[Guest]

Hang a picture on a wall with N nails, such that, if any of the N nails failed, the picture would fall on the floor.

Remarks:

The picture is provided with a sufficiently long hanging thread, which together with the upper frame, forms a single loop.

Only the thread may touch the nails, and the picture is said to have fallen on the floor when the thread is not touching any nail.

Any nail could support the picture without the help of the other nails, if the picture is hung properly on it.

Edited December 14, 2010 by rookie1ja

[Guest]

N = 1

Are we looking for a maximum value for N?

[plasmid]

Make a semicircle of N nails with radius R (its diameter must be larger than the picture frame's top), and drape the string around that semicircle. If any one nail were to fall out (excluding one of the two nails on the edge), then the length of string that would be taken up by the semicircle will decrease by a calculable amount. I think it's R*[4sin(180/2(N-1))-2sin(180/(N-1))] but I haven't checked all that closely to be honest.

The trick is to hang the picture a distance off the floor that is less than half that amount. It could then meet the OP's requirement of falling "to the floor".

If one of the two nails on the edge of the semicircle falls out, then the drop should be even larger as long as the picture frame is significantly smaller than the diameter of the semicircle.

Edited December 14, 2010 by plasmid

[Guest]

Make a semicircle of N nails with radius R (its diameter must be larger than the picture frame's top), and drape the string around that semicircle. If any one nail were to fall out (excluding one of the two nails on the edge), then the length of string that would be taken up by the semicircle will decrease by a calculable amount. I think it's R*[4sin(180/2(N-1))-2sin(180/(N-1))] but I haven't checked all that closely to be honest.

The trick is to hang the picture a distance off the floor that is less than half that amount. It could then meet the OP's requirement of falling "to the floor".

If one of the two nails on the edge of the semicircle falls out, then the drop should be even larger as long as the picture frame is significantly smaller than the diameter of the semicircle.

"the picture is said to have fallen on the floor when the thread is not touching any nail."

In other words, the distance to the floor does not matter: the same effect should be obtained if you take you take out any nail, or all nails, or any combination of them. Both the picture and the whole thread are on the ground then.

Edited December 14, 2010 by Kornrade

[Guest]

N = 1

Are we looking for a maximum value for N?

N =1 is trivial. solve it for N > 1

The final goal should be: either prove that for a certain number of nails, the task is impossible, or find a general solution for any number of nails

But, as in many problems, start simple: N = 2

Edited December 14, 2010 by Kornrade

[Guest]

Pound the first nail into the wall. Bend the remainder such that they can be linked together into a chain between the thread and the nail in the wall.

[Guest]

Use nails with very small heads and angle them such that the hanging thread is held in place but the slightest additional load will pull the thread off the nail head. With this orientation, you could use potentially infinite nails so long as there are no nails directly above or below one another as the load increases on each remaining nail as the first fails followed by others.

Another option would be to use nails with small diameters, hammering them only slightly into the drywall with the heads at a downward angle. The pulling force on each nail should be just under what it would take to remove the nail. In this scenario, one failure again results in an increased load on all other nails, pulling them from the drywall.

Yet another way, would be to never have a nail on the inside of the loop created by the hanging thread and the frame. Weave the 'doubled' thread around an array of nails such that no nail has the thread completely surrounding it and no three consecutive nails form a line. The friction generated by the weaving must be sufficient to keep the thread in place, but weak enough that the removal of one nail anywhere along the array would have the force of gravity overcome the friction; thus, sending the picture to the floor.

[Guest]

Use nails with very small heads and angle them such that the hanging thread is held in place but the slightest additional load will pull the thread off the nail head. With this orientation, you could use potentially infinite nails so long as there are no nails directly above or below one another as the load increases on each remaining nail as the first fails followed by others.

Another option would be to use nails with small diameters, hammering them only slightly into the drywall with the heads at a downward angle. The pulling force on each nail should be just under what it would take to remove the nail. In this scenario, one failure again results in an increased load on all other nails, pulling them from the drywall.

Yet another way, would be to never have a nail on the inside of the loop created by the hanging thread and the frame. Weave the 'doubled' thread around an array of nails such that no nail has the thread completely surrounding it and no three consecutive nails form a line. The friction generated by the weaving must be sufficient to keep the thread in place, but weak enough that the removal of one nail anywhere along the array would have the force of gravity overcome the friction; thus, sending the picture to the floor.

There is no trickery involving the nails, thread, picture, friction or any other particularity of the materials. The solutions should work also in a scenario with long thick nails and without friction.

[k-man]

One nail is above the frame and 1 nail is below. The thread from both sides of the frame is combined and looped around the top nail and then hooked to the bottom nail from the top.

[Guest]

One nail is above the frame and 1 nail is below. The thread from both sides of the frame is combined and looped around the top nail and then hooked to the bottom nail from the top.

I don't know, because I don't see clearly the thread.

But keep in mind that any correct solution for N = 2 has an equivalent solution with both nails next to each other, above the picture.

