No you guys, most shapes do not have a center of mass with this property (but some do)
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I remember thinking to myself "I'm not sure if such a point exists, but if it did then we wouldn't have needed to do all this work, so it probably can't be proven to generally exist".
That might not always be the best approack, but I found myself thinking that, too. Though, then I thought of the triangle case - even though I thought the upper part would actually be bigger haha! - and I figured it wasn't as simple as just picking one point
The Occam’s Razor of giving a shit.
Baste
I'm less surprised that the center of mass isn't always a place where any line cuts the shape in half since there are plenty of irregular shapes so much as I am surprised that it doesn't even work for all shapes which have a circumscribed circle (all vertexes are on a circle) like an equilateral triangle.
I'm a bit surprised it didn't work for all things with an obvious symmetry like n-polygons like the equilateral triangle.
but but but life would be so much easier if it WAS true!
Hi professor
Proof by convenience
"If it makes the problem easier then it must be true, and if enough people on the internet say it then it must also be true"
-Albert Einstein
@@zachstar That Albert guy sounds kinda smart. He should study Quantum Mechanics, I'm sure he'd like them.
* were true
I love how confident all the commenters were that they figured it out
I never see so much 'confidence' (speaking like they know it all) at places were people are really good in math. Most people very good at math dont take themselves seriously, they know they can be fooled and moreover they are more interested in math than their ego. They are happy to talk about math, they dont see it as an opportunity to act smart
It is a little frustrating that a lot of peoples gut reaction is to doubt the video and assume that theyre right. But not all of the comments were framed as "this is easy, iamverysmart" (at least mine wasnt intended to be that way). Have faith in some of us internet randos haha
@@hotdogskid yes true not all people are like this
Having thought about it, most people dont know that they may be something they missed
With experienced you tend to ask yourself more question before concluding sth that has a tricky flavor to be faulty. But someone without experience cannot sense that sth can be tricky or not
Idk if what I wrote makes sense I dont have a very good english
Some people happen to be more humble even in new topics for them. I just think they got humbled elsewhere beforehand
@@holomurphy22 There is also that there is work you don't see.
On this video, one of my comment is quite long, but I wrote a comment twice as long then verified my own claims by attempting to draw a shape meeting the criteria I had... It didn't go well, so I thought a bit more about it and was "ah ah!" and rewrote the whole comment!
As a reader, you only see the end result ^^
But that also means that...I am not wrong there.
That's also why, when we don't have the mean to fully check what we wrote (time constraint or P≠NP), we walk on eggs and add word like "I think, it may, it's possible" and it make us apear much less confident. But in fact, we are highly confident...It's just we don't trust tge result we gave and ask reader to exercise caution. Those cues are often missed and then other people go full steam like 《You're wrong, it should be...》or《It's too convoluted, you didn't even notice...》.
Note that it involves the original author in one way or another too.
@@programaths you will surely appreciate my little caution in my answer to you on the other comment about the CoM, even though Im pretty confident its true but I dont have a formal proof yet in mind
I was indeed a bit surprised you already got a geogebra to show me, lol
Idk if I understood you correctly but I never said you're wrong anywhere
Shortcut on finding the areas at 2:27: The top triangle is similar to the whole triangle, with a ratio of 2/3. So the area ratio is (2/3)^2 = 4/9 (which is not 1/2).
Legend
As for the tetrahedron, using barycenters we can see the ratio is 4/3. (4/3)^3 is 2+10/27 = 2.37... so the volume of the bottom part is 1.37... times bigger than the upper part
It may be fun to mention I didnt use any calculator or done any calculation for 10/27. I knew that 27*37 = 999, so that 1/27=0.037037037... and 1/37=0.027027027... which is kind of cool. Im glad I finally used this knowledge
Big brain guy
You can construct the big triangle using 9 small triangles. it would look like a figure hexagon with 3 triangles placed on every other edge. the center of that hexagon is the centroid of the triangle. If you have magna-tiles you can build this with those triangles. Position the triangle so that the base of it is horizontal, and imagine another horizontal line that passes through that centroid.The triangles will give you a clear path to cut along that line, so cut along that line. Counting the two shapes you get, one of them, the bottom trapezoid, is made of 5 triangles, but the upper mini-triangles, making a triangle figure, is made of only 4 triangles. This is an probably a good easy way for my slow-ass brain to tell that yes, indeed this doesn't work, so the top "half" would have only 0.8 times the area of the bottom "half". So 4 + 5 equals 9 making for the same proportion from OP's comment, 4/9 or 0.444.
@@adam6543 It's a basic triangle similarity theorem. I do agree the use of the theorem is quite ingenious.
