

From equation (3):

 

Case 1: Let, λ = λ₁

 

[P - λ₁I]v₁ = 0

 

\(=> \bigg(1-λ_111-λ_1\bigg)\bigg(\frac{x}{y}\bigg)=0\)

 

\(=>\bigg(\frac{x-xλ_1+y}{x-yλ_1}\bigg)=0 \)

 

From the above equation, we can write,

 

x - yλ₁ = 0

 

=> x = yλ₁

 

Let, y = 1 then the value of x = λ₁.

 

\(v_1=\bigg(\frac{λ_1}{1}\bigg)\)

 

\([P-λ_2I]v_2=0\)

 

\(=>\big(1- λ_21\ 1- λ_2\big)\big(\frac{x}{y}\big)=0\)

 

\(=>\bigg(\frac{x-x λ_2+y}{x-y λ_2}\bigg)=0\)

 

From the above equation, we can write,

 

x - yλ₂ = 0

 

=> x = yλ₂

 

Let, y = 1 then the value of x = λ₂.

 

\(v_2=\bigg(\frac{λ_2}{1}\bigg)\)

 

Therefore, we can conclude that,

 

\(N = \big(v_1,v_2\big)= \big(λ_1λ_21\ 1\big)\)

 

\(N^{-1}=\frac{1}{det\ det(λ_1λ_21\ 1)}\)\(adj(λ_1λ_2\ 1\ 1)\)

 

\(=> N^{-1}=\frac{1}{λ_1-λ_2}\)\((1-λ_2 -1\ λ_1)\)

 

\(=> N^{-1}=\frac{1}{\frac{1\ +\ \sqrt5}2\ -\ \frac{1\ -\ \sqrt5}2}(1\ -\ λ_2\ -\ 1\ λ_1)\)

 

\(=>N^{-1}=\frac{1}{\sqrt{5}}(1 -λ_2 -1λ_1)\)

 

\(D=N^{-1}.P.N\)

 

\(=>D=\frac{1}{\sqrt{5}}\big(1-λ _2-1λ _1\big).(1\ 1\ 1\ 0).\big(λ _1λ _2\ 1\ 1\big)\)

 

\(=>D=\frac{1}{\sqrt{5}}\big(1-λ _2-1λ _1\big).\big(λ_1+1λ_2+1λ_1λ _2\big)\)

 

\(=>D=\frac{1}{\sqrt{5}}\big(λ_1+1-λ_1λ_2-λ^2_2\)

 

\(+λ _2+1λ^2_1-λ _2-1-λ _2-1+λ _1λ_2\big)\)

 

\(=>D=\frac{1}{\sqrt{5}}\bigg(λ_1+1-\frac{1+\sqrt{5}}{5}.\frac{1-\sqrt{5}}{5}0\ 0-λ _2-1\)

 

\(+\frac{1+\sqrt{5}}{5}.\frac{1-\sqrt{5}}{5}\bigg)\)

 

Since, λ²- λ - 1 = 0

 

\(=>D=\frac{1}{\sqrt{5}}\big(λ_1+1+100-λ _2-1-1\big)\)

 

\(=>D=\frac{1}{\sqrt{5}}\bigg(\frac{1+\sqrt{5}}{2}+200-\frac{1-\sqrt{5}}{2}-2\bigg)\)

 

\(=>D=\frac{\sqrt{5}}{5}\bigg(\frac{5+\sqrt{5}}{2}0\ 0\frac{-5+\sqrt{5}}{2}\bigg)\)

 

\(=>D=\bigg(\frac{5\sqrt{5}+5}{10}0\ 0\frac{-5\sqrt{5}+5}{10}\bigg)\)

 

\(=>D=\bigg(\frac{\sqrt{5}+1}{2}0\ 0\frac{-\sqrt{5}+1}{2}\bigg)\)

 

=> D = (λ₁ 0 0 λ₂)

 

Now, we know that,

 

P = N . D . N^-1

 

=> P^n-1 = (N . D . N^-1)^n-1

 

=> P^n-1= N . D . N^-1N . D . N^-1N . D . N^-1N . D . N^-1

 

…………(upto n-1 times)N.D.N^-1

 

=> P^n-1 = N . D . D . D . D…………(upto n-1 times)D . N^-1

 

=> P^n-1 = N . D^n-1. N^-1

 

\(∴D^{n-1}=\bigg(λ^{n-1}_1\ 0\ 0λ^{n-1}_2\bigg)\)

 

Now, initially we have represented n^th term in terms of first two terms as:

 

\(\bigg(\frac{A_n}{A_n-1}\bigg)=P^{n-1}\bigg(\frac{A_1}{A_0}\bigg)=P^{n-1}\bigg(\frac{1}{0}\bigg)\)

 

\(\bigg(\frac{A_n}{A_n-1}\bigg)=N.D^{n-1}.N^{-1}\bigg(\frac{1}{0}\bigg)\)

 

\(=>\bigg(\frac{A_n}{A_n-1}\bigg)=\big(λ _1λ _2\ 1\ 1\big).\bigg(λ ^{n-1}_{1}\ 0\ 0λ ^{n-1}_{2}\bigg)\)

 

\(\cdot\frac{1}{\sqrt{5}}\big(1-λ _2-1λ _1\big)\bigg(\frac{1}{0}\bigg)\)

 

\(=>\bigg(\frac{A_n}{A_n-1}\bigg)=\frac{1}{\sqrt{5}}\bigg(\frac{λ^n_1-λ ^n_2}{λ ^{n-1}_1-λ ^{n-1}_2}\bigg)\)

