Waste less time on Facebook — follow Brilliant.
×

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
3 years, 5 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

Click here for the image

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 \( \phi \approx 1.618\) Lim Zi Heng · 3 years, 5 months ago

Log in to reply

@Lim Zi Heng I attach the image already Lim Zi Heng · 3 years, 4 months ago

Log in to reply

@Lim Zi Heng I will attach an image after I know how to attach an image Lim Zi Heng · 3 years, 5 months ago

Log in to reply

@Lim Zi Heng Is there another way? Bob Krueger · 3 years, 5 months ago

Log in to reply

@Bob Krueger 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 \[a = \phi \]. Harman Kahlon · 3 years, 4 months ago

Log in to reply

@Harman Kahlon Right. This is the way I first discovered it. Bob Krueger · 3 years, 4 months ago

Log in to reply

@Bob Krueger this is the way I think Lim Zi Heng · 3 years, 5 months ago

Log in to reply

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 Ayush Chowdhury · 3 years, 4 months ago

Log in to reply

@Ayush Chowdhury Very interesting indeed. Bob Krueger · 3 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...