Thanks for shouting STEMerch EU out, Zach
Hello, I'm early
And thank you for making this hilarious watch/clock
@@simsim4910 Np :D
Fan of yours :)
man i dont know what is happening in math anymore. everything looks complex in this video XD
Huh... So I've done a lot of simulation stuff in the past using boring filled solids as approximations. My immediate reaction to the premise was surprise, but the centroid doesn't just measure the volume, but the volume AND distance. A large lump of volume near the centroid can be balanced out by a small amount of volume far from it. Derp. Good to be humbled and remember the basics every once in a while. >_
Yup, turning effect of the force, torque.
Yip, I'm fairly what's going on is conflation of the center of mass, where the integral of distance x mass will always be the same on either side of any intersecting dividing line or plane, versus the total mas on either side of said dividing line or plane being the same.
Given that the first is true, the second should only be true when the shape is 180deg rotationally symmetric.
I was one of those people who commented something like this. For some reason it seems intuitive to me that a "magical point" would always exist, but my intuition was wrong. Super interesting!
Every shape has a centre of mass. Any plane through that balances. If you say that the cost of area or volume is its distance from the line, each side has equal cost. The top of the triangle has slightly less area, but the area it does have is further from the line.
@@donaldhobson8873 Yeah. In a certain sense, the term "center of mass" is a bit misleading in that regard. It's really more like a "center of torque", I think.
Same here - in my case I was conflating it with the physical center of mass, which is the same as the center of gravity. Turns out that doesn't hold when you're talking about area instead of weight distribution. I even double checked my thinking by Googling the center of mass and rotation, and thought I found nothing but confirmations on my theory! Then again I also thought I was the only one to comment about it based on a quick scroll through the comments of the original video, so I guess I double learned the same lesson on this one.
@@batlrar I'm not entirely sure what you mean. It doesn't hold for the "physical" center of mass or the center of gravity either.
@@EpicMathTime I think unconsciously the idea (at least for me) was that the "center" of a line cuts it in half, so the "center of mass" of a polytope must also cut it in half. Clearly thats not the case now that ive thought about it a little more
People who commented on other video: "this seems easy just do this, why make a video. "
Those same people this episode: "I totally agree, good thing I never said that last episode" *deletes comment*
Was wondering why there were no comments until I realised this was just posted. Darn, I can't read other people's insights until after I've watched it...
Lol
Jade bishop: suffering from success.
It's really good that you addressed this! Very often I read a neat proof but if I didn't try to solve it myself I'd have no clue why they chose to use that idea, and why other easier ones don't work.
Thanks for pointing out I was being dumb! You made me less dumb now. I didn't comment it, but I definitely thought it. My intuition was the same as everyone else's: "If I can balance a shape on a line, there's equal mass (and thus area) on each side". Cool, now try that with a spoon.
I think the simplest way of seeing how this doesn’t work is an L shape, when the centre of gravity is outside the shape.
Or any shape with the centre of mass outside.
It doesn’t cut the area into half because the centre of mass depends on the torque/moments instead of the distribution of mass.
For those who are still confused, the triangle balances across the horizontal line because the center of mass of the top triangle is farther away from the line than the center of mass of the bottom trapezoid. So long story short leverage is weird.
I like that -- it's a good intuitive summary. *Balancing* cares about distance, but area/volume doesn't.
One could probably use that to make a nice easy counterexample. Maybe a square next to and a long skinny rectangle with the same area, showing that the center of mass isn't at the point where they meet (which it'd have to be for the line to cut the area in half).
@@pfeilspitze Yeah, basically, when cutting through the center of mass, the moment of rotation of the two halves have to be the same, but their mass doesn’t.
For a good counter example, imagine a barbell with two weights of different sizes
###---------#; the balancing point is going to be closer to the left but there will still be 3 weights on the left and 1 on the right.
Trying to balance it between the second and third weights will make it tip to the right, because the right weight sticks out farther and has a greater moment of rotation.
⬆️##^#---------#⬇️
@@KnakuanaRka Well, assuming it would balance at around the between point of ###－ and －－－－－－－－# it'd be because they have the same mass on both sides, right? IDK i think that it would balance based on mass and also the triangle *wouldn't* balance on that horizontal line, but it would balance on that slightly lower one that Zach Star mentions at the end.
No, quite simply, the masses don’t need to be equal the same on either side of a solid (or thin object). What needs to be same are the moments (integral *). You can have a ‘solid’ that is a 1kg, and a 10kg particle, exactly 11 m apart, and you will find that centre of mass is on the line joining them, 1m from the 10kg weight and 10m from the 1kg weight. For a simple case, consider the axis (of rotation) to be a line perpendicular to this system, of course passing through the centre of mass. One side obviously has ten times as much mass as the other, yet the moments are equal (1 kg*10 m = 10 kg*1 m).
The moduli space of oriented planes in R^3 is homeomorphic to S^2 × R in the obvious way. This maps into I^3, whose coordinates represent the portion of mass of each object on the positive side of the plane. This map extends to the compactification S^3, where one pole at negative infinity maps to (0, 0, 0) and the other pole at positive infinity maps to (1, 1, 1). It's now a trivial exercise in topology to conclude that there is a point that maps to (1/2, 1/2, 1/2). Hint: Excising this point leaves a target space with homotopy type S^2. Hint 2: Consider what happens when you reverse the orientation of a plane.
