Mathologer - 2018-07-28
NEW (Christmas 2019). Two ways to support Mathologer Mathologer Patreon: https://www.patreon.com/mathologer Mathologer PayPal: paypal.me/mathologer (see the Patreon page for details) This video is about the absolutely wonderful wobbly table theorem. A special case of this theorem became well-known in 2014 when Numberphile dedicated a video to it: A wobbling square table can often be fixed by turning it on the spot. Today I'll show you why and to what extent this trick works, not only for square tables but also general rectangular ones. I'll also let you in on the interesting history of this theorem and I'll tell you how a couple of friends and I turned the ingenious heuristic argument for why stabilising-by- turning should work into a proper mathematical theorem. Here is a link to a preprint of the article by Bill Baritompa, Rainer Löwen, Marty Ross and me that I refer to in the video: http://arxiv.org/abs/math/0511490 . This preprint is pretty close to the printed article that appeared in the Mathematial Intelligencer and which lives behind a pay wall, Math. Intell. 29(2), 49-58 (2007). The argument that I show you in this video is somewhat different from the one Bill, Rainer, Marty and I used in our paper. It's a mix of what we do in our paper and the original argument by Miodrag Novacovic as presented by Martin Gardner in his Mathematical Games column. Here is a link to the 2014 Numberphile video on table turning featuring the prominent German mathematician Matthias Kreck http://youtu.be/OuF-WB7mD6k And this is an article by Andre Martin that features an alternative proof for why stabilising-by-turning works for square tables on continuous grounds that are not too steep: http://arxiv.org/abs/math-ph/0510065 Thank you very much to Danil for his Russian subtitles and Marty for his help with getting the draft of the script for this video just right. Enjoy!
Annoyed that it's not called the stable table theorem >:(
If the table wasn't wobbly in the first place, you wouldn't even know about the theorem.
Ian Allen Fair point...
the unstable table theorem?
Novel Wobble Theorem
why not the fix-the-unstable-table-by-turning-it-if-the-floor-is-sloped-less-than-thirty-five-point-two-six-degrees theorem XD
Don't lie, that circle was rotating too!
It's a joke. Even if the circle were rotating along with the square, it wouldn't matter because of the inherent properties of a circle.
Ding dong ding.
JustOneAsbesto I can't really read sarcasm from text...
Alright, fair enough, buddy.
I don't really think there's much to work on. If you had said it in person, i may have gotten the hint, but reading a string of letters is not enough so read the tone of a message.
JustOneAsbesto I guess I could have seen the ! And the thumbs up maybe.
Turning the tables with DJ Mathologer
* sees unofficial Linux mascot in thumbnail, instantly upvotes *
Jonadab the Unsightly One, that's what I knew as well.
Found the redditor
FYI :
https://web.archive.org/web/20040401161253/http://www.linux.org/info/logos.html
https://en.wikipedia.org/wiki/Tux_(mascot)
Scrolling down . . . Saw Tux . . . Clicked Instantly . . .
Same there
In typography and computer graphics, a “hugging rectangle” is called a minimum bounding box.
In regard to the question posed about other shapes where all hugging rectangles are squares:
Clearly a square would fit this criteria, as the minimized square is a repeat of itself, and both sides would grow identically as you moved to the corners. Logically this would also apply to any other regular polygon with a number of sides which is a multiple of 4, since all sides would climb to a tip and descend to a flat synchronously. The circle shown can be approximated as a case of above as the number of sides goes to infinity.
Helgefan: If Im not mistaken, as you keep the radius of the circles constant but shrink the equilateral triangle used to construct the Reuleaux triangle you still have a shape of constant width, it just approaches a circle as the size of the triangle approaches zero. Ill have to think about that though. The Reuleaux triangle is a special case of how to construct shapes of constant width.
Yeah, you can add to the radius as much as you want though the bit at the corners is a separate arc. As the equilateral triangle shrinks, the construction approaches a circle constructed of 6 equal length arc segments. So at least for shapes of constant width constructed from triangles, what I said holds for a transformation to a circle.
htomerif Oh I see what you mean now. Interesting idea! I wonder what such a curve midway between a square and a Reuleaux triangle would look like if that really works as well!
