The fix-the-wobbly-table theorem
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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: 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 • Fix a Wobbly Table (wi...
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: 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!
The wobbling tables at work have always gotten the best of me. Now the tables have turned.
A+
This comment is more clever than the theory.
We don't deserve this comment
David Fan quick, hit em when their back's turned!
do you mean a turntable?
Hey that totally worked! Fixed my wobbly table that's been wobbling for years. Thanks Mathologer!
two most important figures in math over centuries leonhard euler and homer simpson
In typography and computer graphics, a “hugging rectangle” is called a minimum bounding box.
This is my all time favorite Numberphile topic to bring up in daily small talk and now even better due to learning it is a 180 degree turn for recantular tables! Such a great video. From a guy that grasps only about 40% of what you discuss I still love it all. Solid editing and realistic enthusiasm with unpretentious delivery, Thanks!
:)
Turning the tables with DJ Mathologer
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
Don't lie, that circle was rotating too!
Yeah, I can see why that would be confusing. But the points never exit the sphere created with center at the center of the square, and radius of half the diameter of the square.
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.
You don't know history ahead of its time. I may dare say, however, that you and 3brown are really together with Gardner in creating the next generation of mathematicians, physicists and so forth. I was just beginning college by the time I discovered this sort of education on YTB, however, I think it still had some profound impact on my interests and to deepen my studies, that's for sure.
Annoyed that it's not called the stable table theorem >:(
Yeah, because now we basically have "The wobbly table theorem states that no such table exists*" (*assuming nice tables and sensible floors)
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?
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
THANK YOU MATHOLOGGER!!! I always wanted to watch a video about the wobbly table theorem!
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.
Galdo145 let me add to that, any shape that has 90 degree rotational symetry or two orthogonal lines of symmetry
And also any object of constant width such as the Reuleaux triangle
I solved it without the logo...
cayge sinnett 90° rotational symmetry works, but 2 orthogonal lines of symmetry doesn't. An ellipse has two orthogonal lines of symmetry, but there is only one hugging rectangle that is actually a square for any ellipse other than a circle.
It seems people have figured out the same 2 types of shapes that I did though: shapes with 90° symmetry and shapes of constant width. I'm curious to see if there are any others, though. Surely there are, but I can't think of any.
Reuleaux triangle also has all hugging rectangles as squares.... & all other many many types of fixed width shapes as well
The proof at the end is AMAZING!!
Hi Mathologer, a friend recently linked one of your videos for me, and I liked it so much that I have now binged-watched all your videos. I'm quite impressed overall, and I think math education (particularly K-12 in the U.S.) could take some notes. As for this video, as soon as you presented the original square table problem, I immediately thought, "this is just an application of the Intermediate Value Theorem," and almost commented before watching the rest of the video :P. I will have to rewatch it to clarify some of the details of the rectangular table case, though.
Amazing, really amazing visual proof. Thank you very much.
Brilliant video, Burkard; the Intermediate Value Theorem is beautiful!
You simplify math. Amazing. Keep it up
The reuleaux triangle also has this property of having all it's friendly hugging rectangles being squares (which is why it can roll correctly)
That was my first thought as well. Any shape of constant width will do that.
Excellent video. Thanks!
Thank you once again BP.
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 .. ♥
I loved Gardner, and Scientific American. I read every magazine my highschool library had, and then rescued my favorites when they were being thrown out.
Thank you for the video! All of you friends awesome!
very good presentation
One of the best math lectors I ever saw!
I live in Spain where many coffee bars have tables in the street so I've come across lots of wobbly tables and a significant proportion of my coffee ends up in the saucer. This video may well transform my breakfast experience. Thanks.
Wonderful. Thank you so much.
All set of points with convex hulls of constant width have infinitely many "hugging squares". Since the shape can "roll" freely through two parallel lines and maintain a single point of tangency with each, then for every orientation there are 4 lines which are perpendicular to each other, are all tangent to the shape, and are the same distance apart (sorry for the informal wording, but you get it)
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.
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 livelogic.blogspot.com/2019/09/a-concrete-way-to-see-what-computable.html
I've been using this theorum ever since I saw the original video. It's a handy Cafe trick.
