A surprising topological proof - Why you can always cut three objects in half with a single plane

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  • Опубліковано 8 чер 2024
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КОМЕНТАРІ • 1 тис.

  • @zachstar
    @zachstar  3 роки тому +143

    Shoutout to Jon from the youtube channel Epic Math Time (link here: ua-cam.com/channels/isjF-Un7hf9lsMhoStF3OQ.html ) for the borsuk-ulam animations in the beginning of this video. Came out great and if you haven't seen his channel I definitely recommend subscribing for cool math videos and animations.
    Also sorry about the huge delay since the last video. Starting just a few weeks ago things in my city have finally opened up without restrictions so I've been going out again for the first time in over a year. Still working on videos during the week but at least for the summer they might come out a little slower while try to have a bit of a social life and make up for time lost in 2020 lol. Hope you guys enjoy!

    • @lombre9149
      @lombre9149 3 роки тому

      I swear to god if anyone tries to call you a time traveller
      How do you not know this by now

    • @masternobody1896
      @masternobody1896 3 роки тому +1

      lol about losing my brain cells

    • @masternobody1896
      @masternobody1896 3 роки тому +1

      nice video

    • @antx1026
      @antx1026 2 роки тому

      Hey Zach… I’m about to begin taking some Upper-level math courses. I’ve done the whole calculus series and some differential equations. Not linear algebra or discrete math yet. I’m kind of excited for Real Analysis and Abstract Algebra, even though I don’t really know what they are. I was hoping you can tell me which upper-level courses I should take at the University of Nevada Las Vegas (UNLV) for my B.S. in Mathematical Sciences. I’ve watched some videos on some of the courses but I have no idea what they’re really talking about. And I just don’t know which of the courses are the most important in current mathematical research or which ones would benefit me the most in terms of employment. It’s crazy how much math there is. I can’t learn all of them in college.

  • @p11111
    @p11111 3 роки тому +2483

    In other words a fruit ninja can evenly split multiple fruit in midair with just one flick of the sword

    • @p11111
      @p11111 3 роки тому +42

      @stillFLiP I guess 3 is better than none? 😊

    • @felhomaly
      @felhomaly 3 роки тому +51

      The solutions depend on the velocity of his sword, even in case of 3 fruits.

    • @yonatanbeer3475
      @yonatanbeer3475 3 роки тому +18

      @stillFLiP up to two

    • @thomasrad5202
      @thomasrad5202 3 роки тому +3

      @stillFLiP 4 dimensional plane, whatever that looks like

    • @tanned_cosines_
      @tanned_cosines_ 3 роки тому +3

      @@thomasrad5202 yeah 😂😂
      plane by name is a 3D structure

  • @edsanville
    @edsanville 3 роки тому +320

    So, you're saying there's definitely a plane that divides my sandwich, Gary Coleman's head, and the Andromeda Galaxy exactly in half? Very interesting to know...

    • @Fidder492
      @Fidder492 3 роки тому +25

      hot dang, you're right! lmao

    • @corvusmonedulas4895
      @corvusmonedulas4895 3 роки тому

      I said, hop in!

    • @jasonreed7522
      @jasonreed7522 3 роки тому +33

      Well considering how all those objects are moving you would need to define at a moment in time, or define the olane as a function of time. Which would be continuous through time as all the inpit variables are continous over time.

    • @ViratKohli-jj3wj
      @ViratKohli-jj3wj 3 роки тому +3

      @@jasonreed7522 hmmm right

    • @kensandale243
      @kensandale243 3 роки тому +4

      "So, you're saying there's definitely a plane that divides my sandwich, Gary Coleman's head, and the Andromeda Galaxy exactly in half?"
      No. Gary Coleman doesn't have a head.

  • @DrTrefor
    @DrTrefor 3 роки тому +792

    So cool! And I love that it generalizes. Want to cut an n component n dimensional “hyper sandwich” each component exactly in half? You just need an n-1 dimensional hyperknife to cut them all:D As an algebraic topologist myself, this theorem is always my favorite for sharing with students

    • @ozargaman6148
      @ozargaman6148 3 роки тому +2

      Can you help me with something I don't understand? If we take the example of (3, 2), (-3, -2), why does that plane have to go through (0, 0)? Can't it go through (3, -2), (-3, 2) and then move to (-3, -2), (3, 2) or something like that? Like can't they switch on one axis and then on another one?

    • @anonymous_4276
      @anonymous_4276 3 роки тому +9

      @@ozargaman6148 I'm no expert but from what I understand, according to Borsuk Ulam theorem, if you map each point on the sphere to a point on the 2D plane such that the mapping is continuous, at least one pair of antipodal points on the sphere get mapped to the same point. In the case of the sphere, we are mapping each point on the sphere to a point in 2D plane. But from the nature of the mapping in this video, we can also see that if a point on the sphere is mapped to (a,b) in 2D, the diametrically opposite point must be mapped to (-a,-b). But according to Borsuk Ulam theorem there must be at least one pair of diametrically opposite points on the sphere being mapped to the same point in 2D. So a point on the sphere is mapped to a point (a,b) in 2D such that a=-a and b=-b => (a,b)=(0,0). But since the x coordinate of these points in 2D space represents the distances between first and second plane and the y coordinate represents the distance between the second and third plane, the existence of one point on the sphere which can be mapped to (0,0) implies that there exists at least one plane which cuts all the three objects in half.

