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tech / sci.math / Re: daily problem: golden rectangles

SubjectAuthor
* daily problem: golden rectanglesRichD
+* Re: daily problem: golden rectanglesChris M. Thomasson
|`* Re: daily problem: golden rectanglesRichD
| +- Re: daily problem: golden rectanglesChris M. Thomasson
| `* Re: daily problem: golden rectanglesChris M. Thomasson
|  `- Re: daily problem: golden rectanglesChris M. Thomasson
`* Re: daily problem: golden rectanglessobriquet
 +- Re: daily problem: golden rectanglesChris M. Thomasson
 +* Re: daily problem: golden rectanglesRichD
 |+- Re: daily problem: golden rectanglesChris M. Thomasson
 |`- Re: daily problem: golden rectanglessobriquet
 +* Re: daily problem: golden rectanglesChris M. Thomasson
 |`* Re: daily problem: golden rectanglessobriquet
 | +- Re: daily problem: golden rectanglesChris M. Thomasson
 | +- Re: daily problem: golden rectanglesChris M. Thomasson
 | `* Re: daily problem: golden rectanglesChris M. Thomasson
 |  `- Re: daily problem: golden rectanglesRichD
 `* Re: daily problem: golden rectanglessobriquet
  `* Re: daily problem: golden rectanglesRichD
   +- Re: daily problem: golden rectanglesChris M. Thomasson
   `* Re: daily problem: golden rectanglessobriquet
    +* Re: daily problem: golden rectanglesFromTheRafters
    |`* Re: daily problem: golden rectanglesRichD
    | `- Re: daily problem: golden rectanglesRichD
    +* Re: daily problem: golden rectanglesMike Terry
    |`* Re: daily problem: golden rectanglesRichD
    | `- Re: daily problem: golden rectanglessobriquet
    `- Re: daily problem: golden rectanglesChris M. Thomasson

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daily problem: golden rectangles

<777436c0-b865-451c-86a7-0b9e875b320bn@googlegroups.com>

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Subject: daily problem: golden rectangles
From: r_delane...@yahoo.com (RichD)
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 by: RichD - Sat, 23 Apr 2022 23:45 UTC

Pull out your graph paper for this one -

We're going to extend the golden ratio, from line
segment to 2-D; a golden rectangle.

Draw a rectangle of height a, width a + b
Call it R[1]. (large, for clarity)
b/a = r ~ .6

To be precise, it must satisfy the golden ratio:
r = 1/(1 + r)

Next, from R[1], chop off a square with sides a,
using the side a. This leaves a new rectangle, R[2],
with sides b and a.
Iterate: from R[2], chop off a square of side b,
leaving R[3], with sides a-b and b.

If you do this right, you construct an endless cw
whirl of golden rectangles (or maybe ccw, depending
how you started).

I'm sure we can count on Chris Thomasson to
code some pretty pictures.

Now, algebra tells us that r is irrational, trivially.
But that's a cheat. Problem: show that r is irrational,
using geometry and logic only, as Euclid would have
(maybe he did, I don't know). i.e. assume r is rational,
derive a contradiction.

It isn't too hard. Use your drawing for inspiration.

--
Rich

Re: daily problem: golden rectangles

<t424mn$jqc$1@dont-email.me>

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From: chris.m....@gmail.com (Chris M. Thomasson)
Newsgroups: sci.math
Subject: Re: daily problem: golden rectangles
Date: Sat, 23 Apr 2022 17:13:08 -0700
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 by: Chris M. Thomasson - Sun, 24 Apr 2022 00:13 UTC

On 4/23/2022 4:45 PM, RichD wrote:
> Pull out your graph paper for this one -
>
> We're going to extend the golden ratio, from line
> segment to 2-D; a golden rectangle.
>
> Draw a rectangle of height a, width a + b
> Call it R[1]. (large, for clarity)
> b/a = r ~ .6
>
> To be precise, it must satisfy the golden ratio:
> r = 1/(1 + r)
>
> Next, from R[1], chop off a square with sides a,
> using the side a. This leaves a new rectangle, R[2],
> with sides b and a.
> Iterate: from R[2], chop off a square of side b,
> leaving R[3], with sides a-b and b.
>
> If you do this right, you construct an endless cw
> whirl of golden rectangles (or maybe ccw, depending
> how you started).
>
> I'm sure we can count on Chris Thomasson to
> code some pretty pictures.
>
> Now, algebra tells us that r is irrational, trivially.
> But that's a cheat. Problem: show that r is irrational,
> using geometry and logic only, as Euclid would have
> (maybe he did, I don't know). i.e. assume r is rational,
> derive a contradiction.
>
> It isn't too hard. Use your drawing for inspiration.

Here is what I coded up so far:
________________________
// Test for RichD over on sci.math...
namespace ct_golden
{

void golden_v0(
ct::plot::cairo::plot_2d& plot,
unsigned long ri,
unsigned long rn,
glm::vec2 p0,
glm::vec2 p1
) {
if (ri >= rn) return;

glm::vec2 dif = p1 - p0;
glm::vec2 perp = { -dif.y, dif.x };

float ratio_0 = 1.618033988749f;
float ratio = 1.f / ratio_0;

glm::vec2 g0 = p0;
glm::vec2 g1 = p1;
glm::vec2 g2 = g1 + perp * ratio;
glm::vec2 g3 = g0 + perp * ratio;

plot.line(g0, g1, CT_RGBF(1, 0, 0), 5.f);
plot.line(g1, g2, CT_RGBF(0, 1, 0), 5.f);
plot.line(g2, g3, CT_RGBF(0, 0, 1), 5.f);
plot.line(g3, g0, CT_RGBF(1, 1, 0), 5.f);

// test
float length_0 = glm::length(dif);
float length_1 = glm::length(g2 - g1);

float length_ratio = length_0 / length_1;

std::cout << "ri[" << ri << "] = " << length_ratio << "\n";

golden_v0(plot, ri + 1, rn, g1, g2);
}

void manifest(ct::plot::cairo::plot_2d& plot)
{
std::cout << "\nct_golden()\n";
std::cout << "__________________________________________\n\n";

golden_v0(plot, 0, 7, { -1, 0 }, { 1, 0 });
}
} ________________________

Here is the output:

https://i.ibb.co/9WcDZC6/ct-golden.png

________________________
Hello World

canvas_width: 1920
canvas_height: 1080

ct_golden()
__________________________________________

ri[0] = 1.6180340051651001
ri[1] = 1.6180340051651001
ri[2] = 1.618033766746521
ri[3] = 1.618033766746521
ri[4] = 1.6180340051651001
ri[5] = 1.6180338859558105
ri[6] = 1.6180342435836792
________________________

this is just a test. Now, did I get what you have in mind? Infinite
golden rectangles from a single line? I can use any line...

