A window problem

[FORTRANS]

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.

[avcaballero]

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):

(x-a)^2 + (y-a)^2 = a^2

Isolating the y:

y = +/- sqrt(a^2-(x-a)^2) + a

What interests us is:

y = 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], so we would have the area equivalent to "x+2y". Thus perhaps we could already have the third equation:

x + 2y = a^2 (1 - pi/4)

[mineiro]

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}

[avcaballero]

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 

[Gunther]

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.

[FORTRANS]

Hi,

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.

[Gunther]

Steve,

   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.

[avcaballero]

Very clever, Gunther  , 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.

[Gunther]

caballero,

Very clever, Gunther  , thank you.

thank you. But I had some help. Matthäus (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.

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.

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.

[HSE]

Hi Gunther!

  Fantastic, more easy than the linear equation system from the russian book I posted before.

[avcaballero]

>  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

[Gunther]

HSE,

  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.

[Gunther]

caballero,

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.

[mineiro]

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).

[Gunther]

mineiro,

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.