# The Golden Ratio: Beginning the Geometry of Here is the previous post concerning the Golden Ratio. For a collection of all the posts concerning the Golden Ratio, click #GoldenRatio below.

Now that we know what the golden ratio is, and why people like it, let's look at it in a geometrical sense, as the greeks would have. One definition of the golden ratio is the one that started this series, based on the golden rectangle. Another definition is as follows:

Divide a line segment into two parts. The Golden Ratio is the ratio of the larger part to the whole, when that ratio is also equal to the ratio of the smaller part to the larger part.

Basically, imagine the line segment pictured above. When we set the two proportions equal to each other we get

$\frac{a+b}{a} = \frac{a}{b} = \phi$

Solving this like we did in the first post will indeed yield $\frac{1+\sqrt{5}}{2}$.

Since this post is kind of short too, I'll add in a teaser. If someone can effectively answer my question before this evening (in the next ten hours), I'll put up the tomorrow's post tonight.

Here's the question: What are the lengths of the diagonals of a regular unit pentagon? Prove it.

See the full proof in here Note by Bob Krueger
6 years, 1 month ago

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ABCDE is a regular pentagon with side length 1

AD,AC & EB is the diagonals of the pentagon

AD intersect EB at point F,AC intersect EB at point G

AF=EF=y, so $\bigtriangleup AEF$ is a isosceles triangle

$\bigtriangleup DEF$ is a isoceles triangle, so $ED=FD=1$

So,the diagonals$= 1+y$

$\angle FAE=\angle AEF=\angle EDF=36^\circ$

$\angle AED=\angle AFE=108^\circ$

$\bigtriangleup AFE \sim \bigtriangleup ADE$ So that, $\frac{1+y}{1}=\frac{1}{y}$

$\frac{a+b}{a}=\frac{a}{b}=\phi$

By substituting $a=1, b=y$

We get, $\frac{1+y}{1}=\frac{1}{y}=\phi$

The length of the diagonals is $\phi \approx 1.618$

- 6 years, 1 month ago

Is there another way?

- 6 years, 1 month ago

An alternate way is to apply cosine rule in one of the triangles.

If a is the length of the diagonal, use cosine rule to get $a^2 = 2.618$approximately equal to $1 + \phi$ and from here recall that $1 + \phi = \phi^{2}$it is easy to get $a = \phi$.

- 6 years ago

Right. This is the way I first discovered it.

- 6 years ago

this is the way I think

- 6 years, 1 month ago

I will attach an image after I know how to attach an image

- 6 years, 1 month ago

- 6 years ago

it is very interesting to know that the equivalent resistance of thiscircuit where all the resistances are 1 ohm is GOLDEN RATIO. I think this is because of the existence of the ratio in the radical form

- 6 years ago

Very interesting indeed.

- 6 years ago