Tricky Box Problem

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lati0s
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Tricky Box Problem

Postby lati0s » Wed Apr 20, 2011 12:42 am UTC

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?

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Re: Tricky Box Problem

Postby jestingrabbit » Wed Apr 20, 2011 4:55 am UTC

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).
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Re: Tricky Box Problem

Postby krucifi » Wed Apr 20, 2011 9:46 am UTC

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:
its impossible no? Any a+b+c that is greater than x+y+z also means a greater volume. i did try think about perhaps an angled box inside but that wouldn't do it either.
for any |a| + |b| + |c| > |x| + |y| + |z|
then a.b.c > x.y.z
V= L.B.H for any cuboid
if V1 is greater than V2 then V1 can never fit inside V2.
and as L.B.H of V1 is a.b.c and L.B.H if V2 is x.y.z
this is sufficient enough proof methinks
i apologize for the crappy way i have explained it lol
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Re: Tricky Box Problem

Postby jaap » Wed Apr 20, 2011 9:51 am UTC

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

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Re: Tricky Box Problem

Postby sfwc » Wed Apr 20, 2011 10:55 am UTC

Here's my attempt at a solution:
Spoiler:
It isn't ever possible.

Suppose we have an a*b*c cuboid A which fits inside an x*y*z cuboid X. The main diagonal of A fits inside X, so is at most as long as the main diagonal of X, and so a2 + b2 + c2 [imath]\leq[/imath] x2 + y2 + z2. For each point p on the surface of A, draw a ray through p perpendicular to the surface of A and away from A. Let f(p) be the point of intersection of this ray with the surface of X. This gives a map f from the surface of A to a subset S of the surface of X (this subset consists of 6 chunks, 1 for each face of A). This map is locally linear, and each linear part has determinant greater than 1. So the surface area of A is at most the area of S, which is at most the surface area of X. Thus ab + bc + ca [imath]\leq[/imath] xy + yz + zx. So (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) [imath]\leq[/imath] x2 + y2 + z2 + 2(xy + yz + zx) = (x + y + z)2 and so a + b + c [imath]\leq[/imath] x + y + z.

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Re: Tricky Box Problem

Postby cjr22 » Wed Apr 20, 2011 10:56 am UTC

Spoiler:
Think about the length of the longest diagonal. This is length [imath]\sqrt{( a^{2} + b^{2} + c^{2} )}[/imath] in the first box and [imath]\sqrt{( x^{2} + y^{2} + z^{2} )}[/imath] in the second. To fit, the first value must be less than the second. If a+b+c is constant, the diagonal length is minimised when [imath]a \gg b[/imath] and [imath]a \gg c[/imath], in which case the diagonal length and a+b+c are approximately equal to a. If x+y+z is constant, the diagonal length is maximised when x = y = z, in which case the diagonal length is [imath]\sqrt{3}[/imath]x and x+y+z=3x.
So, from the problem definition a > 3x, but to fit we need a < [imath]\sqrt{3}[/imath]x, which is a contradiction, so it is not possible.


Arggh, ninja'd

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Re: Tricky Box Problem

Postby redrogue » Wed Apr 20, 2011 4:26 pm UTC

Just use negative numbers for the dimensions on the x, y, z box. A 2x2x2 box will totally fit inside an inside-out 3x3x3 box. :)
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Re: Tricky Box Problem

Postby Markus__1 » Wed Apr 20, 2011 6:53 pm UTC

Spoiler:
Sort sides so that [imath]a \le b \le c[/imath] and [imath]x \le y \le z[/imath].
Geometrically this orients both boxes in the same way.

Now if [imath]a + b + c > x + y + z[/imath], that means at least one of the sides must be larger than its equivalent, i.e. [imath]a > x[/imath] or [imath]b > y[/imath] or [imath]c > z[/imath] (or two or even all three).
Geometrically that larger side sticks out of the other box.


Sorting the sides makes it easy to see that turning the box around does not help the situation.

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Re: Tricky Box Problem

Postby douglasm » Wed Apr 20, 2011 8:02 pm UTC

cjr22 wrote:
Spoiler:
Think about the length of the longest diagonal. This is length [imath]\sqrt{( a^{2} + b^{2} + c^{2} )}[/imath] in the first box and [imath]\sqrt{( x^{2} + y^{2} + z^{2} )}[/imath] in the second. To fit, the first value must be less than the second. If a+b+c is constant, the diagonal length is minimised when [imath]a \gg b[/imath] and [imath]a \gg c[/imath], in which case the diagonal length and a+b+c are approximately equal to a. If x+y+z is constant, the diagonal length is maximised when x = y = z, in which case the diagonal length is [imath]\sqrt{3}[/imath]x and x+y+z=3x.
So, from the problem definition a > 3x, but to fit we need a < [imath]\sqrt{3}[/imath]x, which is a contradiction, so it is not possible.


Arggh, ninja'd

You got your minimized and maximized cases backwards.

