Mathologer - 2020-07-25
Bit of a mystery Mathologer today with the title of the video not giving away much. Anyway it all starts with the quest for equilateral triangles in square grids and by the end of it we find ourselves once more in the realms of irrationality. This video contains some extra gorgeous visual proofs that hardly anybody seems to know about.
0:00 Intro
0:47 First puzzle
2:24 Second puzzle
3:50 Edward Lucas
4:41 Equilateral triangles
13:15 3d & 3rd puzzle
19:52 30 45 60
29:31 Credits
Here are links to/references of some of the things I mention in the video:
Joel Hamkin's blog posts that inspired this video:
http://jdh.hamkins.org/no-regular-polygons-in-the-integer-lattice/
http://jdh.hamkins.org/no-regular-polygons-in-the-hexagonal-lattice/
There is also a whole chapter about all this and much more related maths in his new book
https://www.amazon.com/Proof-Mathematics-Joel-David-Hamkins/dp/0262539799
Here is another really good article which includes a nice characterisation of the triangles that can be found in square grids plus a very good survey of relevant results:
Michael J. Beeson, Triangles with Vertices on Lattice Points, The American Mathematical Monthly 99 (1992), 243-252, https://www.jstor.org/stable/2325060?seq=1
Scherrer's and Hadwinger's articles:
Scherrer, Willy, Die Einlagerung eines regulären Vielecks in ein Gitter, Elemente der Mathematik 1 (1946), 97-98.
https://tinyurl.com/y45p64t7
https://gdz.sub.uni-goettingen.de/id/PPN378850199_0001?tify={%22pages%22:[101]}
Hadwiger, Hugo Über die rationalen Hauptwinkel der Goniometrie, Elemente der Mathematik 1 (1946), 98-100.
https://tinyurl.com/yx98kkqt
https://gdz.sub.uni-goettingen.de/id/PPN378850199_0001?tify={%22pages%22:[102],%22view%22:%22info%22}
Another, nice paper on rational (and algebraic) cosines
https://arxiv.org/pdf/1006.2938.pdf
Here is a solution to the first puzzle (one way to find the general formula):
https://nrich.maths.org/657/solution
The music in this video is by Chris Haugen, Fresh Fallen Snow (playing in the video) and Morning Mandolin (for the credits)
A couple of remarks:
1. Probably the simplest way to deduce the sin and tan parts of the rational trig ratio theorem is to realise that they follow from the cos part via the trigonometric identities: sin(x)=cos(90-x) and tan^2(x) = (1-cos(2x))/(1+cos(2x)). Note that the second identity implies that if tan(x) is rational, then cos(2x) is rational (if tan(x)=c/d, then tan^2(x)=c^2/d^2=C/D and cos(2x)=(D-C)/(D+C)).
2. Bug report.
a) Here I redefine cos(120◦) = 1.
https://youtu.be/sDfzCIWpS7Q?t=1362
Remarkable :(
b) This transition to the good stuff I clearly did not think through properly.
https://youtu.be/sDfzCIWpS7Q?t=1018
It's possible to make this work for all regular n-gons. There is only one complication that occurs for n's that are of the form 2 * odd. For the corresponding regular n-gons, if you pick up the edges in the order that they appear around the n-gon and assemble them into a star, things close up into (n/2)-stars. For all other n, things work exactly as I showed in the video. Having said that you can also assemble the edges of one of the exceptions into stars. Have a look at this https://imgur.com/68A3fEe and you'll get the idea. Anyway lots more nice side puzzles to be explored here if you are interested :)
Enjoy!
Burkard
Two ways to support Mathologer
Mathologer Patreon: https://www.patreon.com/mathologer
Mathologer PayPal: paypal.me/mathologer
(see the Patreon page for details)
14. Sep. 2021: Thank you very much Michael Didenko for your Russian subtitles.
That “hexagon exists in a cubic lattice” is why two sorts of crystal lattices in chemistry are identical. I can’t remember which ones, but I think it’s hexagonal and either face-centred-cubic or body-centred-cubic.