This is what happens when a core audience is comprised of engineers instead of physicists! Love your vids btw
More like physicists instead of mathematicians
I love how artists are bullied by architets, wich are bullied by engineers, wich are bullied by physicists wich are likely to be bullied by Mario 64 speedruners
I acctually kinda forgot the distance property, so no it's not because there are too few physicists here ;)
Sick burn, dude. *Engineers wipe away their tears with 100 dollar bills*
why physicists not mathematicians? This is way more a mathematical question than a physical one. There are literally no forces or particles or anything that would indicate physics other than maybe "mass", but that concept is not uncommon in maths. This is geometry and not really physics.
A line intersecting the center of mass does not cut the shape into two parts of equal area. Instead, the invariant is going to be based on the area weighted by its position relative to the center of mass.
This is just the odd number rule of symmetry.
The odd number rule states that all polygons with odd number points/line segments will only have the same number as the number or points for symmetry.
Irregular polygons are weird
I tried to understand but I really couldn't. What do mean by "have the same number as", the number of what?
I also have no clue what this means, and I can't find anything about an odd number rule of symmetry online :/
@@jonasdaverio9369 I think he just meant Polygons with an odd number of sides will have the same number of lines cutting it in half as it has sides. So a triangle has 3 different lines through the center of mass cutting it in half, a pentagon has 5, a heptagon has 7 and so on. And for Polygons with even number of sides every line through the center of mass will split it in half.
@@jonasdaverio9369 Vertexes/Sides
I don't know about weird. I would say irregular.
you should also explain WHY the center of mass does not have to split volume evenly. I believe it is about distance; a bunch of area (mass) close to the center can be balanced by a smaller area that is further away on the opposite side. Just like a heavy guy close to the fulcrum of a seesaw can be balanced by a lighter kid far away at the other end.
I've actually come to this conclusion for 3d space by asking a somewhat similar question regarding fluid containers: I was looking for some rule or algorithm by which I could quickly find the half-way point where a container is half-full. I started with the known case of a cylinder, since half-way up the cylinder is half, halfway deep is half, and halfway into the breadth is half, since all those planes intersect at right angles, I figured that for all possible orientations of a half-full cylinder, the fluid level would be tangential to a single point, being the center of volume for the cylinder.
I continued that line of reasoning, thinking of what the easiest way to find the half-way point in a cup was. I started by thinking of cups as cones, and immediately came to the conclusion you stated upon discovery that cones have infinite orientations where a line plane through the center of mass would have been above the half-way point in the container.
My question is if there is a name for the shape that's described as the region where all planes that bisect a polyhedral area in half are tangential to this shape.
Literally realized this seconds after posting this comment, but the plane that bisects all three objects will simultaneously be tangential to some part of the surface of each of their respective 'half-oids'
The equilateral triangle's case is a particularly easy one to disprove using simple geometry. You can prove that there are five congruent equilateral triangles on the bottom and four on the top. No trig or irrational numbers necessary. It's quite beautiful how you can prove that 4 to 5 ratio with two different methods
Nice one
For that property to work you would need an object who's points mirror on it's centre of mass (ball, square, 6 sided dice, etc). Think of it as a see-saw, a father playing with his son and trying to balance things would try to move up the see-saw, his mass doesn't shrink, but it is balanced at some point.
This just seems so intuitive, but our brains are very good at... optimism i guess, when i was doing the questions for the math olympiad where i live i always had to double check a pattern by actively searching bad scenarios because otherwise i'd just filter out the best case scenarios.
An interesting observation I derived from this - while it is true not all shapes have a dot where any line splits the shape in half, you could derive a sort of "loop", where for any given point the tangent splits the given shape in half.
Because making this without brute-forcing a bunch of points is beyond me, I'm not quite sure how this loop will look, but im sure that's a brilliant idea for another video!
It's not the area that is cut in half, it's the integral xdA over the area of the object, where dA is a small area segment and x is the distance coordinate of the small area segment perpendicular to the line that cuts through the centroid.
This integral is actually the torque on the object generated by a uniform gravitational field, and it is always zero around the centroid (or center of mass in the case where we have variable density), which is why a gravity will never cause a free falling object to start rotating by itself.
It holds for shapes with 180 deg rotational symmetry around some point. when you're rotating the plane you ad the area that the plane passed at the one side of the point and subtracting at the other, only if they're the same the areas won't change after rotation
I almost mistook this video as saying there is NO line that bisects the perimeter of an eqtri. I was gonna have a cow. I almost mentioned euclidea's theta 1 lvl for easy reference - then I realized I was tripping. Good video. Thanks.
