@@standupmathsplease do this for higher dimensions. This explains so much. The projection from 3d to 2D explains the 4D to 3D projections I have seen and was beautiful.
16:42 just to spell out what Grant is saying here, if you calculate V - E + F for a polyhedron that has a hole in it (e.g. if you approximated the surface of a torus with plane faces), then you won't get 2. Instead, you'll get 2 - 2g, where g is the number of holes. So this is a way to formalize the notion of "holes" (since you can just count them via vertices, edges, faces) and prove that the number of holes is invariant with respect to continuous deformations.
6:30 Funny that you demonstrated a simulation of the polyhedra being projected onto a plane, when in fact, due to the nature of them being rendered on a computer, and displayed on a flat screen, they were already being projected onto a plane, just by us looking at them.
The United States education system uses "y = mx + b" for the equation of lines. Also, big fan of the "technically correct if you're a topologist" entries.
@@carolinecowley427 We do, and it's really not that confusing at all. Variables get reused all over the place, it's not any weirder here than when it happens elsewhere. We just don't think about it.
@@carolinecowley427 The distinction I was taught was that "b" was the y-intercept, while "B" was the coefficient of the term with exponent 1. Since they were different looking, "They're different Bs, so they're different values" was easy to accept.
@@sphaera2520, with a, b and c there’s clear pattern. Meanwhile, if you’re using m and b for linea, what’s the clear pattern for going to higher order polynomials?
Why you get lines and not just planes: For any polyhedron with only triangular faces, you have the additional relation 3F=2E (each face touches three edges, and each edge touches two faces). The intersection of V-E+F=2 and 3F=2E gives a line that contains all polyhedra with triangular faces. It just so happened that the only polyhedra Matt used in his visualization were triangle-faced polyhedra and their duals (which satisfy 3V=2E, giving the other line). There are lots of polyhedra that don't lie on either line that just didn't get drawn - but the triangle-faced ones and their duals are definitely quite common! (In particular, every platonic solid or its dual is triangle-faced)
And, just to make one final point clear: The reason that the Triangular-faced objects satisfy 3V=2E while their duals satisfy 3F=2E is that the switch between duals swaps the number of vertecies and number of faces. This also explains why the tetrahedron and square pyramid (any pyramid really) is on the line of symmetry between the groups: That line is F=V, since mirroring at that line is how you swap number of faces and number of vertices. A pyramid is always on that line since pyramids are self-dual! So they must have identical face-count and vertex-count! This brings up the question: are there other self-dual polyhedra? I don't know and I really shouldn't get into the Geometry Wikipedia rabbithole at 3 in the morning.
OK, so the two diverging lines are because of triangle shenanigans, but what about the center line? Are all duals, even non-triangular ones, at reflections of each other across that center line? And if there is a universal center line, what IS the center line? I would think it’s V=F, since duals swap faces and vertices; is that correct?
@@mathcookie8224 That's exactly right: the center line is the V=F line, and every dual is given by reflection across that line because dual corresponds to swapping the V and F coordinates.
@@walterkipferl6729 Good clarification! Also worth noting that the number of vertices on each face turns into the number of faces touching each vertex in the dual. So while one line contains all the "every face is a triangle" polyhedra (tetrahedron, octahedron, icosahedron, etc), its reflection contains all the "exactly three faces meet at each vertex" polyhedra (tetrahedron, cube, dodecahedron, etc). And yes, there are many other self-dual polyhedra that can be easily found in the geometry wikipedia rabbithole.
Also, there are polyhedra that satisfy V=F but are NOT self-dual. For example, you can start with a cube and draw two new edges coming out of one of the vertices. The result has 8 faces (4 squares and 4 triangles), 14 edges, and 8 vertices. The dual has 8 faces (5 triangles, two quadrilaterals, and a pentagon), 14 edges, and 8 vertices. These are clearly not the same, so you get two distinct polyhedra occupying the same point (8,14,8) on the V=F line.
Ask any good programmer and they'll tell you there's no such thing as two - the only numbers are zero, one and infinity. Two is just a special case of infinity. 😁
I've not seen the livestream, but I'm pretty sure that's why they did it that way. And I can say with a relatively high certainty that the conversation went like this: "Make a tetrahedron with 3 blue and 1 brown face!" "That's a great idea. But wait, we don't have any brown tiles. We'll use gold, it's close enough." *makes the tetrahedron* "Here we have it. 3 Blue, 1 Gold"
I feel like the easiest shortcut to understanding the "why" of the symmetry of duals is that a dual is very much by definition what you get if you swap the things being counted by two of our three variables for one another (while keeping the thing counted by the third constant...)