Galdo145 s
htomerif Yeah but your conjecture was not that there is a continuum of shapes from a constant-width shape to a circle.. you said youd guess there is a continuum from a constant-width to a 90 symmetry shape and the things in between, which are neither, will still have a hugging square. That was the conjecture. So continuously shrinking the triangle of circle centers for the reuleaux triangle accomplishes nothing proves nothing. I was interested and with you up until the conjecture which does not seem intuitive or even likely to me but I dont know could be. But the 90 symm is certain and certain that you could find shapes deviating from that by changing the shape in the parts that dont hit the square, and even the 90symm convex hull defining each family of shapes seems right.
"which also applies to non square tables"
except triangle/three legged tables, which do not and can not wobble.
well in the 3 dimensional region at least.
Your logo of mathologer is the answer
That's a square 😎
Love from India🙂 :)
Logo is a copy of Helsinki metro logo
I once tried to apply the wobbly table theorem but didn't realize the floor had a slight kind of step to it. A discontinuity that wouldn't let me stabilize the table, so I had to find other means.
(Also, roughly half of all wobbly tables in my lifetime are due to uneven leg lengths, not just uneven floors, so it's been very inconsistent for me)
Eric Vilas is this a joke? I really can't tell.
not really, just an actual thing that happened. Whenever a table is wobbly I try to see if I can fix it by rotating it. It just so happened that the "continuous function" part of the proof didn't apply that time
Eric Vilas you must have seen a lot of wobbly tables then!
Eric Vilas I feel your pain. Nearly all wobbly furniture I've encountered had uneven leg lengths. Too bad there's no solution for shoddy craftsmanship.
More like 99%. It is almost always the table not the floor. Rotating almost never works
14:35 – Is that angle, 35.26° equal to arcsin(1 / sqrt(3)) ... and why?
Yes, that angle is the angle between a diagonal of a cube and one of sides of the cube that ends in this diagonal. Have a look at our paper liked in from the description for why exactly this angle pops up in this context.
If a table's legs are not on a circle, then when the table is rotated each leg follows a different path. Therefore, a certain angle of rotation for one leg doesn't correspond to the same position on the plane for another leg at the same angle. Which makes the intermediate value theorem invalid
The legs of any rectangular table will sweep out a circle when rotated.
except that, in general, if you start a rectangular table at a given position, most of the circle defined by its feet will not be in contact with the floor, so if you rotated the table so that its feet traced that circle, you'd have to pass the table legs through the floor and/or have the able hovering off the ground...
Although saying that the intermediate value theorem isn't valid doesn't guarantee that there isn't a way to turn the table. Just that this method of proof doesn't work. You'd still have to construct a counterexample.
For example, you could have that 3 of the legs (say A, B and C) lie on one circle (called oABC) and the fourth (D) lies on a different circle (or rather on a concentric circle with a different radius, called oD). Now imagine your floor is a series of circular ripples, with the centre of oABC at the centre of the ripples. Suppose oABC coincides with a peak of the ripples and oD lies in one of the troughs. Then however you rotate the table about this axis, A, B and C will always be on the floor and D floating.
Now I guess the question would be can you construct a table with a 'centre' such that oABC exists? What does the 'centre' of the table mean? The geocentre?
Although as I say this I realise that the half-hexagon is stabilised by turning it around a point other than its centre, namely its edge. So maybe this is all just a ramble that misses the point... :p
andymcl92 You are absolutely right, just saying the intermediate value theorem doesn't work doesn't prove there isn't a way to turn the table. However, actually inventing a proof is way beyond me.
Yeah you could think about circular grooves of constant elevation/depth that keep the table wobbling at any angle of rotation.
If your donut rolls off the table, you'll wish you had the wobble back.
If it rolls off the table it becomes a D’oh!-nut
D’oh
Doug Rosengard Homer would eat it anyway.
If your donut rolls off the table, then why did you put it down on its side? I feel like it's more of a personal problem than how the table's arranged.
The wobble won't stop the roll.
I’d imagine all shapes of constant width would have hugging rectangles always in the shape of squares. The English 50p coin for instance. Would mean there is an infinite number of shapes which have nothing but hugging squares.
2:15 "I've cut it off twice and it's still too short!"
18:20
So, what you're saying it, for any continuous typology, there's a circle at some altitude with at least four prechosen points that intersect both the surface of the terrain and the circle.
5:00 Треугольник Рёло? Другие фигуры постоянной ширины?
Shapes of constant width should have only hugging squares, right?