    • @Bentu
      @Bentu 3 роки тому +1

      @@ozargaman6148 the coordinates are just the distances of the planes (that cut the objects in half) from each other, with the negative sign being the indicator of the direction of a perpendicular line to the plane. the plane doesn't go through (0, 0), we get it by taking the tangents to the points on the sphere that get mapped to (0, 0). since the distances from each of the planes is zero and the slope (gradient) of the planes is the same, we know that all the planes must be the same one

    • @pbj4184
      @pbj4184 3 роки тому +1

      @@ozargaman6148 I am just as much a noob as you but from what I gathered, here is the explanation-
      Borsuk-Ulam says atleast one pair of antipodal points on the n-sphere corresponds to the same output in the range set of the function, if the condition of isomorphism is maintained.
      So let's say our mapping function is f:S->P (S for n-sphere, P for n-Plane) and ' is a functionon the n-sphere that maps each point to its antipodal point
      Seeing that opposite pointing normal vectors to parallel lines will always be -ve of each other, we can say that there is atleast one pair of antipodal points A and A' such that
      f(A)=f(A') (that is, same output) ----(1)
      But we see from the demostration in the video that for any arbitrary pair of antipodal points p and p',
      f(p)= -ve f(p') ----(2)
      Letting p=A (or by symmetry of antipodality, p=A')
      we find from (1) and (2) that-
      f(A)=f(A')
      f(A)= -ve f(A')
      Now here is where my noobness shines in all its full glory. Assuming (very carelessly) that you can "add" outputs of topological mappings like vectors of the same vector space,
      after a linear equation-like addition of both results, we get-
      f(A)+f(A)= f(A') + (- f(A'))= "0"
      => 2*f(A)= "0"
      => f(A)= "0"
      => f(A')= "0" as f(A)=f(A')
      I put the zero in quotes because it is the "null vector" of the n-Plane in each case (Is there a concept of null vector in topology? Idk)
      Since f(A)=f(A')="0", the Borsuk-Ulam pair has to be the null vector which is why it must pass through (0,0) and not any other points because only the null vector is its own additive inverse when the additive identity is the null vector
      I hope people who actually know this stuff rigorously can correct me (Or just post your own explanation if my explanation is beyond salvageable)

    • @woody442
      @woody442 3 роки тому

      Is it true, that the n-1 dimensional hyperknife (by definition has always zero curvature?) is always the hyperspace of dimension n-1, that contains all the centres of mass of all n objects?

  • @johnchessant3012
    @johnchessant3012 3 роки тому +590

    3b1b's video said anyone talking about Borsuk-Ulam is required by law to give the example that there exist antipodal points on Earth that have the same temperature and pressure. I see you're a law-abiding citizen. :D

    • @alexismandelias
      @alexismandelias 3 роки тому +26

      Got flashbacks from 3b1b when I saw this part of the video

    • @Myrslokstok
      @Myrslokstok 3 роки тому +15

      Probably one of the few real math things I know.

    • @zachstar
      @zachstar  3 роки тому +138

      Honestly I forgot that he said that, but it does feel illegal.

    • @winged777
      @winged777 3 роки тому +2

      @@zachstar I follow the math, easy enough thanks to how well you broke things down - great job! But aren't pressure/temperature discontinuous, making the Earth example not work? Or are they just continuous "enough", or at a large enough scale? I can shine a magnifying glass onto a point on a sidewalk to produce a small "discontinuity" but zoom out far enough and I guess it's unnoticeable. I get that it's not the point here, just seems like it's treated seriously so I'm curious.

    • @kilian8250
      @kilian8250 3 роки тому +8

      Trinity Dickinson of course it’s not continuous in a mathematical sense, since the universe is inherently discrete (there is vacuum between particles), but we could approximate the temperature and pressure VERY well with a continuous function if we ”fill in the gaps”. So for all practical purposes, the theorem holds in this case (but mathematically, no).

  • @PapaFlammy69
    @PapaFlammy69 3 роки тому +594

    Didn't know about this theorem, very interesting :)
    And you got that Applaud button? :D

    • @telecorpse1957
      @telecorpse1957 3 роки тому +10

      3B1B also did an amazing video featuring Borsuk-Ulam:
      ua-cam.com/video/yuVqxCSsE7c/v-deo.html

    • @IustinThe_Human
      @IustinThe_Human 3 роки тому +47

      an engineer teaching a math teacher about topology

    • @maxcl3474
      @maxcl3474 3 роки тому +8

      hi papa flammy

    • @Tony-cm8lg
      @Tony-cm8lg 3 роки тому +16

      There is no way you didn’t know the Borsuk Ulam theorem, you know soo much maths

    • @user_2793
      @user_2793 3 роки тому +8

      @@Tony-cm8lg Math is very, very vast though

  • @manuelgiron4232
    @manuelgiron4232 3 роки тому +77

    Zach Star- professional actor AND professional nerd

  • @j.vonhogen9650
    @j.vonhogen9650 3 роки тому +52

    Amazing how you managed to explain all this so effortlessly in such a short video (I'm sure you worked hard on this, but at least you made it look effortlessly)! I'm really impressed by your talent as an educator. Thanks a lot!

  • @omlett6482
    @omlett6482 3 роки тому +52

    VSauce viewers on the explanation of the Borsuk-Ulam theorem: Hey, i've seen that one before

    • @frodododo
      @frodododo 3 роки тому +3

      Yeah, i saw him talking about borsak-ulum and thought 'hmm, that seems familiar' then as soon as he busts out the temperature-pressure model it clicked

    • @nanamacapagal8342
      @nanamacapagal8342 2 роки тому +1

      3b1b even said it was an unwritten law to use the 2 antipodes example when talking about borsuk-ulam

  • @lexscarlet
    @lexscarlet 3 роки тому +211

    Brother I don't know if you wrote the curriculum, method, script, whatever it's called, for this lesson yourself but this is like one-to-one continuous home run on successful teaching strategies. Repetition with slight variation and difference in expectation (for the negative signed value in the second exercise), slowly moving to the 50/50 point, etc.
    This would def be a stealth sub and like on my kids' UA-cam feed. Have em speaking spinnorial representation by grade 3.

    • @h00db01i
      @h00db01i 3 роки тому +1

      disgusting

    • @johnkeegann
      @johnkeegann 3 роки тому +1

      @@h00db01i huh

    • @koraptd6085
      @koraptd6085 3 роки тому +4

      @@h00db01i disgustingly genius indeed

    • @h00db01i
      @h00db01i 3 роки тому

      @@koraptd6085 geniuses are disgusting, as you can not reach their level no matter how hard you try. parents on the other hand are my pet peeve; fuck em. kill your masters, with success - with gusto.

