aa r X i v : . [ m a t h . HO ] A p r Balancing non-rectangular tables
Antoine Gournay
Abstract
Balancing square and rectangular tables by rotation has been ainteresting way to illustrate the intermediate value theorem. Theaim of this note is to show that the balancing act but with non-rectangular tables can be a nice application of the ergodic theorem(or more generally, invariant measures).
If you have tried to set up a pick-nick table or used a stepladder, then you are certainlyaware how of the wobbly table problem. How can you make this object stop to wobble(without using extra tools)? Well, if you have a square or even rectangular table, then youare in luck. It turns out that rotating the table along some axis will suffice to eventuallystabilise it. Unfortunately, many offices are now full of tables which have a trapezoidalshape. What now?Unfortunately, the arguments for square or rectangular table hinge crucially on thesymmetry of the object. This note will hopefully convince you, that you should have nofear of buying tables which are less symmetric.Here is a more detailed description of this problem, which was originally made publicin [5] and [6]. You have some table (or other object) with 4 legs and you find yourself onsome terrain. Find conditions on the legs and quite unrestrictive conditions on the terrain,so that turning the table around some axis can make it stop to wobble. Note that thetable may not be level, it will just stop to wobble.The original arguments of [5] and [6] cover square tables on a continuous terrain, butis presented in an abstract setting which overlooks the rigidity of the problem. A morerealistic setup is covered in [2] and [9]. The terrain needs to be Lipschitz (with a Lipschitzconstant bounded above). One of the very first work which relates to this problem is [4];the reader is directed to [2] for an extensive historical overview.The main result of this note is to discuss an extension to cover cyclic quadrilaterals(a conjecture raised in [9]). The main narrative of the proof is presented in §2. However,much like the first presentations of this problem in [5] and [6], this narrative overlookssome technicalities. A closer look at the problem is then discussed in §3. The completehypothesis are as follows: − the quadrilateral formed by the ends of legs is cyclic − its diagonals are of equal length. − the angle (from the centre of the excircle) which support the diagonal is rational1 the angle which brings one diagonal on the other is also rational.Then the table can be stabilised by rotating. (As in [9] or [2], the terrain needs to beLipschitz.)A typical example of a non-rectangular table satisfying the conditions of the theoremwhich the reader might have in his office has the following shape: take an hexagon and cutalong two opposite vertices. The resulting quadrilateral is a symmetric trapezoid (with 3equal sides). Let us start by some simple results, reductions, assumptions and notations: • First off, it seems the very least to demand that your table be stable if the terrain isperfectly flat. Hence the ends of the legs must be on the same plane P . The end ofthe legs will be denoted A , B , C and D . • The axis around which you rotate should be perpendicular to the plane P (you don’twant to turn your table upside down!). • Also the legs should all run along the same circle ( i.e. the quadrilateral
ABCD iscyclic) otherwise stabilisation by rotation is not possible (the terrain could be highalong the circle described by one leg and low along the circle described by anotherleg). This is a necessary condition raised in [9]. Z will denote the centre of this circle. • This note stays in a fairly idealised setup, Physical problems, such as “the terrainis going through the [legs of the] table”, “the table will turn over since its centre ofmass is ill-placed” or “the legs have a thickness”, will be ignored. • The whole problem boils is about the four feet of the table. The actual shape of thetable is not really important. • The table starts in the air. There is some basic position corresponding to the angle ◦ . This means that, in this basic position, the leg A makes an angle of ◦ as seemfrom the excentre Z . The angles θ B , θ C and θ D are the angles which make the otherlegs when in this basic position. (For convenience θ A = 0 .) • As you turn your table, all the angles of the legs are changed by the same number.(The table is rigid after all.)
The next step is to look at possible touchdowns for your table. As you rotate the tableconsider the following way to put it down. First, you could put the legs A and C on theground (and completely ignore where B and D are). If B and D are above the terrain,then the table wobbles, if not, then this was not completely legal, but it is still useful to2hink about it (as some sort of negative wobbling). [2] discusses the touchdown much morecarefully.Next, you could do the same thing with B and D touching down first. Note that thesetouchdowns should happen on the same curve (some distorted circle), otherwise there isno hope to conclude. Assume for now that this is the case, see [2] and §3.1 for a correctdescription of this processConsider X to be the intersection point of the diagonals AC and BD . If we make atouchdown with AC , the coordinates of X do not change when the table wobbles. Likewisewith BD . This gives us two height functions: h AC ( θ ) is the height ( z -coordinate) of X after a AC -touchdown, where θ is an angle of rotation (of the table). And likewise for h BD . Lemma 2.1.