 

\(∴A_n=\frac{1}{\sqrt{5}}\big(λ^n_1-λ ^n_2\big)\)

 

\(=>A_n=\frac{1}{\sqrt{5}}\big(\frac{1+\sqrt{5}}{2}\big)^n-(\frac{1-\sqrt{5}}{2}\big)^n\)

 

\(∴A_{n-1}=\frac{1}{\sqrt{5}}\big(λ^{n-1}_1-λ^{n-1}_2\big)\)

 

\(=>A_{n-1}=\frac{1}{\sqrt{5}}\big(\frac{1+\sqrt{5}}{2}\big)^{n-1}-(\frac{1-\sqrt{5}}{2}\big)^{n-1}\)

 

Lastly, the limiting ratio of two consecutive terms are given below:

 

\(\frac{A_n}{A_{n-1}}\)

 

\(\frac{\frac{1}{\sqrt{5}},(λ^n_1-λ^n_2)}{\frac{1}{\sqrt{5}},(λ^{n-1}_1-λ^{n-1}_2)}\)

 

\(\frac{\lambda_1^n\left(1-\frac{\lambda_2^n}{\lambda_1^n}\right)}{\lambda_1^{n-1}\left(\frac{\lambda_2^{n-1}}{\lambda_1^{n-1}}\right)}\)

 

\(\frac{\lambda_1 \left(1-\frac{\lambda_2^n}{\lambda_1^n}\right)}{\left(1-\frac{\lambda_2^{n-1}}{\lambda_1^{n-1}}\right)}\)

 

\(\frac{\left(1\frac{\lambda_2^n}{\lambda_1^n}\right)}{\left(1-\frac{\lambda_2^{n-1}}{\lambda_1^{n-1}}\right)}\)

 

\(=\lambda_1=\frac{1+\sqrt{5}}{2}=1.61803399\)

 

Thus, we can state that the ratio between any two term of a Fibonacci Sequence is 1.618. This is also known as Golden Ratio.

Phi (φ) or Golden Ratio is the only number in this number system which follows a definite relationship with the Pythagoras Theorem. The relationship of phi states that, if phi is added with 1, the resultant value will be square of phi.

 

φ + 1 = 2…………(5)

 

If we construct a pyramid considering it follows the golden ratio, co-relating with equation (11), we can write, the base of the triangle will be 2 units because, two triangles are joined back to back to form a pyramid, the height of the triangle will be  units and the slant height will be  units. Such a triangle, will have a height of 1.272 units. In such case, the ratio of height to base will be:

 

Ratio of height to base = \(\frac{\sqrt{φ}}{2}=\frac{1.272}{2}\)= 0.636

 

The Height of the great pyramid of Giza is 146.5 metres and the base of the great pyramid of Giza is 230.4 metres. In this case, the ratio of height to base is:

 

Ratio of height to base = \(\frac{146.5}{230.4}\) = 0.635

 

Percentage error = \(\frac{0.636-0.635}{0.636}\times\)100 = 0.157 %

 

Furthermore, from the writings of Herodotus, it was proposed that, the area of each face of the pyramid was equal to the area of a square produced by the height of the pyramid.

Considering the image with notations as, r be the distance between the centre of the pyramid and a side of the pyramid with square base which is equal to the half of length of side of the base a. h be the height of the pyramid, s be the slant height of the pyramid, and e be the length of edge of the pyramid.

 

Following the writings of Herodotus, we can formulate:

 

\(\frac{1}{2}\times a\times s=h^2\)

 

\(=>\frac{1}{2}\times(2r)\times s=h^2\)

 

\(=>r×s= h^2…………(6)\)

 

From the Pythagoras Theorem, we can state that,

 

\(r^2+h^2=s^2\)

 

\(=>h^2=s^2-1\)

 

From (6), we can write,

 

\(=>r×s=s^2-r^2\)

 

For r =1

 

\(=>s=s^2-1\)

 

\(=>s^2-s-1=0…………(7)\)

 

Equation (13) is the exact same equation of the golden ratio, equation (4) thus, we can state that, the construction of The Great Pyramid of Giza also followed the Golden Ratio.

In this IA, we have explored the value of Golden Ratio using Fibonacci Sequence. In the course of the derivation, we have come across a characteristic equation whose root is the value of golden Ratio. There are several theories based on the construction of The Great Pyramid of Giza. We have taken references of one hypothetical proposal that was found in the writings of Herodotus and one quantitative measure to prove that the construction of the pyramid obeys the golden ratio. In reference with the quantitative data, the ratio of height to base of the great pyramid of Giza has an error percentage of 0.157% as compared to that of the pyramid that is considered to be following the Golden ratio. The height to base ratio of a pyramid following golden ratio is 0.636 whereas the height to base ratio of the Great Pyramid of Giza is 0.635. On the other hand, following the findings of Herodotus, we can conclude that, if the radius of the pyramid (half of the side of base of pyramid) is 1, it forms the characteristic equation of the golden ratio that was initially found in the process of derivation of the golden ratio. So, in a nutshell, the construction of the great pyramid of Giza does follow the golden ratio.

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• Dunlap, Richard A. The golden ratio and Fibonacci numbers. World Scientific, 1997.
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• https://www.goldennumber.net/phi-pi-great-pyramid-egypt/#:~:text=According%20to%20Wikipedia%2C%20the%20Great,significant%20decimal%20places%20of%20accuracy.
• Fischler, Roger. "What did Herodotus really say? or how to build (a theory of) the Great Pyramid." Environment and Planning B: Planning and Design 6.1 (1979): 89-93.