For pictures, please draw a thicker thread and use colour if necessary to avoid confusion

[Guest]

It seems to me that if you put one nail above the other, that if the top nail failed then the nail below would fail also because the bottom nail would not be able to "catch" the wire, therefore bending and failing also. It is either that or some complicated math thing which I know nothing about.

[Molly Mae]

You could create a vertical line such that the bight of the thread will "slolam" around each nail on the way up, wrap around the top nail, and "slolam" back down the opposite side of each nail. The bight could then be secured on the first nail beneath the beginning of the thread. I'll draw something up in mspaint shortly.

It's more in the way the bight of the thread is secured than the layout of the nails, though.

[Guest]

It seems to me that it is impossible for N>1. Proving it is a bit tricky...

If you forget about the picture frame for a moment, we have a simple loop. It just happens that part of it is rigid. My understanding of the OP is that the picture hangs if there is at least one nail inside the loop, and falls if there is no nail inside it. In order for it to fall by removing any nail, we have a contradiction that all nails must simultaneously be inside and outside the loop. This could be overcome if there were a way of wrapping the loop around the points to form a second closed loop, but there is no way of doing so without either tying a knot in the ends of the original loop (which is pointless), or hooking part of it over one nail.

I don't think knots help, because any knot can be reduced to a braided loop, and we're essentially back to where we started.

[Guest]

It is very clearly impossible without involving gimmickry.

[Molly Mae]

It seems to me that it is impossible for N>1. Proving it is a bit tricky...

If you forget about the picture frame for a moment, we have a simple loop. It just happens that part of it is rigid. My understanding of the OP is that the picture hangs if there is at least one nail inside the loop, and falls if there is no nail inside it. In order for it to fall by removing any nail, we have a contradiction that all nails must simultaneously be inside and outside the loop. This could be overcome if there were a way of wrapping the loop around the points to form a second closed loop, but there is no way of doing so without either tying a knot in the ends of the original loop (which is pointless), or hooking part of it over one nail.

I don't think knots help, because any knot can be reduced to a braided loop, and we're essentially back to where we started.

I had thought that it was considered hanging when at least one nail was inside the loop. When I read back, though, it merely says that the thread must be touching at least one nail.

[Guest]

One nail on each edge of the picture, securing the thread to the frame, one in the middle on the wall. Remove the wall nail and down it comes. Remove either nail on the corner of the picture and the thread is no longer secure and the picture falls.

[Guest]

Disregard last transmission.

Take the loop, stretch it out, and fold it in half around one nail. Put the other nail through the two lobes of the free ends.

[Molly Mae]

But I've also discovered that my method only works if the lowest nail is removed unless the bight folds under the crossing between each nail (I'm sure that's confusing) and enough slack is given in order to allow the frame to gain enough downward momentum to pull free the other slip-folds at each thread intersection. It would be impossible to calculate, though, without mass of the frame and maximum tension of the thread. As those two are not provided, I will disregard this approach.

However, if I'm feeling creative tonight, I'll make a video and upload it.

[Guest]

Can the nails also be holding up the wall? Does your solution involve the wall falling down? Can I use my anti-gravity nails?

[Guest]

Disregard last transmission.

Take the loop, stretch it out, and fold it in half around one nail. Put the other nail through the two lobes of the free ends.

If I'm following your description correctly, (I might have mis-understood) what you're describing is in diagram A.

I tried this with some pens, a notepad and an elastic band. When I removed the bottom nail (pen) I got what I've drawn in diagram B.

[Guest]

I will say this again, it is very easy to show that this puzzle is unsolvable given that we are not using a gimmick of some sort (trick nails, no gravity etc..)

In order for the painting to hang on the wall, AT LEAST 1 nail must be within the closed loop. It therefore stands that when N > 1, you will always be able to choose a nail which leaves at least one nail in the loop, leaving the painting on the wall.

The only case for which this problem is solvable is N = 1.

Stop banging your heads against a wall!

Edited December 14, 2010 by Ring Ring

[wolfgang]

we can hang the picture to a middle nail carried on two other nails.

[Guest]

we can hang the picture to a middle nail carried on two other nails.

That seems the only likely explanation: nailing a nail into the head of the preceeding nail, and hanging the picture on the last, topmost, nail. This is sort of a "trick nail" as you normally cannot nail into a head of a nail.

[Guest]

from the original text:

..."if any of the N nails failed, the picture would fall on the floor."

..."Any nail could support the picture without the help of the other nails"

These two statements are self-contradictory, no?

Given the latter statement, the former can never be true...(i.e there is NO way to "hang a picture on a wall with N nails, such that, if any of the N nails failed, the picture would fall on the floor."

...unless I am missing/misinterpreting something...

I've got to go with Ring Ring and say that this is unsolveable for N <>1...

Smoo

[superprismatic]

Yes, it can be done for any N. Here's my drawing for N=2. Things get pretty messy as N increases.

[superprismatic]

One nail is above the frame and 1 nail is below. The thread from both sides of the frame is combined and looped around the top nail and then hooked to the bottom nail from the top.

Please tell me how to put an attached image into a spoiler like you did here.

I tried to do it for my last post but I couldn't figure it out. Thanks.