Mathematical Table Turning Revisited

arXiv:math/0511490v1 [math.HO] 19 Nov 2005

by Bill Baritompa, Rainer Löwen, Burkard Polster, and Marty Ross

June 28, 2018

Abstract
We investigate under which conditions a rectangular table can be placed with all four feet on a
ground described by a function R2 → R.

We start by considering highly idealized tables that are just simple rectangles. 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.

An old, simple, beautiful, intuitive and applicable, but not very well known argument guarantees
that a square table can be balanced on any ground that is not “too wild”, by turning it on the
spot. In the main part of this paper we turn this intuitive argument into a mathematical theorem.
More precisely, our theorem deals with rectangular tables each consisting of a solid rectangle as top
and four line segments
 of equal length as legs. We prove that if the ground does not rise by more
1
than arctan √2 ≈ 35.26◦ between any two of its points, and if the legs of the table are at least
half as long as its diagonals, then the table can be balanced anywhere on the ground, without any
part of it digging into the ground, by turning the table on the spot. This significantly improves on
related results recently reported on in [16] and [21] by also dealing with tables that are not square,
optimizing the allowable “wobblyness” of the ground, giving minimal leg lengths that ensure that
the table won’t run into the ground, and providing (hopefully) a more accessible proof.

Finally, we give a summary of related earlier results, prove a number of related results for tables
of shapes other than rectangles, and give some advice on using our results in real life.

1
1 %&$#!!!

You sit down at a table and notice that it is wobbling, because it is standing on a surface that is
not quite even. What to do? Curse, yes, of course. Apart from that, it seems that the only quick
fix to this problem is to wedge something under one of the feet of the table to stabilise it. However,
there is another simple approach to solving this annoying problem. Just turn the table on the spot!
More often than not, you will find a position in which all four legs of the table are touching the
ground. This may seem somewhat counterintuitive. So, why and under what conditions does this
trick work?

2 Balancing Mathematical Tables—a Matter of Existence

In the mathematical analysis of the problem, we will first assume that the ground is the graph of
a function g : R2 → R, and that a mathematical table consists of the four vertices of a rectangle of
diameter 2 whose center is on the z-axis. What we are then interested in is determining for which
choices of the function g can a mathematical table be balanced locally: that is, when can a table be
moved such that its center remains on the z-axis, and all its vertices end up on the ground.

We first observe that it is not always possible to balance a mathematical table locally. Consider,
for example, the reflectively symmetric function of the angle θ about the z-axis with

2 if 0 ≤ θ < π2 or π ≤ θ < 3π

g(θ) = 2 ,
1 otherwise.

Figure 1: On this discontinuous ground a square mathematical table cannot be balanced locally.

So, the ground consists of four quadrants, two at height 1 and two at height 2; see Figure 1. It
is not hard to see that a square mathematical table cannot be balanced locally on such a clifflike
piece of ground. On the other hand, we can prove the following theorem:

2
Theorem 1 (Balancing Mathematical Tables) A mathematical table can always be balanced
locally, as long as the ground function g is continuous.

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. Note that since our mathematical table has diagonals of length 2, its four
vertices will be on the unit sphere iff the centers of the table and the sphere coincide. Choose as
the continuous function the vertical distance from the ground,

f : S2 → R : (x, y, z) 7→ z − g(x, y).

Note that here and in everything that follows the vertical distance of a point in space from the
ground is really a signed vertical distance; depending on whether the point is above, on or below
the ground its vertical distance is positive, zero or negative, respectively. Now, we are guaranteed
a position of our rectangle with center at the origin such that all its vertices are the same vertical
distance from the ground. This means that we can balance our mathematical table locally by
translating it this distance in the vertical direction. Easy!

3 Balancing Real Tables...by Turning the Tables

So, one of our highly idealized tables can be balanced locally on any continuous ground. However,
being an existence result, Theorem 1 is less applicable to our real-life balancing act than it appears
at first glance. Here are two problems that seem worth pondering:

1. Mathematical vs. Real Tables. A real table consists of four legs and a table top; our theorem
only tells us that we can balance the four endpoints of the legs of this real table. However,
balancing the whole real table in this position may be physically impossible, since the table
top or other parts of the legs may run into the ground.
To deal with this complication, we define a real table to consist of a solid rectangle with
diameters of length 2 as top, and four line segments of equal length as legs. These legs are
attached to the top at right angles, as shown in Figure 2. The end points of the legs of a real
table form its associated mathematical table. We say that a real table is balanced locally if its
associated mathematical table is balanced locally, and if no point of the real table is below
the ground.