Another usefull solid to think about to get an intuition on this is the following:
Imagine a barbell with three weights on one side and one on the other. Assume the bar is of negligible volume and mass.
|||----|
Finding the center of mass is easy. It's like balancing a seesaw. There's three times more mass on one end so the other end must be three times further from the fulcrum. So the center of mass is 3/4 of the way along the bar towards the side with three weights.
|||-*---|
But obviously with four weights, half of the volume is when you have two weights on each side
||*|----|
So center of mass =/= center of volume.
My physics textbook for 9th grade had an explanation of an experiment for finding the center of mass of any comparably flat object(eg a key which was used as test subject there) and it involved kind of pinning the object on some point, waiting until it stopped swinging and drawing a line through the pinned point straight down on the object itself and repeating this process one more time with a different pinned point and it stated that no matter what point you choose the perpendicular containing this point will also contain the center of mass
Great explanatory follow up! Thanks!
Bro this channel is amazing. Im so glad I got recommended this channel.
For those confused how this is possible, the long story short is leverage.
yup. the parts of the shape far away from the center skew it more than those close to it.
dude my brain was melting until I saw your comment and remembered that was a thing
thanks
The more interesting question is: How good is it to approximate the plane that cuts the objects in half by the plane that goes through the center of masses
I suppose you could define a surface that a plane at a given slope cutting the object in half would lie tangent to.
It depends on the shapes used, but you could construct shapes such that an arbitrarily large amount of mass lies on one side of that plane...
Incredibly, this question is equivalent to saying: how well does the median approximate the mean? To see why you take the volume and do a double integral parallel to xy plane to get a 1D function, the median of this function is the point at which the plane splits the shape in half and the mean is the plane that includes the center of mass :)
I want to see what shape all lines that cut the triangle in half would trace out
@@farissaadat4437 Haha, that's an awesome insight, and makes it abundantly clear why it's a bit dumb. Also goes to show how mindblowing the original theorem is.
I think there's a big difference between a line which cuts the shape in half (resulting two shapes that have the same area) and a line that has the property of the following: if the shape is supported on such a line, the shape would be in equilibrium. This statement holds for any line through the center of mass right?
Yes. Because equilibrium takes into consideration not just the division of mass on the two sides, but the torque generated by gravity on the partial masses of the two sides. The side which weighs less (e.g. the small triangle in the video) will have a barycentre further from the barycentre of the overall shape than the side with more mass (e.g. the trapezoid at the bottom), to satisfy:
τ1=τ2 -> R1*m1*F=R2*m2*F -> R2=R1*(m1/m2)
So since m1/m2 in this case equals 4/5 we will have R2=(4/5)R1.
but but but life would be so much easier if it WAS true!
2.2K
Zach Star
Knasterbart
Knasterbart
1 month ago
I love how confident all the commenters were that they figured it out
845
Flammable Maths
Flammable Maths
1 month ago
Thanks for shouting STEMerch EU out, Zach _<
112
Jade Bishop
1 month ago (edited)
Was wondering why there were no comments until I realised this was just posted. Darn, I can't read other people's insights until after I've watched it...
243
LZH1703
3 weeks ago (edited)
I think the simplest way of seeing how this doesn’t work is an L shape, when the centre of gravity is outside the shape.
Or any shape with the centre of mass outside.
It doesn’t cut the area into half because the centre of mass depends on the torque/moments instead of the distribution of mass.
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mha hart. mah sole
4 weeks ago
However, a cool fact is that all convex shapes have a point in their interior whereby any lime drawn through it will cut the shape into two pieces, each with area at least 4/9ths of the shapes original area.
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Christopher Gibbons
1 month ago
For those who are still confused, the triangle balances across the horizontal line because the center of mass of the top triangle is farther away from the line than the center of mass of the bottom trapezoid. So long story short leverage is weird.
95
John Chessant
1 month ago
It's really good that you addressed this! Very often I read a neat proof but if I didn't try to solve it myself I'd have no clue why they chose to use that idea, and why other easier ones don't work.
45
Heph
3 weeks ago
People who commented on other video: "this seems easy just do this, why make a video. "
Those same people this episode: "I totally agree, good thing I never said that last episode" deletes comment
15
Vereinigte Religionen der Ente
1 month ago
This is just the odd number rule of symmetry.
The odd number rule states that all polygons with odd number points/line segments will only have the same number as the number or points for symmetry.
Irregular polygons are weird
82
jucom
4 weeks ago (edited)
This just seems so intuitive, but our brains are very good at... optimism i guess, when i was doing the questions for the math olympiad where i live i always had to double check a pattern by actively searching bad scenarios because otherwise i'd just filter out the best case scenarios.
9
Wafi Marzouq Mohammad
1 month ago
For those confused how this is possible, the long story short is leverage.
44
Sean Sdahl
1 month ago
The more interesting question is: How good is it to approximate the plane that cuts the objects in half by the plane that goes through the center of masses
34
cmilkau
4 weeks ago
There's another useful "center" that helped me prove a fixed point theorem. Every nonempty set in a compact metric space has a minimum-radius "circumsphere".
