Welcome to another Mathologer video.
Today's video is about the absolutely wonderful wobbly table theorem. Some of
you may already be familiar with a very pretty special case of this theorem
which became quite well known a few years ago when Numberphile dedicated a
nice video to it. This special case of the theorem runs as follows: take a
square table like the one over there. At the moment this table is hovering above the
floor - a bit unusual behavior for a table but perhaps it's due to the ghostly
presence at our seance. Anyway the feet of our spooky table are blue and the points
directly below the four feet are marked in green. The ghost then disappears and the
table plunks onto the surface. Chances are that it will wobble, right? Not
because the ghost is still playing games but just because the floor is uneven. How
do we stabilize the table? Well, usually people will wedge a napkin or whatever
under one of the wobbling feet, like this. However, the wobbly table theorem says
there's another method. The theorem says that if the floor is not too crazily
uneven, then you can stabilize the square table simply by rotating the table on
the spot, like in this example. So just rotate it and eventually all four feet
will be touching the ground. Super neat isn't it? And this actually works very
well in practice. Try it on your next cafe trip for example. And this is only a
small piece of some nice and mostly unknown table turning maths which also
applies to non square tables. Well let's see. Now before we get going I have to
make two very important disclaimers. First when I'm talking about a square
table I always mean that the feet of the table form a square. Of course, that's not
always true, even if the table surface is square. Often enough real-life wobbling
is not caused by the uneven floor but by the table having feet of unequal lengths.
For example, over there we've got a table on a nice flat surface but one of the
legs has been gnawed at by a beaver or a ghost or whatever. Anyway
one leg is shorter so even though the table is standing on a flat ground
obviously there's no way of shifting the table around to make it stable. Well we
could turn it over with its legs in the air but that would be cheating. So, again,
by a square table we mean the feet of the table form a square. Second
disclaimer. By having stabilized a table we do not mean that the tabletop is
necessarily horizontal. Usually the stabilized table will still slope in
some manner consistent with the uneven floor. Of course such a slope can also be
annoying but definitely not as annoying as a wobble and typically in real life a
slight unevenness in the ground will translate into a slight and not too
annoying slope of the tabletop. With those two disclaimer stored away it's
time to have some serious fun turning the tables. Now to do that
we'll start with Homer Simpson. Of course it's never the wrong time to introduce
some Simpsons into the discussion. Recall that my last video was all about Fourier
series and how to draw complicated pictures like this Homer drawing with
epicycles. I'll announce the winner of the epicycle competition at the end of
this video. For now let me reCYCLE :) Homer as a warm-up exercise for our
table turning. Let's begin by framing Homer in a rectangle like this. I call
this sort of rectangle a hugging rectangle because Homer touches all four
sides of the rectangle. Of course there are infinitely many hugging rectangles, one
for each possible orientation. Here are a few more. There's one and there's another one
and one more. Now the aspect ratios of Homer's hugging rectangles will vary but
it turns out that one of them is special. One of the hugging rectangles is a
perfect square. And that was not luck. It turns out that any picture will have a
hugging square. Not obvious at all but it's true. We'll give the idea of a proof in
a minute but first here a couple more examples to illustrate. Apple, Apple, Apple,
and here is its square. Linux penguin, and here's its square. Fish ... square. Just a few
random points ... square. Now it can happen that the shape has more than one hugging
square. For example, all hugging rectangles of circles are squares.
Question for you: Are there any other shapes all of whose hugging rectangles
are squares? Leave your answers in the comments.
Here's a hint. Take a close look at the Mathologer homepage. Okay here's the gist
of a proof that every picture has a hugging square. We start with any hugging
rectangle as in the example over there and color pairs of opposite sides red
and blue. Now let's rotate this rectangle through an angle of 90 degrees so that
at every stage we have a hugging rectangle. Alright the key observation
is that the dimensions of the hugging rectangle will change slowly as we
rotate with no sudden jumps. Using maths jargon we would say that the dimensions
change continuously as we rotate. Now having rotated through 90 degrees we're
back where we started, with the same hugging rectangle. Let's double check. Hmmm, yep exactly the same right? Well not quite. Have a closer look. What has
happened is that the red and blue sides have swapped places. That may not seem
important but it's the key to capturing our hugging square. Okay back to the
beginning zero degrees. Let's record the difference between the red and blue
lengths as we rotate. To begin red is longer than blue, so the starting
difference will be what? Positive negative or zero? Well positive, of course!
