Tricky Box Problem
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Tricky Box Problem
Determine with proof whether it is ever possible to fit an a by b box inside of an x by y by z box where a+b+c>x+y+z?
 jestingrabbit
 Factoids are just Datas that haven't grown up yet
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Re: Tricky Box Problem
Do you mean an [imath]a\times b\times c[/imath] box inside an [imath]x\times y\times z[/imath] box? (you dropped a "by c" in the problem statement).
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Re: Tricky Box Problem
jestingrabbit wrote:Do you mean an [imath]a\times b\times c[/imath] box inside an [imath]x\times y\times z[/imath] box? (you dropped a "by c" in the problem statement).
assuming thats the real puzzle
Spoiler:
Yes, my name is the negative frequency of radioactive decay, of the initial speed of light's radius.
And on the eighth day God created Irony.
But on the ninth day Satan was all like, "Nuh uh!"
And ironically made Alanis Morrisette his minion.
Re: Tricky Box Problem
krucifi wrote:Any a+b+c that is greater than x+y+z also means a greater volume.
No it doesn't.
1+1+8 > 3+3+3
1*1*8 < 3*3*3
Re: Tricky Box Problem
Here's my attempt at a solution:
Spoiler:
Re: Tricky Box Problem
Spoiler:
Arggh, ninja'd
Re: Tricky Box Problem
Just use negative numbers for the dimensions on the x, y, z box. A 2x2x2 box will totally fit inside an insideout 3x3x3 box.
Is 'no' your answer to this question?
Re: Tricky Box Problem
Spoiler:
Sorting the sides makes it easy to see that turning the box around does not help the situation.
Re: Tricky Box Problem
cjr22 wrote:Spoiler:
Arggh, ninja'd
You got your minimized and maximized cases backwards.
 jestingrabbit
 Factoids are just Datas that haven't grown up yet
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Re: Tricky Box Problem
Markus__1 wrote:Sorting the sides makes it easy to see that turning the box around does not help the situation.
But it doesn't explain why putting the box in diagonally doesn't help. Also, please spoiler solutions when in the logic puzzles subforum.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Re: Tricky Box Problem
jestingrabbit wrote:... please spoiler solutions when in the logic puzzles subforum.
Sorry, I noticed the moment I pressed submit  just that little bit too late
 John Citizen
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Re: Tricky Box Problem
Spoiler:
"Yields falsehood when preceded by its quotation"
yields falsehood when preceded by its quotation.
yields falsehood when preceded by its quotation.
Re: Tricky Box Problem
John Citizen wrote:Spoiler:
You just state this without proof.
It is possible for example easy to fit a 1x1x6 box inside a 5x5x5 box without 'folding', despite the fact that 6>5.
You have to somehow use the fact that a+b+c>d+e+f (which my counterexample did not satisfy) and you did not use this in any way. It is not enough to show one example that doesn't work and to then just state that this must for some reason hold for all possible examples. You have to prove it.
 John Citizen
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Re: Tricky Box Problem
Ah yes, I was assuming that
Spoiler:
"Yields falsehood when preceded by its quotation"
yields falsehood when preceded by its quotation.
yields falsehood when preceded by its quotation.
 John Citizen
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 Joined: Fri Jan 21, 2011 12:25 pm UTC
Re: Tricky Box Problem
Actually, I think that I may have worked it out:
I think that's right.
Spoiler:
I think that's right.
"Yields falsehood when preceded by its quotation"
yields falsehood when preceded by its quotation.
yields falsehood when preceded by its quotation.
 John Citizen
 Posts: 23
 Joined: Fri Jan 21, 2011 12:25 pm UTC
Re: Tricky Box Problem
Cpt. Red wrote:Spoiler:
Spoiler:
"Yields falsehood when preceded by its quotation"
yields falsehood when preceded by its quotation.
yields falsehood when preceded by its quotation.
 skeptical scientist
 closedminded spiritualist
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Re: Tricky Box Problem
I read a really nice solution to this puzzle online somewhere:
Skimming the earlier proofs, it looks like sfwc's proof works, and none of the other alleged proofs are actually proofs. Am I right?
Spoiler:
Skimming the earlier proofs, it looks like sfwc's proof works, and none of the other alleged proofs are actually proofs. Am I right?
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.
"With math, all things are possible." —Rebecca Watson
"With math, all things are possible." —Rebecca Watson
 t1mm01994
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Re: Tricky Box Problem
Yep, you are in fact right. You are the 2nd one to post an actual proof and not one that has hands waving.
