20070321, 12:53  #1 
"Lucan"
Dec 2006
England
194A_{16} Posts 
Rocky table
Since Mally has gone off in a huff, I shall put this teaser to you.
A four legged table typically rocks when placed on an unlevel floor. However, with two simple assumptions, you can rectify the problem by rotating the table through an angle <90 degrees. How come? and what are the assumptions? David 
20070321, 13:01  #2  
Nov 2003
2^{2}×5×373 Posts 
Quote:
The height of the floor at each point is uniformly random. [the interval does not matter] 

20070321, 13:11  #3 
Sep 2006
Brussels, Belgium
3135_{8} Posts 
Round table, the four legs are inside the table top.
You tilt the table, it rests on three points only and thus does not rock anymore. 
20070321, 13:24  #4 
"Lucan"
Dec 2006
England
194A_{16} Posts 
Neither of these were the answers I had in mind.
But at least I got some instantaneous response 
20070321, 13:29  #5 
"Lucan"
Dec 2006
England
6474_{10} Posts 

20070321, 13:31  #6 
"Lucan"
Dec 2006
England
2·3·13·83 Posts 

20070321, 13:36  #7  
Jun 2005
565_{8} Posts 
Quote:
Uniformly? random? interval? Can you explain what you mean please? The only thing I can imagine you might be meaning is that the floor say of my office being discribed by the rectangle [0,3]x[0,5], for every point in that space, there is an identically and independently distributed (iid) (uniform) U[0,b] random variable, where b would be the end of the interval you are referring to. That would be a hell of a floor, and it is not obvious (at least for me) why you should get the table fixed. Yours, H. 

20070321, 14:08  #8 
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
When I first told my good friend (now a professor at Princeton)
of my discovery, he went into deep thoght for a minute then replied in excitement "Yes. And we can make it rigorous". (That was about 35 years ago). He didn't go into the details though. David Last fiddled with by davieddy on 20070321 at 14:18 
20070321, 14:12  #9  
Nov 2003
2^{2}·5·373 Posts 
Quote:
[a,b]. The values of a and b do not matter. Rotate the table using one leg as a pivot. Any three legs must be coplanar. That there exists *some* rotation where the 4th point is in the same plane follows from e.g.: The Ham Sandwich Theorem, or Browder's Fixed Point Theorem 

20070321, 15:32  #10 
Jun 2005
373 Posts 
So, for the sake of understanding:
is it like I described, space $R^2$, for any $x\inR^2$ you take a U[0,1] r.v., all iid? This floor would be as I said highly discontinuous. Or do you mean by "uniformly random" "arbitrary, but continuous". In this case, I can vaguely figure how it should work with what you cited (I think you meant Brouwer's fixed point theorem). 
20070321, 16:52  #11  
Nov 2003
2^{2}×5×373 Posts 
Quote:


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