Hole in a Cube
Is it possible to cut a hole in a cube in such a way that the larger cube can be passed through the hole ?
Labels: mathschallenge
posted by ToM at
9:07 AM
Monday, August 17, 2009
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32 Comments:
Yes. Do I win?
Let me make sure I am understanding this correctly: the question is whether or not a hole can be cut into a cube that is large enough for the cube to fit through, correct?
Wouldn't this be akin to eating your own face? Other than cutting the hole, and then flipping the cube inside out though it, I dont see any other way.
Yes, the question is poorly phrased. But that makes it cuter.
Imagine two identical cubes. Make a hole in one, push the other cube through the hole. That's what's being suggested. If not, my head will hurt too much bending the space-time.
It depends on the material, and how you make the cut. If the material is stretchy it could be easy. You may have heard of cutting a playing card in a zig zag pattern to make a hole big enough to walk through. In truth, that card is a 3 dimensional object, similiar to a cube.
If you can cut similar zig zag and the material does not break you can unfold it, and pass a large cube through it.
Ragknot, stretching is cheating.
This problem reminds me that it is possible to square the circle. I can't find a link. But the method involves cutting the circle into some 10s of millions of extremely complex shapes. Amazingly the pieces only need to be slid (i.e. not rotated) to achieve it. It is an exact solution.
cube of solid --> Cube of gas
see where i'm going?
PQ.
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Sure, when you can bend the cube (not necessarely stretching).
Or when you pass them with speeds close to c (speed of light). Because in relativistic physics you don't have rigid bodies.
[Careful here, simple length contraction won't work here, as physics has to be true for both inertial frames.]
Oh and Chris, I think what you're talking about is a special case of the Banach-Tarsky paradox. Where you can cut a sphere in (i think at least 5) crazy shapes and put them together, you get two mathematical identical spheres (also identical to the first one)
greetz
Hi tha b.:.H
SR wouldn't help. The transverse dimensions of the hole aren't affected by length contraction.
Nope, not confused with Banach-Tarski, but it turns out to be closely linked. So I guess you're pretty much right.
I've found a link for the squaring the circle: http://www.maa.org/mathland/mathtrek_11_01_04.html
You will probably find it helpful to find a cube and twirl it about
to see the truth in my early statements. This proof is far from
rigourous, but should satisfy you that it is correct.
Without loss of generality, assume a unit cube.
Largest diagonal in cube = Sqrt(3) (3D Pyhagoras).
Now look at the cube from one corner, so that the opposite corner is
directly below it. You should see a hexagon. Draw in all the
diameters on this plane figure. So you now have 6 equilateral
triangles - making a sort of jewel pattern.
I'll use approx values to reduce eyestrain and typing. The corner
diameters of the hexagon have lengths Sqrt(3). So each equilateral
triangle has side Sqrt(3)/2 = 0.866 (3SF). Now draw a point along
one of the diagonals, 0.5 from the centre of the hexagon. 0.5 is
the half width of a face of the normally oriented cube. So this
point will be 0.366 from the corner of the hexagon. Draw a
perpendicular from that point to intersect an edge of the hexagon.
If the length of that line is greater than 0.5 we are done. And we
are because the length is tan(60 deg) * .366 = .634 (approx).
The exact distance is Sqrt(3)((Sqrt(3)/2-1/2) = 3/2 - Sqrt(3)/2
That is fortunate for me, as I guessed the result after mentally
reorienting a cube.
To Chris. That link u posted works for me. and yes stretching is considered cheating. i remember my wood shop teacher teaching me something to do with cutting a piece of wood (thats y i had woodshop) with some kind of a bit. u put the bit in a drill press put it at ether 200 or 2000 rpm and it cuts a squre for u. who knows what question really asks for.
ps. iam a different chris( notice how one is C(caps.) Chris and another is chris
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On the SR. At normal speeds the whole cube cannot be completely contained inside the holed cube.
On the whole cube, if it were travelling very fast towards the holed cube, the holed cube would seem to be very thin, and it would be like flying through a ring, almost none of the whole cube would be inside the holed cube.
Someone on the holed cube would see the whole cube look more like a square and it would easily be contained (in the direction of motion) inside the holed cube.
Try saying that very fast.