However, because of the swapping, at the end of the 90 degree turn red will be
shorter than blue and so the difference will be negative. Now since the hugging
rectangle changes continuously this implies that the difference red minus blue will also
change continuously, ... from positive to negative. And this means that at some
point the difference has to be zero. But the difference being zero means that red
is equal to blue, that the red sides are just as long as the blue sides. And, of
course, that means that at this instant we have a hugging square. And that works
for any picture whatsoever, not just Homer. Super neat, isn't it? Just start
with any hugging rectangle, start rotating and you're guaranteed to come
across the hugging square. What does all this have to do with table turning? Well a very
similar above-then-below argument shows why, as you turn your wobbling square
table through 90 degrees you can expect to come across a stable position with
all four feet touching the ground. Okay mathematical seatbelts on? Then here we
go. Say that square table is on the ground
and it's wobbling with the two blue feet touching the ground and the red feet
moving up and down as the table is rocking back and forth. Now wobble the
table such that the two red feet are exactly the same vertical distance from
the ground. We'll call that an equal hovering position. Now let's rotate the
table 90 degrees, always with the two feet in the air in
an equal hovering position. Here we go. Okay, how have we ended up? Well the table
is exactly in its starting position, except red and blue have swapped. Now
it's the red feet on the ground with the blue feet hovering. But obviously the
swapping is only possible at an instant where both the blue feet and a red feet
are on the ground. We can also graph this as we did for the hugging rectangles.
Let's do it. Here red and blue stand for the vertical distances of the red and
blue feet from the ground. Then just as before this difference function starts
are positive and ends up negative and so should be exactly zero at some
instant along the way. As earlier we concluded both red and blue are the same
at this point. However, since at least one of red and blue is zero at all times, at
this special pink zeroing time both the red and blue distances must be zero. This
means that all four feet are touching the ground and our table has been
stabilized. Fantastic! Okay, so a square table can always be balanced by rotating
it. Always? Really? Well, let's put it like this: I've been using this trick for
decades to stabilize real-life tables and it's never failed me. On the other
hand, it's quite easy to conjure up scenarios in which different parts of
our argument break down and/or stabilizing by rotating won't work.
Second challenge for you today: Try to come up with some setups that foil our
argument. Just to get things going here's something very simple. When we lower the
table over there its top will hit the top of the mountain and the feet will
never reach the ground. n=Now that we have the square table sorted give or take
some finicky details let me tell you a little bit about the history of
mathematical table balancing, the broader scope of the nifty unwobbling-by-turning argument and the part my friends and I played in turning this intuitive
argument into a nailed down mathematical theorem. Okay
hands up who knows this guy. Marty's hand is up. Well out there in YouTube
land not many people are nodding their heads. Marty this is terrible, our hero
has been forgotten. This is Martin Gardner, by far the greatest popularizer
of mathematics of all time. Starting in the 1950s and for a quarter of a century
Gardner wrote the hugely influential Mathematical Games column in Scientific
American. He's also the author of more than 100 books on popular mathematics,
magic tricks, and so on. Gardner inspired more people of my generation to become
mathematicians than anybody else. For example you can
only watch this video because I really got into maths as a result of reading
Gardner's books. And if you enjoy Mathologer videos it's mostly because of
the lessons I've learned from Martin Gardner. He was a master at explaining
real mathematics as simply as possible, and no simpler. In fact much of popular
mathematics in books or on YouTube or wherever have their origin in a Martin
Gardner column. Hexaflexagons, Conway's Game of Life, Dragon fractals, public key
cryptography, wobbly tables. Yes table turning. In the May 1973 issue of his
Mathematical Games column Gardner challenged his readers to discover a
version of our heuristic argument for why stabilizing a square table by
turning should work. He then supplied an answer to his puzzle in his next column.