Regarding shapes with all hugging rectangles being squares: Looking at the Mathologer homepage, the Mathologer icon shows the M-shape (a square by itself) hugged by a square. All the hugging rectangles of this square will be squares, but with different side lengths depending on the rotation angle of the hugging rectangle. The side length variation repeats every 360/4 degrees. This also applies for all regular polygons with an even number (2N) of sides. The side length variation decreases with increasing number of sides and repeats every 360/(2N) degrees. The circle is just the extreme case with 2N approaching infinity and the side length variation approaching zero. Other shapes with hugging squares are curves of constant width (Gleichdicke), as the Reuleaux triangle or Reuleaux polygons in general, e.g. the shapes of the British 20 and 50 pence coins. The center of the hugging square will wobble when rotated around such a shape. Back in the 1970s, I had a book written by Martin Gardner (Mathematischer Karneval) and if I remember correctly, he described there the "super ellipses" which were used to design the layout of public places. By choosing the two axes of the super ellipses to be equal, these shapes could be specialized into "super circles". Just as with the square shape (which could be seen as an extreme case of the super circle), all the hugging rectangles of such super circles would be squares, with a variation in side length repeating every 90 degrees of rotation.
Great answer :)
For the hugging square puzzle, Rouleaux "polygons" (I don't know how else to call them) are another possibility. Of course, regular squares also have only hugging squares, as opposed to hugging rectangles.
For the hugging squares question, All hugging rectangles of any shape who's width has 90 degree rotational "symmetry" will be squares. These shapes are defined so that, when taking the width for the shape from any angle, rotating the shape by 90 degrees will result in the same width. This includes all shapes of constant width and all shapes with 90 degree rotational symmetry, but I also theorize that there should be shapes which have neither property but still have exclusively hugging squares.
Shapes with N straight sides where N=4(2^x) - where x is an integer - have hugging rectangles that are all square (shapes include perfect squares, octogons, 16, 32, etc sides). This is because from every point in these objects, you can draw a line to its opposite point and that lines perpendicular bisector of equal length will contacts two other extremity points (four equal radii if you will) All four of these end-points are now the midpoints of each side of our perfect hugging square.
At first sight I thought it was MacLane (which I had no idea what he was doing in your video), then I remembered it was Gardner. Extremely nice proof! I can see where you used all the main assumptions you had.
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.
I’m a simple man, I see a maths video with the Linux mascot (tux) in the thumbnail, I click
Brilliant
Well yes! All hugging rectangles of a square are squares. What a hint. In fact, all the hugging rectangles for just the four vertices of a square are hugging squares. In fact, any concave shape may as well not have concavity at all for these purposes - the hugging rectangles for the letter A, for instance, are the same as the hugging rectangles for the triangle formed by getting rid of the horizontal line and putting a line connecting the two bottom-most points. All concave shapes are identical to a corresponding convex shape, here. If that convex shape is a square, or a regular octagon, etc., or a circle, then the hugging rectangles are all squares. I don't know if this covers every case though!
4:34 That makes sense. If you rotate a hugging rectangle 90°, the aspect ratio will swap, and as it smoothly transitions from one to the other, it must become a square at least once along the way.
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
you make me laugh like a baby its so much fun when you realize whats happening before he says it :)
Bro this is awesome
any shape that has 4 points along its edge that create the 4 corners of a square whose area that extends past that square must have a 90 degree rotational symmetry to have a square hugging rectangle at any rotation of the rectangle. this can also be scaled up to any hugging regular polygon, the general rule is that a shape must have n points along its edge that create the vertexes of a regular polygon, and the area that extends outside that polygon must have a rotational symmetry of 360/n degrees to have a hugging regular polygon at any rotation about the shape.
You, Grant Sanderson, Salman Khan, Matt Parker and James Grime are Martin Gardeners for us.
1. Stabilizing the table by rotating wont work anymore as soon as the ground doesnt change continously anymore because then there could be gaps in the graph. 2. And I believe a reauleaux triangle would have a hugging square at all times.
