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

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

lol about losing my brain cells

nice video

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.

@stillFLiP I guess 3 is better than none? 😊

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

@stillFLiP up to two

@stillFLiP 4 dimensional plane, whatever that looks like

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

hot dang, you're right! lmao

I said, hop in!

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.

@@jasonreed7522 hmmm right

"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.

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?

@@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.

@@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

@@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)

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?

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

Probably one of the few real math things I know.

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

@@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.

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).

3B1B also did an amazing video featuring Borsuk-Ulam: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-yuVqxCSsE7c.html

an engineer teaching a math teacher about topology

hi papa flammy

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

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

disgusting

@@h00db01i huh

@@h00db01i disgustingly genius indeed

@@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.

@@h00db01i sure buddy

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

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

Lol

Ah, the intermediate value theorem, my old friend.

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.

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).

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

@@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?

@@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

centroid or inertia

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.

@@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.

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

@@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.

@@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.

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

No because it can't go through the center

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

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.

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

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.

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

@@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.

@@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).

@@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.

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

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.

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

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

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)

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?

@@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

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

@@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 🤔

@@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)

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.

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.

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

Not necessarily. CoV not always cuts it in half

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

offset the plane in such a way that you get more

@@kijete big brain

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.

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?

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

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

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

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.

@@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.

@@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.

Found the engineer

​@@heiswatching Oops :D

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.

Thought of the exact samething

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

@@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.

@@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.

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

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

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

@@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.

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

Proof?

@@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.

@@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

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

_"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.

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.

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

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

Try splitting an equilateral triangle in half using the centroid You would have realized the proof doesn’t work. A line parallel to a side cuts the triangle into 4/9 and 5/9.

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

Zach Star already made a video about this claim. The summary is that not all shapes have centers of masses through which all lines/planes divide equally by volume. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-IJumRmwYsN4.html