Also there’s both tetrahedra and octohedra within a cubic lattice, which tesselate with each other in 3D space.
The way I’d look at that initial problem, finding equilateral triangles in a cubic lattice, is that all points in a cubic lattice are either 1 or sqrt(2) from their neighbours, and an equilateral triangle needs a sqrt(3) in there. Plus or minus an inverting scale-factor. But on a cubic lattice, the distance between diagonally opposite points is sqrt(3). Not exactly rigorous, but intuitive to me.
This is also why the two obvious ways to pack spheres are the same.
You mean this is not all hypothetical and pointless mental bubblegum, but I might stumble into actual real life goo like dark matter playing with this?!
For the ones I think you are thinking of, not quite. They are not entirely identical, but have significant similarities.
Face centred cubic is quite similar to hexagonal close packed.
They both contain a hexagonal arrangement of atoms (like the hexagon shown, if you remove some of the atoms to make it face centred cubic). The difference is the shifting between layers.
When you go from one layer of hexagons to the next there are 2 ways to shift.
Hexagonal close packed shifts back and forth (i.e. it shifts one way, then the other) to give a layer arrangement of ABABAB...
Face centred cubic shifts the same way continually to get ABCABCABC...
This makes them different structures.
The other thing you might have been thinking about are the less symmetric ones being equivalent. For cubic lattices, there is primitive, body centred and face centred, and these are distinct.
But for tetragonal (where the cube has been stretched along one axis) there is only primitive and body centred. The face centred system is equivalent to the body centred one.
As for the size of the grid, for the square lattice you get sqrt(j^2+j^2), not just sqrt(2). But the same kind of argument might hold. There is no way to make the sqrt(3) or scaled version of the grid sqrt(j^2+k^2) as that would require j^2+k^2=3.
You're the Bob Ross of mathematics.
@Mathologer
I made a formula for the square problem (using only pen and paper):
(1/6)*(-(n^4)+5*(n^3)+4*(n^2)-2*n)
Here are two links to the two pictures of my proof if someone bothers:
(I did not bother writing an detailed explanation on how I got to where I was in the beginning because it was just meant for myself, but its based on two key observations. The first observation is that for a n*n grid, it is exactly n different types of squares that can be fitted in the n*n grid that needs at least a n*n grid to exist (can easily be observed geometrically). The second observation is that a n*n grid can fit exactly (x-n+1)^2 times in a x*x grid when x is bigger or equal to n.)
https://ibb.co/BnPNXpj
https://ibb.co/C5s6C7k
Wow. That is high praise.
Except there's no happy little mistakes. All mistakes are punished by false
@ReznoV Vazileski v
This is a happy accident, right? WRONG.
Now this is quality entertainment.
The proof were quite elegant and unique.
Well, after that one, I (probably irrationally) suppose mathematicians tend to avoid too much sun exposure cos tan is a sin.
@SenselessUsername you're*
Lmao
lel , you've now got yourself a t-shirt
@Pranav Limaye *your
@Pseudotaco he has edited the comment since then, yes. "Your" is the appropriate word now
Mathologer often uses the expression “mathematical spidey sense”. He is right. Math is not invented or discovered,; it is sensed.
Intuitionists agree, and 'sensed' is as good or better word as 'intuited'.
I certainly feel that way about geometric proofs, but not so much other maths!
Hmm, you missed the point Charles, driven by idealism
Hi... I hope nobody here eats "Peanuts"... ;-))
@ZEE IBIT I admit my comment was deliberately designed to be provocative by making a categorical statement about a controversial topic. Good grief! There is a special place in heaven where God will reveal the truth to mathematicians.
The regular polygons you can fit inside a 3D grid are also the regular polygons you can use to tile a surface. Coincidence?
No. Tiling is basically just shifting. Since the whole shift argument proved that only triangles, squares, and hexagons work and that nothing else does, that also means that only triangles, squares, and hexagons can tile the plane and nothing else does.
