The wobbling tables at work have always gotten the best of me. Now the tables have turned.

@bobthornton8282

5 жыл бұрын

A+

@pluto8404

5 жыл бұрын

This comment is more clever than the theory.

@nonachyourbusiness1164

5 жыл бұрын

We don't deserve this comment

@lordtachanka903

5 жыл бұрын

David Fan quick, hit em when their back's turned!

@Blox117

5 жыл бұрын

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!

@Mathologer

5 жыл бұрын

:)

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)

@zeldajerk

5 жыл бұрын

Eric Vilas is this a joke? I really can't tell.

@ericvilas

5 жыл бұрын

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

@zeldajerk

5 жыл бұрын

Eric Vilas you must have seen a lot of wobbly tables then!

@wilddogspam

5 жыл бұрын

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.

@ibperson7765

5 жыл бұрын

More like 99%. It is almost always the table not the floor. Rotating almost never works

Turning the tables with DJ Mathologer

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.

Don't lie, that circle was rotating too!

@zeldajerk

5 жыл бұрын

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.

@JustOneAsbesto

5 жыл бұрын

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.

@zeldajerk

5 жыл бұрын

JustOneAsbesto I can't really read sarcasm from text...

@JustOneAsbesto

5 жыл бұрын

Alright, fair enough, buddy.

@zeldajerk

5 жыл бұрын

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.

Annoyed that it's not called the stable table theorem >:(

@andymcl92

5 жыл бұрын

Yeah, because now we basically have "The wobbly table theorem states that no such table exists*" (*assuming nice tables and sensible floors)

@ianallen738

5 жыл бұрын

If the table wasn't wobbly in the first place, you wouldn't even know about the theorem.

@andymcl92

5 жыл бұрын

Ian Allen Fair point...

@tsvtsvtsv

5 жыл бұрын

the unstable table theorem?

@arkanon8661

5 жыл бұрын

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

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.

@caygesinnett6474

5 жыл бұрын

Galdo145 let me add to that, any shape that has 90 degree rotational symetry or two orthogonal lines of symmetry

@DanielHesslow

5 жыл бұрын

And also any object of constant width such as the Reuleaux triangle

@alexwang982

5 жыл бұрын

I solved it without the logo...

@plasmaballin

5 жыл бұрын

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.

@plasmaballin

5 жыл бұрын

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.

THANK YOU MATHOLOGGER!!! I always wanted to watch a video about the wobbly table theorem!

Amazing, really amazing visual proof. Thank you very much.

The proof at the end is AMAZING!!

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)

@bretterry8356

3 жыл бұрын

That was my first thought as well. Any shape of constant width will do that.

Thank you for the video! All of you friends awesome!

Brilliant video, Burkard; the Intermediate Value Theorem is beautiful!

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.

Reuleaux triangle also has all hugging rectangles as squares.... & all other many many types of fixed width shapes as well

Excellent video. Thanks!

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.

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 .. ♥

One of the best math lectors I ever saw!

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.

You, Grant Sanderson, Salman Khan, Matt Parker and James Grime are Martin Gardeners for us.

very good presentation

Wonderful. Thank you so much.

I've been using this theorum ever since I saw the original video. It's a handy Cafe trick.

I’m a simple man, I see a maths video with the Linux mascot (tux) in the thumbnail, I click

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, ....

@iangrant9675

4 жыл бұрын

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

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.

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.

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.

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.

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!

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.

@Mathologer

5 жыл бұрын

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 :)

@zeldajerk

5 жыл бұрын

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.

@Mathologer

5 жыл бұрын

If the theorem "could be wrong", it would not be a theorem :)

@zeldajerk

5 жыл бұрын

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...

@zeldajerk

5 жыл бұрын

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?

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.

@Mathologer

5 жыл бұрын

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 :)

@rijumatiwallis7597

5 жыл бұрын

Mathologer Thanks, I will check out the Mandelbrot video again.

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.

@DFPercush

5 жыл бұрын

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

@DFPercush

5 жыл бұрын

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....

@soppelpost1573

5 жыл бұрын

A regular triangle have a hugging squares at 15° and 45° rotation.