2. Balancing by Turning. A second problem with our analysis so far is that Theorem 1, while
guaranteeing a balancing position, provides no practical method for finding it. After all,
although we restrict the center of the table to the z-axis, there are still four degrees of
freedom to play with when we are actually trying to find a balancing position.
The following rough argument indicates how, by turning a table on the spot in a certain way,
we should be able to locate a balancing position, as long as we are dealing with a square table
and a ground that is not “too crazy”.

3
 Balancing a Square Table by Turning–the Intermediate Value Theorem in Action
Consider a wobbling square table. We wobble the table until two opposite vertices of the
associated mathematical table are on the ground, and the other two vertices are the same
vertical distance above the ground; see the left diagram in Figure 2. Let’s call this position of
the table its initial position. By pushing down on the table, we can make the hovering vertices
touch the ground and, in doing so, we have shoved the “touching” vertices that same vertical
distance into the ground. We call this new position of the table its end position; see the right
diagram in Figure 2. Starting in the initial position, we now rotate the table around the z-axis;
in doing so, we ensure that at all times the center of the mathematical table is on the z-axis,
that the same pair of vertices as in the initial position are touching the ground, and that the
other two vertices are an equal vertical distance from the ground. Eventually, we will arrive
at the end position. So, we started out with two vertices hovering above the ground, and we
finished with the same vertices shoved below the ground. Furthermore, the vertical distance of
the hovering vertices depends continuously on the rotation angle. Hence, by the Intermediate
Value Theorem, somewhere during the rotation these vertices are also touching the ground: that
is, the table has been balanced locally.

Figure 2: The initial position of the table on the left and its end position on the right. In both
positions, two opposite legs are on the ground and the other two are at equal vertical distance from
the ground (above the ground in the initial position and below the ground in the end position).
Also, in both positions the center of the associated mathematical table is on the z-axis. Also check
out the Quicktime movie at www.maths.monash.edu.au/∼bpolster/table.mov to see what happens
when we rotate around the z-axis.
Unlike most other real-world applications of the Intermediate Value Theorem, it seems that this
neat argument is not as well-known as it deserves. We have not been able to pinpoint its origin,
but from personal experience we know that the argument has been around for at least thirty five
years and that people keep rediscovering it. In terms of proper references in which variations of
the argument explicitly appear, we are only aware of [7], [8], [9] (Chapter 6, Problem 6), [11], [13],
[16], [20], [21] and [22]; the earliest reference in this list, [7], is Martin Gardner’s Mathematical
Games column in the May 1973 issue of Scientific American. Note that an essential ingredient of
the argument is the simple fact that a quarter-turn around its centre takes a square into itself—to
move the table from the initial position to the end position takes roughly a quarter-turn around the

4
z-axis. Closely related well-documented quarter-turn arguments date back almost a century; see,
for example, Emch’s proof that any oval contains the vertices of a square in [3] or [17], Section 4.
At any rate, we definitely do not claim to have invented this argument.

At first glance, the above argument appears reasonable and, if true, would provide a foolproof
method for balancing a square table locally by turning. However, for arbitrary continuous ground
functions, it appears just about impossible to turn this intuitive argument into a rigorous proof. In
particular, it seems very difficult to suitably model the rotating action, so that the vertical distance
of the hovering vertices depends continuously upon the rotation angle, and such that we can always
be sure to finish in the end position.

As a second problem, it is easy to construct continuous grounds on which real tables cannot be
balanced locally. For example, consider a real square table with short legs, together with a wedge-
shaped ground made up of two steep half-planes meeting in a ridge along the x-axis. Then it is
clear that the solid table top hitting the ground will prevent the table from being balanced locally
on this ridge.

By restricting ourselves to grounds that are not too wild, we can prove that balancing locally by
turning really works.

Theorem 2 (Balancing Real Tables) Suppose the ground is described by a Lipschitz continuous
function1 with Lipschitz constant less than or equal to √12 . Then a real table with ratio

length short side

r=
length long side

can be balanced locally on this ground by turning if its legs have length greater than or equal to √ 1 .
1+r 2

1
Since 0 < r ≤ 1, the maximum of √1+r 2
in this range is 1, while all our tables have diagonals of
length 2. Thus we conclude that any real table whose legs are at least half as long as its diagonal
can be balanced locally by turning on any “good” ground. If we are dealing with a square table,
then this table can definitely be balanced locally by turning if its legs are at least half as long as
its sides.

Because of the half-turn symmetry of rectangles, we can be sure to reach a balancing position of a
rectangular table whilst turning it 180 degrees on the spot. As we indicated earlier, to balance a
square table we never have to turn it much more than 90 degrees.

