Box and string

[ceptimus]

Imagine a rectangular box. The dimensions of the box are 2 units x 1 unit x 1 unit. Firmly attached to one corner of the box is a length of string. The other end of the string (when swung every which way) will reach any point of the box. If the string were any shorter then this would not be possible.

To the nearest one thousandth of a unit, what is the length of the string?

[Tanda]

Is a unit equal to an inch??

[Moxy]

2.449 units

I think.

:scratch:

[Crumb]

Isn't that :sqroot:3 Units?

[Crumb]

Oh nm... I am stuck in 2 dimensions...

ok is it :sqroot:5 units?

[ceptimus]

Is a unit equal to an inch??It can be, if you like Tanda. It doesn't really matter. If you assume the units are inches, then the answer will be in inches. If you assume that the units are feet, or metres or whatever, then the answer will still be the same number of those units.

[Moxy]

2.449 = :sqroot:6

[ceptimus]

No one has the right answer yet. I should make it clear that the string is on the outside of the box. It has to be long enough so that it can wrap around the outside in order to touch the point farthest away from the corner it is attached to.

[Moxy]

Hm. Am I wrong to assume the string is on the inside of the box?

If it's outside, it would have to be 3.236

[Joshua Adams]

2.449 units

I think.

:scratch:I was thinking the same thing... If you imagine the string as a diagonal (i.e., attached inside the box), then the distance to the opposite inside corner is just the vector sum of the sides of the box, :sqroot:(22 + 12 + 12) = :sqroot:(6) = 2.449.

But, it might be outside the box as well. In which case, to reach the most distant inside corner it would need to be at least the vector sum of the above diagonal plus the height of the box (imagine the string travelling up one side and then turning downward into the box, and going across to the other corner). Thus, :sqroot:(12 + 6) = :sqroot:(7) = 2.646.


Have I overanalyzed it?

Edit, Xpost. Where'd your new answer come from, Moxy?

[ceptimus]

Hm. Am I wrong to assume the string is on the inside of the box?

If it's outside, it would have to be 3.236That is 1 + sqrt(5), I think. But it's not the correct answer.

[Ari]

guess 3.414213562u

[Moxy]

Edit, Xpost. Where'd your new answer come from, Moxy?

If it was ouside, it would still have to reach the opposite corner, and the shortest way would be across the diagonal of the largest side (:sqroot:5) plus the length of one of the shorter sides (1).

[ceptimus]

<snip>

. Thus, :sqroot:(12 + 6) = :sqroot:(7) = 2.646.


Have I overanalyzed it?

Edit, Xpost. Where'd your new answer come from, Moxy?Still not right. :nope:

[Joshua Adams]

2.449 units

I think.

<snip>

. Thus, :sqroot:(12 + 6) = :sqroot:(7) = 2.646.


Have I overanalyzed it?

Edit, Xpost. Where'd your new answer come from, Moxy?Still not right. :nope:Yeah, I just realized that can't be right... the string is travelling along two dimensions, so it wouldn't be that vector length. It'd just be the square root of 6 plus 1, or 3.449, if my other assumptions were correct.

[ceptimus]

I'm going to bed now, so I'll just post a clue. It is not that big a spoiler, just a clue to get you thinking along the right lines, and an upper and lower limit that you can test your answers against.

How did you decide which point on the box is farthest away from the corner that the string is attached to?

The string needs to be longer than 2.8 units, but shorter than 3.0 units.

[Julie]

You peoples and your fancy :sqroot: This kinda math was way I neved done gradumated from Hi Skool!

[Dragar]

I think Ari has it.

Edit: But not in light of ceptimus post...

[Ari]

Ha, yeah my guess is way off. I went the simple but long way to what I think/thought/think is the farthest point.

[Crumb]

Is the string attached inside or outside the box? :scratch:

[Dragar]

Edit: Nevermind.

[Moxy]

Shit...

ok, I have seen the answer, and I still don't believe it. I think I'm going to have to find a box and some string to test experimentally.