https://i.ibb.co/fDdbM74/ct-golden-p1.png

void manifest(ct::plot::cairo::plot_2d& plot)
{ std::cout << "\nct_golden()\n";
std::cout << "__________________________________________\n\n";

golden_v0(plot, 0, 7, { -1, 0 }, { 0, 1 });
golden_v0(plot, 0, 7, { 0, 1 }, { 1, 0 });
}

Re: daily problem: golden rectangles

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Subject: Re: daily problem: golden rectangles
From: dohduh...@yahoo.com (sobriquet)
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 by: sobriquet - Sun, 24 Apr 2022 19:53 UTC

On Sunday, April 24, 2022 at 1:45:54 AM UTC+2, RichD wrote:
> Pull out your graph paper for this one -
>
> We're going to extend the golden ratio, from line
> segment to 2-D; a golden rectangle.
>
> Draw a rectangle of height a, width a + b
> Call it R[1]. (large, for clarity)
> b/a = r ~ .6
>
> To be precise, it must satisfy the golden ratio:
> r = 1/(1 + r)
>
> Next, from R[1], chop off a square with sides a,
> using the side a. This leaves a new rectangle, R[2],
> with sides b and a.
> Iterate: from R[2], chop off a square of side b,
> leaving R[3], with sides a-b and b.
>
> If you do this right, you construct an endless cw
> whirl of golden rectangles (or maybe ccw, depending
> how you started).
>
> I'm sure we can count on Chris Thomasson to
> code some pretty pictures.
>
> Now, algebra tells us that r is irrational, trivially.
> But that's a cheat. Problem: show that r is irrational,
> using geometry and logic only, as Euclid would have
> (maybe he did, I don't know). i.e. assume r is rational,
> derive a contradiction.
>
> It isn't too hard. Use your drawing for inspiration.
>
> --
> Rich

Some initial geometric sketches:
https://www.desmos.com/calculator/lr7o4i0uat

Re: daily problem: golden rectangles

<t44c2b$5uo$1@dont-email.me>

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From: chris.m....@gmail.com (Chris M. Thomasson)
Newsgroups: sci.math
Subject: Re: daily problem: golden rectangles
Date: Sun, 24 Apr 2022 13:31:03 -0700
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 by: Chris M. Thomasson - Sun, 24 Apr 2022 20:31 UTC

On 4/24/2022 12:53 PM, sobriquet wrote:
> On Sunday, April 24, 2022 at 1:45:54 AM UTC+2, RichD wrote:
>> Pull out your graph paper for this one -
>>
>> We're going to extend the golden ratio, from line
>> segment to 2-D; a golden rectangle.
>>
>> Draw a rectangle of height a, width a + b
>> Call it R[1]. (large, for clarity)
>> b/a = r ~ .6
>>
>> To be precise, it must satisfy the golden ratio:
>> r = 1/(1 + r)
>>
>> Next, from R[1], chop off a square with sides a,
>> using the side a. This leaves a new rectangle, R[2],
>> with sides b and a.
>> Iterate: from R[2], chop off a square of side b,
>> leaving R[3], with sides a-b and b.
>>
>> If you do this right, you construct an endless cw
>> whirl of golden rectangles (or maybe ccw, depending
>> how you started).
>>
>> I'm sure we can count on Chris Thomasson to
>> code some pretty pictures.
>>
>> Now, algebra tells us that r is irrational, trivially.
>> But that's a cheat. Problem: show that r is irrational,
>> using geometry and logic only, as Euclid would have
>> (maybe he did, I don't know). i.e. assume r is rational,
>> derive a contradiction.
>>
>> It isn't too hard. Use your drawing for inspiration.
>>
>> --
>> Rich
>
> Some initial geometric sketches:
> https://www.desmos.com/calculator/lr7o4i0uat

Nice! I like the circles within the golden rectangle. Also, its nice how
the rectangle is inscribed within the unit circle. Well done. Now, I
want to put circles inside of the squares in the infinite rectangle.
Let's see here...

Here is a try starting with circles in green within the squares as the
iteration progresses:

https://i.ibb.co/M9bktgL/ct-golden-p2.png

(note: the circle in white is unit centered at point (0, 0), just used
as a guide...)
__________________________
void golden_v0(
ct::plot::cairo::plot_2d& plot,
unsigned long ri,
unsigned long rn,
glm::vec2 p0,
glm::vec2 p1
) {
if (ri >= rn) return;

// The basics... ;^)
glm::vec2 dif = p1 - p0;
glm::vec2 perp = { -dif.y, dif.x };

// Our scale
float ratio_0 = 1.618033988749f;
float ratio = 1.f / ratio_0;

// The golden rectangle
glm::vec2 g0 = p0;
glm::vec2 g1 = p1;
glm::vec2 g2 = g1 + perp * ratio;
glm::vec2 g3 = g0 + perp * ratio;

// internal circle for the square
glm::vec2 cg0 = g3;
glm::vec2 cg1 = p0 + dif * ratio;
glm::vec2 cgdif = cg1 - cg0;
glm::vec2 cgcenter = cg0 + cgdif / 2.f;
float radius = glm::length(g2 - g1) / 2.f;

// Render
plot.circle(cgcenter, radius, CT_RGBF(0, 1, 0), 4.f);
plot.line(g0, g1, CT_RGBF(1, 0, 0), 5.f);
plot.line(g1, g2, CT_RGBF(0, 1, 0), 5.f);
plot.line(g2, g3, CT_RGBF(0, 0, 1), 5.f);
plot.line(g3, g0, CT_RGBF(1, 1, 0), 5.f);

// Test
float length_0 = glm::length(dif);
float length_1 = glm::length(g2 - g1);
float length_ratio = length_0 / length_1;

std::cout << "ri[" << ri << "] = " << length_ratio << "\n";

golden_v0(plot, ri + 1, rn, g1, g2);
} __________________________

Humm... Now to inscribe the golden rectangle within the unit circle.

