Tag: golden ratio

Posted on

The Metallic Ratios: Beyond the Golden Ratio

Many people are familiar with the Golden Ratio (φ), but it is just one member of a broader family known as the Metallic Ratios. These ratios describe a recursive relationship between the sides of a rectangle. Given a rectangle with side lengths A and B (B > A), the Metallic Ratios satisfy the equation:

A / B = (B – n × A) / A

where n is a fixed integer.

For the Golden Ratio (n = 1):
A / B = (B – A) / A
which leads to:
φ² – φ – 1 = 0
solving this gives:
φ = (1 + √5) / 2 ≈ 1.618…

For the Silver Ratio (n = 2):
A / B = (B – 2A) / A
which leads to:
ψ² – 2ψ – 1 = 0
solving this gives:
ψ = 1 + √2 ≈ 2.414…

Different values of n define other Metallic Ratios. The table below presents some of them.

Curious minds, don’t stop here—explore the fascinating properties of the other Metallic Ratios and see what patterns and surprises you can uncover!

Posted on

Discover the Golden Ellipse

A golden ellipse is one where the axes are in golden proportion, meaning the ratio of the major axis (a) to the minor axis (b) is the golden ratio:
φ = (1 + √5)/2.
To visualize this, draw a golden ellipse along with its inscribed and circumscribed circles: the smallest circle fitting inside the ellipse and the largest circle surrounding it.
Interestingly, the area of the ellipse matches the area of the “annulus” formed between these two circles!
Here’s how it works:
Let a be the semi-major axis and b the semi-minor axis, with a = φb.
The area of the annulus is:
π(a² − b²) = πb²(φ² − 1)
The area of the ellipse is:
πab = πφb²
And as φ² − 1 = φ, then πb²(φ² − 1) = πφb².


Isn’t it fascinating how geometry intertwines with the golden ratio?

Posted on

The Kepler Triangle, Phi and Pi

A “Kepler triangle” is a right triangle having edge lengths in a geometric progression, in which the common ratio is √ϕ, where ϕ represents the golden ratio.
Well, let’s construct a square with side length √ϕ that inscribes a Kepler triangle, that is, a right triangle with edges 1 : √ϕ : ϕ (or approximately 1 : 1.272 : 1.618), as shown in the picture. Draw then the circumcircle of the Kepler triangle (highlighted in orange in the picture) whose diameter is the hypotenuse of the triangle.
Then, the perimeters of the square (4√ϕ≈5.0884) and the circle (πϕ≈5.083) coincide up to an error less than 0.1%. From this, we can get the approximation coincidence π≈4/√ϕ

Kepler triangle, phi and pi