For an outline of the following proof for the special case of square tables, aimed at a very general au-
dience, see [21]. Furthermore, it has just come to our attention that André Martin has also recently
published a proof of this result in the special case
√ of square tables and Lipschitz continuous ground
functions with Lipschitz constant less than 2 − 3. In terms of angles, Martin’s Lipschitz constant
1
Recall that for the ground function g to be Lipschitz continuous means that there exists a k such that the slope of
the line segment connecting any two points on the ground is at most k: |g(P ) − g(Q)| ≤ k|P − Q| for any P, Q ∈ R2 .
The Lipschitz constant of g is then defined to be the optimal (smallest) choice of k. Also, recall that any Lipschitz
continuous function is automatically continuous.

5
corresponds to 15 degrees and ours, which is optimal for local turning, to approximately 35.26
degrees.

Proof. We again start by considering a mathematical table ABCD with diameters of length 2
and centre O. Our approach is to bound the wobblyness of our ground by a suitable Lipschitz
condition such that putting the two opposite vertices A and C on the ground, and wobbling the
table about AC until B and D are at equal vertical distance from the ground, are unique operations.
This ensures that everything in sight moves continuously, as we turn the table on the spot. Following
this, it is easy to conclude that we can balance the table locally by turning it.

Our intuition tells us that to successfully place the four corners, we need four degrees of freedom,
four separate motions of the table. Putting our intuition into effect, we approach our balancing act
as a succession of four Intermediate Value Theorem (IVT) arguments, taking one dimension at a
time.

FIRST VERTEX: A

Start out with the table hovering horizontally above the ground so that OA lies above the positive
x-axis, and lower the table until A touches the ground.

SECOND VERTEX: C

We now show that since the ground function g is continuous, A can be slid along the ground,
towards the z-axis (with O sliding up or down the z-axis), so that C also touches the ground. To
do this consider the function
D(t) = |(t, 0, g(t, 0)) − (−t, 0, g(−t, 0))|2 = 4t2 + (g(t, 0) − g(−t, 0))2 .
Since our table has diagonals of length 2, we want a value of t ≤ 22 = 1 such that D(t) = 22 = 4.
Since D is continuous, D(0) = 0 and D(1) ≥ 4, this follows trivially from IVT.

UNIQUENESS OF C

Assuming that g is Lipschitz with Lip(g) ≤ 1, we show that the above positioning of C on the ground
is unique. This follows from the fact that the function D is strictly monotonic; this can be seen by
differentiating D, noting that |g(t, 0)−g(−t, 0)| ≤ 2t with equality only if the function g(t, 0) is linear
with slope ±1 in the interval under consideration. (Lipschitzness is enough for this differentiation
argument to work, but a direct algebraic argument is also easy).

EQUAL HOVERING POSITION

We now rotate the table through an angle θ ∈ [− π2 , π2 ] about the diagonal AC. We choose the
direction so that rotating the table through the angle − π2 brings the table into a vertical position
with B lying above AC. We want to prove the existence of a θ for which the points B and D are at
an equal vertical distance from the ground: we call such a position an equal hovering position. To
show that there is such a special position, we first choose θ = − π2 . The table is now vertical, with B
above AC and D below AC. Since the segments AB and BC are orthogonal, one of the slopes2 of
2
We emphasize that the slope of a line in space is always nonnegative.

6
these segments will be greater than or equal to 1. Hence, since Lip(g) ≤ 1 and since both A and C
are on the ground, we conclude that B is above or on the ground; similarly, we conclude that D is
below or on the ground. If we now rotate the table about AC until θ = π2 , then B is below or on
the ground and D is above or on the ground. Now, a straightforward application of IVT guarantees
a value of θ for which B and D are an equal vertical distance from the ground.

UNIQUENESS OF THE EQUAL HOVERING POSITION

We now fix k ≤ 1 and take the ground to have Lipschitz constant at most k. We show there exists
a choice of k which guarantees the uniqueness of the hovering position.

Take A and C to be touching the ground as above, with AC then inclined at an angle φ. In
the following, we sometimes need to express the various objects as functions of θ, the rotation
angle about AC (when assuming the inclination angle φ to be fixed, which is the case when we
are referring to a particular ground); then, for example, AB would be expressed as AB(θ). At
other times, we need to express the objects as functions of φ and θ (when we are not referring to
a particular ground); AB would then be expressed as AB(φ, θ). Here φ ∈ [− π4 , π4 ] and θ ∈ [− π2 , π2 ].

We first note that for any equal hovering position the slopes of both AB and BC must be at most k
in magnitude. To see this, suppose AB has slope greater than k. Then, clearly, B is either above
or below the ground. Since CD is parallel to AB, it has the same slope as AB; further, if B is
higher than A, then D is lower than C, and vice versa. Therefore, if B is above the ground, then D
is below the ground, and vice versa. It follows that equal hovering is impossible.