Re: daily problem: golden rectangles

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Subject: Re: daily problem: golden rectangles
From: r_delane...@yahoo.com (RichD)
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 by: RichD - Sun, 24 Apr 2022 21:18 UTC

April 23, Chris M. Thomasson wrote:
>> Pull out your graph paper for this one -
>> We're going to extend the golden ratio, from line
>> segment to 2-D; a golden rectangle.
>> Draw a rectangle of height a, width a + b
>> Call it R[1]. (large, for clarity)
>> b/a = r ~ .6
>> To be precise, it must satisfy the golden ratio:
>> r = 1/(1 + r)
>> Next, from R[1], chop off a square with sides a,
>> using the side a. This leaves a new rectangle, R[2],
>> with sides b and a.
>> Iterate: from R[2], chop off a square of side b,
>> leaving R[3], with sides a-b and b.
>> If you do this right, you construct an endless cw
>> whirl of golden rectangles (or maybe ccw, depending
>> how you started).
>
> namespace ct_golden
> { void golden_v0(
> ct::plot::cairo::plot_2d& plot,
> unsigned long ri,
> unsigned long rn, )
> https://i.ibb.co/9WcDZC6/ct-golden.png

Well done, Grasshoppa!
Except your construction is ccw, opposite of mine.

You drew six rectangles, right?

> ct_golden()
> this is just a test. Now, did I get what you have in mind? Infinite
> golden rectangles from a single line? I can use any line...
>
> https://i.ibb.co/fDdbM74/ct-golden-p1.png

That one is overkill.

What initial values did you use for a and b?

--
Rich

Re: daily problem: golden rectangles

<t44fj5$vhn$1@dont-email.me>

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From: chris.m....@gmail.com (Chris M. Thomasson)
Newsgroups: sci.math
Subject: Re: daily problem: golden rectangles
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 by: Chris M. Thomasson - Sun, 24 Apr 2022 21:31 UTC

On 4/24/2022 2:18 PM, RichD wrote:
> April 23, Chris M. Thomasson wrote:
>>> Pull out your graph paper for this one -
>>> We're going to extend the golden ratio, from line
>>> segment to 2-D; a golden rectangle.
>>> Draw a rectangle of height a, width a + b
>>> Call it R[1]. (large, for clarity)
>>> b/a = r ~ .6
>>> To be precise, it must satisfy the golden ratio:
>>> r = 1/(1 + r)
>>> Next, from R[1], chop off a square with sides a,
>>> using the side a. This leaves a new rectangle, R[2],
>>> with sides b and a.
>>> Iterate: from R[2], chop off a square of side b,
>>> leaving R[3], with sides a-b and b.
>>> If you do this right, you construct an endless cw
>>> whirl of golden rectangles (or maybe ccw, depending
>>> how you started).
>>
>> namespace ct_golden
>> { void golden_v0(
>> ct::plot::cairo::plot_2d& plot,
>> unsigned long ri,
>> unsigned long rn, )
>> https://i.ibb.co/9WcDZC6/ct-golden.png
>
> Well done, Grasshoppa!
> Except your construction is ccw, opposite of mine.
>
> You drew six rectangles, right?
>
>> ct_golden()
>> this is just a test. Now, did I get what you have in mind? Infinite
>> golden rectangles from a single line? I can use any line...
>>
>> https://i.ibb.co/fDdbM74/ct-golden-p1.png
>
> That one is overkill.
>
> What initial values did you use for a and b?

Well, I am using two points to define a line. Wrt the "overkill" result,
I used:

(the last two parameters for the golden_v0 function are 2, 2-ary vectors
(x, y), p0 and p1. The two 2d points the define a line in in 2d space
from which infinite golden rectangles can be generated:
____________________________
golden_v0(plot, 0, 7, { -1, 0 }, { 0, 1 });
golden_v0(plot, 0, 7, { 0, 1 }, { 1, 0 });
____________________________

Fwiw, check this out, "fractalizing" it, so to speak...:

https://i.ibb.co/XbHMQ40/ct-golden-p3.png

;^)

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Subject: Re: daily problem: golden rectangles
From: r_delane...@yahoo.com (RichD)
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 by: RichD - Sun, 24 Apr 2022 21:31 UTC

On April 24, sobriquet wrote:
>> We're going to extend the golden ratio, from line
>> segment to 2-D; a golden rectangle.
>> Draw a rectangle of height a, width a + b
>> Call it R[1]
>> b/a = r ~ .6
>> To be precise, it must satisfy the golden ratio:
>> r = 1/(1 + r)
>> Next, from R[1], chop off a square with sides a,
>> using the side a. This leaves a new rectangle, R[2],
>> with sides b and a.
>> If you do this right, you construct an endless cw
>> whirl of golden rectangles
>
> Some initial geometric sketches:
> https://www.desmos.com/calculator/lr7o4i0uat

You drew two rectangles, it looks right.
The dimensions are unclear. What's a and b, in the
initial rectangle?

I suggest you re-jigger the x and y co-ordinates, negative
numbers aren't helpful in geometry.

The circles look cool, but I don't see the connection to
the golden ratio.

This Desmos doodad, is public domain, or what?
Looks like fun.
I don't have any good tools for geometric constructions.

--
Rich

Re: daily problem: golden rectangles

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From: chris.m....@gmail.com (Chris M. Thomasson)
Newsgroups: sci.math
Subject: Re: daily problem: golden rectangles
Date: Sun, 24 Apr 2022 15:14:34 -0700
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 by: Chris M. Thomasson - Sun, 24 Apr 2022 22:14 UTC

On 4/24/2022 2:31 PM, RichD wrote:
> On April 24, sobriquet wrote:
>>> We're going to extend the golden ratio, from line
>>> segment to 2-D; a golden rectangle.
>>> Draw a rectangle of height a, width a + b
>>> Call it R[1]
>>> b/a = r ~ .6
>>> To be precise, it must satisfy the golden ratio:
>>> r = 1/(1 + r)
>>> Next, from R[1], chop off a square with sides a,
>>> using the side a. This leaves a new rectangle, R[2],
>>> with sides b and a.
>>> If you do this right, you construct an endless cw
>>> whirl of golden rectangles
>>
>> Some initial geometric sketches:
>> https://www.desmos.com/calculator/lr7o4i0uat
>
> You drew two rectangles, it looks right.
> The dimensions are unclear. What's a and b, in the
> initial rectangle?
>
> I suggest you re-jigger the x and y co-ordinates, negative
> numbers aren't helpful in geometry.
>
> The circles look cool, but I don't see the connection to
> the golden ratio.
>
> This Desmos doodad, is public domain, or what?
> Looks like fun.
> I don't have any good tools for geometric constructions.

