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Solve the brick problem/perfect box problem?

Posted 9 months ago
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The problem of finding such a cuboid is also called the brick problem, diagonals problem, perfect box problem, perfect cuboid problem, or rational cuboid problem.

Not 100% how to put this so I will get strat to the point I have a solution for this problem but am not sure who to show need help I am on facebook.

I find it very hard to write as I have Irlen syndrome and dyslexia I have tested it on graph paper and works. I am struggling to put into word.

so the question is where is the best place to go? I was going to post on here is this a no no?

6 Replies
Posted 9 months ago

I have a solution for the perfect cube I do not nowhere to go. I live in England Cornwall Liskeard. The small diagonal is in 4 circles and 1 small like so. The long diagonal is 8 sphere then a small sphere in the middle which can be measured it is very beautiful I have put a picture for you. thank for any help Aaron Cattell

enter image description here

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Posted 9 months ago

I was thinking how to know what the small circles and small sphere = so I did this code is

(20^2 + 20^2)^(1/2)

this gives you 28.28427124746190097603377448419396157139343750753896146353...

so 28 - two LR = 20 That gives you 8 the small circle SD

small diagonal

Now I will try the small sphere the long diagonal so I did this code

(20^2 + 20^2+ 20^2)^(1/2)

that give you
34.64101615137754587054892683011744733885610507620761256111...

so 34 - two LR = 20 That gives you 14 the small sphere SD

have not tried long diagonal yet waiting for a delivery of different size sphere? to check.

long diagonal

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Posted 9 months ago

why no comments?

Posted 5 months ago

I have tested 3d and works 14mm is correct

Posted 5 months ago

Hi Aaron

Have you read this, it should point you in the right direction or give you an idea which way to go next.

https://en.wikipedia.org/wiki/Euler_brick

Posted 5 months ago

I have been looking at that the only way it works is with transcendental numbers will add a photo

AB=C

AD=E

BD=F

BE=G

Diophantine equation

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