Second, let tangentB(θ) and tangentD(θ) be the tangent vectors to the semi-circles swept out by
the points B(θ) and D(θ), and let vertB(θ) and vertD(θ) be respectively the vertical distances of B
and D to the ground. Note that we have an equal hovering position iff vertB(θ) − vertD(θ) = 0.
It is easy to see that in the θ-interval where the slope of tangentB(θ) is greater than or equal to k,
then vertB(θ) is strictly decreasing. And also, since tangentB(θ) = −tangentD(θ), vertD(θ) is
strictly increasing in this interval. Thus, vertB(θ) − vertD(θ) is strictly decreasing.

Now, let’s choose

k = min max{slopeAB(φ, θ), slopeBC(φ, θ), slopetangentB(φ, θ)}.

φ,θ

Of course, slopeAB(φ, θ), slopeBC(φ, θ), and slopetangentB(φ, θ) denote the slopes of AB, BC
and the tangent vector at B, respectively, and the minimum is taken over all choices of φ ∈ [− π4 , π4 ]
and θ ∈ [− π2 , π2 ]. Also, because of compactness, the minimum above is actually achieved.

Given this choice of k, we shall show that in the interval where equal hovering is possible the slope
of the tangent is at least k. So, in this interval, vertB(θ) − vertD(θ) is strictly decreasing and thus
the equal hovering position must be unique.

We first show that k = √12 . Note that the vectors AB(φ, θ), BC(φ, θ), and tangentB(φ, θ) are
mutually orthogonal. If we then write (0, 0, 1) in terms of this orthogonal frame and take norms,
it immediately follows that
1 = sin2 β1 + sin2 β2 + sin2 β3 ,

7
where β1 , β2 and β3 are the angles the three vectors make with the xy-plane. Therefore, at least
one of the sin2 βj is at least 13 , and thus the vertical slope (= | tan βj |) of the corresponding vector
must be at least √12 . It follows that k ≥ √12 .

To demonstrate the minimum k = √1 is achieved, we show any table can be oriented in the critical
2
position, with all three slopes equal to √12 , and with associated tilt angle φ between − π4 and π4 . To
do this, consider the tripod formed from three edges of a cube tilted to have vertical diagonal shown
in the left diagram in Figure 3. These edges are mutually orthogonal, and one easily calculates that
the slopes of all three edges are √12 . Notice that every table is similar to one of the grey rectangles,
shown in the right diagram, created by moving the point A′ from P to B ′ . Furthermore, it is
clear that the slope of the diagonal A′ C ′ is less than the slope of B ′ C ′ , which is equal to √12 ,
guaranteeing the tilt angle φ is in the desired range. By scaling and translating the rectangle
suitably, and relabelling the vertices A′ , B ′ , and C ′ as A, B, and C, respectively, we arrive at the
desired orientation of our table.

Figure 3: Three mutually orthogonal segments of length 1 balanced on the xy-plane. Then (grey)
rectangles of any shape can be fitted as indicated.

It remains to show that k = √1 implies that the slope of the tangent is at least k in the interval
2
where equal hovering is possible. Note that in this interval, slopeAB and slopeBC are at most √1 .
2
Then the equation 1 = sin2 β1 + sin2 β2 + sin2 β3 implies that slopetangentB is at least √1 .
2

CONTINUITY OF A, B, C, AND D

All of the above calculations were performed with OA projecting to the positive x-axis. We now
consider rotating the table about the z-axis (while of course being willing to tilt the table as we
rotate). So let γ be the angle the projection of OA makes with the positive x-axis. By our Lipschitz
hypotheis, for any γ there is a unique equal hovering position (with the projection of OA making
the angle γ, and tilting the table around AC an angle between − π2 and π2 in a fixed direction).

We need to show that the positions of the four vertices A, B, C and D, of the equally hovering
table are continuous functions of γ. To do this, consider a sequence γn → γ, with corresponding
corner positions An , Bn , Cn , and Dn and A, B, C, and D. We want to show An → A, Bn → B,
Cn → C, and Dn → D. By compactness, we can take a subsequence so that the corners converge
to something: An → A∗ , Bn → B ∗ , etc. But, by continuity of everything in sight, A∗ and C ∗ are
touching the ground B ∗ and D ∗ are an equal vertical distance from the ground, all four points are

8
corners of the kind of table we are considering, with the projection of OA∗ making an angle γ with
the positive x-axis. By uniqueness, we must have A∗ = A, B ∗ = B, C ∗ = C, and D ∗ = D, as
desired.

