Rocky table

[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

[R.D. Silverman]

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]

[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.

[davieddy]

Neither of these were the answers I had in mind.
But at least I got some instantaneous response

[davieddy]

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.

[davieddy]

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.

[hhh]

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.

[davieddy]

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]

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

[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).

[R.D. Silverman]

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.