    • @Actualshard
      @Actualshard 3 роки тому +6

      @@h00db01i sure buddy

  • @highgroundproductions8590
    @highgroundproductions8590 3 роки тому +47

    Absolute giga chad studying algebraic topology by yourself. Now I must eventually learn algebraic topology because as a physicist I cannot let an engineer be better at math than me.

  • @smalin
    @smalin 3 роки тому +140

    Is there such a thing as a “center of volume” that’s analogous to center of mass (center of gravity)?

    • @DavidSharpMSc
      @DavidSharpMSc 3 роки тому +47

      Yeah, such ideas are used a lot in naval architecture: the “centroid of volume” for the underwater volume of the hull is the point at which the buoyancy force is said to act. Ship stability is often determined by the interaction of the upward buoyancy force acting at the centroid of volume vs the gravity force acting downward at the centre of gravity (mass).

    • @forloop7713
      @forloop7713 3 роки тому +65

      Center of volume is the center of mass if the density is uniform

    • @smalin
      @smalin 3 роки тому +28

      @@forloop7713 Okay, if that's the case, is it true that any plane that contains an object's center of volume divides its volume into two halves of equal volume?

    • @Vizaru
      @Vizaru 3 роки тому +22

      @@smalin that's correct!
      if you were to find any plane that cut an objects volume in half, and that plane didn't go through its perported center of volume, then you wouldn't actually be going through the true center of volume.
      just as center of mass can be conceptualized as the point where mass is always balanced on either side, the same is the case where volume is always balanced on either side of the center of volume

    • @gabwiel
      @gabwiel 3 роки тому +1

      centroid or inertia

  • @anthonycannet1305
    @anthonycannet1305 2 роки тому +10

    For an easier way to understand how the theorem works, if the two points have values that are opposite of each other, and the values change continuously, then in order to get from the positive point to the negative one, you must have crossed zero somewhere. You might not know where but you know the place exists.

    • @himmelsdemon
      @himmelsdemon 11 місяців тому +1

      Ah, the intermediate value theorem, my old friend.

  • @thisisnotmyrealname628
    @thisisnotmyrealname628 3 роки тому +22

    "Aahhhh yeeeeaaaaaahhhh🤤"
    -Me, realising where the proof was going

    • @fangzhang9376
      @fangzhang9376 3 роки тому +1

      I thought he was going to use the volume of object 2 on the "positive side" when he divided object 1 in half. The actual proof feels more symmetrical though and I like that.

  • @georgecantu856
    @georgecantu856 2 роки тому +2

    “You may have seen numberphile’s video, but here we’re going to actually prove it”
    Big flex and why I subscribed 👌🏼

  • @tarunyadav3567
    @tarunyadav3567 3 роки тому +49

    To everyone thinking you can take CoM of all three and draw a plane through them; That plane doesn't necessarily divide the objects in half.

    • @JamesChurchill
      @JamesChurchill 3 роки тому +11

      Only if the object has a non-uniform density. But since we're talking about volume, you instead use the CoM of an identically shaped uniform mass instead, and any plane going through that *will* divide the volume in half.

    • @Flutesrock8900
      @Flutesrock8900 3 роки тому +36

      @@JamesChurchill Short answer: no.
      Not so short answer: the center of mass of a cone of uniform density with height *h* and base of radius *r* will be at h/4 over the center of the base. A plane parallel to the base going through that center of mass will split the cone into a cone with 27/64 the volume of the original cone. Clearly, not *any* plane going through the center of mass will divide the volume in half.

    • @tarunyadav3567
      @tarunyadav3567 3 роки тому +7

      @@JamesChurchill no not true. just imagine a uniform cone or tetrahedron.

    • @quacking.duck.3243
      @quacking.duck.3243 3 роки тому +2

      @@Flutesrock8900 I wonder how would the cone balance if you put an imaginary pin on its CoM then, but I guess it has to do with the fact that the cone portions away from the centre have a greater moment of inertia.

    • @fangzhang9376
      @fangzhang9376 3 роки тому +6

      @@quacking.duck.3243 Yeah. There's an old riddle about balancing a carrot by hanging it from a rope and cutting it where the rope was. The thicker half would be heavier for exactly this reason.

  • @dylanholloway8017
    @dylanholloway8017 3 роки тому +5

    You're the kind of channel in which I like your videos without having watched then yet :)

  • @laineylain
    @laineylain Рік тому +8

    does that mean you can cut 4 objects in 4D with a 3D plane? since 3D can have 3 shapes, 2D can have 2, and 1D can have 1?

    • @ivarangquist9184
      @ivarangquist9184 Рік тому +1

      I believe this is the general statement of the ham sandwich theorem.

    • @DageLV
      @DageLV 11 місяців тому

      4th dimension is considered time. Humans barely can even understand what the hell is 4d object, like imagine 4d cube, you've probably seen it, but that's inaccurate representation. that cube has to have every corner at 90 degrees and all faces same size, you cant even wrap your mind around it
      But you probs have played games. All models are made of triangles. Those 3 triangles are connected with a flat plane, always. That proves this quite simply.

  • @rohanshah6178
    @rohanshah6178 3 роки тому +1

    What an ingenious proof!!! Was just blown away by the cleverness and creativity of the argument. Loved it so much. Thank you so much for sharing it

  • @bane937
    @bane937 3 роки тому

    Zach this is amazing! I really like the concept of you pickig up some problem/exercise from one of those mad textbooks, and solving it step by step, with some fine choice of abstractions.Thanks for sharing and putting in so much work ✌🏼✌🏼

  • @missinglegs
    @missinglegs 2 роки тому +4

    That's... Such a fascinatingly weird Theorem... It's so convenient for seemingly no reason

  • @yohangross5518
    @yohangross5518 3 роки тому +5

    Just wow, beautiful proof

  • @Benny_Blue
    @Benny_Blue 3 роки тому +2

    Very well done on the Borsuk-Ulam section from 1:14 to 3:40 - it stands well as an interesting segment on its own! And honestly, that’s probably the best you can strive for with those video sections of “We need to step aside and learn this really quick.”

