It is a common calculus / analysis exercice to give a square lawn table, with legs at the corners, and ask to prove that it is always possible to rotate it so that it stands stably (not necessarily in perfect horizontal), as long as the wobblyness comes from the uneven lawn, not from the legs of the table itself.

The proof, which assumes the lawn is suitably nice, uses the intermediate value theorem by rotating the table $90^\circ$ while keeping three legs firmly on the ground and noting that since all that's happened is that the diagonals of the table swapped places, the fourth leg must have passed through the ground at some point.

What about a rectangular table? This has the complication that the table only has $180^\circ$ rotational symmetry, so there is no way to rotate it so as to have the diagonals swap position nicely. So, how can we prove that if we keep three legs along the ground as we turn the table around, the last leg must touch the ground at some point? Or is there some different approach that works?

We are, of course, making the same assumptions as in the square case about the niceness of the lawn and that the table stands stably on an even floor.

  • $\begingroup$ @AlexR Why should that work? If the upper left leg doesn't touch the ground, and you turn the table $180^\circ$, then it has become the lower right leg. But it will still be above the ground, since you can just wobble the table and have the upper left leg up in the air again. Thus we don't have any leverage to use the intermediate value theorem. $\endgroup$ – Arthur Nov 7 '16 at 20:49
  • $\begingroup$ Hmm, that's right. Sorry, my bad. $\endgroup$ – AlexR Nov 7 '16 at 20:59
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    $\begingroup$ Hope you agree with the additional tags. If not, you're free to remove them again. $\endgroup$ – Han de Bruijn Nov 11 '16 at 12:30
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    $\begingroup$ how nice is "suitably nice"? I have seen a paper by Matschke about the squarepeg problem ams.org/notices/201404/rnoti-p346.pdf and he references a paper by Fenn about the table theorem blms.oxfordjournals.org/content/2/1/73.extract, and google search on Fenn table theorem brings more results, e.g. Mark D. Meyerson , Remarks on Fenn's the table theorem'' and Zaks' the chair theorem''. projecteuclid.org/euclid.pjm/1102711107 ..there seem to be different versions with somewhat different assumptions, and I find "the same assumptions" in your question a bit vague. $\endgroup$ – Mirko Nov 15 '16 at 4:39
  • $\begingroup$ @Mirko To me it seems like it should be "the restriction of the garden to any given circle is the graph of a continuous function", or perhaps, "Any assumption that makes the standard proof for square tables actually work" $\endgroup$ – Arthur Nov 15 '16 at 8:46

It seems the answer is known, so I will just provide a reference and some comments. A google search with the terms Fenn's table theorem rectangle brings up a paper by:
Bill Baritompa, Rainer Löwen, Burkard Polster, and Marty Ross, titled
Mathematical Table Turning Revisited,

From their introduction: We prove that given any rectangle, any continuous ground and any point on the ground, the rectangle can be positioned such that all its vertices are on the ground and its center is on the vertical through the distinguished point. This is a mathematical existence result and does not provide a practical way of actually finding a balancing position.

Towards a proof, they define "a mathematical table" using a rectangle, as follows.
In the mathematical analysis of the problem, we will first assume that the ground is the graph of a function $g : \Bbb R^2 \to \Bbb R$, and that a mathematical table consists of the four vertices of a rectangle of diameter $2$ whose center is on the $z$-axis.

They state further down: This result is a seemingly undocumented corollary of a theorem by Livesay [15], which can be phrased as follows: For any continuous function $f$ defined on the unit sphere, we can position a given mathematical table with all its vertices on the sphere such that $f$ takes on the same value at all four vertices.

[15] Livesay, George R. On a theorem of F. J. Dyson. Ann. of Math. 59 (1954), 227–229.

The following is Theorem 3 from Livesay [15]
(where $S_2=S^2$ is the sphere, and $E_1=\Bbb R$ is the real line).
Let $f:S_2\to E_1$ be continuous, $0 < \theta \le 90^\circ$.
Then there exist two diameters of $S_2$ subtending the angle $\theta$, such that the four end points $y_1, y_2, y_3, y_4$ of these diameters satisfy $f(y_i) = f(y_j), i, j = 1,...,4$.

Going back to the paper by Baritompa, Löwen, Polster, and Ross, they use
Livesay's Theorem as follows. Given any continuous ground function $g : \Bbb R^2 \to \Bbb R$ and any "mathematical table" (a rectangle with diagonals of length $2$) they construct $f:S^2\to\Bbb R$ by $f(x,y,z)=z-g(x,y)$. Then apply Livesay's Theorem to position the given mathematical table with all its vertices on the sphere and such that $f$ takes on the same value at all four vertices.

The picture (as I interpret it) is that the table is the given rectangle (assuming without loss of generality that the diagonals have length $2$), and the above proof positions the vertices of the table on the unit sphere in such a way that if we assume that the legs are vertical, then the table is perfectly balanced on the given ground $g$. This was confusing a bit initially, since I am used to think that the legs ought to be perpendicular to the table (and the table need not be horizontal, so the legs would not need to be vertical). But this is not really a problem (the authors don't seem to comment on it, but it appears to be a triviality ... and actually they do comment later in their paper about mathematical vs real tables), once we balance the table (not necessarily horizontal) so that the legs a vertical, then we could "move" the table so that the bottom of each leg remains fixed on the ground, while the table and the top of each leg move until the legs become perpendicular to the table. (I was a bit lost for a while to see exactly what was proved in which paper and how the result follows, so now that I think I figured it, I will include my interpretation here, for convenience.)

I indicated in a comment that this subject also related to the inscribed square problem, so I include a few more links (about both the squarepeg problem, and about versions of the table theorem), in case someone might find them useful.

A Survey on the Square Peg Problem, by Benjamin Matschke,
Notices of the AMS Volume 61, Number 4, p.346-352.
In particular note Conjecture 13 (Table problem on $S^2$) there:
Suppose $x_1,x_2,x_3,x_4\in S^2\subset\Bbb R^3$ are the vertices of a square that is inscribed in the standard $2$-sphere, and let $h : S^2\to\Bbb R$ be a smooth function.
Then there exists a rotation $\rho\in SO(3)$ such that $h(\rho(x_1))=h(\rho(x_2))=h(\rho(x_3))=h(\rho(x_4))$.
So far this result has been proven only when $x_1,x_2,x_3,x_4$ lie on a great circle (see Dyson).

Balancing acts, by Mark Meyerson,
Topology proceedings, Volume 6, 1981, pages 59-75

Remarks on Fenn's "the table theorem" and Zaks' "the chair theorem", by Mark D. Meyerson,
Pacific Journal of Mathematics, Volume 110, Number 1 (1984), 167-169,

The Table Theorem, by Roger Fenn,
Bull. London Math. Soc. (1970) 2 (1): 73-76, doi: 10.1112/blms/2.1.73

Dyson, F. J. Continuous functions defined on spheres.
Ann. of Math. 54 (1951), 534–536.


Let points A,B and C be constantly on the floor and D in the air, then center E of the rectangle has its lowest point for various positions of ABC. Lets call that position $A_1B_1C_1D_1E_1$

If we rotate rectangle so A get in the place of $B_1$ then E is bellow $E_1$ (cos A and C are on the floor and $B_1$, $D_1$ not) and we have contradiction.

  • $\begingroup$ naaah, wrong answer, but then also the proof for square doesn't stand $\endgroup$ – Djura Marinkov Nov 15 '16 at 19:50

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