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Re: Tricky Box Problem

Postby jestingrabbit » Thu Apr 21, 2011 10:12 am UTC

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.

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Re: Tricky Box Problem

Postby Markus__1 » Thu Apr 21, 2011 2:52 pm UTC

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 :oops:

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Re: Tricky Box Problem

Postby John Citizen » Sun Apr 24, 2011 11:12 am UTC

Spoiler:
let
V1=abc
V2=def
S1=a+b+c
S2=d+e+f

abc<def
a+b+c>d+e+f
It would appear that this is possible (eg. a=b=1, c=63, d=e=f=4), this only applies if folding is allowed. If not, then each dimension must be smaller; therefore:

a<d
b<e
c<f

Thus a+b+c<d+e+f
and the premise is false.
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Re: Tricky Box Problem

Postby jaap » Sun Apr 24, 2011 11:49 am UTC

John Citizen wrote:
Spoiler:
abc<def
a+b+c>d+e+f
It would appear that this is possible (eg. a=b=1, c=63, d=e=f=4), this only applies if folding is allowed. If not, then each dimension must be smaller

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.

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John Citizen
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Re: Tricky Box Problem

Postby John Citizen » Mon Apr 25, 2011 7:45 am UTC

Ah yes, I was assuming that
Spoiler:
the boxes were lying flat, rather than one diagonal. If one is diagonal, the diagonal of the smaller box must be smaller than the diagonal of the larger box (ie. [imath]\sqrt{a^2+b^2+c^2} < \sqrt{d^2+e^2+f^2}[/imath]).

So we have:

[math]\sqrt{a^2+b^2+c^2} < \sqrt{d^2+e^2+f^2}[/math]
and

[math]a+b+c > d+e+f[/math]
where abc defines the smaller box.

Squaring both sides of the first inequation:
[imath]a^2+b^2+c^2<d^2+e^2+f^2[/imath]
Squaring the second:
[imath]a^2+b^2+c^2+2ab+2ac+2bc>d^2+e^2+f^2+2de+2df+2ef[/imath]
Therefore:
[imath]ab+ac+bc>de+df+ef[/imath]
and
[imath]2(ab+ac+bc-[de+df+ef])>d^2+e^2+f^2-(a^2+b^2+c^2)[/imath]


I'm not quite sure where to go from there.
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Re: Tricky Box Problem

Postby John Citizen » Mon Apr 25, 2011 8:53 am UTC

Actually, I think that I may have worked it out:
Spoiler:
For the smaller box, we want to maximise y = a+b+c for a given volume v=abc.
b=v/(ac)
y=a+v/(ac)+c
[imath]\frac{\partial y}{\partial a}=1-\frac{v}{a^2c}[/imath]
As a tends to 0, [imath]\frac{\partial y}{\partial a}[/imath] tends to infinity , therefore b also approaches infinity (c is constant).
[imath]\frac{\partial y}{\partial c}=1-\frac{v}{c^2a}[/imath]
As c tends to 0, [imath]\frac{\partial y}{\partial c}[/imath] tends to infinity, therefore b also approaches infinity (a is constant).
Therefore the maximal a+b+c is when b is large and a and c are small.
[imath]b>a=c[/imath]

For the larger box, we want to minimise y = d+e+f for a given volume v=def.
e=v/df
y=d+v/(df)+f
[imath]\frac{\partial y}{\partial d}=1-\frac{v}{d^2f}[/imath]
When [imath]d=\sqrt{\frac{v}{f}}[/imath], y is stationary (local min.)
[imath]\frac{\partial y}{\partial f}=1-\frac{v}{f^2d}[/imath]
When [imath]f=\sqrt{\frac{v}{d}}[/imath], y is stationary (local min.)
e=v/(df), therefore [imath]f=\sqrt{ef}[/imath] and f=e
e=v/(df), therefore [imath]d=\sqrt{df}[/imath] and d=e=f
Therefore the minimal d+e+f is when d=e=f.

For abc to fit in def, [imath]\sqrt{a^2+b^2+c^2}<\sqrt{d^2+e^2+f^2}[/imath]
Substituting the maximal and minimal solutions:
[imath]b=\sqrt{3}d[/imath] as a and c approach zero.

From my earlier spoiler:
[imath]ab+ac+bc>de+df+ef[/imath]
[imath]ab+ac+bc>3d^2[/imath]

As a and c approach zero, the entire LHS approaches zero. Therefore, the original statement is false.

I think that's right.
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Re: Tricky Box Problem

Postby Cpt. Red » Wed Apr 27, 2011 8:22 am UTC

Spoiler:
The box [imath]0\times50\times1[/imath] fits inside a [imath]1\times1\times1[/imath] box so yes it is possible if you allow the box to have a volume of zero.

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John Citizen
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Re: Tricky Box Problem

Postby John Citizen » Wed Apr 27, 2011 8:55 am UTC

Cpt. Red wrote:
Spoiler:
The box [imath]0\times50\times1[/imath] fits inside a [imath]1\times1\times1[/imath] box so yes it is possible if you allow the box to have a volume of zero.