The dual line is easy to explain. One shape and its dual are reflections of each other along the line. That is because when making a dual shape, each Vertex becomes a Face, each Face becomes a Vertex, and each Edges just changes orientation. So reflections of the line is just swapping the V and F.
LOL In germany schoolsystem is a mess. It is "länder"-specific, so in saxony you have other standards than in bavaria for example 😂 One other big thing i think are the axes. I heard in some regions at school they label the axes x1, x2 and x3. We always labeled them x, y and z (probably mathematicly x1,x2,x3 makes more sense but maybe it's easier to get confused too idk ‾\°°/‾)
From my experience in the Netherlands we use "y = ax + b". Nice and clear that we use the first two available letters for unknown parameters, so I thought everyone did. Then I saw you use "m" and I just felt sorry for 14-year-olds learning Newton for the first time.
I learnt both in the UK, (m,c, and a,b). I don't actually remember when but ax+b turned up later, possibly at uni, and I wouldn't go for it naturally. I do prefer it though.
In france we also use ax+b and for polynomials you just add new letters in alphabetic order e.g ax²+bx+c or ax³+bx²+cx+d. I logical and it old itself up when integrating and derivating.
Given how prevalent TI-80-something graphing calculators are in the US, I'm surprised we haven't seen a shift from y = mx + b to y = ax + b, since that's how those calculators have always presented it.
In his memoir, mathematician Goro Shimura says that he once set an exam question for a student who was trying to transfer from another university, which went something like this: Find the equation of the line in the plane that passes through the points (1,5) and (1,2). He wanted to see if the student would blindly use the formula y = mx + c. The student fell into the trap and then complained about being tricked.
m+c results in being both 2 and 5, which is impossible for the equation. But if you think about it a little, or even plot them, you see the obvious solution with m=oo.
@@garr_inc Or you use the generalized formula for a line ax+by+c=0. Yes, the constants are equivalent up to a non-zero scalar multiplier, but it's symmetric in the variables and can represent any line without infinities. It can also represent lines at infinity, which is nifty. See projective geometry.
Here in Switzerland we used a multiplicity of letters for the line: ax+by+c=0 or y=ax+b or y=px+q or y=mx+h or y=px+h were all things i encountered in my education. I believe the goal was to teach us that the letters didn't really matter. Also, since Swiss education is very decentralised and each teacher can more or less choose the material they want to use i wouldn't be surprised if elsewhere in Switzerland they would use completely different letters.
Sweden uses y =kx + m, though I think that’s just because k-value (Swedish: K-värde) sounds better in Swedish than a lot of alternatives I’ve seen here
@@pyramear5414 "Constant" is spelled "konstant" in Swedish so here it really should be "y = kx + k". Guess it would be a bit crippling for our mathematicians...
In Germany we have Different Letters vor y=mx+b ==>(m,b). So we also use (m,n),(a,b),(p,q),(m,k). In A-levels it's common using m for the pitch. It depends on the teacher and also the schoolbooks they use.
im so proud of myself, i knew nothing about this before the video, never even thought about arranging any polyhedra or anything, and when you were saying "well,, what different ways can we arrange them" i said... "i bet the euler characteristic is what makes it a plane"
In Austria (not Australia) we typically use f(x) = kx+d for linear functions. I assumed this was the same in Germany but as other comments have shown me it isn't! Very interesting
I would have been tempted to submit my favourite shape: 7 triangles making up a torus, but that would have been disqualified as it has Euler characteristic 0. (7 vertices, 14 edges, 7 faces), and hence, not on the plane. I remember tinkering with an early version of Mathematica for hours to get an R^3 embedable 7-triangle torus. But as an ex-topologist, I do agree with the "off the scale" submissions. Two sides faces, vertices with just two edges, or multiple edges between pairs of vertices, nothing wrong with that. As for the proof of the Euler characteristic being a constant (for planar graphs), instead of starting with a spanning tree, you can start with just a single vertex (V = 1, E = 0, F = 1), then add edges one by one, in such a way the graph remains connected. Each edge either adds a new vertex (in which case, V := V + 1, E := E + 1), or connects two existing vertices, adding a face (in which case E: = E + 1, F := F + 1). In either case, V - E + F remains constant.
I once made a graph of how to transform between polyhedronae using simply moves like: "corner cutting", "edge cutting", "vertex expanding". All very nice when animated.
13:00 In Hungary, in 5-6th grade, we learn it like "y=ax+b" but later, in high school (9th grade and up) we use "y=mx+c". We often use 'm' as slope, and 'c' as a constant, for moving the graph up and down.
y = mx + c for Australia, however I use y = zx + c for my physics classes as m is for mass, and we do a lot of topics where you are trying to solve for mass from a gradient of an experiment and students writing m = f(m) is problematic. Z doesn't get used (no 3d vectors at high school) in any equations in our formula book so that's our side step!