If the table wobbles then h BD ( θ ) = h AC ( θ ) .Proof. If the table wobbles then one of the pair BD or AC can be pushed further down.As a consequence, the height of the centre will differ.The coordinates of X are a convex combinations of the coordinates of the legs. Moreprecisely, it’s a small exercise with vectors that if τ CXCA ∈ ]0 , and µ = DXDB ∈ ]0 , , then −−→ OX = τ −→ OA + (1 − τ ) −−→ OC = µ −−→ OB + (1 − µ ) −−→ OD In particular, this holds for the height coordinate: h AC ( θ ) = τ h A ( θ ) + (1 − τ ) h C ( θ ) and h BD ( θ ) = µh B ( θ ) + (1 − µ ) h D ( θ ) where h E ( θ ) gives the height of (the end of) leg E after a touchdown after a rotation [ofthe table] by the angle θ and E ∈ { A, B, C, D } . Since every leg can be brought at theposition of the other leg by a rotation, the functions h A , h B , h C and h D are all identicalup to a translation (see §3.1 and §3.2 for details). Theorem 2.2.
Assume h E are continuous, then there is an angle θ so that the table doesnot wobble.Proof. Since h E is measurable, let R h E = H (the average height). By assumption h E iscontinuous, hence so are h BD and h AC . Let h ∆ = h BD − h AC .Assume the table cannot be stabilised, then, by Lemma 2.1, h ∆ is never 0. Since h ∆ is continuous, there is an ε > so that,either ∀ θ, h ∆ ( θ ) > ε or ∀ θ, h ∆ ( θ ) < − ε Without loss of generality, we may assume the first holds. Let θ be an irrational angle.Then ∀ N, N N X i =1 h ( i · θ ) > ε
3n the other hand N N X i =1 h ( i · θ ) = 1 N N X i =1 (cid:18) µh B ( i · θ ) + (1 − µ ) h D ( i · θ ) − τ h A ( i · θ ) − (1 − τ ) h C ( i · θ ) (cid:19) By the equidistribution theorem or the ergodic theorem (see, among many possibilities [1,Appendix 1], [8, Section 23.10], [10, Exercise 2.2.12] and [11]) lim N →∞ N N X i =1 h ∆ ( i · θ ) = Z h ∆ = µ Z h B + (1 − µ ) Z h D − τ Z h A − (1 − τ ) Z h C But the right-hand side is just µH + (1 − µ ) H − τ H + (1 − τ ) H = H − H = 0 . Hence lim N →∞ N N X i =1 h ∆ ( i · θ ) = 0 > ε a contradiction. A tool from [2, §3] is to look only at the touchdown with respect to AC and then considerhow far the vertices BD are in the air. This is done so that both vertices B and D areequally far from the ground and is called the “equal hovering position”.Let us sketch how to apply Theorem 2.2 in this context. Consider the average (overall angles) of h AC , R h AC . Since the average of h BD is equal to that of h AC , the equalhovering position is on average 0. But the equal hovering position is continuously defined,hence if the average is , it should be 0 somewhere. When the equal hovering position is0, then the table does not wobble.However there is a hidden hypothesis in this argument. Namely, if the two diagonalsdo not have the same length, then it could be that h BD takes on different value than h AC . Variations in the steepness of terrain could make the pair of vertices B and D lingerdifferent time in different regions. This leads to the following Assumption 3.1.
The length of the diagonals of
ABCD are equal.
