I had the idea for this program just before Christmas 2022. After I went to the Christmas market with some students, we visited a church. When I saw the patterns in the leaded windows, I remembered that I had read about those in old
books years ago. For more details, see the file mw.pdf. You should definitely read that, otherwise it's not clear what it's about.
The software is written with PB 3.5 for DOS. However, the dirty calculation work is done by a subroutine in assembly language. The ZIP archive contains the source code and the executable program. The application should run under quite
a few configurations. As an example here is the output under DOSBox 0.74-3:
With this program you can calculate the values for a tracery on leaded
glass windows. The values of the partial areas depend only on the size
of the square side a. It is therefore irrelevant which system of units
is used. If a is specified in centimeters or inches, then the results
are either square centimeters or square inches. The same is true for
meters and yards.
Please enter the size for sqare side a as valid floating point number,
for example 1.5: 2.5
x = .7989
y = .2712
z = 1.9697
Please, press any key to end the application...
However, the program also runs under plain DOS, as DOS emulation or as a virtual machine. The only restrictions are, as a minimum requirement, the presence of the 80386/87 team. That's checked by the software.
Test results from other members would be very nice. I would also welcome some feedback on the readability of the document mw.pdf. Thank you in advance.
Hi Gunther,
Your PDF read just fine. Your English is certainly much better than my DE.
With this program you can calculate the values for a tracery on leaded
glass windows. The values of the partial areas depend only on the size
of the square side a. It is therefore irrelevant which system of units
is used. If a is specified in centimeters or inches, then the results
are either square centimeters or square inches. The same is true for
meters and yards.
Please enter the size for sqare side a as valid floating point number,
for example 1.5: 3.0
x = 1.1504
y = .3905
z = 2.8363
Please, press any key to end the application...
Seems to work here, Window 7 32 bit... :thumbsup:
The PDF is fine. :thumbsup:
Steve,
wow, an immediate feedback. Thank you. :thumbsup:
Quote from: hutch-- on December 29, 2022, 02:40:22 PM
Your PDF read just fine.
That's good to know. It is always such a thing if you are not a native speaker.
Quote from: hutch-- on December 29, 2022, 02:40:22 PM
Your English is certainly much better than my DE.
I can say little about your German because I don't know it. But if you have time next year, feel free to visit me here. After that, your German will definitely be much better.
After all, we would have some important things to discuss in private. I find Berlin especially beautiful in May and June. The city is full of green trees and you can sit in a
sidewalk cafe on every corner and people watch. Just let me know in time when you will come. You can take lodging in my house.
zedd151,
Quote from: zedd151 on December 29, 2022, 02:58:48 PM
Seems to work here, Window 7 32 bit... :thumbsup:
The PDF is fine. :thumbsup:
thank you for the feedback. I had forgotten that Win 7 (32 bit) has a DOS emulation.
Quote from: Gunther on December 29, 2022, 03:05:23 PM
I had forgotten that Win 7 (32 bit) has a DOS emulation.
And I had forgotten that Windows 7 (64 bit) does not. :biggrin: So, I quickly installed 32 bit version for this test, on a spare partition. :tongue:
As a side note... I had to google 'tracery' as I had not heard that term before, iirc. You just added to my vocabulary. :biggrin: And to think I thought it was a typo at first.
zedd151,
Quote from: zedd151 on December 29, 2022, 03:08:47 PM
And I had forgotten that Windows 7 (64 bit) does not. :biggrin: So, I quickly installed 32 bit version for this test, on a spare partition. :tongue:
that's a pretty big effort for a short test run. Thank you. :thumbsup: You could, of course, use other options. For small tests or old DOS games DOSBox is a good alternative.
It's a 32 bit Windows application. The DOS program plays in the sandbox, so to speak, and cannot do any damage. That already has its charm.
A virtual machine with VirtualBox is another good option. However, you then have to rely on FreeDOS. This has to do with the data exchange between host and guest. For very
special cases I can start a pure DOS from an USB stick.
Quote from: zedd151 on December 29, 2022, 03:08:47 PM
As a side note... I had to google 'tracery' as I had not heard that term before, iirc. You just added to my vocabulary. :biggrin: And to think I thought it was a typo at first.