I inscribed the infinite golden rectangles from a line thing I coded up
in the unit circle:

https://i.ibb.co/VxGTXny/ct-p4.png

Re: daily problem: golden rectangles

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Subject: Re: daily problem: golden rectangles
Date: Sun, 24 Apr 2022 15:15:33 -0700
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 by: Chris M. Thomasson - Sun, 24 Apr 2022 22:15 UTC

On 4/24/2022 12:53 PM, sobriquet wrote:
> On Sunday, April 24, 2022 at 1:45:54 AM UTC+2, RichD wrote:
>> Pull out your graph paper for this one -
>>
>> We're going to extend the golden ratio, from line
>> segment to 2-D; a golden rectangle.
>>
>> Draw a rectangle of height a, width a + b
>> Call it R[1]. (large, for clarity)
>> b/a = r ~ .6
>>
>> To be precise, it must satisfy the golden ratio:
>> r = 1/(1 + r)
>>
>> Next, from R[1], chop off a square with sides a,
>> using the side a. This leaves a new rectangle, R[2],
>> with sides b and a.
>> Iterate: from R[2], chop off a square of side b,
>> leaving R[3], with sides a-b and b.
>>
>> If you do this right, you construct an endless cw
>> whirl of golden rectangles (or maybe ccw, depending
>> how you started).
>>
>> I'm sure we can count on Chris Thomasson to
>> code some pretty pictures.
>>
>> Now, algebra tells us that r is irrational, trivially.
>> But that's a cheat. Problem: show that r is irrational,
>> using geometry and logic only, as Euclid would have
>> (maybe he did, I don't know). i.e. assume r is rational,
>> derive a contradiction.
>>
>> It isn't too hard. Use your drawing for inspiration.
[...]
>
> Some initial geometric sketches:
> https://www.desmos.com/calculator/lr7o4i0uat

Check these experiments out:

https://i.ibb.co/8mTnfYp/ct-p1.png

https://i.ibb.co/JqWGmzW/ct-p2.png

https://i.ibb.co/pJm7Y7z/ct-p3.png

https://i.ibb.co/VxGTXny/ct-p4.png

Re: daily problem: golden rectangles

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Subject: Re: daily problem: golden rectangles
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 by: Chris M. Thomasson - Sun, 24 Apr 2022 22:28 UTC

On 4/24/2022 2:18 PM, RichD wrote:
> April 23, Chris M. Thomasson wrote:
>>> Pull out your graph paper for this one -
>>> We're going to extend the golden ratio, from line
>>> segment to 2-D; a golden rectangle.
>>> Draw a rectangle of height a, width a + b
>>> Call it R[1]. (large, for clarity)
>>> b/a = r ~ .6
>>> To be precise, it must satisfy the golden ratio:
>>> r = 1/(1 + r)
>>> Next, from R[1], chop off a square with sides a,
>>> using the side a. This leaves a new rectangle, R[2],
>>> with sides b and a.
>>> Iterate: from R[2], chop off a square of side b,
>>> leaving R[3], with sides a-b and b.
>>> If you do this right, you construct an endless cw
>>> whirl of golden rectangles (or maybe ccw, depending
>>> how you started).
>>
>> namespace ct_golden
>> { void golden_v0(
>> ct::plot::cairo::plot_2d& plot,
>> unsigned long ri,
>> unsigned long rn, )
>> https://i.ibb.co/9WcDZC6/ct-golden.png
>
> Well done, Grasshoppa!
> Except your construction is ccw, opposite of mine.
>
> You drew six rectangles, right?

Actually I drew 7 of them. Count again... ;^)

>
>> ct_golden()
>> this is just a test. Now, did I get what you have in mind? Infinite
>> golden rectangles from a single line? I can use any line...
>>
>> https://i.ibb.co/fDdbM74/ct-golden-p1.png
>
> That one is overkill.
>
> What initial values did you use for a and b?
>
> --
> Rich
>

Re: daily problem: golden rectangles

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Subject: Re: daily problem: golden rectangles
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 by: Chris M. Thomasson - Sun, 24 Apr 2022 22:35 UTC

On 4/24/2022 3:28 PM, Chris M. Thomasson wrote:
> On 4/24/2022 2:18 PM, RichD wrote:
>>   April 23, Chris M. Thomasson wrote:
>>>> Pull out your graph paper for this one -
>>>> We're going to extend the golden ratio, from line
>>>> segment to 2-D; a golden rectangle.
>>>> Draw a rectangle of height a, width a + b
>>>> Call it R[1]. (large, for clarity)
>>>> b/a = r ~ .6
>>>> To be precise, it must satisfy the golden ratio:
>>>> r = 1/(1 + r)
>>>> Next, from R[1], chop off a square with sides a,
>>>> using the side a. This leaves a new rectangle, R[2],
>>>> with sides b and a.
>>>> Iterate: from R[2], chop off a square of side b,
>>>> leaving R[3], with sides a-b and b.
>>>> If you do this right, you construct an endless cw
>>>> whirl of golden rectangles (or maybe ccw, depending
>>>> how you started).
>>>
>>> namespace ct_golden
>>> { void golden_v0(
>>> ct::plot::cairo::plot_2d& plot,
>>> unsigned long ri,
>>> unsigned long rn, )
>>> https://i.ibb.co/9WcDZC6/ct-golden.png
>>
>> Well done, Grasshoppa!
>> Except your construction is  ccw, opposite of mine.
>>
>> You drew six rectangles, right?
>
> Actually I drew 7 of them. Count again... ;^)
[...]

