Phi and Fibonacci in Kepler and Golden Triangles

Creating a Triangle based on Phi (or Pythagoras meets Fibonacci):

Pythagoras discovered that a right triangle with sides of length a and b and a hypotenuse of length c has the following relationship:

a² + b² = c²

A foundational equality of phi has a similar structure:

1 + Phi = Phi2

( 1+ 1.618… = 2.618… )

By taking the square root of each term in this equality, we have the dimensions of a triangle, known as a Kepler Triangle, a right triangle based on this phi equality, where:

SideLength squared
per above
Length,
or square root
Length divided
by phi so c = 1
a111 / Phi
bPhi√ Phi1 / √ Phi
cPhi2Phi1

 

This triangle is illustrated below.  It has an angle of 51.83° (or 51°50′), which has a cosine of 0.618 or phi.

Golden Triangle based on Phi (1.618 0339 ...), or Golden Ratio relationships

Although difficult to prove due to deterioration through the ages, this angle is believed by some to have been used by the Egyptians in the construction of the Great Pyramid of Cheops.

Other triangles with Golden Ratio proportions can be created with a Phi (1.618 0339 …) to 1 relationship of the base and sides of triangles:

Golden Triangles based on the Golden Ratio with Phi (1.618 0339 ...) to 1 relationships

The isosceles triangle above on the right with a base of 1 two equal sides of Phi is known as a Golden Triangle.  These familiar triangles are found embodied in pentagrams and Penrose tiles.

 

Pentragram showing golden triangles based on phi (1.618 0339 ...), the golden ratio

Penrose tiles showing golden triangles based on phi (1.618 0339 ...), the golden ratio

Creating a Triangle based on Fibonacci numbers

No three successive numbers in the Fibonacci series can be used to create a right triangle.  Marty Stange, however, contributed the following relationship in January 2007:  Every successive series of four Fibonacci numbers can be used to create a right triangle, with the base and hypotenuse being determined by the second and third numbers, and the other side being the square root of the product of the first and fourth numbers.  The table below shows how this relationship works:

 

Fibonacci Series

The Fibonacci Triangle

b’

a

c

b”

b’xb”

a² + b’xb”

= c²

0

1

1

2

1

0

1

1

1

2

3

1

3

4

1

2

3

5

4

5

9

2

3

5

8

9

16

25

3

5

8

13

25

39

64

5

8

13

21

64

105

169

8

13

21

34

169

272

441

13

21

34

55

441

715

1,156

21

34

55

89

1,156

1,869

3,025

34

55

89

144

3,025

4,896

7,921

55

89

144

233

7,921

12,815

20,736

89

144

233

377

20,736

33,553

54,289

144

233

377

610

54,289

87,840

142,129

Fibonacci triangles based on relationship from Marty Stange

 

Thus for the illustration highlighted in gold, Stange’s Treatise on Fibonacci Triangles reveals that a triangle with sides of 5 and the square root of 39 (e.g., 3 x 13) will produce a right triangle with a hypotenuse of 8.

As greater numbers in the series are used, the triangle approaches the proportions of the phi-based Kepler Triangle above, with a ratio of the hypotenuse to the base of Phi, or 1.618…

Comments

  1. Ken Landin says

    What a brilliant analysis by Stange. I loved your beautifully done illustrations, and how clear your descriptions were.
    Anyone interested in phi should look up how using the Fibonacci numbers can create a series of rectangles that approach the phi ratio.

  2. says

    CORRECT REPETITION
    ————————
    A very interesting topic.

    My decoding Plato’s Timaeus “MOST BEAUTIFUL TRIANGLE” shows that Kepler / Magirus Triangle is a similar triangle, “not the same” and” not as beautiful” , but constituent to that of Plato’s:
    http://www.stefanides.gr/Html/quadrature.htm
    This orthogonal scalene triangle has all its sides in ratio T and scalene angle ArcTan [ T ] , T=SQRT[Phi].
    Its hypotenuse is T^3, its bigger side is T^2 and its smaller is T^1.

    { http://www.stefanides.gr/Html/gmr.htm
    Shows the ruller and compass structure of T [ = sqrt(Phi) ] and Θ [ = ArcTan(T) ] ,
    via known construction of Phi [ = T^2].
    It is based on a fourth order equation : T^4 – T^2 -1 =0 }
    For quadrature of circle, [ as proved imposibility by transcendentals ] is obtained , however, by the irrational value of 4/[SQRT(Phi)] , following the equation : [ Πσχ ]^4 + { [ Πσχ ]^2}*4^2 – 4^4 =0
    Ref: http://www.stefanides.gr/Html/piquad.htm
    And an AutoCad drawing:
    http://www.stefanides.gr/pdf/D=5,083FOUR_1.pdf

    © Copyright 1999 P. C. Stefanides.

    Regards,

    Panagiotis Stefanides

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