Non-continuous can definitely be an issue. Also shapes of constant width are of course supernice in terms of hugging rectangles as their hugging rectangles are not only square but even better all squares are of the same size, just like with circles :)
It could be possible that the difference between heights of the legs could be non-continuous at 0 while motion remains continuous. Is it possible the theorem could be wrong? Let me explain. The reds could have an infinitely small height while the blues are at zero. Then in the next moment in time, the reds jump from infinitely small to zero, while the blues jumped from 0 to infinitely small. Now we're in the negatives without actually touching zero. All we have to do is assume that a distance can go continuously from infinitely small to zero, and under that assumtion, the theorem is wrong.
If the theorem "could be wrong", it would not be a theorem :)
Did you consider what I had to say after that? I am completely okay with being wrong about this as long as I understand why the theorem is correct. I mean no disrespect by trying to refute the theorem. Maybe I'm more a scientist than mathematician...
Also, to correct myself, what I am implying is that It is mis-named and should be called the fix-the-wobbly-table conjecture. I guess I would be undermining all of mathematics if im right, but is that even likely?
I can't wait to encounter my next wobbly table so I can try this out!
Any shape which has four points on its linies that form a square (which do not intersect with the shape itself) produce square hugging rectangles at any angle. You could also rephrase that as: Any concave shape of which its four most outer points from the center form a square always have a hugging square at any angle.
4:00 "hugging rectangle". Often in computer graphics we call these things "bounding boxes". "Axis-aligned bounding box" or AABB for the straight up and down rectangle that you started with.
10:08 If we assume that every part of the table except for the four points at the bottom of the legx are intangible (meaning they can go through the ground), we can avoid exceptions like the cylinder there, since those aren't really part of the math problem. Then, the only exceptions will be non-continuous surfaces. First of all, many attempts to make a non-continuous surface don't work. Even if there is a cliff where the height goes up suddenly, it is still continuous, you just have to change the x-axis of the function from being the angle you rotated it to the distance you have traveled, so that you can include the continuous change in the height as you drag the leg up the cliff. You can also account for ledges this way. Note that this will, however, lead to some possibilities that don't work in real life, like the leg resting halfway up a perfectly vertical part of the ground (technically touching the ground, but practically, still not able to prevent a wobble) or the leg having to go through a ledge to get to ground underneath it. In order to make a non-continuous surface, the surface has to actually have some sort of gap in it.
"...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.
Although it's visually clear that rotating a hugging rectangle is a continuous transformation, is there a way we can show that mathematically? My guess is that it would be based on the formal definition of the "hugging rectangle" function.
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, ...
The fact that a Rouleax triangle has constant-size hugging squares has a practical application -- those triangular drills that drill square holes.
Hello, I'm a big fan of Mathologer videos, thanks for sharing all these wonderful insights. I have a request: Burkard please could you do something on chaotic dynamical systems, the transition to chaos and it's links to fractal geometry. I studied all that many years ago but I don't think I ever really got the essence of it.
One of my earlier videos is about the Mandelbrot set and the nice connection between the logistic map and the Mandelbrot set. If you have not seen that one yet ... Definitely a lot more nice videos to be done in terms of dynamical systems :)
Mathologer Thanks, I will check out the Mandelbrot video again.
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 KZhead? 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!
For the rectangle case, you can also do a contraction along one axis of the table and the surface to reduce the rectangular case to the square case.
Well, the problem with this approach is that if we apply the contraction along one axis to everything in sight the rotation around the axis turns into something that is no longer a rotation. Or if you only contract the table, then we are talking about a completely different problem :)
I did not know of this theorem, however I have already corrected a wobbly table at a local bar by turning the table.
Tabled! I'm floored in your hippopotamus hypotenuse! OMG 😂!
Amazing video 😍✌❤️❤️
Answer to question posed at 5:00 Any shape with a constant width would have a hugging square in any orientation. For example, a Reuleaux triange would have a hugging square in any orientation. Proof for 17:49 Imagine the ground is a shallow right cone with a large enough base, pointed upward (like a large anthill). The table has one leg positioned outside of the Circle made by the other three legs (any table without symmetry will satisfy this condition for at least one leg). The table is positioned above the ground so that the Circle is parallel to the base of the cone and the cone's axis passes through the center of the Circle. The one leg will never be at the same level as the other three legs, even if you rotate it 360°.