BALANCING POSITION

With the uniqueness of the equal hovering position for a given γ, and with the continuous depen-
dence of this position of the table upon γ, we conclude that the distance that B and D are hovering
above the ground is also a continuous function of γ. If we are dealing with a square table, we can
now finish the proof using IVT one more time, as described in the intuitive table turning argument
presented at the beginning of this section.

For a general rectangle, let the initial position be an equal hovering position for which the z-
coordinate of the center of the table is a minimum, and let the end position be an equal hovering
positions for which the z-coordinate of the center of the table is a maximum. Note that the
hovering vertices in the initial position must be on or below the ground: if not, we could create
a lower equal hovering position, contradicting minimality, by pushing vertically down on the table
until the hovering vertices touch the ground. Similarly, in the end position the hovering vertices are
on or above the ground. Now, IVT can be applied to guarantee that among all the equal hovering
positions there is at least one balancing position.

BALANCING REAL TABLES

To balance a real table of side lengths ratio r, we determine a balancing position of the associated
1
mathematical table, as described above. We now show that legs of length at least √1+r 2
guarantee
that, balanced in this position, none of the points of the real table are below the ground. We give
the complete argument for a square table, and then describe how things have to be modified to
give the result for arbitrary rectangular tables. In the following, we will refer to the four vertices
of the table top as A′ , B ′ , C ′ and D ′ , corresponding to the vertices A, B, C and D, respectively.

We first convince ourselves that no matter how long are the legs of our table, no part of a leg of the
balanced table will be below the ground. Let’s consider the orthogonal tripod consisting of AB,
AD and the leg at A. Since the Lipschitz constant of our ground is at most √12 , the slopes of AB
and AD are less than or equal to this value; thus, arguing as above, we see that the leg must have
slope at least √12 . This implies that no leg of our balanced table will dip below the ground.3

It remains to choose the length of the legs such that no point of the table top of our balanced table
will ever be below the ground. First, fix the length of the legs and consider the inverted solid
circular cone, whose vertex is one of the vertices of our mathematical table, whose symmetry axis
is vertical, and whose slope is √12 . Intersecting this cone with the plane in which the table top
lies gives a conic section which is either an ellipse, a parabola or a hyperbola.4 Note that since we
intersect the plane with a solid cone this conic section will be “filled in”. We can be sure that a
point in this plane is not below the ground if it is contained in the conic section. Therefore, what
3
For certain grounds with Lipschitz constant √12 it is possible that a leg of a balanced table may lie along the
ground.
4
Parabolas and hyperbolas can occur since the plane that the table top is contained in can have maximum slope
greater than √12 .

9
we want to show is that the union of the four conic sections associated with the four vertices of
our mathematical table contains the whole table top. It is clear that the four conic sections are
congruent and that any two of them can be brought into coincidence via a translation. Furthermore,
given any point of one of these conic sections, this point and the respective points in the other three
conic sections form a square that is congruent to our table top. Finally, since the legs have slope
of at least √12 , the end point of a leg of our table on the plane is contained in the conic section
associated with the other end point of this leg.

To show that we need legs of length at least √1 , consider a special ground with Lipschitz con-
2
stant √1 . This ground coincides with the xy-plane outside the unit circle, and above the unit circle
2
it is the surface of the cone with vertex (0, 0, √12 ) and base the unit circle. Since the diagonals of
our table are of length 2, the mathematical table will balance locally on this ground iff its vertices
are on the unit circle. This means that the length of the legs have to be at least as long as the cone
is high if we want to ensure that no point of the table top is below the ground; it follows that we
have to choose the length of our legs to be at least √12 . If the length of the legs is equal to √12 and
the table is balanced on this ground, then the four conic sections are circles that intersect in the
center of the table top as shown in Figure 4. As you can see, the table top is indeed contained in
the union of these four circles. If we make the legs longer, the circles will overlap more. If we make
the legs shorter, the circles will no longer overlap in the middle.

Figure 4: The four conic sections and the table top in the case of a square table with legs of
length √12 that is balanced horizontally.

Now, consider any ground, and take the legs to be of length √12 ; clearly, if we can show that this
table does not dip below the ground, then the same is true for any table with longer legs. When
we tilt the table away from the horizontal position, the intersection pattern of the conic sections
gets more complicated. The critical observation is that tilting the table results in the conic sections
getting larger: it can be shown that each conic section contains a copy of one of the circles in
Figure 4.5 Since, given any point of one of these conic sections, it and the corresponding points
in the other three conic sections form a square that is congruent to our table top, the union of
these conic sections will contain a possible translated image of the union of the circles, that we
encountered before; see Figure 5. So, in a way our previous picture has just grown a little bit and
been translated. (Note, however, it is not immediate that the conic sections together cover the
table top rather than some translation of the table top). Using this fact and the simple possible
convex shapes of the conic sections that we are dealing with, we can conclude that no matter how
we tilt the table, the union of these conic sections will always be a simply connected domain. This
5
To see this note that what we are looking at are the possible intersections of a given cone with planes that are a
fixed distance from the vertex of the cone.

