## [sub-project] Perfect cuboid

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Zak
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### Re: [sub-project] Perfect cuboid

So we have g>1, and we haven't sqr(2)/2, sqr(2)/3, thus we can't say that we have integer value in denominator of fraction. And all the squares of all integer values are integer, but we have an irrational sqr(2), why?

fwjmath
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### Re: [sub-project] Perfect cuboid

Zak hat geschrieben:So we have g>1, and we haven't sqr(2)/2, sqr(2)/3, thus we can't say that we have integer value in denominator of fraction. And all the squares of all integer values are integer, but we have an irrational sqr(2), why?
Who told you that d,e,f, as face diagonals in situ, form a cuboid? If you only take the length, then who told you the cuboid with d,e,f as sides has an integral body diagonal?

I suggest you first ask a math teacher in a middle school before posting these.

Zak
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### Re: [sub-project] Perfect cuboid

Because d, e, f are also mutually orthogonal like a, b, c. Thus such cuboid d,e,f may to exist for any cuboid a,b,c.
Try:
a=240 b=117 c=44 g=sqr(73225)
d=267 e=244 f=125 h=sqr(146450)
h/g=sqrt(2), g and h are not integer, and we will always have h/g=sqr(2) for all cubouds.
g may be integer in this expression when it is equal 1: h/g=sqrt(2)/1, but we have always g>1.

PS
So we searching integer h in cuboid d,e,f, bigger, and bigger, but we always have smaller cuboid a,b,c, early checked.

x3mEn
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### Re: [sub-project] Perfect cuboid

Why did you decide that (d, e, f) are mutually orthogonal?

Of course, you always can build a cuboid having 3 integer lengths, but why did you decide that the space diagonal of such cuboid should be integer?

Zak
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### Re: [sub-project] Perfect cuboid

x3mEn hat geschrieben: Of course, you always can build a cuboid having 3 integer lengths, but why did you decide that the space diagonal of such cuboid should be integer?
Why not?
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x3mEn
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### Re: [sub-project] Perfect cuboid

Zak hat geschrieben: Why not?
Because, it's a middle school task, to prove that an angle between each pair of face diagonals in your picture is acute.

Zak
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### Re: [sub-project] Perfect cuboid

You think that a perfect cuboid exists in large values. How do you imagine it? It's not the only one, and, for example, if you add 8 such cuboids, you'll get a perfect cuboid with edges 2a, 2b, 2c and a space diagonal twice as big: 2g, etc. So there must be a lot of perfect cuboids, but we didn't find them ... You can, for example, take some edge, expressed in large values, per unit, and then it turns out that the perfect cuboid must exist in smaller values. Or will it cease to be perfect from this? ))) And it does not have to be a convenient brick with edges that are not very different in length, it is quite possible that it has two sides as the legs of the smallest Pythagorean triple: a = 3, b = 4, and the third side c
rushes together with the spatial diagonal to infinity, and we will rather have not a brick, but a long beam, a straight line. The properties of numbers are repeated, no matter how large these numbers are, the Pythagorean triples can be absolutely different numbers. But in small numbers we can not find the perfect cuboid ... So, I have much more confidence that it does not exist, than in that it exists somewhere... That is my parametrization for perfect cuboid, so I created an algorythm based on this formulas whith checking m and n. I started to check this by using several computers from value of g=8999999999999976 down to yours processing value from statistics diagramm.
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x3mEn
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### Re: [sub-project] Perfect cuboid

Zak hat geschrieben:it is quite possible that it has two sides as the legs of the smallest Pythagorean triple: a = 3, b = 4, and the third side crushes together with the spatial diagonal to infinity, and we will rather have not a brick, but a long beam, a straight line.
If you know the smallest edge a, there's also known a higher limit for the 3rd edge c. As far as the hypotenuse has to be integer and also it should be greater than the biggest edge at least by 1.
(c+1)^2 = a^2 + c^2
2*c + 1 = a^2
c = (a^2 - 1)/2
So, for known a=3
max(c) = (3^2-1)/2 = 4
So, it's a stupid enough idea for a=3 to search the 3rd edge over 4.

Zak
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### Re: [sub-project] Perfect cuboid

O.K.
x, f - are integer
g=sqr(f^2+x^2-2*f*x*cos(α)) where angle (α) - obtuse angle and angle (α) can't equal 120 degrees for all cases with each of three face diagonals, thus expression 2*f*x*cos(α) haven't integger value at least once of three cases and the space diagonal g will not integger.
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Zak
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### Re: [sub-project] Perfect cuboid

We also can to consider the triangle d,e,f with acute angles and we can find cosinuses of them analogically...

And finally we have three Pyphagorean triples, which may be described by two integer values of m and n, and if we inevitably have at least one edge even, thus all the parameters: a,b,c,d,e,f,g,m1,n1,m2,n2,m3,n3 we have also even, and we don't have odd space diagonal 4k+1:
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x3mEn
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### Re: [sub-project] Perfect cuboid

We've successfully completed the 0-10T range verification. As was expected, we found a lot of cuboids there:
54'761 edge cuboids
94'501 face cuboids
228'198 imaginary cuboids
1'132'380 twilight cuboids
A total of 1'509'840 different cuboids.
The most fortunate were users year02 and krzyszp who snatched the jackpot and found
7'241 edge cuboids
11'341 face cuboids
27'719 imaginary cuboids
138'903 twilight cuboids
thanks to just a one, the very first task.

x3mEn
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### Re: [sub-project] Perfect cuboid

Perfect Cuboid 1st Batch finish
Now we are close to 1st Batch ending. For those who's interested, what's next, we inform, after complete validation of 0-2^50 range we are going to publish a new app version which will be goaled to Perfect cuboids search only. Thus it's 7 times faster than the current is, so we expect that the next goal — 2^51 will be achieved quick enough.
Stay tuned!