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 2007-03-21, 12:53 #1 davieddy     "Lucan" Dec 2006 England 2·3·13·83 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
2007-03-21, 13:01   #2
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by davieddy 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
I can do better: ONE simple assumption:

The height of the floor at each point is uniformly random. [the interval
does not matter]

 2007-03-21, 13:11 #3 S485122     Sep 2006 Brussels, Belgium 32×181 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.
 2007-03-21, 13:24 #4 davieddy     "Lucan" Dec 2006 England 2×3×13×83 Posts Neither of these were the answers I had in mind. But at least I got some instantaneous response
2007-03-21, 13:29   #5
davieddy

"Lucan"
Dec 2006
England

647410 Posts

Quote:
 Originally Posted by S485122 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.
Until you try to cut the steak you're eating.

2007-03-21, 13:31   #6
davieddy

"Lucan"
Dec 2006
England

194A16 Posts

Quote:
 Originally Posted by R.D. Silverman I can do better: ONE simple assumption: The height of the floor at each point is uniformly random. [the interval does not matter]
I'm not sure I dined at that restaurant.

2007-03-21, 13:36   #7
hhh

Jun 2005

373 Posts

Quote:
 Originally Posted by R.D. Silverman I can do better: ONE simple assumption: The height of the floor at each point is uniformly random. [the interval does not matter]
Is that you, Mr. Silverman?

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.

 2007-03-21, 14:08 #8 davieddy     "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 2007-03-21 at 14:18
2007-03-21, 14:12   #9
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by hhh Is that you, Mr. Silverman? 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.
The height of the floor is a uniform random variable taken over some interval
[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 co-planar. 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

 2007-03-21, 15:32 #10 hhh     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).
2007-03-21, 16:52   #11
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by hhh 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).
Highly discontinuous? Indeed!! It is nowhere continuous.

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