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 davieddy 2007-03-21 12:53

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

 R.D. Silverman 2007-03-21 13:01

[QUOTE=davieddy;101613]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[/QUOTE]

I can do better: ONE simple assumption:

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

 S485122 2007-03-21 13:11

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.

 davieddy 2007-03-21 13:24

But at least I got some instantaneous response:smile:

 davieddy 2007-03-21 13:29

[quote=S485122;101615]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.[/quote]

Until you try to cut the steak you're eating.

 davieddy 2007-03-21 13:31

[quote=R.D. Silverman;101614]I can do better: ONE simple assumption:

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

I'm not sure I dined at that restaurant.

 hhh 2007-03-21 13:36

[QUOTE=R.D. Silverman;101614]I can do better: ONE simple assumption:

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

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.

 davieddy 2007-03-21 14:08

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

 R.D. Silverman 2007-03-21 14:12

[QUOTE=hhh;101620]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.[/QUOTE]

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

 hhh 2007-03-21 15:32

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 [URL="http://mathworld.wolfram.com/BrouwerFixedPointTheorem.html"]Brouwer's fixed point theorem[/URL]).

 R.D. Silverman 2007-03-21 16:52

[QUOTE=hhh;101631]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 [URL="http://mathworld.wolfram.com/BrouwerFixedPointTheorem.html"]Brouwer's fixed point theorem[/URL]).[/QUOTE]

Highly discontinuous? Indeed!! It is nowhere continuous.

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