2:10: I tried to stop the wobble on my mini-table with the trick, I was like "Don't really work...", and then, I realized that a feet is a bit broken: lenghts are unequal ! Thanks for the info, it really helped me with a 2 year-old problem 5:12: A square ^^ 5:46: It remembers me Borsuk-Ulam theorem 16:18 now I know why xD 18:53: Why 180° ? I think maybe there is a better answer. I propose you the biggest (obtuse) angle made by the diagonals at the center (which is 90° for a square): it moves the wobbling/red pair at the initial position of the stable/blue pair (it do the trick I think ;) )
When we minimise to find the contradiction we have to be sure that we are minimising over ALL possible equal hovering positions of the table to be sure that when we construct the new equal hovering position it is one in the set we have been minimising over :)
Thx Prof for the reply :D ! I understand the contradiction proof of a stable position and the fact that the contradiction is not in set, I totally agree (how I couldn't?) and it's always amazing math with you guys :D ; however, I think we can prove it directly with the intermediate value theorem and optimize the angle needed (less than 90°). Since the rotation of any of the 2 angles (the vertex angles of the isosceles triangles, not just the obtuse one in fact, the acute one too) between the diagonals at the center of the rectangle WILL "always automatically provide us the 2nd crucial position" (19:19), we can then use the intermediate value theorem as for the square, no ? I got something wrong on this one Prof ? In fact, with a rotation of any on this 2 angles, The 2 red feets that were hovering at the start ends up on the ground: they reach the initial stable position of the blue pair, therefore since we got continuity and the red feets reach a vertical distance of 0 (and the blue distance can't be negative) , we got our stable position as an intermediate value, and so no need to prove it by contradiction. It gives us 2 optimized angles, less than 180° for the obtuse one and less than 90° for the acute one (some trigo and we get the values)
I have been doing this rotation trick for decades. Not sure I can remember even how I cam across it. But I encountered a table yesterday with on leg about a half in longer than all the others ! Oh man - had me frustrated.
@12:50 Gardner's last sentence is correct. Because his argument was different from yours. He required 3 feet A, B, and C to touch the floor all the time and check the behavior of the last one D. That works for rectangular table too.
Well, it may appear so at first glance, but maybe try the rotating with a physical object like an iphone and be very clear about what the start and end positions of your rotations are. There is really nothing really obvious to prevent A,B,C to be on the ground at all times and D in the air. If it was really that easy Gardner would not have bothered with restricting himself to talking about square tables :)
You're right, I didn't think about this carefully enough. Thank you!
I thought about constant width shapes, like the ruleaux triangle, all of whose hugging rectangles, like the circle's, are also squares, and similarly to the circle, all of the edge sizes of the squares are equal - to the width of the shape
shapes who's hugging rectangles are squares: 1. all shapes that have 90°/n; n€N rotational symmetry, including the square 2. shapes of constant width, including the circle (because parallel tangents to them are always the same distance apart)
concerning the hugging squares question: it seems intuitive that any shape with a 90 degree rotational symmetry will work. Or even more generally: all shapes whose convex hull has a 90 degree rotational symmetry. Did I get them all?
0:00 Video Starts 1:53 Disclamers 1. Table legs need to bof equald lenghts 2. Table is "stable" in the sense that all parts of the tableis touching the ground . In real life this is not a big deal as it is hardly noticeable so the application of this model to real life still holds good 3:48 "Hugging rectangles"/ Intermediate value theorem 6:10 Using calculus to find hugging rectangles of different shapes 7:48 Simulation of the model for squares 8:47 Using "Hugging rectangels" principle to find points of stability 12:10 The different models of table turning , Miordrag's and Martiez Krieg "Bier gartne guy" 14:28 The assumptions (Lipschitz continous and leg lenghts and stuff) 16:15 Intermediate value theorem 17:05 Possibility of other shapes of stablisation 17:30 The reason why it needs to be on a circle 18:00 The full showcase of the model 19:07 Rectangular case 19:34 Proof by contradiction
For those of you who may not have taken calculus, this is an application of the mean value theorem. The floor has to be continuous/differentiable for the rotating to work, which means more concretely that this might not work if the floor has holes, a step, or the table is near a corner. In practice, this rule usually works because a flat-looking floor has only a small curvature, and most tables are chosen to fit in a room comfortably.