Well, the word is not so often used and probably comes from ancient times, but it is quite common for these circumstances. There are: branch traceries, frost traceries or
plate traceries, for example. Anyway, my paper at least helped expand your vocabulary. :cool:
By the way, have you already found the linear equation system to solve the problem?
Quote from: Gunther on December 30, 2022, 12:36:50 AMthat's a pretty big effort for a short test run.
Actually not a big deal. Takes ten minutes sometimes a little less (win 7) to reinstall from a .wim image :cool: I generally leave one partition available for OS changes when needed. Either Windows xp 32 bit, windows 7 32 bit and 64, or windows 10 64 bit. :tongue: Windows xp reinstalls very quick ~5 minutes, windows 10 maybe 20 minutes.
Quote
By the way, have you already found the linear equation ...
A little beyond my skill set and pay grade. :biggrin:
Hi,
Quote from: Gunther on December 29, 2022, 02:31:20 PM
Test results from other members would be very nice. I would also welcome some feedback on the readability of the document mw.pdf. Thank you in advance.
The program runs with a Win2000 NTVDM. The PDF reads quite well.
Amusingly, the problem looks like calculus might be easier to set up
rather than a geometric solution. (Well maybe, maybe not.)
Regards,
Steve N.
great :thumbsup:
works in Dosbox
frost traceries?like if swedish ice hotel artist carves those shapes in the ice?
me often want to use formulas to draw things in code ,expected some image drawn by curves?
zedd151
Quote from: zedd151 on December 30, 2022, 02:36:41 AM
A little beyond my skill set and pay grade. :biggrin: [/font]
You're understating the case. It's really not that difficult.
Steve,
Quote from: FORTRANS on December 30, 2022, 02:46:52 AM
The program runs with a Win2000 NTVDM.
Good to know. Thank you. :thumbsup:
Quote from: FORTRANS on December 30, 2022, 02:46:52 AM
The PDF reads quite well.
Amusingly, the problem looks like calculus might be easier to set up
rather than a geometric solution. (Well maybe, maybe not.)
Here I must clear up a misunderstanding. The drawings, or rather sketches, are only intended to help you set up the equations. There are 3 unknowns x, y and z. Therefore, we need 3 equations
to determine their values. From the picture in the PDF document you can immediately identify 2 equations. For the necessary third equation, further considerations and an additional sketch are
required. Then you have the linear equation system, which is easily solvable.
I think that there is no geometric solution for this question, since we are looking for partial areas. Of course, your remark is correct that integral calculus can be used to solve it. But this could be
associated with certain difficulties. We have quarter circles. In Cartesian coordinates, roots would then appear in the integrand. A transformation to plane polar coordinates could help. To determine
the integration limits, the 4 circle equations must be set up and the corresponding intersection points calculated. This is not exactly elementary either. Finally, the 3 integrals must be calculated.
In this time, the linear equation system has long been solved and the results of both calculation methods will match.
My view of the things is like this: Use integral calculus where it is necessary. However, if there are simpler and equivalent alternatives, they should be applied. We should also remember that: In the
Renaissance, when this task was first given, the ancient builders did not yet know about the infinitesimal calculus. Nevertheless, the practical problem was successfully solved.
daydreamer,
Quote from: daydreamer on December 30, 2022, 03:44:54 AM
great :thumbsup:
works in Dosbox
Thank you for testing.
Quote from: daydreamer on December 30, 2022, 03:44:54 AM
frost traceries?like if swedish ice hotel artist carves those shapes in the ice?
me often want to use formulas to draw things in code ,expected some image drawn by curves?
I was thinking more of ice flowers and snowflakes. Fractal curves - like the Koch curve - are probably better suited to describe such shapes.
I only have the solution handwritten so far and have yet to convert it to an electronically readable form. I am currently tutoring 8th grade learners at a high school
in Brandenburg, Germany. In the second half of the school year (April, May), linear equation systems will be taught. I will set this task as a longer term homework
project. That's why I won't publish the solution here. Students are very creative when it comes to searches in the Internet. But who is interested in the solution,
can send me a PM with a mail address. As soon as the solution is available as a file, I will transmit it this way.
Quote from: Gunther on December 30, 2022, 05:06:39 AM
... It's really not that difficult ...
