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bonanova
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* takes a quick glance at the puzzle title, hoping no one notices, ;) and then confidently replies *

Three.

I think it is possible

if the smaller cube is just slightly smaller than the bigger one, and the hole is from 1 corner drill through to the opposite corner (in 3 D manner), with the diameter of the edge of the bigger cube.

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I think it is possible

if the smaller cube is just slightly smaller than the bigger one, and the hole is from 1 corner drill through to the opposite corner (in 3 D manner), with the diameter of the edge of the bigger cube.

Hi woon,

I don't know how to discuss your answer without spoiling the puzzle, so ...

let me promise to give you a reason later but say for now your solution won't work. ;)

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Was thinking if ...

When, for example, the Length of a side of the smaller cube is x and name this cube A. The Length of a side of the larger cube is x+1 and name this cube B. You drill a hole into A into face FA in such a manner that the four corners of FA lies on the circumference of the circle. This will make the each side of FA a chord in the circle and the diameter of a circle is the longest chord in that circle, a larger size cube should be able to go through it. :D

=>

Hope it is confusing enough. j/k. Let's see what some people have to say about this BS

EDIT: Typo :P

Edited by Ben Law
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Was thinking if ...

When, for example, the Length of a side of the smaller cube is x and name this cube A. The Length of a side of the larger cube is x+1 and name this cube B. You drill a hole into A into face FA in such a manner that the four corners of FA lies on the circumference of the circle. This will make the each side of FA a chord in the circle and the diameter of a circle is the longest chord in that circle, a larger size cube should be able to go through it. :D

=>

Hope it is confusing enough. j/k. Let's see what some people have to say about this BS

EDIT: Typo :P

So ... would that be possible?

Equivalent approach:

Think of the cubes as equal in size.

See if that problem is solvable, and if so, is there any extra clearance?

If so, then the one with the hole can be shrunk very slightly and preserve the solution.

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I think Woon has it, but here's a diagram to make it clearer:

Looked at corner-on, a 1cm cube is equivalent to a hexagon of side ((2/3)^0.5)cm (I think)

You can just about squeeze a 1cm square into that, so a slightly larger cube would also go

post-4017-1223277307_thumbjpg

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I think Woon has it, but here's a diagram to make it clearer:

Looked at corner-on, a 1cm cube is equivalent to a hexagon of side ((2/3)^0.5)cm (I think)

You can just about squeeze a 1cm square into that, so a slightly larger cube would also go

post-4017-1223277307_thumbjpg

Exactly. ;)

he implied a circular hole would be drilled.

It turns out that the inserted cube's face diagonal would have to be the diameter of that drilled hole, and that is exactly the diameter of the inscribed circle of the hexagonal cross section that optimally provides room for the cube to pass, given that it has a preferred angular position. I have a drawing - like yours - that shows the inscribed circle.

The upshot is that for equal size cubes, it passes through, but with zero clearance and 0-thickness walls at some points. Therefore the cube with the hold could not be shrunk to satisfy the puzzle.

If Woon had not used a circular hole, his answer would have been correct.

So ... woon ... if you're reading ... that's why.

-bn

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Exactly. ;)
he implied a circular hole would be drilled.

It turns out that the inserted cube's face diagonal would have to be the diameter of that drilled hole, and that is exactly the diameter of the inscribed circle of the hexagonal cross section that optimally provides room for the cube to pass, given that it has a preferred angular position. I have a drawing - like yours - that shows the inscribed circle.

The upshot is that for equal size cubes, it passes through, but with zero clearance and 0-thickness walls at some points. Therefore the cube with the hold could not be shrunk to satisfy the puzzle.

If Woon had not used a circular hole, his answer would have been correct.

So ... woon ... if you're reading ... that's why.

-bn

Nice. To give Woon credit, though, I think he didn't mean a circular hole, it's just that English is not his first language...;P

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post-1048-1223278323_thumbgif

If you can see the green circle, what you have is three figures, looking along body diagonal of the cube with a hole[black hexagon]

one of the cube axes [red] of the other cube, and

the inscribed circle [green] of the hexagon, for which the red figure is the inscribed square.

If a circular hole is drilled along the cube diagonal, a red cube of the same size as the black cube passes through, with zero clearance.

Also, the walls of the black cube have zero thickness at six points.

So you can't drill a circular hole and use it to pass a cube through a smaller cube.

But if you create a square hole slightly larger than the red cube face, it will clear the black cube walls and permit the red cube to pass, even if you shrink the black cube ever so slightly.

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Exactly. ;)
he implied a circular hole would be drilled.

It turns out that the inserted cube's face diagonal would have to be the diameter of that drilled hole, and that is exactly the diameter of the inscribed circle of the hexagonal cross section that optimally provides room for the cube to pass, given that it has a preferred angular position. I have a drawing - like yours - that shows the inscribed circle.

The upshot is that for equal size cubes, it passes through, but with zero clearance and 0-thickness walls at some points. Therefore the cube with the hold could not be shrunk to satisfy the puzzle.

If Woon had not used a circular hole, his answer would have been correct.

So ... woon ... if you're reading ... that's why.

-bn

I knew what you mean now!

While others are mislook on the 3 D

I keep thinking hole must be circular!

The hole should be as the 2-D shape of the bigger cube, that's it. Right?

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