Ah, it's mentioned at about 16ish minutes in. That'll teach me to comment halfway through the video!
:)
I know the oter shape who's all hugging rectangles are squares, squares! :D Thanks for the info, I directly looked at the channel logo.
:)
Mathologer oh getting a reply from you. Now I can die in peace :)
Anything with 90 symmetry. And shapes modified from such, modified in concave areas that never hit the rectangle. And things with constant width. But anything else? Will he ever tell us?
It looks to me as though you're using a vertical projection of the corners of the table to find the contact points. But as soon as you rotate the the surface of the table about some combination of the X and Y axes ("tilting" the table), your corner projection changes from its original aspect ratio to some other aspect ratio. (In the case of the square table, when you tilt the table, the projection goes from square to rectangle.) I don't know that this causes an insoluble problem for the intermediate value theorem, since all such projections would also be changing continuously, but I'm not sure that the way you've presented the proof here is sufficient to make your case.
It would be great if you did a series on the new winners of the fields medal and tried to explain what they do, especially peter scholze. Great video.
So many nice topics to cover so little time. I've had a look at some of Peter Scholze's mathematics a while back, really beautiful stuff. We'll see :)
the hugging squares will also occur when using a reuleaux polygon because of their constant width
17:41 If the surface has a constant height for each circle that the feet rotate around (e.g. a conical of hemispheric surface), and the table is unbalanced to begin with, which is possible since each circle can be a different height, rotating the surface will change nothing, so the table will always be unbalanced.
Yes, however, imagine that you arrange things in such a way that throughout the rotation the point at which the diagonals of this table intersect are always on the z-axis :)
Mathologer Keep the intersection point completely stationary, and have the shape perpendicular to the z-axis. Rotate it around the z-axis. Trace out the circles that each point makes. There are four possiblilties: 1. Three of the points are on the same circle, and the fourth is not. In this case, have the circle with three of the legs be part of the ground. Have a circle that is directly below the circle with only one point on it also be part of the ground. This ensures that three points will always be on the ground, but the foufth will be above it. The rest of the ground can look like anything as long as those two circles are part of it. 2. One circle has two points on it, and two circles have one. Have the circle with two points and one of the circles with one poing be part of the ground, and then have a circle directly below the remaining one be part of the ground. Like case 1, three points will stay in the ground, but the other will always be above ground. 3. There are four circles, each with one point. Have three circles be part of the ground, and a circle directly below the fourth one be part of the ground. The same argument applies again. 4. There are two circles, each with two points on them. The points that are together on the same circle must also be opposite from each other (meaning that the shape is a parallelogram), because if they were next to eachother, the shape would be an isosceles trapezoid, which can always be inscribed in a circle and therefore doesn't meet the criteria for the shaoes we are considering. Have the outer one be part of the ground. Have a disc with its circumference directly below the inner circle be directly below the ground. This will give us two points that are always on the ground and two points always above it. Having an entire disc of ground below the inner two points ensures that we cannot tilt the entire table and find a spot where both of them hit the ground. If we tilt it until one of them hits the ground, it will hit it somewhere on the disc, and the other one will be above the disc, since it is opposite of the one getting titled down and will therefore be tilted up.
Ah, I see, you were going for the proof that four-legged tables have to have cocircular feet for them to have a chance to always be "balancable" by rotation. My hint was about aiming for one particular point to rotate about to see that those isosceles trapezoid tables actually can be stabilised just like rectangular ones (by rotating around 360 degrees as the table in the Mathologer intro at the very beginning). Anyway, another way to argue that the feet have to be cocircular is that if a table whose feet are by definition all contained in a plane, can also sit with all four feet on a sphere, then the feet have to be contained in the intersection of the plane and the sphere, which is necessarily a circle :)
It's kinda funny how different people understand things differently. I'm so used to Mathologer videos being completely unintelligible until you get a detailed explanation (and even that can leave me mystified most of the time), but the idea that all pictures must have a hugging square seemed immediately obvious to me (for the reasons he explained). Guess I will have to enjoy the feeling while it lasts. :)
At 4:50, aren't those "random dots" the anti-copying constellation used on banknotes?