@Nana Macapagal Well, I see how shifting plays a part in both, but the kind of shifting is different in both cases; in the stuff explored in this video, we shift sides to get new points with integer coordinates, but in tiling, we shift entire shapes as a whole (so there's no scaling going on), and moreover, tiling has nothing to do with integer coordinates.
I think it's just a coincidence of small numbers, because that's exactly why we can tile a plane with equilateral triangles and hexagons in the first place; the internal angle of a regular n-gon is given by π(n-2)/n, and that just so happens to be of the form 2π/k for some integer k when n=3,4,6 and never for any other n > 2, because of how small numbers work.
I was about to say that there are no coincidences in maths, but then I realized that Gödel's incompleteness theorem proved me wrong.
What if instead of square grids we would have penrose aperiodic tilling?
@The Religious Atheists Tiling is very similar to shifting - to "stack" an identical shape next to another one, you just shift the corresponding line segments where they need to be
I think the shirt in this might be one of my top five favorite shirts I've seen him wear so far.
Impossible triangle made of rubik's cubes, perfect hexagram in the middle, and it almost looks 3D.
This shirt is a winner.
Also a tribute to MC Escher.
Steve Spivey,
would there actually be a dodecagon in the middle of a Penrose triangle if they can’t exist without violating Euclidean geometry?
I’m genuinely asking (not trying to be sassy). I won’t pretend that I know any theory behind what would be at the center of a penrose triangle lol
I was more just casually referring to visual sensation of a hexagram in the middle of the shirt, in my original statement.. if that makes sense 😂
I appreciate your response and respect for mathematics.
@Cassie D A hexagram has 6 intersecting lines, arranged like 2 triangles, not just a 6-pointed star shape (which has no intersecting lines).
On his shirt is a negative-space with 12 sides, making a non-convex dodecahedron, or 12-gon, not a hexagram.
Similarly, a pentagram only has 5 (intersecting) sides, but a 5-pointed star is a decagon, it has 10 sides.
Sorry, just the mathematician and word nerd in me coming out. I know, it's a dangerous combination.
https://en.wikipedia.org/wiki/Star_polygon#Interiors
Steve Spivy,
I found this link and I think I'm starting to (maybe) get close to seeing what you are talking about :O
https://www.thingiverse.com/thing:6513 (pics 7 and 8 are the best) / https://etc.usf.edu/clipart/42800/42838/12_42838.htm
Wish I could find some sort of 3D simulation to rotate..
I'm totally wiggin out on trying to find this dodecagon lol
If you have any helpful visual references or links I would be most appreciative.
Also dangerous to the Borg.
This is so freaking cool I got goose bumps! Sad to sat I can't do any math in my head but had no trouble understanding everything. Beautiful presentation.
2:15 Let (a, b), where a≥1 and b≥0, be the lowest side of a square written in vector notation. Precisely, it is the side containing the bottommost (and leftmost in case of a tie) vertex as its left endpoint. This enumerates all possible squares uniquely. There is room for (n - (a+b))^2 such squares in the grid, so the total number of squares is (A2415 on OEIS):
sum[k = 1 to n-1] sum[a+b=k | a≥1, b≥0] (n - k)^2
= sum[k = 1 to n-1] k (n - k)^2
= n^2 (n^2 - 1) / 12
????????????
15:32, it does have a turning property but you must rotate around a cardinal axis. This does mean it is useless for finding other grid points though.
But probably with worse wording :)
But when all you have is just two points, how do you determine where's the axis and what is a grid unit?
@Nikita Kipriyanov we would know that a grid unit is no bigger than the distance between the two points, and would be able to turn perpendicular to one of the two points to find other points. infinitely many 3d grids could be constructed starting from just two points, it would just be up to us how we want to make it. If you are trying to match the two points to the rest of an already existing grid that would not be reasonably possible.
Nikita Kipriyanov I agree with Marcial’s reasoning, I mentioned it is useless for finding more grid points is because you can never be exactly sure where an axis is.