Spoiler:
The former 'box' will not in fact fit inside the latter, because there is no possible way to fit a plane with length 50 and width 1 into a 1x1x1 box. The question isn't about volume, otherwise you'd just get, say, 0.01x50x1 and 1x1x1 as solutions.
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Re: Tricky Box Problem

Postby skeptical scientist » Wed Apr 27, 2011 12:34 pm UTC

I read a really nice solution to this puzzle online somewhere:
Spoiler:
Suppose a box of dimensions a,b,c fits inside a box of dimensions x,y,z. Then we have AXR3, where A and X are rectangular prisms of the given dimensions, not necessarily oriented the same way. Consider the sets Fr(A) and Fr(X) consisting of all points at distance at most r from A and X, respectively. Since AX, we have Fr(A) ⊂ Fr(X) for all r, so the volumes satisfy the inequality
V(Fr(A)) ≤ V(Fr(B)),
for all r.

Now, the set of points of distance at most r from an rectangular prism consists of the prism itself, 6 smaller prisms (one for each face), 12 quarter cylinders (one for each edge), and 8 octants of a sphere (one for each corner). Adding up these volumes, we have
V(Fr(A))=4/3πr3+π(a+b+c)r2+O(r),
and
V(Fr(X))=4/3πr3+π(x+y+z)r2+O(r).
Since the r3 terms are the same, and the terms proportional to r2 dominate the O(r) terms for r large, we must have π(a+b+c)r2 ≤ π(x+y+z)r2 for sufficiently large r, whence a+b+c ≤ x+y+z.


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.

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Re: Tricky Box Problem

Postby t1mm01994 » Wed Apr 27, 2011 1:47 pm UTC

Yep, you are in fact right. You are the 2nd one to post an actual proof and not one that has hands waving.

lati0s
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Re: Tricky Box Problem

Postby lati0s » Wed Apr 27, 2011 10:29 pm UTC

@skeptical scientist, that was the solution I heard when I originally came across this problem

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Re: Tricky Box Problem

Postby Ddanndt » Thu Apr 28, 2011 2:32 pm UTC

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].
God does not care about our mathematical difficulties — He integrates empirically.
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The Chosen One
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Re: Tricky Box Problem

Postby The Chosen One » Mon May 09, 2011 3:29 pm UTC

redrogue wrote:Just use negative numbers for the dimensions on the x, y, z box. A 2x2x2 box will totally fit inside an inside-out 3x3x3 box.


The best part is that it doesn't even have to go "inside" the 3x3x3 because, due to its inside-outedness, 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 inside-out.
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Re: Tricky Box Problem

Postby skeptical scientist » Mon May 09, 2011 6:41 pm UTC

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.

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Andersmith
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Re: Tricky Box Problem

Postby Andersmith » Fri May 27, 2011 10:12 pm UTC

I found the answer:
http://img194.imageshack.us/img194/410/20110518150141.png

willdawg
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Re: Tricky Box Problem

Postby willdawg » Tue May 31, 2011 8:55 am UTC

I've found a simple solution:

Spoiler:
Begin by noting that, for box A (defined by sides a, b, c) to fit inside box X (defined by x, y, z), we must have
[math]a^2 + b^2 + c^2 \le x^2+ y^2 + z^2[/math]
Proving this is trivial: the long-diagonal is the longest strait line contained in any box, and box A cannot contain a strait line which is longer than the longest strait line in X while being enclosed by X. Therefore, the long-diagonal of A must be less than or equal to the long-diagonal of X.

Noting this, let's imagine that [imath]a + b + c > x + y + z[/imath] while A is enclosed by X. Let's begin by rewriting a, b, and c as perturbations from x, y, and z:
[math]a + b + c \to (x + \delta x) + (y + \delta y) + (z + \delta z)[/math]
where
[math]\delta x = x - a, \delta y = y - b, \delta z = z - c[/math]
We therefore require that [imath]\delta x + \delta y + \delta z \ge 0[/imath]. But we also require that
[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]
[math]\to 2\delta x + delta x^2 + 2\delta y + delta y^2 + 2\delta z + delta z^2 \le 0[/math]
[math]\to 2(\delta x + \delta y + \delta z) + (\delta x^2 + \delta y^2 + \delta z^2) \le 0 ???[/math]
But both of the terms on the left hand side are clearly greater than zero, so we have reached a contradiction.
In fact, requiring that [math]a + b + c > x + y + z[/math] implies that the long-diagonal of A is longer than the long-diagonal of X, and therefore A can never be enclosed by X.

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Re: Tricky Box Problem

Postby skeptical scientist » Tue May 31, 2011 9:23 am UTC

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 left-hand-side 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

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Re: Tricky Box Problem

Postby Ambeco » Thu Jun 02, 2011 7:15 pm UTC

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.


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