https://i.ibb.co/g6yxn9H/ct-golden-p4.png

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Subject: Re: daily problem: golden rectangles
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 by: sobriquet - Sun, 24 Apr 2022 23:38 UTC

On Sunday, April 24, 2022 at 11:32:04 PM UTC+2, RichD wrote:
> On April 24, sobriquet wrote:
> >> We're going to extend the golden ratio, from line
> >> segment to 2-D; a golden rectangle.
> >> Draw a rectangle of height a, width a + b
> >> Call it R[1]
> >> b/a = r ~ .6
> >> To be precise, it must satisfy the golden ratio:
> >> r = 1/(1 + r)
> >> Next, from R[1], chop off a square with sides a,
> >> using the side a. This leaves a new rectangle, R[2],
> >> with sides b and a.
> >> If you do this right, you construct an endless cw
> >> whirl of golden rectangles
> >
> > Some initial geometric sketches:
> > https://www.desmos.com/calculator/lr7o4i0uat
> You drew two rectangles, it looks right.
> The dimensions are unclear. What's a and b, in the
> initial rectangle?

https://i.imgur.com/UTeMGGB.png

Line 20 reads:
"rectangular area inscribed in the unit circle p1 x p2 (with proportion of the long side to the short side equal to the golden ratio p0)"

p0, p1 and p2 are defined in lines 3, 13 and 14 respectively.

>
> I suggest you re-jigger the x and y co-ordinates, negative
> numbers aren't helpful in geometry.
>
> The circles look cool, but I don't see the connection to
> the golden ratio.

The circles make it easy to recognize the squares visually.

>
> This Desmos doodad, is public domain, or what?

Yeah, it's really great. Similar to geogebra, but desmos works
more conveniently (the geogebra interface is a bit clumsy).

With desmos you can even create a mathematical construction
and share it with others without having a desmos account (though
a desmos account is free and recommended and would allow you
to save your graphs in your account).

> Looks like fun.
> I don't have any good tools for geometric constructions.
>
> --
> Rich

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Subject: Re: daily problem: golden rectangles
From: dohduh...@yahoo.com (sobriquet)
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 by: sobriquet - Mon, 25 Apr 2022 00:22 UTC

On Monday, April 25, 2022 at 12:15:43 AM UTC+2, Chris M. Thomasson wrote:
> On 4/24/2022 12:53 PM, sobriquet wrote:
> > On Sunday, April 24, 2022 at 1:45:54 AM UTC+2, RichD wrote:
> >> Pull out your graph paper for this one -
> >>
> >> We're going to extend the golden ratio, from line
> >> segment to 2-D; a golden rectangle.
> >>
> >> Draw a rectangle of height a, width a + b
> >> Call it R[1]. (large, for clarity)
> >> b/a = r ~ .6
> >>
> >> To be precise, it must satisfy the golden ratio:
> >> r = 1/(1 + r)
> >>
> >> Next, from R[1], chop off a square with sides a,
> >> using the side a. This leaves a new rectangle, R[2],
> >> with sides b and a.
> >> Iterate: from R[2], chop off a square of side b,
> >> leaving R[3], with sides a-b and b.
> >>
> >> If you do this right, you construct an endless cw
> >> whirl of golden rectangles (or maybe ccw, depending
> >> how you started).
> >>
> >> I'm sure we can count on Chris Thomasson to
> >> code some pretty pictures.
> >>
> >> Now, algebra tells us that r is irrational, trivially.
> >> But that's a cheat. Problem: show that r is irrational,
> >> using geometry and logic only, as Euclid would have
> >> (maybe he did, I don't know). i.e. assume r is rational,
> >> derive a contradiction.
> >>
> >> It isn't too hard. Use your drawing for inspiration.
> [...]
> >
> > Some initial geometric sketches:
> > https://www.desmos.com/calculator/lr7o4i0uat
> Check these experiments out:
>
> https://i.ibb.co/8mTnfYp/ct-p1.png
>
> https://i.ibb.co/JqWGmzW/ct-p2.png
>
> https://i.ibb.co/pJm7Y7z/ct-p3.png
>
> https://i.ibb.co/VxGTXny/ct-p4.png

A bit like this one..

https://www.geogebra.org/m/sdTJfqb7

Re: daily problem: golden rectangles

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From: chris.m....@gmail.com (Chris M. Thomasson)
Newsgroups: sci.math
Subject: Re: daily problem: golden rectangles
Date: Sun, 24 Apr 2022 20:22:57 -0700
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 by: Chris M. Thomasson - Mon, 25 Apr 2022 03:22 UTC

On 4/24/2022 5:22 PM, sobriquet wrote:
> On Monday, April 25, 2022 at 12:15:43 AM UTC+2, Chris M. Thomasson wrote:
>> On 4/24/2022 12:53 PM, sobriquet wrote:
>>> On Sunday, April 24, 2022 at 1:45:54 AM UTC+2, RichD wrote:
>>>> Pull out your graph paper for this one -
>>>>
>>>> We're going to extend the golden ratio, from line
>>>> segment to 2-D; a golden rectangle.
>>>>
>>>> Draw a rectangle of height a, width a + b
>>>> Call it R[1]. (large, for clarity)
>>>> b/a = r ~ .6
>>>>
>>>> To be precise, it must satisfy the golden ratio:
>>>> r = 1/(1 + r)
>>>>
>>>> Next, from R[1], chop off a square with sides a,
>>>> using the side a. This leaves a new rectangle, R[2],
>>>> with sides b and a.
>>>> Iterate: from R[2], chop off a square of side b,
>>>> leaving R[3], with sides a-b and b.
>>>>
>>>> If you do this right, you construct an endless cw
>>>> whirl of golden rectangles (or maybe ccw, depending
>>>> how you started).
>>>>
>>>> I'm sure we can count on Chris Thomasson to
>>>> code some pretty pictures.
>>>>
>>>> Now, algebra tells us that r is irrational, trivially.
>>>> But that's a cheat. Problem: show that r is irrational,
>>>> using geometry and logic only, as Euclid would have
>>>> (maybe he did, I don't know). i.e. assume r is rational,
>>>> derive a contradiction.
>>>>
>>>> It isn't too hard. Use your drawing for inspiration.
>> [...]
>>>
>>> Some initial geometric sketches:
>>> https://www.desmos.com/calculator/lr7o4i0uat
>> Check these experiments out:
>>
>> https://i.ibb.co/8mTnfYp/ct-p1.png
>>
>> https://i.ibb.co/JqWGmzW/ct-p2.png
>>
>> https://i.ibb.co/pJm7Y7z/ct-p3.png
>>
>> https://i.ibb.co/VxGTXny/ct-p4.png
>
> A bit like this one..
>
> https://www.geogebra.org/m/sdTJfqb7