@donielf1074

5 жыл бұрын

Soppel Post He didn’t say they have no hugging squares, he said they won’t always have hugging squares.

@ibperson7765

5 жыл бұрын

Bob Bobson well theres nothing with NO hugging square so it’s unnecessary to say a particular shape has one

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.

I can't wait to encounter my next wobbly table so I can try this out!

Bro this is awesome

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.

@Mathologer

5 жыл бұрын

Great answer :)

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)

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°.

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.

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.

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.

you make me laugh like a baby its so much fun when you realize whats happening before he says it :)

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 ;) )

@Mathologer

5 жыл бұрын

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 :)

@HakoHak

5 жыл бұрын

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)

* _sees unofficial Linux mascot in thumbnail, instantly upvotes_ *

@swedneck

5 жыл бұрын

Isn't tux official?

@OxygenGammling

5 жыл бұрын

thought so aswell

@FourthDerivative

5 жыл бұрын

I saw Tux and I clapped!

@jasondoe2596

5 жыл бұрын

Tim Stahel, yeap, you're right; Wikipedia says Tux is the "official brand character". I was under the impression it started out unofficially, although it was commissioned and immediately endorsed by Linus, which would make it unofficially "official", I guess :) The open source world doesn't care too much about such details - _Linux_ hadn't been registered as a trademark for many many years, which only finally happened because someone else tried to steal it.

@jonadabtheunsightly

5 жыл бұрын

As far as I know, it's semi-official. Torvalds (who is as official a representative as Linux has ever had) expressed approval, in an informal and unofficial venue (a developer's mailing list or somesuch, IIRC). However, this information dates from the late nineties, so it's easily possible that there may be some more recent development of which I am not aware.

Brilliant

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

5 жыл бұрын

:)

@dnzssrl

5 жыл бұрын

Mathologer oh getting a reply from you. Now I can die in peace :)

@ibperson7765

5 жыл бұрын

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?

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.

Thank you for the rectangle proof! Now I can finally fix my wobbly bed :P

@Mathologer

5 жыл бұрын

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, ...

@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.

@Mathologer

5 жыл бұрын

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 :)

@yinq5384

5 жыл бұрын

You're right, I didn't think about this carefully enough. Thank you!

Amazing video 😍✌❤️❤️

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 KZread? 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 ;)

@tetraedri_1834

5 жыл бұрын

Great, I just sent you my contact information to email!

The fact that a Rouleax triangle has constant-size hugging squares has a practical application -- those triangular drills that drill square holes.

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.

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.

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.

@Mathologer

5 жыл бұрын

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 :)

Tabled! I'm floored in your hippopotamus hypotenuse! OMG 😂!

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'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. :)

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.

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?

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?

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.

@Mathologer

5 жыл бұрын

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 :)

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

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!

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.

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.

amazing

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

Your logo of mathologer is the answer That's a square 😎 Love from India🙂 :)

@Kakkos4

5 жыл бұрын

Logo is a copy of Helsinki metro logo

That 35.26 degree angle, from arctan of inverse of root 2, is ~ just 1E-13 more than the angle between the diagonal of a cube & the diagonal of it's side, which is arccos of the ratio of root 2 to root 3. I got this from a math question, but didn't check the work, I assume the answer they gave was correct.

5:00 Треугольник Рёло? Другие фигуры постоянной ширины?

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!!

incredible

14:35 - Is that angle, 35.26° equal to arcsin(1 / sqrt(3)) ... and why?

@Mathologer

5 жыл бұрын

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.

@leif1075

4 жыл бұрын

@@Mathologer I don't understand the proof at the end..isn't it wrong say two of the legs will always be on the ground and two won't be? They're could be a case where only one is touching the ground or indeed one red and one blue..not both of the same color?

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.

@Mathologer

5 жыл бұрын

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 :)

@plasmaballin

5 жыл бұрын

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.

@Mathologer

5 жыл бұрын

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 :)

I did not know of this theorem, however I have already corrected a wobbly table at a local bar by turning the table.

Better explanation than Numberphile's video, thanks!

At 4:50, aren't those "random dots" the anti-copying constellation used on banknotes?

Shapes of constant width should have only hugging squares, right?