10
Figure 5: If the table is not horizontal, the conic sections are larger than the circular (gray) sections
in the horizontal case (left). Their union is simply connected (right).

means that we can be sure that the table top is contained in this union if we can show that the
boundary of the table top is contained in it.

We proceed to show that for all possible positions of our table in space the sides of the table
top never dip below the ground. Clearly, it suffices to show this for one of the sides of the table
top, say A′ B ′ . For this we consider the possible positions of the rectangle with vertices A, A′ , B
and B ′ in space. We start with the rectangle vertical and AB horizontal. Draw lines of slope √12
ending in A and B; see Figure 6 (left). Since the point of intersection of these two lines is not
above A′ B ′ , no point of this segment can be below the ground when the rectangle is positioned in
such a way. Now rotate the rectangle around its center, keeping it in a vertical plane, and keeping
the slope of AB less than or equal to √12 . Again, draw lines of slope √12 ending in A and B; see
Figure 6 (middle). Again, the position of the point at which these two lines intersect tells you
whether A′ B ′ can possibly touch the ground with the rectangle in this position. Since the two lines
always intersect in the same angle, we know that the points of intersection are on a circle segment
through A and B; see Figure 6 (right).

Figure 6: The points of intersection of the two lines of slope √1 for the different possible vertical
2
positionings of the rectangle AA′ BB ′ are on a circle.

Now, tilt the original rectangle around AB. We repeat everything that we have done so far to end
up with another circle segment. However, the apex of this circle segment will the be closer to A′ B ′
than the one we encountered before. In fact, the more we tilt, the closer we will get; see Figure 7.

11
Figure 7: The more we tilt, the closer the apex of the circle segment gets to A′ B ′ . As long as the
apex is below or on A′ B ′ , we can be sure that A′ B ′ does not dip below the ground.

Since the apex of one of these circles is the point closest to A′ B ′ , and since the apex corresponds
to AB being horizontal, we now calculate just how close this apex gets when we tilt around AB
though a maximal possible angle. This maximal possible angle is attained if AD (which is orthog-
onal to AB) has slope √12 . It is a routine exercise to check that in this position the slope of the
line connecting A with the midpoint of A′ B ′ is √1 . This means that in this position the apex will
2
be contained in A′ B ′ . We conclude that if we choose the legs of our square table to be at least √12
long, then the boundary of the table top, and hence also the table top itself, will not dip below the
ground.

For tables that are not square, the same arguments apply up the point where we start tilting the
rectangle AA′ B ′ B around AB. We now have to worry about two different rectangles corresponding
to the longer and shorter sides of the table top. If r is the ratio of the lengths of the sides of the
table, then it is easy to see that the critical length of the legs that we need to avoid running into
the ground is the length that makes the longer of the two rectangles similar to the rectangle that
1
we considered in the square case. This critical length is √1+r 2
. 

4 Other Balancing Acts

Horizontal balancing

When we balance a table locally, the table will usually not end up horizontal, and a beer mug
placed on the table may still be in danger of sliding off. It would be great if we could arrange it
so that the table is not only balanced but also horizontal, maybe by moving the center of the table
off the z-axis and balancing it somewhere else on the ground. Just imagine the ground to be a
tilted plane, and you can see that this will not be possible in general. However, Fenn [4] proved
the following result: If a continuous ground coincides with the xy-plane outside a compact convex
disc and if the ground never dips below the xy-plane inside the disk, then a given square table can
be balanced horizontally such that the center of the table lies above the disk. Let’s call the special
kind of ground described here a Fenn ground and the part of this ground inside the distinguished
compact disk its hill.

The problem of horizontally balancing tables consisting of plane shapes other than squares on Fenn
grounds has also been considered. Here ‘horizontal balancing on a Fenn ground’ means that in the

12
balancing position some interior points of the shape are situated above the hill. It has been shown
by Zaks [28] that a triangular table can be balanced on any Fenn ground. In fact, he showed that
if we start out with a horizontal triangle somewhere in space and mark a point inside the triangle,
then we can balance this triangle on any Fenn ground, with the marked point above the hill, by just
translating the triangle. Fenn also showed that tables with four legs that are not concircular and
those forming regular polygons with more than four legs cannot always be balanced horizontally
on Fenn grounds. Zaks mentions an unpublished proof by L.M. Sonneborn that any polygon table
with more than four legs cannot always be balanced horizontally on Fenn grounds. It is not known
whether any concircular quadrilateral tables other than squares can always be balanced horizontally
on Fenn grounds. See [14], [17], [18], and [19] for further results relating to this line of research.

