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If a four-legged, rectangular table is rickety, it can nearly always be stabilised just by turning it a little. This is very useful in everyday life! Of course it relies on the floor being the source of the ricketiness; if the table's legs are different lengths, it doesn't work (this is why I said 'nearly always').

Here is a quick 'proof': Label the feet A, B, C, D in order, and integrate the function F(A) + F(C) - F(B) - F(D) while the table is turned through 180° (here F is the height of the floor). The result is F - F = 0 (these integrals are over 360°), so the function must be zero somewhere; at this point, the table is stable.

The proof only works if the slope S of the floor is small enough that S2 can be ignored. If not, and if the stable solution is not horizontal, then the feet will not lie on the projection of the circle that we integrated over. So a more complicated argument is required:

Suppose ABCD is a square, and suppose that |F(x)-F(y)| <= S|x-y| everywhere, where S = sin-1(1/√3). Fix a vertical line V. Then ABCD can always be positioned with its centre on V, so that each corner touches the floor.

To see this, consider the two diagonals AC and BD separately. Choose a starting position such that

(i) their endpoints lie on the floor,
(ii) their midpoints are on V,
(iii) they are perpendicular to each other.


We are not requiring that their centres coincide (so we will have to dismantle the table to achieve this). We may suppose that in the starting position, the centre of AB lies above the centre of BD. Now rotate them so that constraints (i)-(iii) remain satisfied, until AC occupies the starting position of BD and vice versa. (The constraint on the slope ensures that we can do this continuously.) Now the centre of AB lies below the centre of BD, so the two centres must have coincided at some time.

(This proof generalises to rectangles, but the maximum slope depends on the rectangle. It does not generalise to cyclic quadrilaterals.)

My first question is this: Is the condition on the slope (or a less stringent slope condition) necessary? Or can we always stabilise the table, whatever the slope of the floor?

My second question (if the answer to the first question is that we can always stabilise the table): Given any partition of $\mathbb R$3 into {L,U} with L bounded above and U bounded below, and a vertical line V, can we always find a unit square whose corners lie in the closures of both L and U, and whose centre lies on V?

Update The paper referenced below by Q.Q.J. answers my first question: a rectangular table can always be stabilised, if the floor function F is continuous. But it can only be stabilised by a smooth rotation if the floor satisfies the slope condition. (I was wrong about different rectangles requiring different slope conditions.)

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# Stable Tables on Fluctuating Floors

If a four-legged, rectangular table is rickety, it can nearly always be stabilised just by turning it a little. This is very useful in everyday life! Of course it relies on the floor being the source of the ricketiness; if the table's legs are different lengths, it doesn't work (this is why I said 'nearly always').

Here is a quick 'proof': Label the feet A, B, C, D in order, and integrate the function F(A) + F(C) - F(B) - F(D) while the table is turned through 180° (here F is the height of the floor). The result is F - F = 0 (these integrals are over 360°), so the function must be zero somewhere; at this point, the table is stable.

The proof only works if the slope S of the floor is small enough that S2 can be ignored. If not, and if the stable solution is not horizontal, then the feet will not lie on the projection of the circle that we integrated over. So a more complicated argument is required:

Suppose ABCD is a square, and suppose that |F(x)-F(y)| <= S|x-y| everywhere, where S = sin-1(1/√3). Fix a vertical line V. Then ABCD can always be positioned with its centre on V, so that each corner touches the floor.

To see this, consider the two diagonals AC and BD separately. Choose a starting position such that

(i) their endpoints lie on the floor,
(ii) their midpoints are on V,
(iii) they are perpendicular to each other.


We are not requiring that their centres coincide (so we will have to dismantle the table to achieve this). We may suppose that in the starting position, the centre of AB lies above the centre of BD. Now rotate them so that constraints (i)-(iii) remain satisfied, until AC occupies the starting position of BD and vice versa. (The constraint on the slope ensures that we can do this continuously.) Now the centre of AB lies below the centre of BD, so the two centres must have coincided at some time.

(This proof generalises to rectangles, but the maximum slope depends on the rectangle. It does not generalise to cyclic quadrilaterals.)

My first question is this: Is the condition on the slope (or a less stringent slope condition) necessary? Or can we always stabilise the table, whatever the slope of the floor?

My second question (if the answer to the first question is that we can always stabilise the table): Given any partition of $\mathbb R$3 into {L,U} with L bounded above and U bounded below, and a vertical line V, can we always find a unit square whose corners lie in the closures of both L and U, and whose centre lies on V?