...no pun intended.
Done with pictures!!!
Ok, after reading what Chris said, I see how it can easily be done.
Here's a picture of a cube 1x1x1.
You see the blue line? It goes from one corner to the opposite corner, striaght thru the center of the cube.
http://1.bp.blogspot.com/_gQkeSWqb63Q/SopEr_xZIXI/AAAAAAAAAFo/J3AX3O8FJ1U/s1600-h/Capture+cube+with+a+line.JPG
Now we can bore a hole along the blue line, The diameter of the hole is larger than 1. I suspect the hole might be as large as 1.7 but that would leave almost nothing left. The length of the blue line is 1.7.
After boring the hole, this is what is left of the cube.
http://2.bp.blogspot.com/_gQkeSWqb63Q/SopEy2qK0RI/AAAAAAAAAFw/cF_44hjRNVU/s1600-h/Capture+cube+with+a+hole.JPG
You can pass a cube larger than 1x1x1 through it.
Did you notice from the cube with a hole, that 6 of the 8 corners of the cube remain, and a small portions of the edges connecting them are left?
You can see both pictures here.
http://ragknot.blogspot.com/
Chris, that's why I said that that the length contraction won't work here, it works only in the direction you are flying, and far more important, when you take the inertial frame of the cube with a hole in it the whole cube will look even bigger.
I pointed out that you have to take SR to get rid of rigid bodies. You make a hole in a cube larger then the half of the other cube. Now pass over the hole with great speed, and that means greater than the speed of sound of the cubes material!
Let's have a force into the direction of the hole (lets say gravity or EM-force for a charged cube or whatever) the atoms passing over the hole (of the whole cube ;)) will fall independently, as the information that they are falling won't have reached the other atoms on top of the holed cube. They can't hold the falling atoms back, because when that information reaches them they will also be falling already.
So in a manner you DO bend, even stretch the cube but there's nothing you can do about it. At v>c_sound that is the physical reality, no matter what material you use.
Hi tha b.:.H. I haven't followed a lot of your chain in the second para - don't bother to explain, I've got a pretty good grasp of SR already (and even some GR using tensors).
However, whichever cube you're moving with, you will say the other is physically smaller in the diretion of motion. I don't usually concern myself with what you actually see (with your eyes), but I'll have a think about that later.
Anonymous with the cube of gas - gotcha. You went over my head - nice one.
tha b.:.H re your 1:23 am post.
Of course no materials are truly rigid even in the Newtonian universe. I know my leg isn't.
The SR rigidity problem is due to the dependence on simultaneity.
I've not really thought about the speed of sound in a solid body. I've always assumed that it is more like a group velocity.
Oops, I oversimplified. Rigidity prob also because of propagation of signal speed. Which is sort of what you were talking about.
Yap, i think you now got the point chris =) speed of sound -> mechanical signal propagation speed ^^
well, GR is pretty complicated, haven't understood it in great depth, especially those Christoffel symbols...
tensoranalysis is tricky but in principle is a good handling of derivatives and coordinate transforms.
If you want a really good and founded introduction check this one out:
http://timms.uni-tuebingen.de/Browser/Browser01.aspx?path=%2fUniversit%c3%a4t+T%c3%bcbingen%2fFakult%c3%a4t+f%c3%bcr+Mathematik+und+Physik%2fPhysikalisches+Institut%2fLecture+Introduction+to+General+Relativity+WiSe+2008-2009%2f
have fun
Hasn't anyone looked at the pictures of gave for a solution above?
I think you could push a 1.5 x 1.5 x 1.5 cube thru a 1 x 1 x 1 cube, if the hole was cut like this.
Ragknot, not sure which picture you mean. I think (with guessed best orientation) that 1.098... is the biggest.
Ragknot. OK, I'll have a think tomorrow. Not sure if the holed cube would be in one piece (with a just below 1.5 hole). I'm too tired too visualise it and I don't have CAD and it's almoost 4am in the UK.
Tilt the big cube and you can
comment posted by:
Vincent Jumper
Actually I mean to say tilt the small cube so one corner is pointing up. Opposite corners on a cube are greater distance than ajacent corners.
Example:
the squareroot of (5 squared + 5 squared) > 5
Comment posted by Vincent Jumper
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