5:07 Of course! Any regular paralelogram is (by regular I mean that all sides are equal and all angles are equal)
"regular parallelogram" = "all sides are equal and all angles are equal" = square ? :)
The proof of contradiction for the Rectangular table works for the highest point as well. Rotate the table until it is highest, mark the rotation and hight. (Let's call that diagonal A) Rotate the table so far, that the other diagonal (Diagonal B) is on the same line (as diagonal A was before), now the feet of diagonal B need to touch the ground, since they are at exactly the same position as the feet of diagonal A before, and the feet of diagonal A are lower, since you chose the point of highest elevation.
A Reuleaux triangle is hugged by squares of the same size in all orientations. If the corners of a table lie on two different concentric circles, as those of a non-rectangular parallelogram are, and you put it on a paraboloid of revolution with its center over the vertex, it will always wobble. The surface {θ∈[0°,90°):0;θ∈[90°,180°):1;θ∈[180°,270°):0;θ∈[270°,360°):1} cannot support a square table centered at the origin without wobbling, but it can support a rectangular table of any other aspect ratio.
incredible
NEW VID!!!!!!!
amazing
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 ;)
Shapes of constant width should have only hugging squares, right?
SwordQuake2 correct
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).
Мне всегда нравилась эта теорема, потому что она имеет важные практические применения в обыденной жизни. Например, я однажды применил её в магазине при покупке стола - он качался, и вращение не помогало его поставить ровно. Аргументируя этим, я заставил продавцов принести другой экземпляр со склада.
I knew the reason for my wobbly table was ghost's gnawing at the legs! Thank you for verifying this, I'm going to collect on a lot of bets!!
The hugging rectangles of a square are all squares, as are all those of any regular polygon with 4X sides. More generally, it's true of any shape with 4-fold radial symmetry.
Is there a refrigerator door theorem? How do I make my refrigerator slightly un-level in a way that always makes the door try to close by itself due to gravity, no matter what position it is open to? My freezer does this, but my refrigerator is opposite and gravity makes the door try to open more once it's open past a certain point. Instead of randomly adjusting all the feet, it would be more fun to figure out the math on how to make this happen. I also have other doors that I would prefer gravity to tend to open, and yet others I would prefer gravity to tend to close, and even others I would prefer to just stay where I put them no matter what position they are in. I think that this should be possible by adjusting the hinges, perhaps with a shim behind either the upper or lower hinges... but again, just randomly doing things is not as good as figuring out the math behind them.
In a similar vein, there are (at least) two points on the surface of the Earth that are exactly opposite of each other, and have the exact same temperature. There are likewise (at least) two points that are exactly opposite of each other and have the same air pressure. Moreover, there are (at least) two points that are exactly opposite of each other and have the same temperature _and_ air pressure.
I don't have the education to substantiate this, but the wobbly table theorem feels very related to the inscribed square conjecture. It feels like somehow linking the curve in the latter to the surface in the former would be fruitful.
Definitely the same circle of ideas with quite a large overlap of references in literature :)
Where does the magical 35.26 angle come from?
5:05 A curve an constant width with a positive or zero padding number
Your logo of mathologer is the answer That's a square 😎 Love from India🙂 :)
Logo is a copy of Helsinki metro logo
Intuitively the square case should also have the leg length constraint. Also what exactly is the flaw in your proof that the maximum slope constraint becomes necessary?
I'm going to take a bit of a 'guesstimate' and say that any shape with 4-fold rotational symmetry has a 'hugging rectangle' which is always square? Great video by the way!
5:00 Треугольник Рёло? Другие фигуры постоянной ширины?
Does the paper include a way to calculate the minimum Lipschitz constant for rectangular tables with legs of arbitrary (equal) length? Many tables aren't as tall as that, but I imagine they still can be balanced this way on most real-world surfaces.
I wonder if it is possible to find out if all legs of the table are the same length? What if I rotate the table first 180 degrees and if it’s wobbling then in another direction as before, we should have defect legs?