A window problem

[HSE]

Gunther,

Here I must clear up a misunderstanding. The drawings, or rather sketches, are only intended to help you set up the equations. ...

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.

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.

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 

[Gunther]

HSE,

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.

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.

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?

Sorry professor, at my level the areas of the square, the circle and the triangle are exactly what means geometry.

And you are able to determine the content of these areas without equations only with circle and ruler?

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 

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.

[FORTRANS]

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.

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.

   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.

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.

   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.

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.

   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.

[avcaballero]

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]

[daydreamer]

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 
Sine curves  rotated 45 degrees to draw image?

[avcaballero]

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

[Gunther]

caballero,

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.

[Gunther]

Steve,

   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.

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.

[HSE]

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:

Code Select Expand

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:

Code Select Expand

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

Adapted to ObjAsm C.2 from an example in ObjAsm32.

[Gunther]

HSE,

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.

For a = 1.0 you obtain gunther's coefficients:

Code Select Expand

Row = [0.1278 ]
Row = [0.0434 ]
Row = [0.3151 ]

The values with which I consistently calculate internally are:

Code Select Expand


; 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:

Code Select Expand


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?

[HSE]

  Are no Caballero equations, but the others.

[Gunther]

HSE,

Are no Caballero equations, but the others.

of course. But my questions still remain.

[HSE]

Gunther,

It doesn't matter which real a you choose.

Obviously, if you choose a=1 equations become:

Code Select Expand

4x + 4y + z = 1
1x + 2y     = 1 - (PI /4.0)
      y     = 1 - (PI/6) - 3^0.5/4

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)

Maybe you have a very simple explanation?

Perhaps at first Trudgen's demo was about complex numbers, but suffer of a severe simplification 

[avcaballero]

This is interesting, though no much time to dedicate for it. 

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 

[daydreamer]

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
I prefer math solution that draws image of window instead