- **Puzzles**
(*https://www.mersenneforum.org/forumdisplay.php?f=18*)

- - **Rocky table**
(*https://www.mersenneforum.org/showthread.php?t=7582*)

Rocky tableSince 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=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] |

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. |

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

[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. |

[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. |

[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. |

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 |

[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 |

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=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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