As mentioned before, [2] contains a detailed description how to realise the touchdown oftwo opposite vertices ( AC or BD ). The focus of this section is to point out what the proofof Theorem 2.2 requires. The various height functions h A , h B , h C and h D need:4 H1). to be well-defined (H2). to be continuous (H3). to differ only by a translation, i.e. h E ( θ ) = h A ( θ + θ E ) Before looking closer at this, let’s point the problem out: while letting the table down,you will need to turn the table according to other axes. This means there is a priori anuncertainty in how you put the table down. As mentioned in [9] these constructions arehighly non-unique.For example, this plays an implicit role in Lemma 2.1. Indeed, on needs to assumethat, for any given angle θ , the touchdown according to AC and BD are done in one go.So you cannot let the table down and see where AC end up, and then let the table downand see where BD end up. You need to let the table down, see which of the two first comeup and then go on to the second.In short, for the proof of Lemma 2.1, a transversality condition is probably necessary,namely that the z -coordinate of X is decreasing while letting the table down. This meansthat for a given rotation angle θ the touchdown of both diagonals must be taken into ac-count simultaneously. To make sure this is possible requires most certainly three additionalassumptions.The first of these assumptions, is that for transversality, the terrain needs to be a C -function with some upper bound on the first derivative. Lipschitz-continuity (with anupper bound on the Lipschitz constant) is however sufficient, see [2] and [9] for details.The second assumption comes form the fact that the height functions h E should onlydepend on the angle (and on θ E ). The touchdown h BD (0) needs to be defined simulta-neously with the touchdown h AC (0) . But then, h AC ( θ B ) , h AC ( θ C ) and h AC ( θ D ) are alsodefined (since the leg A will land where the legs B , C and D did).Repeating this process, one sees that all angles k θ B + k θ C + k θ D (where k , k and k ∈ Z ≥ ) need to be defined at once. In order to avoid an infinite number of choices, thisleads to the following two assumptions: Assumption 3.2.
The rotation bringing one diagonal on the other diagonal is rational.
Assumption 3.3.
The rotation bringing one end of a diagonal on the other end is rational( i.e. the angle supporting the diagonals from the excircle is rational).
Lastly, note that Assumption 3.1 is particularly important for condition (H3) mentionedabove.Note that Assumptions 3.1 and 3.3 are both automatically verified in the case of arectangle (diagonals of equal length is a characteristic feature of rectangles among parallel-ogram; the rotation is 180 ◦ since the excentre lies on the diagonal). Assumption 3.2 mighthowever not hold Remark 3.4.
One might be able to get rid of the rationality assumptions. A standardway to do so would be to consider a sequence of tables with rational angles which tend tothe desired (irrational) table. Since everything is happening over a compact part of R ,the stabilising positions of the rational tables will converge to (at least) one position. Thisposition should stabilise the (irrational) table. ♦ .3 Invariant measures Note that the proof relied on a very specific invariant measure: the uniform measure, whichis invariant under translations.Perhaps one way to get rid of Assumption 3.1 would be to consider a sequence θ i whichtends to some well-chosen measure, rather than just taking this sequence of angles to bemultiples of an irrational angle ( i · θ in the proof).More concretely, let h B ( θ ) = h A ( θ + f B ( θ )) , h C ( θ ) = h A ( θ + f C ( θ )) and h D ( θ ) = h A ( θ + f D ( θ )) . The functions x x + f E ( x ) (where E ∈ { B, C, D } ) generate a monoid (ofcircle maps). If the action of this monoid is amenable, then there is a invariant measure µ (see [7] for generic background on amenability and [10, §2.3 and §3.2] for more specificdetails in the case of groups acting on the circle). Since The convex hull of Dirac massesis weak ∗ dense in the space of means, one can the choose the sequence of angles θ i so that N P Ni =1 δ θ i tends (weak ∗ ) to µ .Consequently, the assumptions presented in §3.1 and §3.2 can perhaps be relaxed.A simple way of satisfying the amenability assumption is to assume that the monoid iscontained in the monoid given by some rational rotations (e.g. coming from Assumptions3.3 and 3.2) as well as some map (which may not be a rational rotation). Since the monoidis then a finite extension of an amenable one (the single map which is not a rational yieldsa monoid isomorphic to N ), it is amenable as well.The author would like to thank Yves de Cornulier [3] for pointing out that [10] essen-tially shows there are not other examples where this monoid is amenable (beside the givenexample where two maps are rational). References [1] V. I. Arnold and A. Avez,
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