Hi Gunther,
If I have some free time, I'll look at it later at some point...
zedd151,
Quote from: zedd151 on December 30, 2022, 05:24:51 AM
If I have some free time, I'll look at it later at some point...
Yes, it's worth a look.
Hi Gunther!
There is a nice geometric solution:
Математика и фокусы: https://www.youtube.com/watch?v=fLRBiPJhmdo
y = 2.7118e-001
z = 1.9697e+000
x = 7.9890e-001
matematika_i_fokusy = 6.2500e+000
gunther = 6.2501e+000
Press any key to continue...
Regards, HSE
HSE,
Quote from: HSE on December 30, 2022, 06:57:49 AM
There is a nice geometric solution:
I think not. The geometry is only used to derive the equations. This is what I have already described above:
Quote from: Gunther on December 30, 2022, 05:06:39 AM
Here I must clear up a misunderstanding. The drawings, or rather sketches, are only intended to help you set up the equations. ...
A strictly geometric solution would mean to determine the area with circle and ruler. How is this supposed to work? I'm glad my students don't know Russian. But why doesn't the presenter
say anything about the history of the exercise and why is he yelling like that? Is that really necessary?
Gunther,
Quote from: Gunther on December 30, 2022, 05:06:39 AM
Here I must clear up a misunderstanding. The drawings, or rather sketches, are only intended to help you set up the equations. ...
:biggrin: Perhaps the oposite apply here: you can forget equations, but never must forget what the problem really is. This is a practical masonery problem about people drawing squares, circles and triangles with compass and ruler. The idea is from Keith Devlin talking about mythical ratios at Stanford.
Quote from: Gunther on December 30, 2022, 09:06:09 AM
A strictly geometric solution would mean to determine the area with circle and ruler.
Sorry professor, at my level the areas of the square, the circle and the triangle are exactly what means geometry. :rolleyes:
Quote from: Gunther on December 30, 2022, 09:06:09 AM
say anything about the history of the exercise and why is he yelling like that?
First idea: he use friendly maners like Putin, and is not a russian history.
Second idea: the video he talk about is from Ukraine, and the woman that listen is an exKGB purging "soft people" in Universities.
Anyway, is not a genius at all. The problem is in the book he have in a hand :biggrin:
HSE,
Quote from: HSE on December 30, 2022, 11:22:10 PM
:biggrin: Perhaps the oposite apply here: you can forget equations, but never must forget what the problem really is.
Maybe you and I just have different perspectives on the same problem. Classical geometric problems are for example: partition of a distance, bisection of an angle or the construction of the tangent to a curve.
This is done with a circle and a ruler.
Quote from: HSE on December 30, 2022, 11:22:10 PM
This is a practical masonery problem about people drawing squares, circles and triangles with compass and ruler.
This exercise isn't about constructing a square, quarter circle, or triangle. Where is the problem to construct a square with side length a? Where is the problem to construct a quarter circle with radius a? Pupils in
the 5th grade can. Even constructing the equilateral triangle is a ridiculous task. That is not what is being asked here. We are looking for areas. This cannot be done without analytical formulas, which is clearly
stated in my paper and in the video you referenced. For this purpose, equations cannot be forgotten; this statement is simply wrong.
Quote from: HSE on December 30, 2022, 11:22:10 PM
The idea is from Keith Devlin talking about mythical ratios at Stanford.
The area of the square, quarter circle or equilateral triangle are anything but mythical. You can read all this already at Euclid. There is nothing left to explore. But what do I know? Perhaps Keith Devlin of Stanford
has found something completely new about this and Euclid was fundamentally wrong? Then all textbooks will have to be rewritten. I haven't heard anything about that yet, but it may be coming soon, who knows?
Quote from: HSE on December 30, 2022, 11:22:10 PM
Sorry professor, at my level the areas of the square, the circle and the triangle are exactly what means geometry. :rolleyes:
And you are able to determine the content of these areas without equations only with circle and ruler?
Quote from: HSE on December 30, 2022, 11:22:10 PM
First idea: he use friendly maners like Putin, and is not a russian history.
Second idea: the video he talk about is from Ukraine, and the woman that listen is an exKGB purging "soft people" in Universities.