Agreed, 2000s high school planar mathematics was y=mx+c. When I got to unii the tutors always used to write z=ax+b. Their reasoning was that m is for mass, c is the speed of light and z is the vertical plane. I still use it z=ax+b now because I've ended up a place where I'm doing calcs with masses and vector-forces and need variables that represent what is actually being input/output.
Spanning tree is a term well known by network engineers. There is a "spanning tree protocol" which ensures your network does not have any loops, independent on how you interconnect everything. The network switches just "figure it out" (if you have loops in your network, everything just breaks down (you can have something called "broadcast storm")
16:00 my artwork is above your hand (but in the background) - made my day to see it make a cameo since it was inspired by watching another of your videos!
good to know that i can always cut a sandwich made of polyhedra plotted by number of faces, edges, and vertices, no matter how many ingredients i add, perfectly in two! also interesting how matt went with the 3 blue 1 brown tetrahedron instead of the parker cube (a 3d solid with parker square faces)
Sweden: in elementary school it was definitely y=kx+m but then in later parts of high school and at university I think ax+b was pretty common to be consistent with polynomials of arbitrary degree (ax^2+bx+c, etc)
I just learned about spanning trees for the first time 2 weeks ago. I thought it was cool but couldn't understand how it would ever be useful. I'm amazed.
Would you really be that surprised if you found out Glen did a bit of math(s) on the side? Cooking, flying planes, video/film making… just another hobby?
Because all shapes are liars, Matt! Had a great time seeing you in LA, by the way! I reference that software engineer joke all the time now and it's glorious. I kind of wish I had that slideshow :)
Here's a nice related result: For a polyhedron (e.g. a cube), at each edge we can define an angular deficit, being 360 degrees minus the angles of all the polygon vertices which meet there. E.g. for the cube, each vertex has three squares, each of which have 90 degree angles. So the deficit is 360 - 3 x 90 = 90. Now calculate this deficit for every vertex of the polygon, and add them up. In the case of the cube, there are eight identical vertices, so the total deficit is 90 x 8 = 720 degrees. Consider a regular triangular prism. Now each vertex has two squares and a triangle, so the vertex deficit is 360 - 2 x 90 - 60 = 120. There are six vertices, and 6 x 120 = 720. For any polyhedron which obeys Euler's polyhedron formula (i.e. no holes) and has plane faces, the answer is always 720 degrees. I leave the proof as an exercise for the student, but leave the hint to use Euler's polyhedron formula. It isn't difficult. I'm pretty sure, but haven't proved, that this extends to continuous surfaces: at every point there is a curvature. Integrate the curvature over the surface, and you'll get 4 pi (720 degrees in radians.) (Assuming your surface is embedded in Euclidian space and is topologically a sphere.)
The equation of a plane is ax + by + cz + d = 0 Euler's formula can be thought of as a plane equation( where a,c = 1 and b = -1 and d = -2) if x,y,z repressent the vertices, edges and faces of a polyhedron. Which is exactly what Matt shows in the video
The dual of the beachball is a sort of puffed up pillow with two nonagonal faces with nine edges, trying to picture the dual of the over-verticed tetrahedron...
It would be an over-faced tetrahedron with the standard four vertices. Each triangular face would have multiple copies bulging above and below the plane of the three vertices it connects to. EDIT: turns out I wasn't quite right about this. Thank you Mr Goomba dude.
I'm honestly confused about how you even define a face in this context. I'm used to assuming that a face is a flat polygon, and the 4 "faces" of that jagged tetrahedron aren't flat
My guess before watching the full video (around 4 minutes): All polyhedra, when squashed, are planar graphs, thus v - e + f = 2 applies and defines a plane.
I think a variation of Euler v-e+f is to include the null face and the whole. Giving -1+v-e+f-1 = 0 in 2d polytopes this works as well -1+v-e+1 for the pentagon is -1+5-5+1 = 0 Also works with all dimensions.
That line at 19:42 reminds me of the elemt table and their isotopes. The further you are away from the line, the more likely it is going to be an unstable isotope.
UK, 1970s, y = mx + c. Obviously c stands for "constant", but I honestly can't remember whether any justification was given for the use of "m", nor what it actually was. Conceptually, I think I would prefer y = a + bx because I like the idea that you start from a fixed point, and THEN add a variable thing. Others here have also pointed out that this generalises more naturally for polynomials (e.g. y = a + bx + cx^2)
@@mehill00 I haven't heard that explanation before, but ax+by+cz=d is a common equation for a plane. And x=x0+ta, y=y0+tb, z=z0+tc is a common parametric representation of a line in 3D space. It's not exactly a 1-to-1 comparison (multiplication instead of addition) but I could see where someone could have associated b with y-intercept and then decided to use a different variable for the slope.