That's a damn good one! I still have not fitted the curves to the
rectangles. Here is a line in white that gives the basic basis for such
a curve:

https://i.ibb.co/ZKFx57P/ct-golden-p5.png

Re: daily problem: golden rectangles

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From: chris.m....@gmail.com (Chris M. Thomasson)
Newsgroups: sci.math
Subject: Re: daily problem: golden rectangles
Date: Sun, 24 Apr 2022 21:01:03 -0700
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 by: Chris M. Thomasson - Mon, 25 Apr 2022 04:01 UTC

On 4/24/2022 5:22 PM, sobriquet wrote:
> On Monday, April 25, 2022 at 12:15:43 AM UTC+2, Chris M. Thomasson wrote:
>> On 4/24/2022 12:53 PM, sobriquet wrote:
>>> On Sunday, April 24, 2022 at 1:45:54 AM UTC+2, RichD wrote:
>>>> Pull out your graph paper for this one -
>>>>
>>>> We're going to extend the golden ratio, from line
>>>> segment to 2-D; a golden rectangle.
>>>>
>>>> Draw a rectangle of height a, width a + b
>>>> Call it R[1]. (large, for clarity)
>>>> b/a = r ~ .6
>>>>
>>>> To be precise, it must satisfy the golden ratio:
>>>> r = 1/(1 + r)
>>>>
>>>> Next, from R[1], chop off a square with sides a,
>>>> using the side a. This leaves a new rectangle, R[2],
>>>> with sides b and a.
>>>> Iterate: from R[2], chop off a square of side b,
>>>> leaving R[3], with sides a-b and b.
>>>>
>>>> If you do this right, you construct an endless cw
>>>> whirl of golden rectangles (or maybe ccw, depending
>>>> how you started).
>>>>
>>>> I'm sure we can count on Chris Thomasson to
>>>> code some pretty pictures.
>>>>
>>>> Now, algebra tells us that r is irrational, trivially.
>>>> But that's a cheat. Problem: show that r is irrational,
>>>> using geometry and logic only, as Euclid would have
>>>> (maybe he did, I don't know). i.e. assume r is rational,
>>>> derive a contradiction.
>>>>
>>>> It isn't too hard. Use your drawing for inspiration.
>> [...]
>>>
>>> Some initial geometric sketches:
>>> https://www.desmos.com/calculator/lr7o4i0uat
>> Check these experiments out:
>>
>> https://i.ibb.co/8mTnfYp/ct-p1.png
>>
>> https://i.ibb.co/JqWGmzW/ct-p2.png
>>
>> https://i.ibb.co/pJm7Y7z/ct-p3.png
>>
>> https://i.ibb.co/VxGTXny/ct-p4.png
>
> A bit like this one..
>
> https://www.geogebra.org/m/sdTJfqb7

I think I got the spiral arcs in there, except I drew full circles
instead of the specific arc segments:

https://i.ibb.co/J7Tcrc8/ct-golden-p6.png

I think the spiral is in here, embedded within the rendering... ;^)

Re: daily problem: golden rectangles

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From: chris.m....@gmail.com (Chris M. Thomasson)
Newsgroups: sci.math
Subject: Re: daily problem: golden rectangles
Date: Sun, 24 Apr 2022 21:21:12 -0700
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 by: Chris M. Thomasson - Mon, 25 Apr 2022 04:21 UTC

On 4/24/2022 5:22 PM, sobriquet wrote:
> On Monday, April 25, 2022 at 12:15:43 AM UTC+2, Chris M. Thomasson wrote:
>> On 4/24/2022 12:53 PM, sobriquet wrote:
>>> On Sunday, April 24, 2022 at 1:45:54 AM UTC+2, RichD wrote:
>>>> Pull out your graph paper for this one -
[...]
>>> Some initial geometric sketches:
>>> https://www.desmos.com/calculator/lr7o4i0uat
>> Check these experiments out:
>>
>> https://i.ibb.co/8mTnfYp/ct-p1.png
>>
>> https://i.ibb.co/JqWGmzW/ct-p2.png
>>
>> https://i.ibb.co/pJm7Y7z/ct-p3.png
>>
>> https://i.ibb.co/VxGTXny/ct-p4.png
>
> A bit like this one..
>
> https://www.geogebra.org/m/sdTJfqb7

I got it!

https://i.ibb.co/BLWJDdv/ct-golden-p7.png

;^)

Re: daily problem: golden rectangles

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Subject: Re: daily problem: golden rectangles
From: r_delane...@yahoo.com (RichD)
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 by: RichD - Mon, 25 Apr 2022 19:20 UTC

On April 24, Chris M. Thomasson wrote:
> I got it!
> https://i.ibb.co/BLWJDdv/ct-golden-p7.png

The golden ratio spiral!

--
Rich

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Subject: Re: daily problem: golden rectangles
From: dohduh...@yahoo.com (sobriquet)
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 by: sobriquet - Tue, 26 Apr 2022 03:05 UTC

On Sunday, April 24, 2022 at 9:53:16 PM UTC+2, sobriquet wrote:
>[..]
> Some initial geometric sketches:
> https://www.desmos.com/calculator/lr7o4i0uat

Here is a version that recursively decomposes a golden rectangle into
squares (use the M slider to control the level of recursion).

https://www.desmos.com/calculator/gf0wq6cwhc

Now the question remains how this recursive decomposition somehow
provides evidence that demonstrates geometrically that the golden ratio
is irrational.