Anyway, is not a genius at all. The problem is in the book he have in a hand :biggrin:
I can't see what Putin or the KGB has to do with this question. But anything is possible these days. There are hardly any things for which Putin isn't to blame. However, I will not discuss on this level any further.
Quote from: Gunther on December 30, 2022, 05:06:39 AM
Quote from: FORTRANS on December 30, 2022, 02:46:52 AM
The PDF reads quite well.
Amusingly, the problem looks like calculus might be easier to set up
rather than a geometric solution. (Well maybe, maybe not.)
Here I must clear up a misunderstanding. The drawings, or rather sketches, are only intended to help you set up the equations. There are 3 unknowns x, y and z. Therefore, we need 3 equations
to determine their values. From the picture in the PDF document you can immediately identify 2 equations.
Well, no I can't see it. It would appear that I need either an
explanation or a rather blunt hint. Sorry to be obtuse.
QuoteFor the necessary third equation, further considerations and an additional sketch are
required. Then you have the linear equation system, which is easily solvable.
I think that there is no geometric solution for this question, since we are looking for partial areas. Of course, your remark is correct that integral calculus can be used to solve it. But this could be
associated with certain difficulties. We have quarter circles. In Cartesian coordinates, roots would then appear in the integrand. A transformation to plane polar coordinates could help. To determine
the integration limits, the 4 circle equations must be set up and the corresponding intersection points calculated. This is not exactly elementary either.
Well I found it easy to calculate the intersection points. I think
it took longer to find some paper and a pen than it did to solve
the equation for the first of the intersection points.
QuoteFinally, the 3 integrals must be calculated.
In this time, the linear equation system has long been solved and the results of both calculation methods will match.
My view of the things is like this: Use integral calculus where it is necessary. However, if there are simpler and equivalent alternatives, they should be applied.
I, of course, agree with you. Always use the easiest solution.
And as I did not see your linear equations, I was left with a
geometric solution or calculus. Using mensuration formulas and
creative partitioning looked tedious at best. Integral calculus
should either be fairly easy, or not easy at all.
QuoteWe should also remember that: In the
Renaissance, when this task was first given, the ancient builders did not yet know about the infinitesimal calculus. Nevertheless, the practical problem was successfully solved.
Of course. If you produce a good explanation, please send
me a copy. In any event, an interesting problem. Thanks for
you posting it. Though I can't see how knowing the areas
would help in cutting the glass for a window anyway. (Sorry
if that is also reasonably evident.)
Regards,
Steve N.
Maybe I'm wrong somewhere?
4x + 4y + z = a^2 [square area]
2x + 2y + z/2 = a^2/2 [area of triangle ABC. I'd say that AC divide by half the areas of x and z]
3x + 2y + z = pi*a^2/4 [One fourth of the area of the circle centered at B and radius a]
Quote from: caballero on December 31, 2022, 10:13:27 PM
Maybe I'm wrong somewhere?
4x + 4y + z = a^2 [square area]
2x + 2y + z/2 = a^2/2 [area of triangle ABC. I'd say that AC divide by half the areas of x and z]
3x + 2y + z = pi*a^2/4 [One fourth of the area of the circle centered at B and radius a]
Seem right :thumbsup:
Sine curves rotated 45 degrees to draw image?
Actually
4x + 4y + z = a^2
2x + 2y + z/2 = a^2/2
are redundants (the same). The second one is just the first one divided on left and right by 2. Need to think on the third equation.
Maybe
a^2 = pi*a^2/4 + x + 2y
caballero,
Quote from: caballero on January 01, 2023, 04:54:20 AM
are redundants (the same). The second one is just the first one divided on left and right by 2. Need to think on the third equation.
that is correct. The second equation is a linear combination of the first. In other words, the second is linearly dependent on the first equation and therefore doesn't add any new information. But you're already on the right path.
You have two equations already. You can find the third equation using a new sketch. Consider a one-third circle and subtract an equilateral triangle from it. The rest is then straightforward.
Steve,
Quote from: FORTRANS on December 31, 2022, 06:01:17 AM
Of course. If you produce a good explanation, please send
me a copy.
