[sub-project] Perfect cuboid

Fehler und Wünsche zum Projekt yoyo@home
Bugs and wishes for the project yoyo@home
Nachricht
Autor
Zak
Prozessor-Polier
Prozessor-Polier
Beiträge: 109
Registriert: 04.11.2017 12:27

Re: [sub-project] Perfect cuboid

#133 Ungelesener Beitrag von Zak » 10.11.2017 22:52

O.K. Thanks. Go on to look for... )))
Dateianhänge
00.gif
00.gif (19.31 KiB) 4546 mal betrachtet
Zuletzt geändert von Zak am 11.11.2017 01:41, insgesamt 1-mal geändert.

Zak
Prozessor-Polier
Prozessor-Polier
Beiträge: 109
Registriert: 04.11.2017 12:27

Re: [sub-project] Perfect cuboid

#134 Ungelesener Beitrag von Zak » 11.11.2017 01:34

Carefully look at the picture. Good luck!

fwjmath
XBOX360-Installer
XBOX360-Installer
Beiträge: 82
Registriert: 19.10.2010 15:26

Re: [sub-project] Perfect cuboid

#135 Ungelesener Beitrag von fwjmath » 11.11.2017 07:25

Zak hat geschrieben:Carefully look at the picture. Good luck!
You think that is a proof? You didn't even explain why your first statement is correct. Yes, both face-diagonals are integers, but that does not mean their ratio is also an integer.

No matter how carefully I look at the picture, I only see ideas from someone that knows no business about perfect cuboid. Or math.

Zak
Prozessor-Polier
Prozessor-Polier
Beiträge: 109
Registriert: 04.11.2017 12:27

Re: [sub-project] Perfect cuboid

#136 Ungelesener Beitrag von Zak » 11.11.2017 07:42

You don't see that DC'/OH - integer, so AD'/OH' must be integer too?

fwjmath
XBOX360-Installer
XBOX360-Installer
Beiträge: 82
Registriert: 19.10.2010 15:26

Re: [sub-project] Perfect cuboid

#137 Ungelesener Beitrag von fwjmath » 11.11.2017 10:03

Zak hat geschrieben:You don't see that DC'/OH - integer, so AD'/OH' must be integer too?
Your implication only shows that AD/OH' is an integer (2, to be precise), but nothing on AD'/OH'.

This is why I said that you have no proof, only wishful thinking, wishfully disregarding an apostrophe. And I am not making a grammar mistake here, because I think it describe something that is invariant.

Zak
Prozessor-Polier
Prozessor-Polier
Beiträge: 109
Registriert: 04.11.2017 12:27

Re: [sub-project] Perfect cuboid

#138 Ungelesener Beitrag von Zak » 11.11.2017 11:32

Of cause AD/OH' is an integer (2) and DC'/OH is an integer (2). So OH we would carry out to the middle of DC' and OH' we would carry out to the middle of distance, wich parallel and equal to edges AB=AB'. So we analogically can to carry out any perpendicular to the middle of any edge or to the midle of any distance wich parallel of choosen edge. Thus and DC'/OO' must be integer and AD/OO' must be integer and AD'/OO' must be integer. But DC'/AD is ever a non-integer and AD'/AD is ever a non-integer, etc and so on.

fwjmath
XBOX360-Installer
XBOX360-Installer
Beiträge: 82
Registriert: 19.10.2010 15:26

Re: [sub-project] Perfect cuboid

#139 Ungelesener Beitrag von fwjmath » 11.11.2017 11:46

Zak hat geschrieben:Of cause AD/OH' is an integer (2) and DC'/OH is an integer (2). So OH we would carry out to the middle of DC' and OH' we would carry out to the middle of distance, wich parallel and equal to edges AB=AB'. So we analogically can to carry out any perpendicular to the middle of any edge or to the midle of any distance wich parallel of choosen edge. Thus and DC'/OO' must be integer and AD/OO' must be integer and AD'/OO' must be integer. But DC'/AD is ever a non-integer and AD'/AD is ever a non-integer, etc and so on.
Please read again before you post.

Zak
Prozessor-Polier
Prozessor-Polier
Beiträge: 109
Registriert: 04.11.2017 12:27

Re: [sub-project] Perfect cuboid

#140 Ungelesener Beitrag von Zak » 11.11.2017 13:17

Is it possible that in Euler brick ABCDA'B'C'D' : OD, OH and HD will be a Pythagorean triple? And OD, O'D, OO' must be a Pythagorean triple too. Is it possible?

Zak
Prozessor-Polier
Prozessor-Polier
Beiträge: 109
Registriert: 04.11.2017 12:27

Re: [sub-project] Perfect cuboid

#141 Ungelesener Beitrag von Zak » 11.11.2017 13:58

Is it possible that same value of the hypotenuse OD forms two different Pythagorean triples?
The Pythagorean triple is a unique combination of the whole integer edges in a right-angeled triangle, and if two different triangles have a common hypotenuse, and one of the legs from each triangle lies in the same plane, then at least the edges of one of them do not constitute a Pythagorean triple.

Zak
Prozessor-Polier
Prozessor-Polier
Beiträge: 109
Registriert: 04.11.2017 12:27

Re: [sub-project] Perfect cuboid

#142 Ungelesener Beitrag von Zak » 11.11.2017 17:22

If the lengths of all three sides of a right triangle are integers, the triangle is said to be a Pythagorean triangle and its side lengths are collectively known as a Pythagorean triple.
So unequal triangles ODH and ODO' can't will be Pythagorean triangles both, because they have common hypotenuse OD. Thus Euler brick never be a perfect (CD/OO'=2).

Zak
Prozessor-Polier
Prozessor-Polier
Beiträge: 109
Registriert: 04.11.2017 12:27

Re: [sub-project] Perfect cuboid

#143 Ungelesener Beitrag von Zak » 11.11.2017 23:09

Sorry, I found example for 3 triangles OAH', ODH, ODO' which have common hypotenuse:
500, 140, 480
500, 176, 468
500, 300, 400
So I confess a false statement

Zak
Prozessor-Polier
Prozessor-Polier
Beiträge: 109
Registriert: 04.11.2017 12:27

Re: [sub-project] Perfect cuboid

#144 Ungelesener Beitrag von Zak » 13.11.2017 00:59

g^2=a^2+b^2+c^2 (first brick)
h^2=d^2+e^2+f^2 (second brick)
h/g=sqr((d^2+e^2+f^2)/(a^2+b^2+c^2))=sqr(2)
a, b,c,d,e,f are integer, but h/g - irrational, so h and g are not integer.

Antworten

Zurück zu „Fehler, Wünsche / Bugs, Wishes“