This can be generalized further, as curves of constant width fulfill a special case of this property where all hugging squares are the same size too. I can't really put the more general case with varying sizes into a sentence, though.
More generally, all squares have only hugging squares :p
It can be shown using a digression from one of the proofs of Pythagoras's Theorem
Does that mean 3-dimensional shapes of constant width such as the Meissner tetrahedron only have hugging cubes? I believe they do! :)
Any shape whose convex hull has order 4 rotational symmetry will also always have square hugging rectangles (order 4k if you insist a regular octagon does not have order 4 rotational symmetry).
I don't know what Order 4 means but I believe any regular polygon with an even number of sides should have only square hugging rectangles. I believe although I haven't tried to do any working out, that something like a pentagon or a circle will have rectangles when there is a flat side opposed by a point and the width is point to point. In even sided polygons the point on the the perimeter which is 180 degrees opposed should have a pair of analogs perpendicular which are of equivalent length and therefore only square hugging rectangles could exist
Not all regular polygons have only square hugging rectangles - consider a regular triangle.
However, any shape which is 90° rotationally symmetric will have all hugging rectangles be square.
Ah, I didn't think about it like that. But if I may add to your post, the length of the sides of a bounding (or hugging) rectangle would oscillate in a pattern of n-phase sine waves (think 3-phase AC current, taking the maximum value, kind of like driving over the top of the waveform) as it spins at constant speed. A circle is like a limit as number of sides goes to infinity, resulting in zero oscillation in the side length. The 90 degree symmetry you mentioned would be like the length of your wheel base, and controls whether your car tilts back and forth as it hits different phases of waves going over it, thus producing a rectangle with a difference in side length, or simply bounces up and down but stays horizontal, meaning the y coordinate of each wheel, aka side length, is equal, forming a square. ... right? :p
Actually, does the length of the car's wheel base even matter? I guess it's just a property of a 3 phase system that no matter what length you choose, as long as that length is constant, you can not trace over the top of the waveform and have both ends remain at the same height. For a 4-phase system, you can. Hmmmm....
A regular triangle have a hugging squares at 15° and 45° rotation.
Soppel Post He didn’t say they have no hugging squares, he said they won’t always have hugging squares.
Bob Bobson well theres nothing with NO hugging square so it’s unnecessary to say a particular shape has one
Omg this theorem is so useful. Whenever I’m holding a seance where the table has an uneven leg and is on sine waves, this helps so much?
Reuleaux triangle also has all hugging rectangles as squares....
& all other many many types of fixed width shapes as well
"One of the (table) legs has been gnawed out by a beaver or a ghost ."
"...Gnawed at by a Beaver, or a Ghost..."? Lol!!! I needed that before bed. This is gonna be the first time I've fallen asleep by exhaustion due to laughing. Good bit sir.
5:05
A curve an constant width with a positive or zero padding number
Thank you for the rectangle proof! Now I can finally fix my wobbly bed :P
Yes, beds, step ladders, all sorts of rectangular stuff can be tamed with this trick. Of course you may end up with your bed stale but not aligned with the walls of your room. Oh well, ...
9:33 but you said yourself that sometimes the issue is that the legs are uneven so surely it has failed sometimes?
"out in the wild", great line. Haha
The proof at the end is AMAZING!!
As a few of you have observed that all hugging rectangles of squares are squares which is nice. What I find extra nice is that a square hugging a square is the diagram at the core of the most popular visual proof for Pythagoras theorem (This also happens to be part of the Mathologer logo :) . https://youtu.be/p-0SOWbzUYI
Thanks for reminding me of this! I once had an idea that this could be a way to get the moral equivalent of the Intermediate Value Theorem in an intuitionistic logical framework. I will think about this a bit harder and try to make the vague idea more concrete, ....
OK, just a note: try defining positive and negative numbers as scalar products of unspecified basis vectors. Negative numbers arise when the vectors point in opposite directions, but the situation is more symmetric because in the usual concrete system there is no way to get -1 as a product of +1s! The idea I have is to a circular definition of a vector space as a Cartesian Closed Category. The idea is then to use some axes with something like Hermitian symmetry, and this would be related to even and odd functions and derivatives. See https://livelogic.blogspot.com/2019/09/a-concrete-way-to-see-what-computable.html
I use this IRL too, it's really effective. Making just one leg shorter on a completely flat surface causes a problem.
