Re: CORRECTION Re: Tiling square with 3 proportional rectangles

Ed Murphy wrote:
> On Sun, 03 Jul 2005 06:19:29 +0000, Peter Renzland wrote:
>
> > In <pan.2005.07.02.21.01.54.251685@xxxxxxxxxxxx> On 2005-07-02 Ed Murphy
> > <emurphy42@xxxxxxxxxxxx> wrote:
> >
> >> On Sat, 02 Jul 2005 07:36:26 +0000, Peter Renzland wrote:
> >>
> >>> BTW, the answer about 5 rectangles (configured as per your
> >>> illustration) should be evident from looking at your two illustrations
> >>> with calculation.
> >>
> >> Would anyone like to run the actual calculations?
> >
> > That "with" was supposed to be "without". (I recommend not wasting any
> > time with calulations -- just look at the nice diagrams and think.)
> >
> >>> The ratio above is the square of the Plastic Number p=1.324717957244746
> >>
> >> Google found this (with some difficulty; most of the top links are
> >> selling various plastic objects with numbers on them)
> >>
> >> http://members.fortunecity.com/templarser/padovan.html
> >>
> >> but is there a simple relationship between the 3-rectangle arrangement
> >> and the triangle arrangement that generates p?
> >
> > I'm not totally sure what you mean. The triangle arrangement generates p
> > by approximation. Just as Phi is the limit ratio of successive Fibonacci
> > terms, f[n+1] = f[n-1] + f[n], p is the ratio of the other morphic
> > sequence: p[n+1] = p[n-2] + p[n-1]
> >
> > The 3 rectangles in a square are a constructive way to produce p(^2).
>
> "Produces the same number" is not a simple relationship, any more than
> e^(i*pi) and 4-5. There are a number of pieces here, but I don't grok
> why they're all related like this:
>
> * phi^2 - phi - 1 = 0 * p^3 - p - 1 = 0
>
> * Ratio of consecutive Fibonacci * Ratio of consecutive Padovan
> numbers tends to phi numbers tends to p
>
> * Spiral of squares leads to phi * Spiral of triangles leads to p
>
> * Spiral of cubes leads to p
>
> * Square + rectangle = similar * Three similar rectangles =
> rectangle leads to phi square leads to p^2
>
> > If you look at your 5-rectangle diagram, you will see 2 large rectangles
> > and a square that make up the large square.
>
> Those are intended to be 2 large rectangles and a square. Building it
> around a diagram with the vertices labelled the same as the 3-rectangle
> diagram (though I adjusted the lengths) was misleading. Sorry about
> that, Chief.
>
> > 2 similar rectangles and a square are dissimilar from 3 similar
> > rectangles. QED.
> >
> > Not only that, but 2 similar rectangles and a square implies a ratio of
> > Phi.
>
> Invalidated by the above. I think. Anyway, let's see:
>
> A-B---C
> | | |
> | D-E-F
> | | | |
> | | G-H
> | | | |
> J-K-L-M
>
> (not bothering to attempt a proportional drawing)
>
> AJ BC DK EF GL
> -- = -- = -- = -- = -- = n
> AB BD DE EG GH
>
> AJ = AB + BC = AB + DE + GH
>
> Without loss of generality, let GH = x = n^4. This avoids negative
> exponents of n.
>
> GH = n^4
> GL = n^5
> EF = n^4
> EG = n^3
> DE = n^4 + n^2
> DK = n^5 + n^3
> BD = 2n^3 + n
> BC = 2n^4 + n^2
> AB = n^4 + 3n^2 + 1
> AJ = n^5 + 3n^3 + n
>
> n^5 + 3n^3 + n = (n^4 + 3n^2 + 1) + (n^4 + n^2) + n^4
> n^5 + 3n^3 + n = 3n^4 + 4n^2 + 1
>
> n^5 - 3n^4 + 3n^3 - 4n^2 + n - 1 = 0
>
> A rough plot over n = 1 to 2 by tenths shows no roots, but I have to get
> offline now - can someone proofread this, please?

Looks correct; n = 2.40220883386...

+---------------+-----------------------+
| 1698 | 2382 |
| | |
| |991 |
| | |
| | |
| +------------+----------+
| | | |
| | |456 |
| | +----------+
| | | 1096 |
|4080 | | |
| | | |
| | | |
| |3089 | |
| | | |
| | |2633 |
| | | |
| | | |
| | | |
| | | |
| | | |
| | 1286 | |
+---------------+------------+----------+



Please reply to drgmayer at hotmail dot com

__/\__
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\ Toronto, Canada
/__ __\
||

.

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