[yguy]

Imagine a rectangular box. The dimensions of the box are 2 units x 1 unit x 1 unit. Firmly attached to one corner of the box is a length of string. The other end of the string (when swung every which way) will reach any point of the box. If the string were any shorter then this would not be possible.

To the nearest one thousandth of a unit, what is the length of the string?

1000.000 units.

Well, you didn't ask for a minimum length, after all. :cool:

Oh, all right. The string is (5^.5) + 1 = 3.236 units long.

OK, that's wrong too. Sue me. :)

[yguy]

I'm going to bed now, so I'll just post a clue. It is not that big a spoiler, just a clue to get you thinking along the right lines, and an upper and lower limit that you can test your answers against.

I don't buy it. I say you get the minimum length by menatlly "unfolding" the box, and that gives you 10^.5=3.1623.

[Dingfod]

2.818, the distance across two 1 x 1 right triangles, the distance to the opposite corner.

[Moxy]

Well, you didn't ask for a minimum length, after all. :cool:

It does say, "If the string were any shorter then this would not be possible."

[quiet bear]

I know enough math to balance my checkbook.

This, I don't need to know.

[Dingfod]

What if you were balancing your checkbook on one corner of a box and the pencil was attached to a string at the farthest point from the checkbook? How long would the string have to be to allow you to correct an error in your checkbook?

[quiet bear]

Then I'd be screwed. I'd have to rob banks to pay bills.

[Ari]

Ding: Easy, you realize that if you only have enough money for one pencil, it obviously isn't your checkbook and you don't need to balance it. :)

[Dragar]

I'm going to bed now, so I'll just post a clue. It is not that big a spoiler, just a clue to get you thinking along the right lines, and an upper and lower limit that you can test your answers against.

I don't buy it. I say you get the minimum length by menatlly "unfolding" the box, and that gives you 10^.5=3.1623.

Try unfolding it in the following way. Leave your string attached to the bottom left of a small face (face A). Now keep that facing the same way, and unfold all the big sides so they lay in a row. So (labelling the big faces B C D E) it will look something like:



A
B C D E
F

F is the small face opposite A.

Now, the opposite corner to the bottom left of A is bottom right of F. But it is equally valid for F to be attached to C (it's just a matter of unfolding it differently). So do that, match up which corner is which (note that F is rotated when attached to C as compared to B!), draw a straight line, and calculate the distance.

I'd suggest drawing a diagram, which includes as many ways of unfolding as possible.

I am finding my diagram very helpful. :yup: But remember that what you have is a 2D map of a 3D space. Sometimes your diagram will seem to outright lie.

[slimshady2357]

2.818, the distance across two 1 x 1 right triangles, the distance to the opposite corner.

After some practical experiments, this seems to me to be the correct answer.

[Dragar]

I have an answer that is slightly larger than that. I feel I have cheated though - I did derive the answer, but only after reading an explanation of what was going on, and then spending a good few hours working out how to draw shortest-paths as straight lines on my paper.

2.851 is my answer.
Once you have done figured out how to draw the multiple shortest paths as straight lines on the same diagram, you can quite quickly trace out the area a string of root(8) would cover on the opposite small face, set up some axes, jot down the equations of four circles, and solve them to reveal the minimum radius where all four circles converge - the sufficient and neccessary condition for the entire small face area to be covered.


Good puzzle Ceptimus! It is very counter-intuitive, and I like things like that. I have learned quite a bit from it and enjoyed working my way through, even if I couldn't figure it all out under my own steam.

[ceptimus]

Sorry, but 2 * sqrt(2) (2.818) is not the answer.

Dragar is just a tiny bit out with his answer - to the nearest one thousandth, the string is actually (warning, exact answer behind spoiler below):

2.850 The exact answer is sqrt(65 / 8) which is about 2.85044

The counter intuitive part of this puzzle is that the farthest point (in terms of distance travelled over the surface) from the corner of the box is not (as one first assumes) the diagonally opposite corner - it is a point on the opposite small face, one quarter of a unit from two of the edges.