Local Balancing of Exotic Mathematical Tables

Taking things to different mathematical extreme, we can consider a table consisting of n ≥ 3

leg points in 3-space together with an additional center point. We then ask whether, given any
continuous ground, it is possible to always balance this table locally, that is, move this configuration
of n + 1 points into a position in which the n leg points are on the ground, and the center is on the
z-axis.

The example of a plane ground shows that the leg points of an always locally balancing table have
to be coplanar. Let’s consider the example of a ground that contains part of a sphere that is large
enough to ensure that all legs of our table end up on this part of the sphere whenever the table is
locally balanced on this ground. Then intersecting this sphere with the plane that the leg points
are contained in gives a circle that all leg points are contained in. Hence the leg points of the table
are concircular. Now, let’s consider a ground that includes part of an ellipsoid which does not
contain a copy of the circumcircle of the leg points of the table; moreover, we choose the ellipsoid
large enough so that all leg points of our table end up on the ellipsoid whenever the table is locally
balanced on this ground. Then intersecting the ellipsoid with the plane containing the leg points
gives an ellipse that is different from the circumcircle of the leg points. However, this is impossible
if the table contains more than four leg points because five points on an ellipse determine this
ellipse uniquely. We conclude that an always locally balancing table must have three or four leg
points and that these points are concircular. Note that requiring concircularity in the case of three
points is not superfluous since we need to exclude the case of three collinear points.

Livesay’s theorem, which made the proof of Theorem 1 so easy, has a counterpart for triangles, due
to Floyd [6]. It is a straightforward exercise to apply this result to prove the following theorem:

Theorem 3 (Balancing Triangular Tables) If the ground function is continuous, a triangular

table whose three leg points are contained in a sphere around its center can be balanced locally.

Of course, one should be able to prove a lot more when it comes to balancing triangular tables!

In the case that the center and the (three or four) leg points of an always locally balancing table
are coplanar, we can say a little bit more about the location of the center point with respect to the
leg points. Begin by balancing the table in the xy-plane and drawing the circles around the center

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that contain leg points; if one of the leg points coincides with the center, then also consider this
point as one of the circles. Now it is easy to see that there cannot be more than two such circles.
Otherwise a ground that coincides with the xy-plane inside the third smallest circle and that lies
above the plane outside this circle would clearly thwart all local balancing efforts. Therefore, if we
want to check whether our favorite set of three or four concircular points is the set of leg points
of a locally balancing table, there are usually very few positions of the center relative to the leg
points which need to be considered.

Perhaps the most natural choice for the center is the center of the circle that the leg points are
contained in. As a corollary to the above theorem for triangles, we conclude that a triangular table
with this natural choice of center is always locally balancing. In the case of four concircular points
with this natural choice of center we do not know whether any tables apart form the rectangular
ones are always locally balancing. However, a result worth mentioning in this context is Theorem 3
in Mayerson’s paper [17] (see also the concluding remarks in Martin’s paper [16]). It can be phrased
as follows: Given a continuous ground and one of these special four-legged tables in the xy-plane,
the table can be rotated in the xy-plane around its center such that in this new position the four
points on the ground above the leg points are coplanar. The quadrilateral formed by the coplanar
points on the ground will be congruent to the table if and only if the plane it is contained in is
horizontal, in which case we have actually found a balancing position for our table. In all other
cases, the quadrilateral on the ground is a deformed version of the table. Still, if the ground is not
too wild, both quadrilaterals will be very similar, and lifting the table up onto the ground should
result in the table not wobbling too much.

Livesay’s theorem is a generalization of a theorem by Dyson [2], which only deals with the square
case. A higher-dimensional counterpart of Dyson’s theorem arises as a special case of results of
Joshi [12], Theorem 2 and Yang [24], Theorem 3: Given a continuous real-valued function defined
on the n-sphere, there are n mutually orthogonal diameters of this sphere such that the function
takes on the same value at all 2n end points of these diameters. Note that the endpoints of n
mutually orthogonal diameters of the n-sphere are the vertices an n-dimensional orthoplex, one of
the regular solids in n-dimensions. (For example, a 1-dimensional orthoplex is just a line segment
and a 3-dimensional orthoplex is an octahedron.) Using the same simple argument as in the case
of Livesay’s theorem, we can prove the following theorem:

Theorem 4 (Balancing Orthoplex-Shaped Tables) An (n − 1)-dimensional orthoplex-shaped

table in Rn can be balanced locally on any ground given by a continuous function g : Rn−1 → R.

For other closely related results see [1], [5], [10], [23], [25], [26], and [27].