Re: daily problem: golden rectangles

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Subject: Re: daily problem: golden rectangles
From: r_delane...@yahoo.com (RichD)
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 by: RichD - Tue, 26 Apr 2022 17:41 UTC

On April 25, sobriquet wrote:
>> Some initial geometric sketches:
>> https://www.desmos.com/calculator/lr7o4i0uat
> Here is a version that recursively decomposes a golden rectangle into
> squares (use the M slider to control the level of recursion).
> https://www.desmos.com/calculator/gf0wq6cwhc

Decomposition into squares, but at the same time, constructing
the sequence of rectangles.
> Now the question remains how this recursive decomposition somehow
> provides evidence that demonstrates geometrically that the golden ratio
> is irrational.

Well stated. The construction is not a proof, but evidence,
a guide to a proof.

hint: Starting from the initial rectangle, with semi-arbitrary
dimensions, consider the sequence of calculated dimensions
of the ensuing rectangles.

--
Rich

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From: chris.m....@gmail.com (Chris M. Thomasson)
Newsgroups: sci.math
Subject: Re: daily problem: golden rectangles
Date: Tue, 26 Apr 2022 12:37:29 -0700
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 by: Chris M. Thomasson - Tue, 26 Apr 2022 19:37 UTC

On 4/26/2022 10:41 AM, RichD wrote:
> On April 25, sobriquet wrote:
>>> Some initial geometric sketches:
>>> https://www.desmos.com/calculator/lr7o4i0uat
>> Here is a version that recursively decomposes a golden rectangle into
>> squares (use the M slider to control the level of recursion).
>> https://www.desmos.com/calculator/gf0wq6cwhc
>
> Decomposition into squares, but at the same time, constructing
> the sequence of rectangles.
>
>> Now the question remains how this recursive decomposition somehow
>> provides evidence that demonstrates geometrically that the golden ratio
>> is irrational.
>
> Well stated. The construction is not a proof, but evidence,
> a guide to a proof.
>
> hint: Starting from the initial rectangle, with semi-arbitrary
> dimensions, consider the sequence of calculated dimensions
> of the ensuing rectangles.

We can build these infinite golden rectangles with a single line... ;^)

Re: daily problem: golden rectangles

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Subject: Re: daily problem: golden rectangles
From: dohduh...@yahoo.com (sobriquet)
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 by: sobriquet - Tue, 26 Apr 2022 20:43 UTC

On Tuesday, April 26, 2022 at 7:41:54 PM UTC+2, RichD wrote:
> On April 25, sobriquet wrote:
> >> Some initial geometric sketches:
> >> https://www.desmos.com/calculator/lr7o4i0uat
> > Here is a version that recursively decomposes a golden rectangle into
> > squares (use the M slider to control the level of recursion).
> > https://www.desmos.com/calculator/gf0wq6cwhc
> Decomposition into squares, but at the same time, constructing
> the sequence of rectangles.
> > Now the question remains how this recursive decomposition somehow
> > provides evidence that demonstrates geometrically that the golden ratio
> > is irrational.
> Well stated. The construction is not a proof, but evidence,
> a guide to a proof.
>
> hint: Starting from the initial rectangle, with semi-arbitrary
> dimensions, consider the sequence of calculated dimensions
> of the ensuing rectangles.
>
>
> --
> Rich

I was thinking along the lines of whether there is some kind of feature that
differentiates between rational proportions, like 1/3 or 2/3 and irrational
proportions like the golden ratio.

https://i.imgur.com/Dn3JMMl.png

Is it just that irrational proportions can only be subdivided into successively smaller
squares while rational proportions can be subdivided into equal parts?

The Fibonacci numbers can be used as successive rational approximations to
the golden ratio and from that you can kind of infer that there is no ultimate
rational proportion (given that there is no biggest Fibonacci number).

https://www.desmos.com/calculator/bwkv9bjtfw

Re: daily problem: golden rectangles

<t49ton$plh$1@dont-email.me>

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From: nom...@afraid.org (FromTheRafters)
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Subject: Re: daily problem: golden rectangles
Date: Tue, 26 Apr 2022 16:03:47 -0700
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 by: FromTheRafters - Tue, 26 Apr 2022 23:03 UTC

sobriquet brought next idea :
> On Tuesday, April 26, 2022 at 7:41:54 PM UTC+2, RichD wrote:
>> On April 25, sobriquet wrote:
>>>> Some initial geometric sketches:
>>>> https://www.desmos.com/calculator/lr7o4i0uat
>>> Here is a version that recursively decomposes a golden rectangle into
>>> squares (use the M slider to control the level of recursion).
>>> https://www.desmos.com/calculator/gf0wq6cwhc
>> Decomposition into squares, but at the same time, constructing
>> the sequence of rectangles.
>>> Now the question remains how this recursive decomposition somehow
>>> provides evidence that demonstrates geometrically that the golden ratio
>>> is irrational.
>> Well stated. The construction is not a proof, but evidence,
>> a guide to a proof.
>>
>> hint: Starting from the initial rectangle, with semi-arbitrary
>> dimensions, consider the sequence of calculated dimensions
>> of the ensuing rectangles.
>>
>>
>> --
>> Rich
>
> I was thinking along the lines of whether there is some kind of feature that
> differentiates between rational proportions, like 1/3 or 2/3 and irrational
> proportions like the golden ratio.
>
> https://i.imgur.com/Dn3JMMl.png
>
> Is it just that irrational proportions can only be subdivided into
> successively smaller squares while rational proportions can be subdivided
> into equal parts?
>
> The Fibonacci numbers can be used as successive rational approximations to
> the golden ratio and from that you can kind of infer that there is no
> ultimate rational proportion (given that there is no biggest Fibonacci
> number).
>
> https://www.desmos.com/calculator/bwkv9bjtfw

Perhaps Continued Fractional Expansions could help a little.