I'll do that. But that will take a while, because I still have to correct tests and exams. The children will soon receive their mid-year report cards.
Quote from: FORTRANS on December 31, 2022, 06:01:17 AM
In any event, an interesting problem. Thanks for
you posting it. Though I can't see how knowing the areas
would help in cutting the glass for a window anyway. (Sorry
if that is also reasonably evident.)
I would hope so. Well, this is a simple cutting problem. For cutting by hand, the areas play a subordinate role. But nowadays the cutting of the glass sheets is done with NC machines. There, the areas and
corresponding related coordinates are very crucial. Quite by the way: In Lyon, a center of the French fashion and clothing industry, up to 2 000 mathematicians work solely on optimizing the cuts.
Quote from: caballero on December 31, 2022, 10:13:27 PM
Maybe I'm wrong somewhere?
4x + 4y + z = a^2 [square area]
2x + 2y + z/2 = a^2/2 [area of triangle ABC. I'd say that AC divide by half the areas of x and z]
3x + 2y + z = pi*a^2/4 [One fourth of the area of the circle centered at B and radius a]
4x + 4y + z = a^2
1x + 2y = a^2 * (1 - (PI /4.0))
y = a^2 * (1 - (PI/6) - 3^0.5/4)
For a = 2.5:
New A Matrix of complex terms:
[esi].dRows = 3
[esi].dColumns = 3
Mtx Type: MXI_SQUARE
Row = [4.0000 4.0000 1.0000 ]
Row = [1.0000 2.0000 0.0000 ]
Row = [0.0000 1.0000 0.0000 ]
──────────────────────────────────────────────────────────────────────
Matrix B of complex terms:
[esi].dRows = 3
[esi].dColumns = 1
Mtx Type: MXI_COL_VECTOR
Row = [6.2500 ]
Row = [1.3413 ]
Row = [0.2712 ]
──────────────────────────────────────────────────────────────────────
Inverse Matrix A:
[esi].dRows = 3
[esi].dColumns = 3
Mtx Type: MXI_SQUARE
Row = [0.0000 1.0000 -2.0000 ]
Row = [0.0000 0.0000 1.0000 ]
Row = [1.0000 -4.0000 4.0000 ]
──────────────────────────────────────────────────────────────────────
Complex Solution [ F ; G ]
(Column 1 is Real, Column 2 is imaginary):
[esi].dRows = 3
[esi].dColumns = 1
Mtx Type: MXI_COL_VECTOR
Row = [0.7989 ]
Row = [0.2712 ]
Row = [1.9697 ]
For a = 1.0 you obtain gunther's coefficients:
Row = [0.1278 ]
Row = [0.0434 ]
Row = [0.3151 ]Solved with an interesting piece of code: Matrix.inc (Jaymeson Trudgen @ March 2004, also known as NaN). Not so new :biggrin:
Adapted to ObjAsm C.2 from an example in ObjAsm32.
HSE,
Quote from: HSE on January 02, 2023, 07:36:39 AM
Quote from: caballero on December 31, 2022, 10:13:27 PM
Maybe I'm wrong somewhere?
4x + 4y + z = a^2 [square area]
2x + 2y + z/2 = a^2/2 [area of triangle ABC. I'd say that AC divide by half the areas of x and z]
3x + 2y + z = pi*a^2/4 [One fourth of the area of the circle centered at B and radius a]
[/code]
this system of linear equations isn't sufficient for solving because it contains linear dependent equations. I've already explained that in this post (http://masm32.com/board/index.php?topic=10571.msg117060#msg117060).
Quote from: HSE on January 02, 2023, 07:36:39 AM
For a = 1.0 you obtain gunther's coefficients:Row = [0.1278 ]
Row = [0.0434 ]
Row = [0.3151 ]
The values with which I consistently calculate internally are:
; Factors for x, y, z
fx REAL10 0.127824791583588083
fy REAL10 0.0433885225094818035
fz REAL10 0.315146743627720453
Everyone can read and check this in the source code of mw.asm. These values result from a combination of the following constants:
1, -1, sqrt(3), sqrt(3)/2, sqrt(3)/4, Pi/3, Pi/6, Pi/12
on the right side of the solution vector. It doesn't matter which real a you choose. That's the trick.