However, if you get the ghost to gnaw one of the adjacent legs to be shorter, then the feet will make a rectangle again and problem solved ;)
I figured this out on my own years ago while trying to orient my scale to have all four feet firmly on my non-level floor. I always found that I could eventually find a rotation where all four feet were on the floor.
The final one was an Awesome Proof!! And any regular polygon — any hugging rectangle around a regular is itself a square since the triangles formed between the 2 shapes are all congruent by AAS .. ♥
9:24 regular polygon shapes
YES there are other shapes with infinitely many hugging squares : the solids of constant width
like the reuleaux rotor, for example
you make me laugh like a baby its so much fun when you realize whats happening before he says it :)
11:04 Gardner was born in 1914 not 1918.
Thank you very much! I must say that the epicycle challenge was very fun and inspiring, so much that I ended up spending whole weekend with it :D. And I did 2nd to 4th iteration, not the first 3, since the first one would've been quite boring ;).
However, I don't feel quite comfortable to give my contact information publicly here in the comment section. Is there a way to send a private message through Google+ or YouTube? If not, then I can leave my email here, until I get confirmation that you've received it.
And while I'm at it, maybe it's finally time to say this out loud: your videos have been fantastic so far! I started as a silent lurker, until I finally decided to subscribe to your channel. You definitely are up there with other top math-content creators, and often give a new perspective to familiar concepts. Keep up the great work, I'm always looking forward to your videos ;)
Great, I just sent you my contact information to email!
It's been a while since I watched the Numberphile video, but I seem to remember that in their video they mis-attributed the explanation to the Mean Value Theorem rather than the Intermediate Value Theorem.
It did bother me that he didn't use either of those terms in the video.
(edit) He actually did.....
What is proof that dimensions of hugging rectangle change continuously while it turns 90 degrees?
It's hard to explain without pictures, but it's definitely true. Think of it this way: the bounding line can never jump inwards as you rotate it because the point it was previously contacting will always be in the way. Perhaps, then, it might need to jump outward in order to avoid a new point that comes into its path. But then you could just make that new point your point of contact (and your previous point of contact would never be in the way because you've rotated past it), so such outward jumps are unnecessary.
Good question. My intuition says to me that it's obvious, but I can't think of a sensible proof immediately...
So I'll have a think :)
I guess what @trogdorstrngbd said is the basis for the proof. As you turn, the only ways the lengths could be non-continuous is if the rectangle were to jump in or out. Since each edge of the rectangle is tangent to its contact point, an inward jump would not be possible is it would now cut through beneath the previous contact point. So now we just have to eliminate outward jumps.
As trogdorstrngbd said, an outward jump would only be if a new point came in to the path of the edge. But this would happen continuously as the pine is just rotating and shifting in a continuous manner. As this happens, the edge would now be tangential to both the old and the new contact points, and as it continues to rotate it would simply leave the old contact point and attach to the new one. Therefore there would be no outward jump. Since the rectangle can't jump out of in, it must move in a continuous manner.
Does that make sense (both mathematical sense and is it easy enough to follow)?
@2:15 curse those Ghost Beavers and their hunger for table legs.
I never realised you looked like homer, skinnier though !! Everything makes sens now :D
Maybe I'm missing something but when you turn the square table, around what point do you do it?
Amazing, really amazing visual proof. Thank you very much.
Your logo :P
A square's hugging rectangles are always squares!
Exactly, and all of these squares hugging squares "prove" Pythagoras theorem :)
I think that would also hold at least for any other regular polygons with a number of sides that is a power of two...
I knew this already. We were taught our tables in primary school.
Hugging rectangle is a square: SQUARE
If there happens to be an infinitely steep cliff in the floor, the function will not be continuous and might not become zero.
THANK YOU MATHOLOGGER!!! I always wanted to watch a video about the wobbly table theorem!
David Fan - 2018-07-28
The wobbling tables at work have always gotten the best of me.
Now the tables have turned.
Logan Revesby - 2018-07-30
Cannot fault this comment. Damn.
Non' AchYourbusiness - 2018-10-13
We don't deserve this comment
Lord Tachanka - 2018-11-23
David Fan quick, hit em when their back's turned!
Blox117 - 2018-12-10
do you mean a turntable?
Jace Wright - 2019-01-31
You just earned a sub.