  • @TheFootballPlaya
    @TheFootballPlaya 3 роки тому +1

    I love your videos man. been watching them for years. they are great content. I really appreciate all that you do.

  • @TheGoldenFluzzleBuff
    @TheGoldenFluzzleBuff 3 роки тому +27

    Question: Can you cut an avocado perfectly in half using a spoon? Aka, can a curved plane perfectly cut in half an object if there is another object in the middle of it that cannot be penetrated?

    • @edwardfanboy
      @edwardfanboy 3 роки тому

      No. What if the whole object cannot be penetrated - the avocado is all stone?

    • @acidgiraffes
      @acidgiraffes 3 роки тому

      No because it can't go through the center

    • @TheGoldenFluzzleBuff
      @TheGoldenFluzzleBuff 3 роки тому

      @@acidgiraffes I mean like the fruit part of the avocado, excluding the pit

    • @jasonreed7522
      @jasonreed7522 3 роки тому +7

      Vocab adjustment:
      "Planes" are inherently flat, you would instead be defining a "surface" that cuts through 1 object without cutting a second object located entirely withing the first object.
      Meaning cutting only the flesh of an avocado and not the pit with a surface with curvature ≠ 0 in the relevant region of space.
      Also i think this should be entirely possible, but irritating to provide a mathematically sound proof for.

    • @flamencoprof
      @flamencoprof 3 роки тому

      Just don't be so lazy and dreamy. Reach for the knife that is sure to be at hand. :-)

  • @jeroenodb
    @jeroenodb 3 роки тому +6

    A lot of the commenters seem to think that just having the plane go through the three "centres" of the objects would do the trick. But what kind of centre of a volume would that be? The centre of mass will not work as mass that's far away will contribute more to the position of the centre of mass. Take for example a pear. On one side it's long and thin, and on the other it is thick. This will mean that the plane through the centre of mass will have less mass on the skinny side than on the thick side. Maybe you could find a "centre of volume", or in 2D, a centre of area, but these are not guaranteed to exist.

    • @iippari7
      @iippari7 3 роки тому

      I imagine you could find the "centre of volume" by representing the shape as _n_ equidistant points that all lie on the surface of the shape, and taking the average of all points.

    • @Vizaru
      @Vizaru 3 роки тому +1

      right, center of mass only works if you assume uniform density

    • @jeroenodb
      @jeroenodb 3 роки тому +2

      @@Vizaru Even with with uniform density, the centre of mass will not work, except for really symmetric cases, like a sphere or a cube. The reason for this is that a cut through the centre of mass will guarantee that the integral of distance*dV will be equal on both sides. In this way volume farther from the cut plane will "count" more, so you're not guaranteed that the two volumes are the same.

    • @domenkastelic2611
      @domenkastelic2611 3 роки тому +1

      @@jeroenodb It works for any object with uniform density, as any point of dV volume has a mass of dm=dV*density. Thus if the density is uniform, the volume and the mass of a point are directly proportional to each other. In other words - if you cut an object in half through the centre of mass (1/2 of mass on each side) you will also cut the object through the centre of volume (1/2 of volume on each side).

    • @jeroenodb
      @jeroenodb 3 роки тому +1

      @@domenkastelic2611 Almost correct, but it's not true that halve of the mass is on each side of the centre of mass. Say you have a weird 2D shape made of a plate of wood with constant density. If you hold it at one end, and let it rotate due to gravity, the line extending from your hand downwards goes through the centre of mass. Now the object is in equilibrium, which means that the torque on it is zero, so the integral (horizontal distance from line)*g*dm is the same on both sides, to cancel out to zero. Because there's a factor distance in the equation, you're not guaranteed that the mass or the volume is the same on both sides, only that the combined torque due to gravity of the object is the same on both sides. The only time the the volume or mass are the same on both sides, is when by coincidence the average distance from the middle line is the same on both halves.

  • @basspuff514
    @basspuff514 2 роки тому

    Fascinating and surprisingly intuitive! Great video

  • @emperorpingusmathchannel5365
    @emperorpingusmathchannel5365 3 роки тому +1

    Love this channel! Makes me excited for problems!

  • @akshittyagi6482
    @akshittyagi6482 3 роки тому +11

    Zach asking the real questions here...

  • @clevelandHater
    @clevelandHater 3 роки тому +3

    Damn that’s actually pretty neat

  • @takyc7883
    @takyc7883 3 роки тому

    This was SO GOOD! Earned a sub, please do more of these kind of challenging problems. Of course you can also do simpler ones if you would like as well!

  • @oyibechibundu628
    @oyibechibundu628 3 роки тому

    Your explanations are so clear

  • @kahisoerickson1059
    @kahisoerickson1059 3 роки тому +4

    Great analysis of a complex topic. Very simple and understandable

  • @anitagofradump5195
    @anitagofradump5195 3 роки тому +5

    Wait im not sure if im oversimplifying this in my head but isnt this essentially just 3 points makes a plane? Each object has a centre of volume so wouldnt passing the plane through each point give us halves?

    • @half_pixel
      @half_pixel 3 роки тому +1

      Yep, that's another way to prove this! Although, with that approach, you have to first prove that the center of volume exists, i.e. there exists a point such that any plane passing through that point splits the object exactly in half.

    • @fullfungo
      @fullfungo 2 роки тому +2

      Sadly, there is no single point that would cut an object in half

    • @cl0p38
      @cl0p38 2 роки тому +1

      Not necessarily. CoV not always cuts it in half

    • @ValkyRiver
      @ValkyRiver Рік тому

      @@half_pixel No it’s not. For example, a cone doesn’t have this property.

  • @ktroyanos
    @ktroyanos 3 роки тому

    Awesome job on this video. Well done.

  • @TheBigFatVladimir
    @TheBigFatVladimir 3 роки тому

    Thank you for taking the time to make very interesting videos!