For that matter, I can't see how and where complex numbers are supposed to appear in this exercise. For the values we are looking for, x, y and z, we have linear equations throughout. At no point does the
square root of a negative number have to be taken. The complex side length of a real square is very difficult to imagine, as is the complex area.
But hey, what do I know? Maybe you have a very simple explanation with the complicated approach about complex matrices? Who knows?
:biggrin: Are no Caballero equations, but the others.
HSE,
Quote from: HSE on January 02, 2023, 12:03:25 PM
:biggrin: Are no Caballero equations, but the others.
of course. But my questions still remain.
Gunther,
Quote from: Gunther on January 02, 2023, 11:43:49 AM
It doesn't matter which real a you choose.
Obviously, if you choose a=1 equations become:
4x + 4y + z = 1
1x + 2y = 1 - (PI /4.0)
y = 1 - (PI/6) - 3^0.5/4Quote from: Gunther on January 02, 2023, 11:43:49 AM
For that matter, I can't see how and where complex numbers are supposed to appear in this exercise.
Don't worry. Complex numbers don't appear in this exercise. (Anyway a real number is a complex number with zero as imaginary part)
Quote from: Gunther on January 02, 2023, 11:43:49 AM
Maybe you have a very simple explanation?
Perhaps at first Trudgen's demo was about complex numbers, but suffer of a severe simplification :biggrin:
This is interesting, though no much time to dedicate for it. :thumbsup:
I have taken an isosceles triangle with base the lower side of the square and vertex the center of the square. Its area would be (a*a/2)/2 = x+y+z/4, which again leaves us with an equation that is a linear combination of the first. Which leads me to think that to find the third equation we must leave any figure inside the square, perhaps calculating the integral of some curve, not of the circle, but of the one whose area would be x+2y (the similar of 1/x).
> Consider a one-third circle and subtract an equilateral triangle from it
I don't see that
Gunther
This is what I mean,ice hotel with a church, Windows probably not made with nc machines,friction in machine makes it too warm and ice melts
They invite artist and build it every year
https://sv.m.wikipedia.org/wiki/Ishotellet_i_Jukkasj%C3%A4rvi (https://sv.m.wikipedia.org/wiki/Ishotellet_i_Jukkasj%C3%A4rvi)
I prefer math solution that draws image of window instead
Hi,
Just as a side note: I calculated the y and z areas geometrically.
I was looking through my calculus books, but have not made any
progress in that direction.
Cheers,
Steve N.
We only have to find the third equation, not linearly dependent on the other two that we already have. So it occurs to me to set a circle of radius a at coordinates (a, a):
Quote(x-a)^2 + (y-a)^2 = a^2
Isolating the y:
Quotey = +/- sqrt(a^2-(x-a)^2) + a
What interests us is:
Quotey = a - sqrt(a^2-(x-a)^2)
Using the page provided by Mineiro, we could find the definite integral of this function between [0, a] (https://www.wolframalpha.com/input?i2d=true&i=Integrate%5Ba-sqrt%5C%2840%29Power%5Ba%2C2%5D-Power%5B%5C%2840%29x-a%5C%2841%29%2C2%5D%5C%2841%29%2C%7Bx%2C0%2Ca%7D%5D&lang=es), so we would have the area equivalent to "x+2y". Thus perhaps we could already have the third equation:
Quotex + 2y = a^2 (1 - pi/4)
Maybe I thinking wrong, I got these results:
https://www.wolframalpha.com/input?i=z%2B4x%2B4y%3Da%5E2%3Ba%5E2-%28%28%CF%80*a%5E2%29%2F4%29%3Dx%2B2y%3Ba%5E2-2*%28x%2B2y%29%3Dz%2B2x
Input
{z + 4 x + 4 y = a^2, a^2 - 1/4 (π a^2) = x + 2 y, a^2 - 2 (x + 2 y) = z + 2 x}
Wrong try, again I've got an equation that is a lineal combination of the previous two:
4x + 4y + z = a^2
3x + 2y + z = pi*a^2/4
x + 2y = a^2 - pi*a^2/4
The third one is the first minus the second :sad:
I had promised to give the linear equation system to solve the problem. For this purpose I have written a document esyse.pdf which
contains all necessary information and drawings. The archive esyse.zip is under the first post of this thread.
