Can a four-legged table standing on an uneven (but smooth) floor always be turned so that all four legs...

In article <526uoj$kt5@spectrum.tcns.co.uk>, mdecker@tcns.co.uk
(Michael Decker) wrote:

> Yup, I'm having trouble too.  I'm thinking of a table with legs 12"
> apart, with one leg hanging exactly over the middle of a deep hole of
> 11" radius.  Rotating the table around the centre of the table-top
> will never produce four points of contact, as there will always be one
> or more legs over the hole.

mag@ncsa.uiuc.edu (Tom Magliery) wrote:

> You're all just not quite jumping far enough out of your system...  I
> had the same problem until I read one of the postings here, which you
> may have missed.  I don't remember who wrote it.
>
> Where you're going wrong is that you're thinking only of turning the
> table around its perpendicular axis.  Instead of doing this, think of
> rotating it while holding your hands firmly on top of the table above
> two adjacent legs, thereby ensuring that those two legs always remain
> firmly on the floor.

Right--though it would be nice to have a more mathematical-looking
description of this process.  (Keep the center of the table above the
same point, letting the plane of the top swing around?)

> Note that, as someone else pointed out, this only works for a square
> table.  If anyone's given a similarly intuitive explanation for
> non-square tables, I missed it.  Seems like something similar oughta
> work for non-square tables, but you don't have the fact that A,B,C end
> up occupying the exact same 3 points formerly occupied by B,C,D.

I think someone did post a solution that works for rectangular tables.
If you rotate 180 degrees, you'll have exchanged positions of legs.  Two
legs which were 'collectively too short' will now be 'collectively too
long', and there must be some point in between at which they were
'collectively just right'.  (A pair of legs is collectively too long if
the sum of the distances of the top of the legs to their points on the
floor is greater than the sum of the lengths of the legs; and mutatis
mutandis for 'collectively too short' and 'collectively just right'.)

> By the way, in the example described in the quoted part above, if the
> hole is circular, you can touch all four legs to the floor by orienting
> the table so that two of its sides are parallel to a tangent to the
> circle, and leaning the table into the circle until the two legs that
> are inside touch the walls of the hole.

Remember too that the floor is differentiable at all points.  The
'hole' has to have curvy 'edges' (or maybe, *no* edges, I'm not sure how
to describe it).  I think this matters, doesn't it?

-- Jamie Dreier <pl436000@brownvm.brown.edu>

[ Editor: the original 1996 question "Table on a random floor", with a
reply. ]

In article <51qi07$o8d@nntp5.u.washington.edu>,
Mariza Cabral <marizac@u.washington.edu> wrote:
>A kitchen has an uneven floor. There are no "steps", but the
>continuous random waviness of the linoleum is such that when one
>tries to place on it a square table with four legs (of equal
>length), one leg is usually off the floor, causing the table
>to wobble. If one does not mind the table top being on a slant,
>is it always possible to find a place where all four legs are
>firmly on the floor? Or can a floor wave in such a way that no
>such spot is available?

Paint one leg black.

Consider the function F(position) = the length of the black leg
necessary to make the table stand firmly on the floor.  Put the table
down.  If F is less than the length of the other legs, rotate the
table a quarter-turn; F will now be greater than the length of the
other legs (and vice versa).  By continuity, at some angle during the
turn, F was the same as the length of the other legs.

Proof that if the table stands with leg 1 long, it will also stand
with leg 2 short: pivot around legs 3 and 4, so legs 1 and 2 both
shorten.

-- Seth Breidbart <sethb@panix.com>