Re: Tricky Box Problem
@skeptical scientist, that was the solution I heard when I originally came across this problem
Re: Tricky Box Problem
Edit:
I don't think that there's really any need for a proof. Since we are dealing with objects with similar symmetry like rectangular cuboids we just need to place the boxes such that their centre of mass (or symmetry) and the respective planes of symmetry coincide. If we have a configuration such that [imath]a\leq x[/imath] and [imath]b\leq y[/imath] and [imath]c\leq z[/imath] the box ABC fits XYZ and [imath]a+b+c\leq x+y+z[/imath] . Otherwise [imath]a+b+c> x+y+z[/imath].
I don't think that there's really any need for a proof. Since we are dealing with objects with similar symmetry like rectangular cuboids we just need to place the boxes such that their centre of mass (or symmetry) and the respective planes of symmetry coincide. If we have a configuration such that [imath]a\leq x[/imath] and [imath]b\leq y[/imath] and [imath]c\leq z[/imath] the box ABC fits XYZ and [imath]a+b+c\leq x+y+z[/imath] . Otherwise [imath]a+b+c> x+y+z[/imath].
God does not care about our mathematical difficulties — He integrates empirically.
—Albert Einstein
—Albert Einstein
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Re: Tricky Box Problem
redrogue wrote:Just use negative numbers for the dimensions on the x, y, z box. A 2x2x2 box will totally fit inside an insideout 3x3x3 box.
The best part is that it doesn't even have to go "inside" the 3x3x3 because, due to its insideoutedness, its inside volume occurs on the apparent "outside". That is to say, for example, that a 5x5x5 could fit on the inside of a 3x3x3 if it was insideout.
If you ever find yourself asking, "is the answer to the question, "is the answer to the question, "am I the only one?" no?" no?" The answer is still no, but you should probably close all those Logic Puzzles tabs and go to sleep.
 skeptical scientist
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Re: Tricky Box Problem
Ddanndt wrote:Otherwise [imath]a+b+c> x+y+z[/imath].
That part is false, because you can have a configuration with a > x, but b < y and c < z, and in this case, you might have a+b+c < x+y+z. It is even possible to have the smaller box fit inside of the larger box in this case. (For example, a 12x1x1 box just barely fits inside an 8x8x8 box.) However, it is not at all obvious that a+b+c < x+y+z if the smaller box fits in the larger box, and that requires proof.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.
"With math, all things are possible." —Rebecca Watson
"With math, all things are possible." —Rebecca Watson

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Re: Tricky Box Problem
I found the answer:
http://img194.imageshack.us/img194/410/20110518150141.png
http://img194.imageshack.us/img194/410/20110518150141.png
Re: Tricky Box Problem
I've found a simple solution:
Spoiler:
 skeptical scientist
 closedminded spiritualist
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Re: Tricky Box Problem
willdawg wrote:[math](x + \delta x)^2 + (y + \delta y)^2 + (z + \delta z)^2 \le x^2 + y^2 + z^2[/math][math]\to x^2 + 2\delta x + \delta x^2 + y^2 + 2\delta y + \delta y^2 + z^2 + 2\delta z + \delta z^2 \le x^2 + y^2 + z^2[/math]
The second line above is incorrect; [imath]2\delta x[/imath] should be [imath]2x\delta x[/imath], and similarly for the y and z terms.
But now your proof no longer works, because the first term on the lefthandside of[math]2(x\delta x + y\delta y + z\delta z) + (\delta x^2 + \delta y^2 + \delta z^2) \le 0[/math]is no longer clearly positive.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.
"With math, all things are possible." —Rebecca Watson
"With math, all things are possible." —Rebecca Watson
Re: Tricky Box Problem
I'm no math major, but this puzzle seems vaguely similar to Prince Rupert's Cube: http://mathworld.wolfram.com/PrinceRupertsCube.html.
Prince Rupert's cube is the largest cube that can be made to pass through a given cube. In other words, the cube having a side length equal to the side length of the largest hole of a square cross section that can be cut through a unit cube without splitting it into two pieces. However, Prince Rupert's cube doesn't require the "inside" cube to be completely inside all at once, it must simply all pass through.
Based on this, I'd guess (only a guess) that your puzzle can't be done.
Prince Rupert's cube is the largest cube that can be made to pass through a given cube. In other words, the cube having a side length equal to the side length of the largest hole of a square cross section that can be cut through a unit cube without splitting it into two pieces. However, Prince Rupert's cube doesn't require the "inside" cube to be completely inside all at once, it must simply all pass through.
Based on this, I'd guess (only a guess) that your puzzle can't be done.
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