I hope that everything is formulated comprehensible. Some feedback on this would be nice. Thank you.
Hi,
Quote from: Gunther on February 01, 2023, 08:47:46 AM
I hope that everything is formulated comprehensible. Some feedback on this would be nice. Thank you.
Concise, quite understandable, and to the point. As such, a good resolution to
the stated problem. Thank you. If you wanted to, you could put in the decimal
representation of the answers to compare to your code. Though hardly needed.
Regards,
Steve N.
Steve,
Quote from: FORTRANS on February 02, 2023, 03:37:58 AM
Concise, quite understandable, and to the point. As such, a good resolution to
the stated problem. Thank you.
I have you to thank. You took the effort to read and evaluate the document. That shows your interest in the problem. I can't take that for granted - with these few downloads.
That's how it is then. In any case, I have kept my promise.
Very clever, Gunther :thumbsup:, thank you. It didn't occur to me to get out of the box because I couldn't see the triangle. Probably without a ruler and a compass it would never be seen.
Just for curiosity, if anyone wonder, the area of the triangle is calculated by applying the Pythagorean theorem for a right triangle. Since side BD measures a and half of side DE measures a/2, the height of the triangle measures sqrt(3)/2*a. Therefore, the area of this triangle is sqrt(3)*a^2/4. Also it is very useful to see that the black shaded areas are equal so that it is useful for us to subtract the area of the triangle from the third of the circle.
caballero,
Quote from: caballero on February 02, 2023, 05:47:22 AM
Very clever, Gunther :thumbsup:, thank you.
thank you. But I had some help. Matthäus (Mathes) Roriczer (https://en.wikipedia.org/wiki/Mathes_Roriczer) (approximate dates 1435-1495) first set this question. At the same time, he has formulated appropriate solution hints.
I just needed to draw this for everyone clearly to understand and translate it into a halfway readable English. I hope I have succeeded in doing that. That's all I have done. Everything
else is from the old master Roriczer and not from me. I give credit where credit is due. We should look with more humility at the work of our forefathers. We can see further only because
we stand on the shoulders of giants. Some people - even in our forum - seem to have forgotten that. But that's another story perhaps for a new thread when the time is ripe. I think that
you and I agree on this point.
Quote from: caballero on February 02, 2023, 05:47:22 AM
It didn't occur to me to get out of the box because I couldn't see the triangle.
Yes, that's exactly the first trick. Therefore I've clarified it in the second sketch. I drew this by hand because I was too lazy to generate it with PSTricks. But I hope the essential nature
of the matter is still recognizable.
Quote from: caballero on February 02, 2023, 05:47:22 AM
Also it is very useful to see that the black shaded areas are equal so that it is useful for us to subtract the area of the triangle from the third of the circle.
Yes, that's the second trick. After that, the rest is actually pretty trivial and pure hand tool. Choose an appropriate method to solve the linear equation system and solve it. That's all.
The sketches led to the core of the problem. That's often the case. Behind many complicated-sounding mathematical definitions and theorems lie simple geometric facts. Not for nothing
it's said: The language of modern mathematics is geometry.
Hi Gunther!
:thumbsup: Fantastic, more easy than the linear equation system from the russian book I posted before.
> We should look with more humility at the work of our forefathers. Some people - even in our forum - seem to have forgotten that. I think that you and I agree on this point.
Yes, indeed.
A good warm-up exercise would be to prove the Pythagorean theorem without looking at any reference, proven back in ancient Greece, over 2,000 years ago.
You, in Germany, have a wide cast of great mathematicians. For example Gauss, the prince of mathematics.
Another good warm-up exercise would be to find the following sum without looking at any reference:
1 + 2 + 3 + ... + 100
https://www.youtube.com/watch?v=XL52d0XFZVM
HSE,
Quote from: HSE on February 03, 2023, 03:45:59 AM
:thumbsup: Fantastic, more easy than the linear equation system from the russian book I posted before.
Yes, of course. But as I said, all I've done is to illustrate Roriczer's ideas and some translation work. More wasn't necessary.
caballero,
Quote from: caballero on February 03, 2023, 05:51:52 AM
A good warm-up exercise would be to prove the Pythagorean theorem without looking at any reference, proven back in ancient Greece, over 2,000 years ago.
