I recently heard the following puzzle: There are three nails in the wall, and you want to hang a picture by wrapping a wire attached to the picture around the nails so that if any one nail is removed the picture still stays but if any two nails are removed then the picture falls down. An answer, schematically, is given by abca-1b-1c-1 (i.e., wrap it clockwise around each of the three nails in some order, then counterclockwise around each of the three nails in the same order).


This reminded me very much of the [Borromean rings](http://en.wikipedia.org/wiki/Borromean_rings%20%22Borromean%20rings%22), where three rings are linked but when any one ring is removed the other two become unlinked. So I was trying to figure out if there might be some way to transform one situation into the other. My first instinct was to put the rings in S3 and have one of them pass through ∞, but that isn't really right. What seems to be tripping me up is that with the picture there's an extra object (the wire) that doesn't show up with the Borromean rings, but I have a vague idea that perhaps we could change the former situation by saying that we make the loop in the complement of three unlinked rings, and then perhaps "pulling really hard on the wire" would somehow thread the rings together. Maybe my issue is that what's *really* going on with the picture is just that we're making a loop in the complement of three points in plane, and I'm confounding phenomena of different dimensions...


Does anyone have any ideas?



relationship between borromean rings and hanging-a-picture-from-three-nails puzzle?
This question is related to this post and its answers:
[Collapsible group words](https://mathoverflow.net/questions/15316/collapsible-group-words)


You may think of the three nails as giving a 3-punctured
plane, which has fundamental group a rank 3 free group. 
An element of this group may be thought of as pushing a point
around in the surface. If one takes the trace of this motion
in time, you get a braid with 4 strands, three strands of which
are straight. This represents an element of the pure braid group, an
example of the Birman exact sequence. Closing up the
braid, you get a four component link. Your condition implies 
that removing any two of the last three strands gives the trivial link, so it is 
a kind of second order Brunnian condition. 


