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O.K. Thanks. Go on to look for... )))

Carefully look at the picture. Good luck!

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.

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

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.

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.

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.

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?

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.

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).

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

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.