  • @VaradMahashabde
    @VaradMahashabde 3 роки тому +39

    Can we please just call Borsuk-Ulam as the multidimensional intermediate value theorem

    • @xavierstanton8146
      @xavierstanton8146 3 роки тому +9

      I'd say otherwise.
      The Borsuk-Ulam Theorem states that for any continuous mapping (which we will call f) of an n-sphere to R^n, there exists some point p on the n-sphere such that f(p)=f(-p).
      The Intermediate Value Theorem just states that for a function continuous on an interval
      [a,b], for all M in (f(a), f(b)), there exists a c in (a,b) such that f(c)=M.
      You could generalize the Intermediate Value Theorem to multivariate functions, but such generalizations don't tell you about where the points in the input will be mapped. So the Borsuk-Ulam Theorem isn't really a multidimensional Intermediate Value Theorem.

  • @factsheet4930
    @factsheet4930 3 роки тому +13

    My mind: "Wait but what if one of Ai is unmeasurable!"
    Also my mind: "Oh yeah compact sets in R^3 are closed hence borel measurable"

    • @martinshoosterman
      @martinshoosterman 3 роки тому +3

      If you substitute measure for outer measure, it might actually still be true for bounded sets.

    • @factsheet4930
      @factsheet4930 3 роки тому +2

      @@martinshoosterman I feel like you could still make a pathological and weirdly disjointed set that's also none measurable as counter example, that or the proof is just identical 🤔

    • @martinshoosterman
      @martinshoosterman 3 роки тому +2

      @@factsheet4930 the only thing that could possibly go wrong is that the position of the line that cuts the set in half might not be continuous as you rotate the line. Everything else should actually be fine because outer measure behaves nicely when your set is cut by measurable sets. (Even if the original set itself is non measurable, and in this case the set defining the cut is a half plane and therefore measurable)

  • @aashsyed1277
    @aashsyed1277 3 роки тому

    Very very nice!
    Best visualisation!

  • @reneejones6330
    @reneejones6330 2 роки тому

    Thank you. I love seeing the more rigorous version of the proof.

  • @hotdogskid
    @hotdogskid 3 роки тому +3

    Interesting video! You never fail to get me excited about math :) Question: Couldn't you find each center and draw a line/plane through all of them? Two points always define a line, three always define a plane, etc.
    Edit: Just watched the followup vid and saw these replies, it makes sense now that there isn't always a "magical center" where all lines through that point would cut the shape in half, counter to my initial intuition. Fascinating!

    • @kinacu_287
      @kinacu_287 3 роки тому

      I think that there might be combinations in which this line doesn't include the center of one of the figures(or both)

    • @UnCavi
      @UnCavi 3 роки тому

      Yes, you could, but this construction says nothing about the volumes of the parts you cut. Could be not half-half

    • @chrisg3030
      @chrisg3030 3 роки тому

      I was thinking that just as there exists a 2d plane that can any cut three finite 3d objects in half, so a 1d line can cut any two finite 2d objects in half. For the latter I visualize two plane figures A and B of any finite size and shape each being cut in half by any of an infinite number of lines intersecting at a central point. One such line in A must be colinear with one in B. The two central points define a line, and that line cuts both figures in half. (Also there exists a 0d point that cuts any one finite 1d line in half)

    • @1vader
      @1vader 2 роки тому

      Yeah, that's what I immediately thought of (although admittedly I saw a similar puzzle a few weeks ago and it took my a while longer to come up with this back then). This seems like a much simpler solution. Every 2d object has a center of mass and every line through that point cuts the area in half and the same applies to 3d objects, planes and volume. Two points can always be connected by a line and 3 points always uniquely construct a plane so there trivially exists such a line or plane. The only assumption (besides the 2 points -> line/3 points -> plane thing) is that objects have a center of mass and every line/plane through that half's their area/volume which seems fairly trivial. Although I guess that could be proven using the theorem from the video?

    • @UnCavi
      @UnCavi 2 роки тому

      @@1vader This is false. Counterexample: take an isosceles triangle and a line that passes thorugh the center of mass parallel to one of the edges

  • @marz.6102
    @marz.6102 3 роки тому +8

    Me when splitting food for the siblings

    • @kijete
      @kijete 2 роки тому

      offset the plane in such a way that you get more

    • @marz.6102
      @marz.6102 2 роки тому

      @@kijete big brain

  • @TheNetkrot
    @TheNetkrot 3 роки тому

    this is great, thanks this was the clearest explanation I have seen about this. I am not studying topology but it might happen later on. When I read about this problem I actually thought the solution would be focused on center of gravity of each object and the plane would be constructed through these points.

  • @whydontiknowthat
    @whydontiknowthat 3 роки тому

    Hey Zach! I’ve been watching your videos for a few years and I must say that I love the way that you’ve grown as a presenter, both in the depth and complexity of the topics you cover and the clarity of your presentation, to the point where now I feel that your content rivals some of the best math content creators on the site, like 3blue1brown, Mathologer, and Numberphile to name a few. You have a relatability that makes you easy to root for, at least to me, since it looks like we did our math undergrad at around the same time. It’s been a pleasure to see your transformation over the years, I love your content, and I can’t wait to see what you’ll do next!

  • @user-mi5xq8zj7u
    @user-mi5xq8zj7u 3 роки тому +26

    That seems pretty intuitive on its face though. Just treat each object as a point, it’s central point that is. Now you have three dots…which makes a plane.
    Edit: it doesn’t have to be the central point as in a symmetrical shape.
    Of course, I might be wrong

    • @alejandrocoria
      @alejandrocoria 3 роки тому +6

      Not all planes at all angles that pass through the center point will cut the object right in half. If you choose the plane that passes through the 3 central points, it will have an arbitrary angle that does not necessarily cut them in half.

    • @samuelwerley528
      @samuelwerley528 3 роки тому +1

      How do you define a central point on irregular shapes though? Can you always find a point where all planes that pass through that point cut a shape's volume in half?

    • @user-mi5xq8zj7u
      @user-mi5xq8zj7u 3 роки тому +3

      @@samuelwerley528 “cut them all in half by volume”, that can be done at any angle, right?

