Golden Ratio?

Toast · 2006-11-26 19:25:20

So I was bored and decided to stuff around in paint (yes i know, I was pretty bored), I started with a circle, added a midpoint, then made all kinds of links and connections to come up with what is attached.
Then I noticed that the rectangles sorta looked familiar. Are they the golden ratio???

Last edited by Toast (2006-11-27 02:25:16)

mathsyperson · 2006-11-26 22:53:44

Ooh, very interesting. Let's see if we can work out the lengths of that highlighted rectangle.

If we say that the circle has a radius of 1, then its equation would be x² + y² = 1.

The upper-left corner of the rectangle lies on the circle and on a line drawn from the centre at 45° to the horizontal. From standard trigonometric triangles, this means that the co-ordinates of that point are (-1/√2, 1/√2), which in turn means that the length of the rectangle is 1/√2.

The lower-left corner has the same x co-ordinate as the upper-left corner, and it also lies on the line y = 1+x. We know that at that point, x = -1/√2, so that means that y = 1 - 1/√2.

So the difference in y-values of the upper-left and lower-left corners of the rectangle is given by (1/√2) - (1 - 1/√2) = √2 - 1.

We have now worked out that the highlighted rectangle has dimensions of 1/√2 and (√2 - 1).

To find the ratio of these, we need to divide one by the other.

(1/√2)/(√2 - 1) = 1/(2-√2) = (2+√2)/2 = 1+1/√2 ≈ 1.70

So, unless I've made a mistake somewhere, that rectangle isn't golden. Just quite close to it.

Very nice diagram though. It's amazing what you can produce with a simple drawing tool and a bit of boredom.

luca-deltodesco · 2006-11-27 03:49:35

but it begs the question... if you keep seperating it into smaller rectangles in the same fashion, will it remain ~1.7, or does it maybe diverge to phi?

Math Is Fun Forum

#1 2006-11-26 19:25:20

Golden Ratio?

#2 2006-11-26 22:53:44

Re: Golden Ratio?

#3 2006-11-27 03:49:35

Re: Golden Ratio?

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