Thanks to everyone who took part. :)

[Moxy]

Yeah, I read that, and honestly, I still don't believe it.

[Moxy]

I think Dragar's answer of 2.851 is saying that you can't round the answer down, because the string would then be too short. You have to round it up.

[ceptimus]

I think Dragar's answer of 2.851 is saying that you can't round the answer down, because the string would then be too short. You have to round it up.Yes, I'm sure you're right. I guess Dragar has the exact answer in terms of the square root, and rounded up, as you suggest.

[Moxy]

Now, in your answer, you say:

(in terms of distance travelled over the surface)

For what reason do you need to travel over the surface?

[ceptimus]

Now, in your answer, you say:

(in terms of distance travelled over the surface)

For what reason do you need to travel over the surface?Well, if it's a real world box, with the string fastened to an outside corner, then the string can't pass through the box. I admit my OP didn't make this clear, but I did clarify it after about the first five posts.

If the string is inside the box, then the opposite corner is the farthest away, but the problem is trivial - the answer being sqrt(6).

[Moxy]

God fucking dammit. The one post in this entire thread I missed.

[Moxy]

Well at least I'm not insane, just illiterate.

[yguy]

Sorry, but 2 * sqrt(2) (2.818) is not the answer.

Dragar is just a tiny bit out with his answer - to the nearest one thousandth, the string is actually (warning, exact answer behind spoiler below):

How do you come to the conclusion that the point most distant from the corner is not the diagonally opposite corner?

[yguy]

Dingfod, you are correct that 2.828 is the shortest distance across the surface to the opposite corner, but I do not see how a string that length could reach all points on the surface of the box.

[Dragar]

I think Dragar's answer of 2.851 is saying that you can't round the answer down, because the string would then be too short. You have to round it up.Yes, I'm sure you're right. I guess Dragar has the exact answer in terms of the square root, and rounded up, as you suggest.

:yup: I bolded the 1 for a reason.

[Moxy]

So now I have to start all over and figure out how the hell to get 2.851. I guess I won't be getting anything done at work today.

[Dragar]

How do you come to the conclusion that the point most distant from the corner is not the diagonally opposite corner?

The best way is geometrically. Start with the opposite corner at (0,0) on your axis, and 'unfold' the box. You may have to play around with how to represent multiple unfoldings at once, and the fact you can 'wrap' around from one corner to another, amongst other things. Then examine the shortest paths (they will be straight lines across two rectangular faces on your unfolded box) to the opposite face. You should find that with a length of root(8) (which is needed to get to the diagonally opposite corner) you can't cover all the face, even using multiple paths.

[ceptimus]

Bonus puzzle (Warning! This is hard):

Okay, we now know that the farthest point from a corner of the box (travelling over the surface) is not the opposite corner, and that we need a string of length 2.851 to reach.

What if the string were attached to some point on the box other than a corner? Might it now need to be even longer, so that it can reach every point on the surface?

What is the longest string required to reach from any point on the box to any other point?

Where is the attachment point that requires the longest string?

[Dragar]

My intuition says that in the middle of one of the faces would require the longest piece of string to reach any point. But we know what intuition is like. ;)

[Dingfod]

Dingfod, you are correct that 2.828 is the shortest distance across the surface to the opposite corner, but I do not see how a string that length could reach all points on the surface of the box.OMG, the correct answer as stated is a whopping .023 longer than mine. If the units were feet, it would be a hair over 1/4" longer than my totally inadequate string. I don't know, it just seemed logical to me that if the string can span the distance between the two most removed corners, it can reach anywhere on the box. But, then again, I was only thinking corners and edges.

[Dingfod]

Bonus puzzle (Warning! This is hard):


What is the longest string required to reach from any point on the box to any other point?

Where is the attachment point that requires the longest string?
3.618, attached in the exact middle of one 1x1 end.

[ceptimus]

:nope:

It's counter intuitive again. The point(s) you're after are neither on a corner, an edge, nor the centre of a face.

[Dingfod]

I'm all hosed up anyway, a 3 unit string would reach any point on the box from any point.