Gardner credits Miodrag Novakovic and Ken Austin as his source for the
wobbly table trick. I actually tracked down Ken Austin about 15 years ago and
he told me that in fact it was his friend Miodrag who discovered the trick
and how to justify it around 1950. Ken only communicated the trick to Martin
Gardner. Miodrag's version of the table turning is actually a bit different from
the one I showed you and is also well worth checking out. Yet another version
of the argument is presented by the prominent German mathematician Matthias
Kreck who is the star of Numberphile's 2014 table turning video, also definitely
check out that one. Now for a surprising twist have a look at the last sentence
of Martin Gardner's write-up: "A similar argument can be applied to wobbly
rectangular tables by giving them 180 degree rotations, Now of course that's
very welcome news since out in the wild most tables are rectangular but not
square in shape. And there are lots of other objects whose feet form rectangles
to which all this applies. For example I personally use stabilizing by rotating
quite often with my trusty stepladder. Anyway why does everybody only go on
about square tables if this all works much more generally for rectangular
tables. Well when you try to adapt this slick argument for square tables to
general rectangular tables you'll find that this is not nearly as
straightforward as Gardener's off-the-cuff comment suggests. Now before
I show you how to extend the argument to rectangles
let me tell you about the almost definitive paper on wobbly tables which
a couple of friends and I published in 2007, in the Mathematical Intelligencer.
There Marty is also taking a bow. This paper was all about making the wobbly
table argument into a nailed down mathematical theorem by hammering out on
which surfaces a perfect rectangular table can be stabilized by turning it on
the spot. One of the main results of our paper is that stabilizing by turning is
guaranteed to work if the ground and the table have the following properties:
First the ground. The ground has to be Lipschitz continuous with associated
angle of at most 35.26 degrees. Whoa that sounds scary
but it's not really that bad. All it means is that the ground is given by a
continuous function and that between any two points on the ground the ground
slopes at an angle of at most 35.26 degrees. Another way of putting this
is to say that when you slide around the cony 3d counterpart of the green wedge,
all of the ground will always be below the wedge. So the scary term Lipschitz
continuous amounts to a maximum slope requirement for our ground. Okay so
ground continuous and not too steep, check!!! And what about the table?
Well obviously in line with our introductory disclaimer the feet should
form a rectangle and we also require a minimum leg lengths. If the legs are at
least half as long as the diagonals of the tabletop then nothing can go wrong
in the way of little hills bumping into the tabletop. So what Bill, Marty, Reiner
and I proved in our paper is that as long as the surface is continues and not
too sloppy and as long as the table legs are sufficiently long a perfect
rectangular table can be stabilized by turning. The heart of this theorem is the
extension of the nice heuristic argument for square tables but to be able to use
this argument we had to do some pretty hard work to make sure that our
conditions guarantee that the table always rotates as nicely as suggested by
the animations that I showed you earlier. The idea underlying the proof and many
similar proofs is the so-called Intermediate Value Theorem. This is a
mathematically rigorous version of it the intuitively obvious fact captured
by our positive-to-negative diagram, the fact that if a continuous function
changes from positive to negative it will be 0 somewhere along the way. In
fact, in the proof we use the intermediate value theorem not just once
but about a zillion times and for those of you keen on the gory details of the
tricky proof and some of the other nice results in our paper I'll provide a link
in the description. Now before we look at the proof for rectangular tables, here is
an interesting open problem. Is there any non-rectangular four-legged table which
can always be stabilized by turning. For example, what about the half-hexagon
table here. So is it possible to stabilize this or some other mutant
table by rotating? Let me just tell you about one neat observation. It turns out
that it is important whether or not the feet of a table lie on a circle. Well
that's true for our half-hexagon table and, of course, it's also true for all
rectangular tables. Why is this circle business important? Well, it turns out
that if the feet of a table are not on a circle, then there is definitely a not
too sloppy surface on which this table will always wobble. Last challenge for
you today, prove this non-concircular- feet-implies-doomed-to-wobble theorem.