@vapenation7061

5 жыл бұрын

SwordQuake2 correct

@Marci124

5 жыл бұрын

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.

@atrumluminarium

5 жыл бұрын

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

@helgefan8994

5 жыл бұрын

Does that mean 3-dimensional shapes of constant width such as the Meissner tetrahedron only have hugging cubes? I believe they do! :)

@rmsgrey

5 жыл бұрын

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 have a little silly question about that 35.26° business, if you don't mind. If our floor is more steep does it start to acting like an "almost discontinued" one or does there some other problem appear? And sorry for mistakes, English isn't my first language so I am still learning it

@zeldajerk

5 жыл бұрын

There are no silly questions.

@jonadabtheunsightly

5 жыл бұрын

I think this is related to the minimum leg length. If your table's legs are shorter, you'd have to limit the angle more narrowly. Not sure if the reverse (being able to relax the angle if you require the legs to be longer) also works or not. Intuitively I think it should, but to be sure I'd have to look over the whole proof, to see if there's any additional reason for that particular angle restriction, and I haven't yet done so.

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.

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?

Brilliant video! Thanks. Unfortunately I was a bit disappointed at the end. I guess it is obvious for some reason still it is not obvious to me why, in turning the table, height has to become lower. Please explain.

@Mathologer

5 жыл бұрын

The center of the table can move up and down the z-axis in a much more complicated manner than shown in my animation. But the main point here is that at some point the center will be at a minimum height. There could even be a couple of points in time during the rotation where the same minimum height is reached. Just choose one the position that correspond to one of these points in time and do what I show in the video to come up with the contradiction :)

@christosgeorgiadis7462

5 жыл бұрын

Thanks for answering. Unfortunately, I still don't understand :-( . I follow the argument with no problem up until we are at the minimum height. Then we turn the table to have the other pair of legs at this position, right? I understand that as well. What I don't understand is why, by this last turn, we cannot have an increase of the height. If a decrease in height is necessary then the contradiction is obvious. What is not obvious to me is why the decrease is necessary.

@Mathologer

5 жыл бұрын

Well, remember that to get to the new position we take the diagonal that connects the red points and translate it DOWN. The center of this diagonal coincides first with the center of the original rectangle kzread.info/dash/bejne/k3eeldOfgZmenZM.htmlm43s and after the shift with the center of the rectangle in the new position. Since the height of the center defines the elevation of a position ...

@christosgeorgiadis7462

5 жыл бұрын

It's clear now, thanks a lot!

@MichaelRothwell1

5 жыл бұрын

I had to watch that bit of the video 4 or 5 times before I finally got it. The key is that the blue feet are always on the ground (by assumption), so if we turn the table so the blue diagonal is in the direction the red diagonal is now, since the diagonals are equal, the blue feet will be lower down in the green positions.

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.

the hugging squares will also occur when using a reuleaux polygon because of their constant width

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

@zeldajerk

5 жыл бұрын

The legs points can exit the circle. Therefore the legs ends are some point on the sphere defined at the center of the square with radius half the diagonal of the square. He didn't explain that well in the video.

@zeldajerk

5 жыл бұрын

cayge sinnett edit: radius sqrt (1/2 diagonal^2 + leg length^2)

@Roonasaur

5 жыл бұрын

The legs of any rectangular table will sweep out a circle when rotated.

@rmsgrey

5 жыл бұрын

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...

@andymcl92

5 жыл бұрын

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

Question at 5:00 I believe the Reuleaux triangle is the answer. Some in the comments have said that "a square" is the answer. I guess that works too but that's a bit too obvious.

5:07 Of course! Any regular paralelogram is (by regular I mean that all sides are equal and all angles are equal)

@Mathologer

5 жыл бұрын

"regular parallelogram" = "all sides are equal and all angles are equal" = square ? :)

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.

@Mathologer

3 жыл бұрын

Definitely the same circle of ideas with quite a large overlap of references in literature :)

Great video, could you do a video on knot theory, please.

@Mathologer

5 жыл бұрын

On my list of things to do :)

@juanmonge9391

5 жыл бұрын

Mathologer thanks for answering i'll be waiting for it and keep making this awesome videos!!

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)