Re: daily problem: golden rectangles

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From: news.dea...@darjeeling.plus.com (Mike Terry)
Date: Wed, 27 Apr 2022 02:16:21 +0100
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 by: Mike Terry - Wed, 27 Apr 2022 01:16 UTC

On 26/04/2022 21:43, sobriquet wrote:
> On Tuesday, April 26, 2022 at 7:41:54 PM UTC+2, RichD wrote:
>> On April 25, sobriquet wrote:
>>>> Some initial geometric sketches:
>>>> https://www.desmos.com/calculator/lr7o4i0uat
>>> Here is a version that recursively decomposes a golden rectangle into
>>> squares (use the M slider to control the level of recursion).
>>> https://www.desmos.com/calculator/gf0wq6cwhc
>> Decomposition into squares, but at the same time, constructing
>> the sequence of rectangles.
>>> Now the question remains how this recursive decomposition somehow
>>> provides evidence that demonstrates geometrically that the golden ratio
>>> is irrational.
>> Well stated. The construction is not a proof, but evidence,
>> a guide to a proof.
>>
>> hint: Starting from the initial rectangle, with semi-arbitrary
>> dimensions, consider the sequence of calculated dimensions
>> of the ensuing rectangles.
>>
>>
>> --
>> Rich
>
> I was thinking along the lines of whether there is some kind of feature that
> differentiates between rational proportions, like 1/3 or 2/3 and irrational
> proportions like the golden ratio.
>
> https://i.imgur.com/Dn3JMMl.png
>
> Is it just that irrational proportions can only be subdivided into successively smaller
> squares while rational proportions can be subdivided into equal parts?

That's sort of heading in the right direction... If we start with whole number lengths a,b with
a>b, then a-b and b will have the same greatest common divisor (GCD), and as the process continues
taking off more squares, the GCD always remains the same. Since the lengths themselves keep
decreasing, eventually the GCD itself will be reached and the process can't continue because the new
length will become zero.

(This is effectively applying Euclid's algorithm to calculate the GCD.)

The same applies with rational a,b because we can just change the units to make a,b both whole
number lengths.

So if, as with the golden ratio, the process continues indefinitely, that implies the ratio is
irrational.

FromTheRafters mentioned continued fractions as an approach, and that's another way of looking at
it, because they are very closely related to Euclid's algorithm. (And for rationals, the continued
fraction always terminates, while irrationals have an infinite continued fraction, but I think more
people will be familiar with Euclid's algorithm than continued fractions.)

Mike.

>
> The Fibonacci numbers can be used as successive rational approximations to
> the golden ratio and from that you can kind of infer that there is no ultimate
> rational proportion (given that there is no biggest Fibonacci number).
>
> https://www.desmos.com/calculator/bwkv9bjtfw
>

Re: daily problem: golden rectangles

<t4afok$38u$2@dont-email.me>

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From: chris.m....@gmail.com (Chris M. Thomasson)
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Subject: Re: daily problem: golden rectangles
Date: Tue, 26 Apr 2022 21:11:00 -0700
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 by: Chris M. Thomasson - Wed, 27 Apr 2022 04:11 UTC

On 4/26/2022 1:43 PM, sobriquet wrote:
> On Tuesday, April 26, 2022 at 7:41:54 PM UTC+2, RichD wrote:
>> On April 25, sobriquet wrote:
>>>> Some initial geometric sketches:
>>>> https://www.desmos.com/calculator/lr7o4i0uat
>>> Here is a version that recursively decomposes a golden rectangle into
>>> squares (use the M slider to control the level of recursion).
>>> https://www.desmos.com/calculator/gf0wq6cwhc
>> Decomposition into squares, but at the same time, constructing
>> the sequence of rectangles.
>>> Now the question remains how this recursive decomposition somehow
>>> provides evidence that demonstrates geometrically that the golden ratio
>>> is irrational.
>> Well stated. The construction is not a proof, but evidence,
>> a guide to a proof.
>>
>> hint: Starting from the initial rectangle, with semi-arbitrary
>> dimensions, consider the sequence of calculated dimensions
>> of the ensuing rectangles.
>>
>>
>> --
>> Rich
>
> I was thinking along the lines of whether there is some kind of feature that
> differentiates between rational proportions, like 1/3 or 2/3 and irrational
> proportions like the golden ratio.
>
> https://i.imgur.com/Dn3JMMl.png
>
> Is it just that irrational proportions can only be subdivided into successively smaller
> squares while rational proportions can be subdivided into equal parts?
>
> The Fibonacci numbers can be used as successive rational approximations to
> the golden ratio and from that you can kind of infer that there is no ultimate
> rational proportion (given that there is no biggest Fibonacci number).
>
> https://www.desmos.com/calculator/bwkv9bjtfw

Nice! We have the same result wrt:

https://i.ibb.co/VxGTXny/ct-p4.png

Nice!

Re: daily problem: golden rectangles

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Subject: Re: daily problem: golden rectangles
From: r_delane...@yahoo.com (RichD)
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 by: RichD - Wed, 27 Apr 2022 18:27 UTC

On April 26, Mike Terry wrote:
>>>> Here is a version that recursively decomposes a golden rectangle into
>>>> squares
>>>> https://www.desmos.com/calculator/gf0wq6cwhc
>>>> Now the question remains how this recursive decomposition somehow
>>>> provides evidence that demonstrates geometrically that the golden ratio
>>>> is irrational.
>
>>> The construction is not a proof, but evidence, a guide to a proof.
>>> hint: Starting from the initial rectangle, with semi-arbitrary
>>> dimensions, consider the sequence of calculated dimensions
>>> of the ensuing rectangles.
>
>> I was thinking along the lines of whether there is some kind of feature that
>> differentiates between rational proportions, like 1/3 or 2/3 and irrational
>> proportions like the golden ratio.
>> Is it just that irrational proportions can only be subdivided into successively smaller
>> squares while rational proportions can be subdivided into equal parts?
>
> That's sort of heading in the right direction...
> If we start with whole number lengths a,b with a>b, then a-b and b will have
> the same greatest common divisor (GCD), and as the process continues
> taking off more squares, the GCD always remains the same. Since the lengths
> themselves keep decreasing, eventually the GCD itself will be reached and the
> process can't continue because the new length will become zero.
> (This is effectively applying Euclid's algorithm to calculate the GCD.)
> The same applies with rational a,b because we can just change the units to
> make a,b both whole number lengths.
> So if, as with the golden ratio, the process continues indefinitely, that implies the ratio is
> irrational.

That works.

The simplest explanation:
Assume r = b/a is rational.
Start with a rectangle of integer dimensions; like, a and a+b (duh)
i) By construction, the rectangle dimensions continually shrink.
ii) The dimensions are all positive.
iii) By arithmetic, the dimensions of every rectangle are integers.

Then, we see.... <drum roll>... an infinite descending sequence
of positive integers! Which makes the baby Jesus cry.

--
Rich

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