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Topic: AP Cuboid (Read 763 times) |
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Sir Col
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A cuboid has a square base with all the sides having integral length.
Let the total length of the edges be E.
Let the total surface area be S.
Let the volume be V.
E, S, and V are in an arithmetic progression in that order.
Prove that a unique solution exists.
[edit]To clarify that all the sides have integer lengths[/edit]
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| « Last Edit: Nov 4th, 2008, 11:18am by Sir Col » |
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Grimbal
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Re: AP Cuboid
« Reply #1 on: Sep 22nd, 2008, 1:47pm » |
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I guess "with integral length sides" applies to the cuboid, and not only the base?
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ThudnBlunder
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Re: AP Cuboid
« Reply #2 on: Sep 22nd, 2008, 5:36pm » |
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I get
a = 4b(b-2)/(b2 - 8b + 4)
= 48 when b = 8
where b = length of base
a = length of other side
But still got to prove uniqueness.
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Sir Col
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Re: AP Cuboid
« Reply #3 on: Nov 4th, 2008, 11:17am » |
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I'm surprised this hasn't been solved yet. Out of interest, was it too hard for medium or was it just an uninspiring problem?
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Hippo
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Re: AP Cuboid
« Reply #4 on: Nov 4th, 2008, 12:06pm » |
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So we have to solve 4a+4b+4c=E 2ab+2bc+2ac=S abc=V and 2S=E+V where a,b,c are positive integers.
(x-a)(x-b)(x-c)=x3-(a+b+c)x2+(ab+bc+ac)x-abc=0
x3-(E/4)x2+(S/2)x-(V)=0
x3-(E/4)x2+((E+V)/4)x-(V)=0
Hmm, may be not on the path ...
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4ab+4bc+4ca = 4a+4b+4c + abc
hmm, may be later 
Looking on ThudanBlunder's one cuboid means b=c?
So another constraint ...
8ab+4b2=4a+8b+ab2.
What leads to expression of a using b as Thudan made ...
a(8b-4-b2)=(8b-4b2)
a=(8b-4b2)/(8b-4-b2) ... as finished in the next post ...
Or in the original one the roots have diferent multicity ...
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p(x)=x3-(E/4)x2+((E+V)/4)x-(V)
p'(x)=3x2-(E/2)x+(E+V)/4
LCD(p(x),p'(x)) would be (x-b) (otherwise a=b).
r0(x)=3p(x)-xp'(x)=-(E/4)x2+((E+V)/2)x-(3V)
r1(x)=3r0(x)+(E/4)p'(x)=(3/2(E+V)-E2/8 )x-(9V-E(E+V)/16)
so b=(9V-E(E+V)/16)/(3/2(E+V)-E2/8 ) = (16 9V-E2-EV)/(38E+38V-2E2)
but I don't thing this will help ;(
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| « Last Edit: Nov 4th, 2008, 12:56pm by Hippo » |
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teekyman
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Re: AP Cuboid
« Reply #5 on: Nov 4th, 2008, 12:29pm » |
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Continuing Thudan Blunder's line of thought,
a = 4b(b-2)/(b2 - 8b + 4) = 4 + (24b - 16)/(b - 8b + 4).
The second part of that goes to 0 as b gets large, so we only need to check enough values until that second term gets smaller than 1. (which happens at around b = 32). Kind of brute forcey, but I can't think of a cleaner way to prove uniqueness.
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Eigenray
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Re: AP Cuboid
« Reply #6 on: Nov 4th, 2008, 6:10pm » |
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To continue,
| hidden: | b2 - 8b + 4 divides 24b-16. Then it must also divide
(3b-22)*(24b-16) - 72*(b2-8b+4) = 64.
So b2 - 8b + 4 = d, where d | 64, and solving for b shows d+12 is a perfect square. The only possibility is d=4 and b=8. (d=-8 gives (a,b)=(0,2) or (-12,6), which aren't valid.) |
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| « Last Edit: Nov 4th, 2008, 6:13pm by Eigenray » |
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Sir Col
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Re: AP Cuboid
« Reply #7 on: Nov 5th, 2008, 10:43am » |
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Bravo, Eigenray!
By contrast my approach now seems somewhat disappointing:
a = 4b(b - 2)/(b^2 - 8b + 4)
For b > 2 the numerator is positive, so b^2 - 8b + 4 must be positive. By completing the square, (b - 4)^2 >= 12, but as b is integer, (b - 4)^2 must be square, so (b - 4)^2 >= 16 => b >= 8.
Then writing a = 4 + (24b - 16)/(b^2 - 8b + 4) we know that b^2 - 8b + 4 <= 24b - 16.
That is, b^2 - 32b + 20 <= 0, (b - 16)^2 <= 236, and as LHS is square, (b - 16)^2 <= 225 => b <= 31.
Hence 8 <= b <= 30, as we only need to try even values of b, and the only solution in domain is when b = 8.
It is fairly easy to prove that no integer solution exists for a cube, but I'd be interested to know if anyone is able to make progress with a general cuboid measuring a by b by c.
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Eigenray
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Re: AP Cuboid
« Reply #8 on: Nov 5th, 2008, 11:23am » |
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Suppose a,b  c. Suppose c  31. Solving for a,
a = 4(bc-b-c)/(bc-4b-4c+4) = 4 + 4(3b+3c-4)/(bc-4b-4c+4).
b=1 is impossible, so bc-b-c>0, so bc-4b-4c+4>0, so we must have
bc-4b-4c+4 4(3b+3c-4).
Since c 31, this forces
b 4(4c-5)/(c-16) 476/15 < 32,
so b 31, and similarly a 31. On the other hand, if c < 31, then a,b < 31.
So in any case a,b 31, and brute force shows the only solutions are:
(5,17,252)
(5,18,134)
(5,20,75)
(6,11,98)
(6,12,54)
(6,14,32)
(6,18,21)
(8,8,48)
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| « Last Edit: Nov 5th, 2008, 11:33am by Eigenray » |
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