Yes. But there are still many open questions in mathematics to which we should direct our energy. I think that would be sensible.
sir Gunther;
Do you happen to have anything related to Fibonacci? It is in several places, but always hidden.
I'll say what intrigues me:
Several properties, however, little use. And after asking myself about the usability of Fibonacci I'm perplexed, too astonished to realize that it starts from limbo and ends in the wilderness (haphazardly).
https://www.geodose.com/2021/06/golden-section-search-python-application-example.html
In other words, the words esmo (wilderness) and ermo (haphazardly) do not exist in English laguage. So something that starts from nothing and ends with nothing. The fault here is not online translators, but, language limits, its necessary seeds in the air (parable of sowing).
mineiro,
Quote from: mineiro on February 04, 2023, 12:29:05 PM
sir Gunther;
Do you happen to have anything related to Fibonacci? It is in several places, but always hidden.
yes. I've written something about the Fibonacci numbers and the Golden Section. But this is only a compilation of known facts and characteristics. Some I've proved if the proof is easy to understand.
But none of that is new. It's not finished yet. So far there are 12 pages and there will be some more. You still have to be patient. I won't be able to finish the paper before Easter; after that, I'll think
about putting it up for discussion.
Thanks for answering sir Gunther;
I have patience.
I was reading "The Fibonacci Quaterly" in a hope to find some answers. They compiled related Fibonacci things from 1963 to front. Some articles are too advanced to my math knowledge.
Dr Ron Knott have a nice site about, I read all, but stuck in some problems, like fibonacci base (like binary, decimal radix).
Again, thank you.
mineiro,
Quote from: mineiro on February 05, 2023, 11:09:06 AM
Thanks for answering sir Gunther;
no thanks is necessary for that. You asked a reasonable question and you will get a reasonable answer from me.
Quote from: mineiro on February 05, 2023, 11:09:06 AM
I was reading "The Fibonacci Quaterly" in a hope to find some answers.
Hopefully, you won't be disappointed by my paper. As I said, it will not contain any new facts but will only be a compilation of already known things. Maybe this or that proof approach
is new for you, but I cannot and won't promise more.
However, I think the Golden Section is very remarkable. These themes have held a great fascination for academics, scientists, and amateurs since ancient times. The great astronomer
Johannes Kepler (1571 - 1630) expressed his high regard in this way: "
Geometry holds two great treasures: one is the Pythagorean Theorem, the other the Golden Section. We can compare
the first one with a bushel of gold, the second we can call a valuable jewel." That's really very well formulated.
Gunther there is a fibonnacci thread "astronomical number" in soap box with interesting link from rosettacode, showing much faster fibonnacci calculation with help of biginteger and matrices
Golden ratio is used by artist, theory of most appealing pic
daydreamer,
Quote from: daydreamer on February 05, 2023, 11:00:25 PM
Gunther there is a fibonnacci thread "astronomical number" in soap box with interesting link from rosettacode, ...
I know. This is exactly why I started writing my paper, to show the theoretical background. Primarily this is supposed to be for my students, but maybe someone on the forum would be interested.
We'll see when the time comes.
Quote from: Gunther on February 05, 2023, 11:55:46 PM
daydreamer,
Quote from: daydreamer on February 05, 2023, 11:00:25 PM
Gunther there is a fibonnacci thread "astronomical number" in soap box with interesting link from rosettacode, ...
I know. This is exactly why I started writing my paper, to show the theoretical background. Primarily this is supposed to be for my students, but maybe someone on the forum would be interested.
We'll see when the time comes.
Please publish more math. Pdfs
Many programmers are interested in math formulas,ranging from beginner level enter numbers in console mode and calculate formula and print result to old demoscene coders using math skill for example poly proc replace mul in innerloop with add, without that kinda math skill, you are restricted to only replace based on find fewer clock cycles opcode
daydreamer,
Quote from: daydreamer on February 15, 2023, 07:11:37 PM
Please publish more math. Pdfs
i will try it. But the problems and solutions have to be interesting. It's not that simple. In addition, of course, you need time for this. But like I said, I want to try.