    • @user-mi5xq8zj7u
      @user-mi5xq8zj7u 3 роки тому

      @@alejandrocoria I have to think about that, I’m not sure

    • @trangium
      @trangium 3 роки тому +2

      It's good intuition although it's not a rigorous proof

  • @simonwillover4175
    @simonwillover4175 3 роки тому +16

    Wanna become a profficient math professional, or have the same skill as one?
    *Well, you should continue watching these math vidoes, but also read the actual formal mathematic proof along with the video, trying to make sense of the proof as you watch.*
    Now, I know this can be hard to do, so you should ask for some help from others.

  • @srijanyadav4209
    @srijanyadav4209 3 роки тому

    Hey Zach, really Very interesting, keep up the awesome work👍👍👍 plz make more videos on advanced puzzles with clever solution, like the one in which u used vivani's theorem to get 25% chance of forming a triangle..... really love them😍😍

  • @tanelgulerman3073
    @tanelgulerman3073 3 роки тому

    Wonderful explanation

  • @Chondriam
    @Chondriam 3 роки тому +26

    There is an easy way to find the plane: Calculate the center of mass for the 3 objects. Then calculate the plane between that 3 points.

    • @gabrielragum
      @gabrielragum 3 роки тому +12

      Since we're looking for a plane that splits them equally in volume, technically we would need their center of volume, or centroid, or whatever it's called. It would only work with center of mass if the objects have homogeneous mass distribution, I believe.

    • @neelmaniar1454
      @neelmaniar1454 3 роки тому +5

      @@gabrielragum in the case of a uniform mass distribution, is the centre of volume not the same as the centre of mass?
      And in the case of a non-uniform mass distribution, we'd need the centre of mass. Although I'm not sure if any plane through the CoM of any object (uniform or otherwise) splits it in half by mass.

    • @gabrielragum
      @gabrielragum 3 роки тому

      @@neelmaniar1454 Yes, in the case where the mass distribution is uniform, since density is constant, the center of mass and volume are the same. In the case of non constant density, very much like the volume problem, not all ways to slice the object at its center of mass will give the same mass for both slices, but I'm pretty sure there is always at least one way, because mass distribution is continuous, and you can move and rotate the objects/plane.

    • @heiswatching
      @heiswatching 3 роки тому +4

      Found the engineer

    • @gabrielragum
      @gabrielragum 3 роки тому

      ​@@heiswatching Oops :D

  • @tomwells6499
    @tomwells6499 3 роки тому +5

    Three points define a plane - so what's the big deal?
    Three legged tables and stools don't wobble.

    • @daddymike4158
      @daddymike4158 3 роки тому +2

      The video states that the objects must all be cut in half, not just cut

    • @tomwells6499
      @tomwells6499 3 роки тому

      @@daddymike4158 each of the objects have a point in their center - and now we're back to "three points define a plane".

    • @daddymike4158
      @daddymike4158 3 роки тому +2

      @@tomwells6499 True, but not every plane that passes through that point cuts the object in half, only some of those planes do.

    • @Vizaru
      @Vizaru 3 роки тому +1

      @@daddymike4158 if the point in question is the center of volume, then actually every plane cuts perfectly in half. so he is correct is saying that this problem could be accurately conceptualized by simply defining a plane with three points.

    • @ValkyRiver
      @ValkyRiver 2 роки тому

      @@Vizaru This property doesn't exist for an equilateral triangle.

  • @dr.saurabhsinghal
    @dr.saurabhsinghal 3 роки тому

    Wonderful video!

  • @MelodiCat753
    @MelodiCat753 3 роки тому +1

    Beautiful, beautiful video. Intuition is truly the best part of math.

  • @SuperBiologe
    @SuperBiologe 3 роки тому +10

    My brain jumped immediatly to: well if you calculate the center of mass (assuming uniform density) for each body, you get 3 points, which always define a plane. Now i am not 100% sure, but my intuiton says, that if you cut through the center of mass, you should always split any body in half. Of course this would need to be proved.
    Edit: ok i just thought about it some more. (Uniform density) center of mass = center of volume. So if you cut through that with any plane you exactly half the body. In particular, so does the plane defined by the three centers of volume of the three bodys. Therefore it works.
    Lol maybe i should wake up properly before i start commenting on maths stuff

    • @jolusies6836
      @jolusies6836 3 роки тому +1

      Hey @SuperBiologe, I thought exactly the same thing and was wondering if anyone else would agree. My intuition says that it doesn't even have to be the (uniform density) center of mass, just the center of volume where the mass in the two halves could differ, since we only care about the volume. I don't know if this is true but it seems logical to me.

    • @katsuma-csgo8264
      @katsuma-csgo8264 3 роки тому

      Thought of the exact samething

    • @anonymous_4276
      @anonymous_4276 3 роки тому +1

      A plane/line passing through the center of mass need not necessarily cut it in half (by mass/measure/volume).

    • @SuperBiologe
      @SuperBiologe 3 роки тому

      @@jolusies6836 yep, center of volume is what i mean, i just did not know if the term existed in english, thats why i said center of mass assuming uniform density.

    • @jolusies6836
      @jolusies6836 3 роки тому

      @@SuperBiologe I don't know that either, just made the term up to be honnest, but I assume that such a point exists. I have no clue how to calculate it though.

  • @albertmaturanasteinbrugge5678
    @albertmaturanasteinbrugge5678 3 роки тому +4

    Even more basic: if given 3 points anywhere in space you can always connect them and produce a triangle.

    • @holomurphy22
      @holomurphy22 3 роки тому

      Proof?

    • @albertmaturanasteinbrugge5678
      @albertmaturanasteinbrugge5678 3 роки тому

      @@holomurphy22 Take three objects, anything that you can place your gaze upon, imagine you trace a line connecting them together as if they were individual dots. You can do that all day until you prove me wrong, I'll wait.