1
Rafael Freitas
3 weeks ago (edited)
So, if we take say, a 2d perfect homogeneous fluid, and put it inside a triangle, find a line that divides its area in half, we won't be able to rotate that line with respect to the CoM axis, if the total amount of fluid is to be conserved inside the triangle, assuming the line permits no leaks, of course...
So counterintuitive!
Is there any other axis we can use to accomplish continuous rotations?
2
Johnnie Chan
4 weeks ago
A line intersecting the center of mass does not cut the shape into two parts of equal area. Instead, the invariant is going to be based on the area weighted by its position relative to the center of mass.
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Bro you are so much underrated
Cool graph idea. Find the formula for a line that cuts the object in half at a given angle. Then find the point on that line that is closest to the center of that object. Now graph the path that point takes as you change the angle from 0 to pi. What would that look like for a triangle? What about different polyhedrons?
Man. Solved it with simple trig. Love it
Thank you for pointing it out, I did think it was the case and am glad I now know it is not. I can intuitively say that I know the problem was true, but I could not have come up with a true explanation that does not have a circular dependency
Hey mate! I subscribed to your channel I stumbled upon your really interesting video about how expected resuts didn't matched with reality like making parents pay for showing up late to take their kids from kindgergarden actually increased number of such cases, because fine was too low, parents felt that they pay for showing up late and it was ok in their eyes. Could you please make more videos like that? Something about probability (I find that subject really interesting) or touching upon psychology like in that video. I am not asking you to remake your whole channel, I just want to see more videos this from time to time. :)
Suppose you looked at each possible orientation of a line, and found the offset from the center of mass needed to make a line with that orientation bisect the shape, and plotted that point (rephrasing: the point where, if you start at the center of mass, and move perpendicular to the line orientation until the line passing through the point with the given orientation for the line, bisects the shape), and take the resulting curve.
What would this curve look like?
It should be a continuous curve, and I want to say also smooth, but maybe only smooth almost everywhere, idk.
It should also have all the symmetries the original shape has.
The smallest the area bounded by it can be is of course 0, because for the circle (and the square?) the path is a constant point. But what is the maximum area it can be for a shape with a given bound? (E.g. a shape that fits within the unit disk, or, as an alternative question, a convex shape with area at most 1)
There's more cases where it works than just the circle and the square. Any ellipse and parallelogram should work.
First video of yours i saw. I corrected one of my ideas this morning thanks to you.
It would be really cool to see an animation of the set of lines that do cut an equilateral triangle in half, as well as a discussion of which 2D and maybe 3D shapes possess the point property
I was kinda bugged about those peeps saying like it's just that intuitive cus their line of reasoning never really made much sense to me as well. Thanks for clarifying, Mr. Zach!
It is intuitive, but every so often it just happens that obviously true statement is wrong.
I think the property does apply to the example shapes used in your previous video. The shapes had 180 degree rotational symmetry in three perpendicular axes, basically stretched, puffed out cubes. My first thought was the center of mass as well. I think it would have helped your illustration to pick at least one 3D shape that would obviously not work that way.
There's another useful "center" that helped me prove a fixed point theorem. Every nonempty set in a compact metric space has a minimum-radius "circumsphere".
yes yes!! that was awesome. great video dude!!
Awesome! This made me realize I had a wrong intuition about CoM as well. I also wondered what is actually equal on both sides if a plane through the CoM. This quantity includes not just the mass but also its distance from the plane.
It’s the torque or something i guess
@@bcthoburn If you assume uniform gravity along the separation plane and an (arbitrary) anchor point on this plane, then yes.
But you don't need those assumptions, it's basically distance (to the plane, not to the anchor) times density.
I'm less surprised that the center of mass isn't always a place where any line cuts the shape in half since there are plenty of irregular shapes so much as I am surprised that it doesn't even work for all shapes which have a circumscribed circle (all vertexes are on a circle) like an equilateral triangle.
I didn't see the first video. I'm onboard though. For any shape and any angle plane you can move the bisection until both sides have equal volume. And you can join the 3 shapes with a plane.
But figuring out a formula depends very much on the shapes involved.
I think what had people confused is that you use highly symmetrical shapes in your example (with some visible imperfections, sure). But using the center of mass would work for many perfect cylinder / sphere / cube style shapes which are mirrored in all axis
centerpoint theorem guarantees that there is a centerpoint of an object, and every hyperplane going through that object cuts space in such a way that each half-space has at least 1/(d+1) of the volume. So it holds true for 1 dimensional objects.
A lot of people may have jumped to this because there is a similar property the the center of mass does have. It cuts the First Moment of the mass of an object equally in every direction (thats pretty much the definition of the center of mass), but not the mass itself.
learned something new tonight....thank you
The centre of mass of a tetrahedron is at 3/4 of its height, so the volume of the "upper" tetrahedron (considering an horizontal plane through the centre of mass) would be (3/4)^3=27/64, NOT 1/2 of the volume of the whole tetrahedron (since the two are obviously similar. So yeah it doesn't work for a tetrahedron either.