Okay now for the hardcore Mathologer fans let me sketch
how our simple heuristic rotation argument for stabilizing square tables
can be extended to rectangular tables. Okay deep breath. As with square tables
we'll rotate rectangular tables such that at all times they are at an equal hovering
position. So at all times throughout the rotation either the blue pair of feet is
on the ground with the red feet hovering an equal distance above the ground, as
pictured over there, or vice versa. So if we can show that at some instant during
our rotation we are guaranteed to have the blue feet on the ground and there's
some other instant where we're sure that the red feet are on the ground, then we can argue
with continuity as before that somewhere in between all four feet have to be on
the ground. Now an obvious difference between the square and rectangular cases
is that a 90 degree turn won't work for a general rectangular table. We have to
rotate 180 degrees to bring it back to the starting position.
But now the problem is that after a 180 degree turn the same pair of feet are on
the ground as at the start. Let's have a look. There, turn, same feet on the ground.
So unlike in the square case the end position after the half turn is not
distinguished in any way from the starting position.
So our rotation in this case doesn't automatically provide us with that
second crucial position where the two feet that were hovering at the start end
up on the ground. Instead, I'll show you that the crucial position always exists
using a proof by contradiction. So let's assume that the opposite of what we want
to show is true let's assume that somewhere in the universe there's a
rectangular table and a surface on which the table cannot be stabilized by
turning. We're in Fantasyland now and so to keep things as uncluttered as
possible I'll only show you the table and the underlying green points on the
ground. Now if this table cannot be stabilized by
turning, then the blue feet must stay on the ground throughout
the rotation and the red feet will be hovering in the air all the time,
something like that. Alright, blue feet on the ground and red feed hovering all the
time. Let me now show how this assumption leads to a contradiction, which then
finishes the proof, modulo all the finicky details contained in our
Intelligencer paper. Let's highlight the rectangle formed by the feet, and let's
also highlight the z-axis around which we're rotating. That brown line is the z-axis. And let's get rid of the non-essential bits. Now we make sure that
the center of the rectangle will be on the z-axis throughout the rotation, like
that. So the center just wanders up and down the z-axis as we turn. Okay two
important observations. First, since we are rotating through 180 degrees you've
just seen all possible balancing positions of our table, with the center
on the z-axis. There are no other balancing positions like this. Second,
let's call the height of the center of the table in a certain position the
elevation of the table in this position. Now, as we rotate, there will be a
position of the table where the elevation is at a minimum. Let's see, okay.
so it's wandering. Now where's the minimum? Here, okay, did you spot it? Well, there it
is. We're almost ready for the punchline. Notice that if the diagonal connecting
the red points is translated straight down, the diagonal will connect the green
dots underneath. But this means that we can rotate the table into a new balancing
position with the blue feet ending up at those two green points. But this new
balancing position is absolutely impossible. Why because the elevation of
the new balancing position is lower than the supposedly minimal elevation that we
started with and that's the contradiction. Pretty tricky but also
pretty cool don't you think? And that's it for today as far as table
turning is concerned. I've just got one more thing to do before I sign off,
announce the winner of our epicycle competition from last time. Lots of nice
ideas and implementations which I've listed in my comment pinned to the top of
the epicycle video comment section. The one I settled for as the winner is by
Tetrahedri who succeeded in doing something extra special. Their epicycle
system simultaneously draws out three iterations of a space-filling curve how
neat is that? Check out the details of how they did it in the description of
their video. Okay Tetrahedri please send me your contact details via a comment and
I'll send you Marty's and my book. And thanks again to everybody who took part
in this competition. And that's it for today. Happy table-turning.
I once tried this with a table in my backyard over and over and although I totally get the proof it still felt like magic when the feet of the table kept rotating into stable positions.
I'm pretty sure it's just an application of the Intermediate Value Theorem.
Is it also true that any compact n-d shape has a hugging n-d hypercube. In what branch of mathematics would a question like this be considered?
My have the tables turned
Cool.
I recommend the overall work of Keith Devlin if you find this video particularly interesting.
My first thought is an application to the inscribed square problem. Couldn't you define a curve that hugs a plane of a surface, then use this theorem to show that a square table can be stable on it, thus having all 4 verticies of the square base on the curve?
I’ve just done this and it bloody well works 🤣🤣
The Numberphile video on the same topic with Prof. Matthias Kreck is not quite as thorough but is extremely well done and pleasant to listen to.