@@mehill00 He means the more general equation y = m(x - a) + b, which does make the b make a little more since. As he points out, a controls the x axis and b controls the y axis. edit: Never seen a c for the "z axis" but if you wanted to go into a third axis, I guess +cz would do it.
@@kindlin I follow the logic. I was curious whether this was pattern recognition, speculation, or based on some historical knowledge or source. It’s one thing to say this is plausible, perhaps very plausible, and it’s another to say this is the known historical reason.
Since you asked for letters/symbols used in different countries: in Sweden, we use 'k' for slope/gradient and 'm' for intersect. So the line equation would be y=kx+m.
Canadian here (specifically Ontario, if it makes a difference) y=mx+b m means slope, because they said so. b means y intercept, because they said so. Super easy for children to intuit.
I've seen a lot of comments about how the line is defined in the US but a lot of people don't realize that it was recently changed to be 'y=mx+you know the thing'
The way I think about it is V + F = E + C + 1, where C is the number of components. A blank plane has V=0, F=1, E=0, C=0. Adding a vertex adds 1 to V and 1 to C, which keeps the equation true. Adding an edge either connects two components or connects two vertices in the same component. In the first case, it adds 1 to E, and subtracts 1 from C. In the second case, it adds 1 to E and 1 to F. Either way, any addition keeps the equation true.
"Installation" is always a noun, even when it means "the act or process of installing something." Notice the parallelism with e.g. "dinner:" "The installation took three hours;" "Dinner took three hours." (We had clocks, eating which was time-consuming.) Interestingly, -ing words are often both: "driving" is a noun in "Driving is terrible in New York," and a verb in "I can't talk now, I'm driving!"
Here in the czech republic we were always taught that the coefficients of any order polynomial go alphabetically starting from the highest order term, eg: ax^3 + bx^2 + cx + d or ax^2 + bx + c or ax + b
In Sweden se use y=kx+m, seems strange now that I found out that lots of other use m as the derivative… Although, we spell “constant” with a k, so that might be a reason for our choice…
I’m glad my first instinct for the polytopal planar equation was correct. Also, I would like to register a technical addendum. y=mx+c cannot give you the equation of any line, x=6 for example cannot be realized this way, it only gives you all linear functions. I wouldn’t call it a correction, because the video definitely wouldn’t be improved by making the distinction, but it does explain why you have the d value in the plane equation, because d=1 and d=0 are fundamentally different cases as it turns out.
What's up with the captions of this video? We have oiliver, ollier, and olier for Euler. As well as polinomial and plenty other mathematical mispellings..... dual = jewel....
This feels somehow connected to prime numbers. Especially at the end when the central splits off from the square base pyramid. It reminds me of how 2 and 3 branch off into being above and/or below a multiple of 6.
The fact there always is a circle free path through all vertices was new to me. It does not seem obvious, but can probably also be solved be induction.
Denmark does a bit of a mix, which I expect all countries do. We do "y = ax + b" (and for quadratics: y = ax² + bx + c) in primary school, but as you enter university, most tend to move to "y = mx + c"
Huh, that's the exact opposite of how I was taught in the UK, we had y=mx+c up until Year 13 (age 17-18), then it switched to y=ax+b at university. Wasn't until I read the comments on this video that I realised people use so many other options!
Would be funny if someone had sent a polyhedric croissant: basically imagine a croissant turned into ps1 graphics, and the endpoints meet. It doesn't satisfy the euler equation, with an euler characteristic of 3, but its not trivial explaining why (the hole inside is not really a hole like with a polygonal donut, its the shared vertex ruining it all). Polyhedra (1997) by Peter Cromwell explains it so much better than I could in page 209, great book thoroughly recommended
To generalize it. If wi have a sequence of edges linked with vertexs, we have Edges - Vertex = 1. Once you add one more connection point to build a volumed structure you get Edge - Vetexs = 2. What if we go to hyperspace (e.g. add time dimention) and wrap our shape in time as well? We should have Edges - Vertexs = 3, and so on? Isn't it a good topic for a new episode? Reach me if you'd like to igure this out together
During the tree graph proof I kept thinking to myself: "But what does this tree graph have to do with the polyhedron in the first place?" It wasn't until you filled in the missing edges that I, too, "connected the dots" :)
Master class in how not to get to the point even including footage of walking on the sidewalk for no reason at all. I had math teachers like this in school that seemed to pride themselves on being obtuse and fragmented. Then later someone else would explain whatever it was they were teaching in about three sentences and you'd go "Oh, yeah, of course..."