So, if we take say, a 2d perfect homogeneous fluid, and put it inside a triangle, find a line that divides its area in half, we won't be able to rotate that line with respect to the CoM axis, if the total amount of fluid is to be conserved inside the triangle, assuming the line permits no leaks, of course...
So counterintuitive!
Is there any other axis we can use to accomplish continuous rotations?
Your globe looks so cool
There is a point in any object called the centroid or centre of mass but they don't necessarily mean it's inside the object. Take a C for example. But for every object there is a single point inside the volume. No matter how small the object there lies at least one point in space. Any 3 objects will have this and you only need 3 points to make a plane. Easy done
if the object (2d) has 180 degree rotational symmetry then it should have that property. An easy way to see this is that with any given cut, if you rotate the line say a miniscule amount, you must be gaining as much area on one part of the line as you lose on the other. hence the "centroid" must always be the midpoint of this line, implying 180egree symmetry.
A simple way to disprove the 3d case is just with a prism of any 2d shape that fails, as any line can be projected to a plane that (supposedly) should cut the prism in half perfectly
So there must be a formula which describes the height of the center as a function of the angle it makes with the base of the triangle when cutting the triangle exactly in half and it must be root(3)/3 at 30, 90 and 150 degrees. Some sine function probably
I really liked the proof for the triangle that you did with, but actually the first thing I thought of has more of a calculus flavor to it. I was thinking, if you took one of these lines that splits the triangle in two and passes through the "center of mass", then you rotated it by a super small amount around the center, you would be taking off more from one side of the line than the other because the slopes of the edges of the triangle--relative to the line--are different on each side of it. If I'm missing something, though, the reader may feel free to correct me
1:15 Where did you get that floating globe?
All 2D shapes DO have a center of mass & it’s easy to find for any arbitrary shape.
Say, a flat map of the US. Tape it to a piece of cardboard, cut out the shape. Get a pin, a string & a weight attached. Wrap the string around the pin. Push the pin (with string & weight thru the cardboard at ANY point near one edge of the shape & into a wall. Gravity will cause the cardboard shape to rotate until the centroid is directly below the pivot point (the pin). The string also hangs straight down. Make 2 marks on the cardboard directly under the string near the top & bottom og the string. Pul the pin, rotate the cardboard, & repeat the procedure. Join each set of 2 marks with a straight line. Where the two lines cross is the center of mass.
You can prove this by pushing the pin thru the CoM & into the wall. Even with a very “low friction” pin to cardboard interface, the cardboard shape will be stable (not inclined to rotate) in ANY angular orientation.
Oh I see the problem you saw on the argument. I bet mine were in the middle of the bunch BUT look the point of this argument is that the point doesn't need to be at the same spot, if it moves continuously as the plane continuously turns at any moment you'll have 3 points to form a plane and this makes it feel like it obviously works.
Like the 2 points are always inside the 2 first objects and the plane that cross the shape in half even if when turnining you have to ajust them slightly up or down you can see you can move them all across the 3rd shape and will eventually meet with the point that contains the "half plane" inside the 3rd object, still even if it moves slightly, as long as it does so continuously.
I think the best way to illustrate this is with a donut. No point on a donut will allow all lines through it to cut the shape in half, because a donut is a circle without a center, so it lacks that one point on a circle that would have such a property
What if you allow picking a point not on the object though?
No civil engineers commenting I see!
The CoM balances the moments of all opposing sides, not the areas/volumes.
I mean, that’s literally how the CoM is defined; as the balancing point and it should be immediately obvious that the balance point does not necessarily cut an object in half. Take a broom, or rake, slide your fingers together until they balance. They will be much closer to the side of the broom or rake with the head.
Thanks so much for posting this video. I was one of the people who posted on the last video and I was wrong.
I guess people get confused because if you hang an object from any point the centre of mass will fall diectly below that point. It doesn't feel like much of a leap to say that the line through the centre of mass and the pivot must therefore split the mass/area evenly. But if you take a moment to consider it, you realise it makes no sense ;P
It seems as if the center point will move around if you went through many different lines where the 2 halves actually do remain equal, it would be cool to visualize all of those point positions, and see how they move as the dividing line moves.
Can we characterize the shapes such that every line through the center of mass, cuts the area in half?
Or, more generally, characterize the mass distributions such that any line through COM cuts mass in half?
Ngl I had the wrong idea too. I think an even more intuitive example would be to look at the center of mass of two disjoint and unequal sets, and move the smaller set arbitrarily far from the larger one until the perpendicular bisector of the COM lies entirely outside of the larger set. Only confusion could be from having the sets be disjoint, but one could fix it by instead considering an imbalanced dumbbell with an arbitrarily thin, and thus ignorable, crossbar.
