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Topic: Silver Ratio (Read 823 times) |
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Noke Lieu
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Silver Ratio
« on: Sep 15th, 2005, 10:59pm » |
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Preparing teacher notes, extensions etc. and stumbled across the "silver ratio"
It's 1+21/2 to 1
Looking for an elegant way of showing that ratio with geometry. So far, have reduced it to using two circles, one square. Can you better that?
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| « Last Edit: Sep 15th, 2005, 11:06pm by Noke Lieu » |
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Sjoerd Job Postmus
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Re: Silver Ratio
« Reply #1 on: Sep 16th, 2005, 2:03am » |
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Hrm... I could plot it, though...
Get a square...
[ ]
Now, the sides are 1, the diagonals are sqrt(2)
If we now turn the diagonal with as center of rotation being one of the corners of the square, in such a way that it is in a straight line with a corner, we have a line being 1 + sqrt(2). Now, complete the square... ( one circle )
2 squares tghough
Hope that does any good
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Barukh
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Re: Silver Ratio
« Reply #2 on: Sep 16th, 2005, 7:50am » |
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Noke, what exactly do you want to show? Is it OK to have 2 segmens on a line, their lengths in silver ratio?
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SWF
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Re: Silver Ratio
« Reply #3 on: Sep 17th, 2005, 2:54pm » |
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Tessellate the plane with unit squares and octagons of side 1. It is then easy to find a distance between two of the vertices that equals to the silver ratio.
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Noke Lieu
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on Sep 16th, 2005, 7:50am, Barukh wrote:| Noke, what exactly do you want to show? Is it OK to have 2 segmens on a line, their lengths in silver ratio? |
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I reckon that's better... Well, here's my shot.
Just trying to get elegant versions like this.
There is a series of "silver means" such that
0.5(n+(n2 +4)0.5)
when n=1, you get 1.618 (golden ratio)... n=2 you get this. n=3 gets 3.303 and so on.
It's easy to make contrived situations to make these ratios, but elegants ones, that's the trick.
(of course removing one of the circles from this would yeild an elegant solution, but then doesn't have the 1:X disection, just a line of X)
Am making a short series of extension-ideas for teachers. This is just one of the ideas.
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| « Last Edit: Sep 18th, 2005, 11:04pm by Noke Lieu » |
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Barukh
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How about this construction? Point E divides the segment BD in a silver ratio.
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Neelesh
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Re: Silver Ratio
« Reply #6 on: Sep 19th, 2005, 2:08am » |
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on Sep 18th, 2005, 11:38pm, Barukh wrote:| How about this construction? Point E divides the segment BD in a silver ratio. |
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DE/EB = sqrt(2) - 1 if the arc represents portion of circle inscribed in a unit square. Thats not 1+sqrt(2)
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| « Last Edit: Sep 19th, 2005, 2:11am by Neelesh » |
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Barukh
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Re: Silver Ratio
« Reply #7 on: Sep 19th, 2005, 3:32am » |
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on Sep 19th, 2005, 2:08am, Neelesh wrote:| DE/EB = sqrt(2) - 1 if the arc represents portion of circle inscribed in a unit square. Thats not 1+sqrt(2) |
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But EB/DE = 1/([sqrt]2 - 1) = [sqrt]2 + 1.
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Neelesh
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Re: Silver Ratio
« Reply #8 on: Sep 19th, 2005, 4:38am » |
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on Sep 19th, 2005, 3:32am, Barukh wrote:
But EB/DE = 1/([sqrt]2 - 1) = [sqrt]2 + 1.
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Ooops. I did think of that but did some miscalculations 
Sorry, and Thanks to Barukh
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Noke Lieu
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Re: Silver Ratio
« Reply #9 on: Sep 19th, 2005, 4:42pm » |
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*oops*
That's what the "remove one circle" comment was about, but I just looked at the diameter of that arc
Very nice.
so what about (3+[sqrt]13)/2? Haven't got to anything really elegant. been close, using concentric circles but not quite there.
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Barukh
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In the following construction, AB/AC = 3:1. This yields DC/AC = (3+[sqrt]13)/2.
I will let you try to prove it. 
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| « Last Edit: Sep 20th, 2005, 11:32am by Barukh » |
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Sjoerd Job Postmus
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Re: Silver Ratio
« Reply #11 on: Sep 20th, 2005, 12:09pm » |
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Take radius = 6
AC = 2
AO = 3
CO = sqrt(13)
DC = sqrt(13) + 3
DC/AC = (sqrt(13) + 3) / 2
d'uh 
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Sjoerd Job Postmus
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Re: Silver Ratio
« Reply #12 on: Sep 20th, 2005, 12:14pm » |
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Now, I was wondering, what if AC : AB = 1 : 2
Radius = 4
AC = 2
AO = 2
CO = sqrt( = 2sqrt(2)
DC = CO + OD = 2 + 2sqrt(2)
DC / AC = 1 + sqrt(2) = silver ratio
Now, what if it was 1 to 1?
Radius = 2
AC = 2
AO = 1
CO = sqrt(5)
DC = 1 + sqrt(5)
DC / AC = (1 + sqrt(5)) / 2 = golden ratio
Heh, cool... didn't know this before 
----EDIT----
Maybe I can generalyse this
Ratio height : diameter = 1 : n
Ratio height : radius = 2 : n
AC = 2
AO = n
CO = sqrt(2^2 + n^2)
CO = sqrt(n^2 + 4)
CD = n + sqrt(n^2 + 4)
CD : AC = (n + sqrt(n^2 + 4)) / 2
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| « Last Edit: Sep 20th, 2005, 12:39pm by Sjoerd Job Postmus » |
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Noke Lieu
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Re: Silver Ratio
« Reply #13 on: Sep 20th, 2005, 7:23pm » |
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on Sep 20th, 2005, 11:31am, Barukh wrote:| In the following construction, AB/AC = 3:1. |
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Undoubtedly beautiful. How'd you accertain that 3:1? Is that keeping it elegant?
I think I prefer the line with the section idea.
Otherwise, you just draw a right angle triangle, and section off a unit somewhere. Sure you could throw a circle around C radius AC, but that's not so elegant anymore.
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