    • @holomurphy22
      @holomurphy22 3 роки тому

      @@albertmaturanasteinbrugge5678 how do you prove it mathematically? You need first to state the definition of a triangle, which can differ depending on the 'structure' you consider. A common case is affine geometry, where you define a segment between two points A and B to be the set of all the barycenters of A and B with non negative coefficients. Then the triangle defined by 3 points A B C is the union of segments AB BC CA. This definition holds for every triplets of points, and by definition requires exactly 3 points, thus 'proving' your statement (in the case of affine geometry).
      I think we should tell the president we managed to prove such a thing. You got the intuition first I must admit. We should publish as co-autors

    • @albertmaturanasteinbrugge5678
      @albertmaturanasteinbrugge5678 3 роки тому

      @@holomurphy22 Lmao alright sure, let's pretend what I said was groundbreaking.

  • @half_pixel
    @half_pixel 3 роки тому

    Loved the video! Instead of using signed distance, here's an alternative way to assign a number to each point on the circle: Look at the tangent line's slope, then draw the unique line of this slope which cuts object 1 in half. Now use the volume of the portion of object 2 which is on the normal vector's side of the line. (more formally, object 2 intersected with that half of the plane)
    With this definition, the two antipodal points of equal value must correspond to a line that slices object 2 in half, since it's got an equal volume on each side of the line. IMO it's a bit simpler to define/understand, and easier to generalize to higher dimensions.

  • @jamesking2439
    @jamesking2439 3 роки тому

    That's incredible!

  • @BlacksterVFX
    @BlacksterVFX 3 роки тому +5

    When I saw the thumbnail, I thought of an easier proof:
    Any plane going through the COG of an object cuts it in half. Since three points define a plane, it is obvious that the statement is true.

    • @godfreypigott
      @godfreypigott 3 роки тому +7

      _"Any plane going through the COG of an object cuts it in half"_
      Not correct. For example, the centre of gravity of a right circular cone lies one quarter of the way up from the base. A plane through that point parallel to the base divides the cone into two solids whose volumes are in the ratio 9:7.
      The centre of gravity depends not only on volume/mass, but also the distribution of the distances of the volume/mass elements from the centre.

    • @Flutesrock8900
      @Flutesrock8900 3 роки тому +4

      What you said sounds reasonable at first glance, but (as pointed out by Godfrey) is not at all true. This is why mathematicians work so hard on being rigorous in their proofs.

    • @godfreypigott
      @godfreypigott 3 роки тому

      Oops - the ratio of the volumes is 37:27, not 9:7.

  • @sjoerdev
    @sjoerdev 3 роки тому +3

    Just make a triangle from the 3 objects, then expand the triangle to a plane....

    • @ganonfan98
      @ganonfan98 3 роки тому +1

      And which point inside each of the three objects do you choose? The algorithm isn't that simple, I think. ;)

  • @gauravpallod4768
    @gauravpallod4768 3 роки тому

    YES! Even I initially thought that it was the same as the numberphile video. But no!so coool

  • @davidgillies620
    @davidgillies620 Рік тому

    I just love that Borsuk-Ulam works on both the ham sandwich problem and the stolen necklace problem. I can't imagine that revelation not giving me a frisson of joy and satisfaction.

  • @davynolan182
    @davynolan182 3 роки тому

    I demand more videos like this

  • @ShadowSlith789
    @ShadowSlith789 3 роки тому +2

    This makes me think back to geography where any 3 nonlinear points make a circle, which would become the plane in 3d space

  • @ketalesto
    @ketalesto 3 роки тому

    Beautiful!

  • @thanasisrks4944
    @thanasisrks4944 3 роки тому +2

    I would be very interested if you talked about higher time dimensions (if they arera thing that is) because I haven't found much about them online

  • @bryanoconnors7512
    @bryanoconnors7512 3 роки тому +1

    Nice upload. There are a couple of details that are very easy to fill in, but that assumption that the percentages of volume in which a line or a plane divides such a "physical object" vary continuously, does not seem trivial.

  • @Sparkl1ngM1lk
    @Sparkl1ngM1lk 2 роки тому +1

    This is the most interesting and brain teasing yt channel I've ever seen! I find it even better than Vsause or even Vihart!

  • @ethanbove629
    @ethanbove629 3 роки тому

    This is fantastic

  • @christernyqvist3116
    @christernyqvist3116 3 роки тому

    Excellent!

  • @sicapanjesis3987
    @sicapanjesis3987 3 роки тому

    I love how brilliant supports all inspiring math channels...

  • @khalliwalli-obaidfarooqui
    @khalliwalli-obaidfarooqui 3 роки тому

    Nice Sharing, keep it up

  • @snakecodm7249
    @snakecodm7249 2 роки тому

    This is legendary. Awesome
    When i saw them hit 0 on the numb line it all clicked. Great vid broski

  • @JustJanitor
    @JustJanitor 3 роки тому

    This was very interesting

  • @camerongray7767
    @camerongray7767 2 роки тому +1

    That was brilliant

  • @BradleyAndrew_TheVexis
    @BradleyAndrew_TheVexis 3 роки тому

    Very cool!

  • @isbestlizard
    @isbestlizard 10 місяців тому

    what a clever proof! as soon as you guarantee opposite points are negative, you know there's got to be a pair that are 0! this could be super userful for all kinds of things o.o

  • @ReginaldCarey
    @ReginaldCarey 3 роки тому

    Very cool proof

  • @mrl9418
    @mrl9418 3 роки тому

    Bravo!

  • @user-yw1hn7pt4y
    @user-yw1hn7pt4y 3 роки тому

    Very cool video.
    Please correct me if I'm wrong. It doesn't have to be volume, it can also be surface area.

  • @eriktempelman2097
    @eriktempelman2097 3 роки тому

    Epic math, super!

  • @bernardmartinelli268
    @bernardmartinelli268 3 роки тому

    mind spectacularly blown

  • @gsp_admirador
    @gsp_admirador 2 роки тому

    Very interesting

  • @FourthRoot
    @FourthRoot Рік тому

    I suppose this theorem extends into higher dimensions as well.
    One dimension higher and we can imagine temporal 3D objects that pop into existance, wiggle and move around before vanishing, and some sweeping plane that splits the entire field into two regions and sweeps across the field at constant motion. There exists an equation defining that sweeping plane for which the half of the time-integrated volume of each 4D object lies on each side of that sweeping plane. Correct?