Whoah - you’re looking really healthy my dude; good job 👍
I believe that in general, a line would cut shape in half if:
Center of mass of the object is located the same place as the center of mass of a circle (or an ellipse) drawn intersecting the edge of the object.
Well i am an idiot so i might be wrong.
Sorry for not using math jargon as well so it might be confusing
I love this and all I want is to see the faces of the people who thought they had one sentence proofs.
It’s different than the center of mass, but it is possible to find a point of this type in at least 3 distinct planes, if not all of them. Maybe there is still a proof there if you align your choice of planes, but its more complicated than trying the center of mass.
Do someone know which applications Zach use for geometric representation such as the one used to represent plane for Linear Algebra (3Blue1Brown may use the same) ?
Simple proof : take two unequal circles and consider your object to be the sum of both.
As you take the smaller circle farther and farther away, and keep the bigger circle at the origin suppose, then the COM eventually leaves the bigger circle and is in the middles somewhere.
Clearly, a vertical line through this point does not split our "object" in half, as that would imply that both the circles are actually of equal size.
The centre of mass makes it such the area further away has more “weight”.
When cutting area in half, the distance between the cut and other areas does not matter.
An example where this would be obvious is a T shape made from 2 of the same rectangles.
It would be interesting to know what is the set of all shapes for which the property holds. Clearly a subset of it is all shapes with 180 degree rotational symmetry, but I wonder if there are any others, and how to prove it.
You mean the property that any line through the center of mass cuts the area in half, right?
@@EpicMathTime Yes. Also the generalization to higher dimensions (which might be beyond my own ability to prove).
If this video taught you something, it's that any unproven proposition that starts with "clearly" is likely wrong.
@@shmojelfed9664 Indeed if a shape has a central symmetry it verifies the property. For a convex shape, one can show its a necessary condition. For what follows we assume implicitly than the borders of any shape is continuous (quite natural assumption)
Now, hopefully you're familiar with infinitesimal calculus. If so, fix a point P that supposedly verifies the property we look for. Consider a fixed line D that goes through P. And consider the function f(x) which gives the amount of area of the shape between D and a line that makes an angle of x radian with D (Ik there may be an ambiguity but its not important because we only look for the derivative).
Now consider the derivative of this function. I will now get informal as its too tedious and hard to explain properly with my poor english. The line that goes through P has two sides in regards to P. Each side has to verify that the derivative is equal (because it must swipe the same area for each small angle). It implies that each of these lines verify a kind of lever property, because each small fraction of the line swipe an area proportional to its distance from P. Note that holds for each line so that P must be the center of (uniform) mass.
So this is the final answer. A shape verify the property we look for if and only if any lines that goes through the CoM verify this lever property. I dont think we can be more precise with full generality.
We can deduce from this that for convex shapes, the CoM has to be a symmetry center (because each line get to be symmetric in regards to the CoM)
Correct me if I'm wrong
4:28 the ULTIMATE SIMP LEVEL PYRAMID
I wonder what curve you would get if you plotted the area of the two parts of the triangle while rotating the line in a circle? There would be three places the line would cross over itself (since there are 3 lines where the two parts would be equal) but in-between that, one side would be less or more than the other's area. Do you get sine waves? Something else?
Thanks
what happen when you cut thought the center of sierpinsik triangle with a horizontal line?
how much of the mass are on the upper side?
it seems the upper half will be 3/7
This video is solid golld!!!!
For those who obviously don't know the definition of "center of mass" or "center of gravity," its position is
X = (m_1 x_1 + m_2 x_2 +...+ m_N x_N)/(m_1 + m_2 +...+ m_N)
and the same for Y and Z. This has NOTHING to do with volume or area. A support anchored there results in net zero torque and no angular acceleration. The sum of the gravitational TORQUES on the left side of the center of mass exactly cancels the torques on the right side of the center of mass. (And the same for before vs. behind and above vs. below.)
Oh wow, I have a degree in physics and didn't realize this. I always implicitly though the center mass was extra super duper special and every plane through it cut the object in half. What about when the object is hanging from one point? Isn't the center of mass always supposed to be underneath the hanging point? Wait, that's not just mass but also moment of inertia. Nvm.
(Cue people telling me my degree was a waste of time)
So it does not work for a equilateral triangle, but does it not work for any triangle or is there some special case triangle it does work for?
"they all look very... similar" made me chuckle
I shared that baseless intuition about the center of mass until I thought about it and then the difficulty became clear. But the video didn't really show me examples of shapes where it'd feel hard that you could figure out that you can cut them all in half. Like for instance three strings curled up into a mess. It's slightly too hard to keep track off but somewhere between that and the simple shapes there's something that'd show how amazing the proof in the previous video is.
point symmetry any line through the center of mass cuts it in equal areas
Except no
alright, this makes a lot of sense. so much actually that I'm not sure why it wasn't immediately obvious.
however, there is always a plane that can cut the shape in half. as you showed, the triangle can be cut on half if you move the line around. supposedly (i don't see how this isn't the case) this is true no mayter the angle the line is at, even if said line does not have any point in common with other lines of the same property.
this should hold true to the third dimension, no? that any shape CAN be cut in half by a plane, no matter what the orientation of said plane? just move the plane along the perpendicular, and it will eventually be true.
