[sub-project] Perfect cuboid

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Zak
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Re: [sub-project] Perfect cuboid

#109 Ungelesener Beitrag von Zak » 06.11.2017 01:08

x3mEn hat geschrieben:Try d = 1552 e=1886 f=1950 g=2210:
2 * 2210^2 = 1552^2 + 1886^2 + 1950^2
The equation is true, d<e<f<g and d, e, f are even
I have an opinion about you as a fast programmer )))
O.K. guys, I admit that this statement is not correct. Sorry for my crooked English, but it's nice to talk in your good collective.
We with my collegue didn't knew about that it has been proved that Sounderson's parametrization will never give a perfect cuboid, so I wrote vain programm. But we checked that prove is realy works.)))
So I think that if we want to create a fast algorythm, we must omitted some unnecessary cases. And I try to seek them by different approaches and reasoning are not always successful as I would like.

Zak
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Re: [sub-project] Perfect cuboid

#110 Ungelesener Beitrag von Zak » 06.11.2017 13:13

Suppose that the sides of the cuboid a, b, c- the diameters of three intersecting circles, the intersection points of which are the vertices of the cuboid. Then the spatial diagonal is the diameter of the sphere, given by the intersection of these circles, with the center at the center of the cuboid.
On the other hand, the face diagonals d, e, f - also the diameters of three intersecting circles, the intersection points of which are the same vertices of the cuboid. Then the spatial diagonal is also the diameter of the sphere defined by the intersection of these circles, with the center at the center of the cuboid. Consequently, the radius of the sphere in both cases is equal to half the spatial diagonal: r=d/2. What is your opinion:
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Zak
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Re: [sub-project] Perfect cuboid

#111 Ungelesener Beitrag von Zak » 06.11.2017 13:21

So, if I didn't make a mistake, then we have a formula:
g=2sqr(1.5(a^2+b^2+c^2))

x3mEn
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Re: [sub-project] Perfect cuboid

#112 Ungelesener Beitrag von x3mEn » 06.11.2017 14:36

Zak hat geschrieben:the face diagonals d, e, f
Zak hat geschrieben:the radius of the sphere in both cases is equal to half the spatial diagonal: r=d/2
Excuse me, but looks like the spatial diagonal has a different designation: g.
So, the radius of the sphere is equal to half the spatial diagonal: r=g/2.

Zak
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Re: [sub-project] Perfect cuboid

#113 Ungelesener Beitrag von Zak » 06.11.2017 14:38

Oh, I'm sorry, variables are not correct, it will be as: r=g/2, of course
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Re: [sub-project] Perfect cuboid

#114 Ungelesener Beitrag von Zak » 06.11.2017 14:56

So we still have the non-integer value of the spatial diagonal for the integger sides of the cuboid:
g=2sqr(1.5(a^2+b^2+c^2))
Is it true?

x3mEn
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Re: [sub-project] Perfect cuboid

#115 Ungelesener Beitrag von x3mEn » 06.11.2017 15:05

as far as a^2+b^2+c^2=g^2 and g>0, you've proved that 1 = 2 * sqrt(3/2), isn't it?
Zuletzt geändert von x3mEn am 06.11.2017 15:43, insgesamt 1-mal geändert.

Zak
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Re: [sub-project] Perfect cuboid

#116 Ungelesener Beitrag von Zak » 06.11.2017 15:20

It is't. I'm already confused and can not understand exactly where I was wrong...

vasyannyasha
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Re: [sub-project] Perfect cuboid

#117 Ungelesener Beitrag von vasyannyasha » 06.11.2017 16:49

equation g=2*√(S/2)
Where S=(a/2)^2 + (b/2)^2 +...+ (f/2)^
You reduced it to
a^2 + b^2 +...+ f^2
But equation after reducing must looks like
(a^2)/(2^2) + (b^2)/(2^2)+...+ (f^2)/(2^2)=
= (a^2)/4 + (b^2/4) +...+ (f^2)/4=(2(a^2+b^2+c^2))/4=
=(a^2+b^2+c^2)/2
Back to equation with 'g'
g=2*√(((a^2+b^2+c^2)/2)/2)=2*√((a^2+b^2+c^2)/4)=
=2*(1/2)*√(a^2+b^2+c^2)=√(a^2+b^2+c^2)
This equation is true.
P.S Another time think about your equations before posting.Please
Чтобы точно поняли напишу ещё и на русском.
В следующий раз подумайте над вашими уравнениями перед тем как написать на форум.Пожалуйста.
P.S.S
Отвечать на русском не надо.

Zak
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Re: [sub-project] Perfect cuboid

#118 Ungelesener Beitrag von Zak » 06.11.2017 18:17

What is your opinion:
if a - odd then a^2 - odd
else a - even and a^2 - even
So, if we want to get integger value of g, from this formula
g=sqr(d^2/2+e^2/2+f^2/2)=sqr((a^2+b^2)/2+(a^2+c^2)/2+(b^2+c^2)/2)

we have four cases:
1. a^2,b^2,c^2 - odd
2. a^2,b^2,c^2 - even
3. one of squares (of a^2,b^2,c^2) - odd, but two others (of a^2,b^2,c^2) - even
4. two of squares (of a^2,b^2,c^2) - odd, but one remaining (of a^2,b^2,c^2) - even

then everything comes down to two cases:
1. all squares of the face diagonals - even
2. two squares of the face diagonals - odd, but one remaining - even

Right?
Zuletzt geändert von Zak am 06.11.2017 20:04, insgesamt 1-mal geändert.

Zak
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Re: [sub-project] Perfect cuboid

#119 Ungelesener Beitrag von Zak » 06.11.2017 20:03

d^2=a^2+b^2, e^2=a^2+c^2, f^2=b^2+c^2,
We know, that primitive Pythagorean triples consist of two odd squares and one even square...
and both cathetus have a different parity...
Right?
Zuletzt geändert von Zak am 06.11.2017 21:03, insgesamt 2-mal geändert.

vasyannyasha
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Re: [sub-project] Perfect cuboid

#120 Ungelesener Beitrag von vasyannyasha » 06.11.2017 20:05

Zak hat geschrieben:Hello, guys!
I'm from Russia. My name's Zakhar Pekhterev. Your project is very interesting to us. We investigated this with my colleague Natalia Makarova. So we created fast algorytm that allowed to get good result. We varified 156481 Pythagorean triples with calculating values of space diagonals into all possible Euler briks up to value of space diagonal is g=3918370664973947,8649970910420777... The side's values of last verified cuboid are: a=1238639696233380, b=1239555398362281, c=3504697245902160. Last verified Pythagorean triple is w=120561, v=85239, u=85260. So we didn't find at least one integger value of space diagonal. So we have any teqnical dificaltties, but we can get over them if it's necеssary, however, at this time, we decided that here is no further meaning. Good luck for your searching! :wave:
What did you mean under "Last verified Pythagorean triple"?
Do you know about formula for all Pythagorean triple?

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