  • @tehvvisard
    @tehvvisard 2 роки тому

    This is a nice demonstration as it's quite easy to get to grip with the concept of tangent planes.
    Lets say you have a plane intersecting object 1 and 2 but not 3, that could be interpreted as a point in 2D space where X = 0 and Y != 0. Likewise a plane that intersects object 2 and 3 but not 1 would be a point where Y = 0 and X != 0. Got me thinking of how to create a plane that intersects object 1 and 3 but not 2. In my understanding this would be all points in 2D where X = U and Y = -U. Because X = the distant between object 1 and 2 and Y = the distance between object 2 and 3, so for the plane to intersect 1 and 3 the distance must equal = X + Y. This fits even when the plane intersects all objects, as 0 + 0 = 0

  • @jucom756
    @jucom756 2 роки тому +1

    You can find the (name of dimension-1) etc. For things like this by constructing (number of dimensions) independent (name of dimension - 1)'s thrue your objects so that they get cut in half and then you get a singular point where these (name of dimension-1)'s meet (then any (name of dimension - 1) that passes thru this point cuts the object in half), and the points for these (number of dimensions) objects define a (name of dimension-1) that cuts them all in half.

    • @jucom756
      @jucom756 2 роки тому +1

      For example if you draw 2 independent lines for a 2D object that cut it in half the point where they meet is the areal middle and if you get 2 objects their areal middles define a line that cuts both in half

  • @callowaysutton
    @callowaysutton 2 роки тому

    Hey Zach, great video! :D Would it be possible to do a video on the topic of Lie Superalgebras?

  • @sharpfang
    @sharpfang 2 роки тому +1

    What about an iterative approach? Start with a plane through three centers of mass (not perfect but decently close). Pivot the plane around axis coming through two of these points, adjusting 3rd to cut it in half - the two will "spoil" a little, while third makes its shape perfectly cut in half. Affix it, move another point to make its shape cut in half and the prior "spoiled" a bit. Repeat with third of the points - and keep repeating, improving the precision infinitely.

  • @okboing
    @okboing 3 роки тому

    Looking at this, with just the thumbnail itself I can see, intuitively how such a thing is possible

  • @jonathanreal8018
    @jonathanreal8018 3 роки тому

    Since the mapping is continuous, the intermediate value theorem is sufficient to show that given any non zero x,-x pair must have a 0 value between them somewhere on the map.
    Would be interesting to see what it is about compact sets rather than 'physically possibly' that makes ivt insufficient and borsuk-ulam necessary.

  • @keithandnatani7849
    @keithandnatani7849 3 роки тому

    Im glad they figured out the 2

  • @ooos2989
    @ooos2989 2 роки тому +1

    I just check out my topology book, and they prove this as a theorem using Borsul-Ulam and the intermediate value theorem!

  • @matheusreidopedaco
    @matheusreidopedaco Рік тому

    That's so cool.

  • @saimmunir2841
    @saimmunir2841 3 роки тому +1

    I need guidance!!!
    I'm interested in artificial intelligence and neuroscience...
    Which major should I choose? I'm completely new to this... What choices I have regarding AI and neuroscience?!?!

  • @AtzenGaffi
    @AtzenGaffi 3 роки тому

    Is the mapping welldefined? Say we have a disk as object 1 and object 2 is a set around set 1. In which direction points the normal vector?

  • @what76485
    @what76485 3 роки тому +2

    What a convoluted way of saying a triangle is a triangle.

    • @fullfungo
      @fullfungo 2 роки тому

      Bruh, this is not a video about triangles. I don’t even know why you are talking about triangles.

  • @JW-oe6nw
    @JW-oe6nw 3 роки тому

    I don't even understand Topology and I understand this video, you are an amazing teacher!

    • @holomurphy22
      @holomurphy22 3 роки тому +1

      Borsuk Ulam is a topological argument but if you only use its statement then you dont need to understand much if any of topology (its proof requires topological arguments though)

    • @holomurphy22
      @holomurphy22 3 роки тому +1

      But gg on understanding the video though!

  • @threaljo_
    @threaljo_ 3 роки тому +2

    Woah. Did I just understand this proof???!

  • @footlover9416
    @footlover9416 2 роки тому

    lowk geeking out this shit is so cool

  • @Halosty45
    @Halosty45 11 місяців тому

    It's very useful to know this for fantasy novels.

  • @holomurphy22
    @holomurphy22 3 роки тому

    I tried to prove Borsuk-Ulam for dim 2 et 3, and have some doubts about the dim 3 because Im a beginner in topology.
    For dim 2, it's easy with intermediate value theorem. For example for a continuous function f defined on [0,2pi], you have either f(0)=f(pi) which solves the problem, or f(0)>f(pi) (or < but its the same reasoning). Thus the continuous function x -> f(x)-f(pi+x) is positive at x=0 and negative at x=pi (as f(pi+pi)=f(0)). Then use the theorem mentionned above.
    Now for the dim 2 sphere, lets say f is defined on the unit sphere of IR^3 to IR^2 and consider like before the continuous function defined by g(x)=f(x)-f(-x)=(a(x),b(x)). a and b are the coordinate functions in IR^2, they are continuous. Now the idea is to find a 'loop' (a closed curve) where a(x)=0 and apply the dim 2 theorem for b(x).
    To do so, lets take a point y where a(y)>0 (if no such point exists, then a(x)=0 for all x because the mean value on the sphere of a(x) is 0. And then we can take any closed curve). By continuity, the connected component of a(y) in {x, a(x)>x} has an area > 0. Its quite clear that for any x in its boundary, a(x)=0 (because likewise there exists a non trivial area where a(x)

  • @HolyG-sus
    @HolyG-sus 3 роки тому

    Pro man