For those who are still confused: imagine a big circle and a small circle that are far apart, but connected with a very thin ”line” (like a line that has thickness) so it’s all connected. The center of mass shall be somewhere on that line, but if you simply cut it in half through that point, you will end up with a heavy piece and a light piece.
Why would it balance if you lift the whole thing from a point that divides the mass inequally? Because the small mass is further away and thus the torque is the same on both sides.
This proves that the center of mass is not the right point if you want equal area on both sides. It does not prove that an equal-area point does not exist potentially partly inside the big circle.
The video however does prove it because for an equilateral triangle three obvious lines (each starting at a point) determine where the equal-area point would need to be. Then the horizontal line shows that even that point is not working.
Are there different formulas for finding the area of a triangle vs the triangle of a trapezoid?
Now I wanna see the shape made by each line that cuts a polygon in half. I'd imagine it would look like a string art cardioid.
Does there exist a shape with the property of the video( there exist a point where all line passing throught it divides the area in half) that doesn't have a central symmetry?
If you have a central symmetry (shape rotated 180° is no op) then every line going through that center will cut the shape int two surfaces of same area. Simply because those two surfaces can be superposed.
It's possible that these people confused that center with the center of mass as both overlaps when they exists.
I found the same thing. I think we can prove it's a necessary condition if the shape is convex (the border being continuous). Note that it's not a necessary condition if the shape is not convex (hard to describe what I thought of even if its simple, like a disk where I emptied some appropriate parts).
We may prove that its necessary for convex shapes for any point P instead of only the center of mass. The proof I will give here is not formal partly because I've hard times writing in english. To formalize it I think its easier to consider P to be strictly inside the shape.
Idea for proof: We consider there is no central symmetry, such that there is no central symmetry in P. There must thus exist a line D cutting the shape, making a segment going through P such that the P is not in the middle of the segment. The border being continuous, we can make D to rotate with a small enough angle (rotation centered in P) such that the area D went over while rotating is bigger in the longest side of the segment. Thus the area each side of D is not constant while rotating (at least at some point). It thus cannot cut the shape in half for every angle.
@@holomurphy22 To find a point within the shape as such that any line going through it slice the shape in two equal area, you need to have the line sweep the same area.
These swept area can be seen as list of numbers.
So, it would be stating that if you have a list of number, you can cycle it as such one half is always equal to the other half.
For central symmetry, such list looks like:
1,2,2,3,1,2,2,3
If you cycle it:
2,2,3,1,2,2,3,1
Only lists that are made out of a smaller list repeated twice have that property.
That's because you need to get what you lose in your half:
a,z,y,x,Q,x,y,z
When shifted left, the first sub list loses "a". And the second list gains "a"
z,y,x,Q,x,y,z,a
So, Q has to be "a".
I could ignore "x,y,z" and that could have been anything.
But if we try to cycle it once more, it will not work as the equality will not be preserved.
We just showed that the list has to be same sub-list repeated twice.
That means that for convex shapes, center of symmetry is required.
For concave shapes, this is NOT directly the case.
Here is a weird polygon with a line going through its center, you can move L (red dot) and see that the areas stay constant. Though there is no central symmetry, but the swept area is kept. Also, you can see that it is concave. if you look really closer, it is in fact two copies of the same shape at different scale with central symmetry and one is removed from the other.
The geogebra exhibit about it: www.geogebra.org/calculator/f8b7g5mn
We could call this a "central swept area symmetry". That would be nice to have something allowing to generate such shapes. Note that the shape I used is cylomatic for convenience. it doesn't have to.
@@programaths you did not clearly explain how you linked your lists to a problem that is not discrete. Feel free to use formal writing and arguments if possible
I will check your link on my pc later, it doesnt work on phone
@@holomurphy22 The same way you do for integrals. You make the swept angle smaller and smaller.
@@programaths Alright. I think the final rigorous argument should look familiar with my proof, because only same 'radius' length swiping have the same derivative (thanks to continuous borders). Nice way to do it (a bit less straightforward than what I wrote)
Say you were to bisect an equilateral triangle evenly by area using a line segment, and then you gradually rotated that line segment keeping the area the same. What shape would be inscribed within the triangle by the axis of rotation?
Super interesting. I'm ordering something from the STEM store to become a nerdy member of your engineering community. :-)
and how you make this sort of super animation
,any software recommendation?
This is like the effects of gerrymandering in physics
Damn, interesting, very counter-intuitive.
I started watching this at 4:13pm, and at 4:16pm, my brain exploded