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        riddles >> easy >> Silver Ratio
        
(Message started by: Noke Lieu on Sep 15^th, 2005, 10:59pm)

Title: Silver Ratio
Post by Noke Lieu on Sep 15^th, 2005, 10:59pm

Preparing teacher notes, extensions etc. and stumbled across the "silver ratio"
It's 1+2^1/2 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?

Title: Re: Silver Ratio
Post by Sjoerd Job Postmus on Sep 16^th, 2005, 2:03am

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

Title: Re: Silver Ratio
Post by Barukh on Sep 16^th, 2005, 7:50am

Noke, what exactly do you want to show? Is it OK to have 2 segmens on a line, their lengths in silver ratio?

Title: Re: Silver Ratio
Post by SWF on Sep 17^th, 2005, 2:54pm

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.

Title: Re: Silver Ratio
Post by Noke Lieu on Sep 18^th, 2005, 5:12pm

on 09/16/05 at 07:50:05, 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?

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.

Title: Re: Silver Ratio
Post by Barukh on Sep 18^th, 2005, 11:38pm

How about this construction? Point E divides the segment BD in a silver ratio.

Title: Re: Silver Ratio
Post by Neelesh on Sep 19^th, 2005, 2:08am

on 09/18/05 at 23:38:23, Barukh wrote:

How about this construction? Point E divides the segment BD in a silver ratio.

DE/EB = sqrt(2) - 1 if the arc represents portion of circle inscribed in a unit square. Thats not 1+sqrt(2)

Title: Re: Silver Ratio
Post by Barukh on Sep 19^th, 2005, 3:32am

on 09/19/05 at 02:08:09, Neelesh wrote:

DE/EB = sqrt(2) - 1 if the arc represents portion of circle inscribed in a unit square. Thats not 1+sqrt(2)

But EB/DE = 1/([sqrt]2 - 1) = [sqrt]2 + 1.

Title: Re: Silver Ratio
Post by Neelesh on Sep 19^th, 2005, 4:38am

on 09/19/05 at 03:32:34, Barukh wrote:

But EB/DE = 1/([sqrt]2 - 1) = [sqrt]2 + 1.

Ooops. I did think of that but did some miscalculations :-[

Sorry, and Thanks to Barukh

Title: Re: Silver Ratio
Post by Noke Lieu on Sep 19^th, 2005, 4:42pm

*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.

Title: Re: Silver Ratio
Post by Barukh on Sep 20^th, 2005, 11:31am

In the following construction, AB/AC = 3:1. This yields DC/AC = (3+[sqrt]13)/2.

I will let you try to prove it. :D

Title: Re: Silver Ratio
Post by Sjoerd Job Postmus on Sep 20^th, 2005, 12:09pm

Take radius = 6

AC = 2
AO = 3
CO = sqrt(13)

DC = sqrt(13) + 3
DC/AC = (sqrt(13) + 3) / 2
d'uh ;)

Title: Re: Silver Ratio
Post by Sjoerd Job Postmus on Sep 20^th, 2005, 12:14pm

Now, I was wondering, what if AC : AB = 1 : 2

Radius = 4

AC = 2
AO = 2
CO = sqrt(8) = 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

Title: Re: Silver Ratio
Post by Noke Lieu on Sep 20^th, 2005, 7:23pm

on 09/20/05 at 11:31:17, Barukh wrote:

In the following construction, AB/AC = 3:1.

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.