Balance Everywhere

Imagine a square table with diameter of length 2 suspended horizontally high above some ground,
with its center on the z-axis. Rotate it a certain angle about the z-axis, release it, and let it drop to
the ground. It is easy to identify continuous grounds such that all four leg of the table will hit the
ground simultaneously, no matter what release angle you choose. Of course, any horizontal plane
will do, and so will any ground that contains a vertical translate of the unit circle. We leave it as an

14
exercise for the reader to construct a ground that is not of this type but admits horizontal balancing
for any angle. Also, the reader may wish to convince themselves that the following is true: we are
dealing with a ground as in Theorem 2. If the center of the table has the same z-coordinate in all
its equal hovering positions (positions in which A and C touch the ground and B and D are at
equal vertical distance from the ground), then in fact the table is balanced in all these positions.

5 Some Practical Advice

Short Legs and Tiled Floors

Note that if you shorten one of the legs of a real-life square table, this table will wobble if you set it
down on the plane, and no turning or tilting will fix this problem. In real life rectangular tables the
ends of whose legs do not form a perfect rectangle are not uncommon and, as our simple example
shows, those uneven legs may conspire to make our anti-wobble tactics fail.

Considering our example of a discontinuous ground at the beginning of this article, it should be
clear that a wobbling table on a tiled floor may also defy our table turning efforts.

How to Turn Tables in Practice

In practice, it does not seem to matter how exactly you turn your table on the spot, as long as
you turn roughly around the center of the table. Notice that you needn’t actually establish the
equal hovering: as you rotate towards the correct balancing position, there will be less and less
wobble-room until, at the correct rotation, the balancing position is forced. With a square table,
you can even go for a little bit of a journey, sliding the table around in your (continuous) backyard.
As long as you aim to get back to your starting position, incorporating a quarter turn in your
overall movement, you can expect to find a balancing position.

References

[1] de Mira Fernandes, A. Funzioni continue sopra una superficie sferica. Portugaliae Math 4
(1943), 69–72.

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

[3] Emch A. Some properties of closed convex curves in a plane, Amer. J. Math. XXXV (1913),
407–412.

[4] Fenn, Roger. The table theorem. Bull. London Math. Soc. 2 (1970), 73–76.

[5] Fenn, Roger. Some applications of the width and breadth of a closed curve to the two-
dimensional sphere. J. London Math. Soc. 10 (1975), 219–222.

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 [6] Floyd, E. E. Real-valued mappings of spheres. Proc. Amer. Math. Soc. 6 (1955), 957–959.

[7] Gardner, Martin. Mathematical Games column in Scientific American (May 1973), page 104.

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[10] Hadwiger, H. Ein Satz über stetige Funktionen auf der Kugelfläche. Arch. Math. 11 (1960),
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[11] Hunziker, Markus. The Wobbly Table Problem. In Summation Vol. 7 (April 2005), 5–7
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[12] Joshi, Kapil D. A non-symmetric generalization of the Borsuk-Ulam theorem. Fund. Math. 80
(1973), 13–33.

[13] Kraft, Hanspeter. The wobbly garden table. J. Biol. Phys. Chem. 1 (2001), 95–96.

[14] Kronheimer, E. H. and Kronheimer, P. B. The tripos problem. J. London Math. Soc. 24 (1981),
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[15] Livesay, George R. On a theorem of F. J. Dyson. Ann. of Math. 59 (1954), 227–229.

[16] Martin, André. On the stability of four feet tables. http://arxiv.org/abs/math-ph/0510065

[17] Meyerson, Mark D. Balancing acts. The Proceedings of the 1981 Topology Conference (Blacks-
burg, Va., 1981). Topology Proc. 6 (1981), 59–75.

[18] Meyerson, Mark D. Convexity and the table theorem. Pacific J. Math. 97 (1981), 167–169.

[19] Meyerson, Mark D. Remarks on Fenn’s “the table theorem” and Zaks’ “the chair theorem”.
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[20] Polster, Burkard; Ross, Marty and QED (the cat). Table Turning Mathsnack in Vincu-
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[22] Vinculum, Editorial Board. Mathematical inquiry—from a snack to a meal. Vinculum,

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[23] Yamabe, Hidehiko and Yujobô, Zuiman. On the continuous function defined on a sphere. Osaka
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[25] Yang, Chung-Tao. On theorems of Borsuk-Ulam, Kakutani-Yamabe-Yujobô and Dyson. II.
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[26] Yang, Chung-Tao. Continuous functions from spheres to euclidean spaces. Ann of Math. 62
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[27] Yang, Chung-Tao. On maps from spheres to euclidean spaces. Amer. J. Math. 79 (1957),
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