@Aidan Christensen @Nikita Kipriyanov I think this can be done without a defined axis. If you rewrite the "rotation" rule to be "take n+1 points as long as n points align with existing grid points without scaling than the other point is also on the grid" where n is the number of dimensions.
The "equilateral triangles" theorem shows also an important poligon property: the only regular poligons with a 90° rotational symmetry are squares
idk why but i love proofs that use recursion/infinity to cause contradictions. just like in the triangular squares vid.
It's because they re beautiful
Another fun but elementary observation: the sines of the "nice" angles 0⁰,30⁰,45⁰,60⁰,90⁰ are √0/4, √1/4, √2/4, √3/4, √4/4. I could never remember what their values were until I noticed that!
@Richard Smith Then you can use parentheses.
@UC9w-TsWWijQSNZ3rkn0L_Dw I think that you would have a hard time, or that it is impossible, to get 60º angle in a square grid as much as it is impossible to have a equilateral triangle.
And it would not make any sense if you could have 30º by any means because then you would be able to do the process 2 times and get a 60º angle (i think).
And in 3D you do 1²+1²+1²=3.
Haha, I used to remember that cos is X and sin is Y, imagine a circle with r = 1, center 0;0, draw a line from center with aporox. required angle, till intersection with a circle line, project this on X for cos and on Y for sin, approximate, remember to put square root of two or three into nom or denom if approximate value does not look precise, calc. this approx to find out if I out nom / denom properly, find an answer.
For some reason imaging and calculating this was faster and easier than just remembering the sin / cos values!
Do it the opposite way and you get the cosines.
Vinícius Moretti Divide the sines by the cosines and you get the tangents.
Beautiful proofs! At the start I feared it was just going to be "Slopes in the grid are rational but tan 60 is sqrt(3), irrational"
One bit that you probably know but I'll say it anyways for the group: The way you found a triangle in 3d is a special case of a more general construction for simplexes. You can always find a regular N-simplex in an N+1 dimensional grid by labeling one point as the origin and taking (1,0,0...), (0,1,0...), (0,0,1...), etc as your vertexes. For instance, here are the vertexes of a regular tetrahedron in 4-space: {(1,0,0,0), (0,1,0,0),(0,0,1,0),(0,0,0,1)} By a symmetry argument, all sides are the same length, all faces are congruent, etc.
Nice video! Q: can a /3-D/ equal-sided simplex (e.g. tetrahedron) fit, like the 2-D equilateral triangle does?
Had to pause so many times just to truly appreciate the beauty of this visual proof. So much to reflect on.
There's a simple process to get the first step answer without having to count all of the boxes. The pattern is made up of 4×4 rows of boxes that equal 16 boxes in total. Divide 16 in half to get 8. Divide 8 in half to get 4. Divide 4 in half to get 2. 16+8+4+2=30
@acetate909I don't think there is a relation to powers of two but it's a nice outcome. I've found the general formula for an n×n square to be (n-1)n²(n+1)/12, it's a surprisingly nice looking formula.
It's just coincidence. Take a 9x9 grid. (Your method only works for grids of length (2^n)+1).
Then 64+32+16+8+4+2 = 126 and 64+49+25+16+4+2+1 = 168.
Don't claim to have found something, if you haven't checked it on one (better 3) example(s).
@Faris Saadat I think this formula isn't correct. n=1 ; (1-1) x (other part) = 0 but a 1x1 square has 1 square.
The correct formula is Sum 1;n (k²) = n(n+1)(2n+1)/6
@Teumii, when it is 1×1 then there is only one grid point, so you get no squares. You need four different points to form a square as the video has shown. I am sure the formula I have written is correct because I have checked OEIS and one of the definitions for the formula is the same as the square counting one, you can see it here: https://oeis.org/A002415.
@Faris Saadat Ok, i've got it, you're including the tilted ones.
I found the same formula except i used a different definition of a nxn square (a nxn square is made of n² 1x1 squares)
For the 2d equilateral, Wouldn't it be a lot faster to just prove that sqrt(5) is irrational and that any conceivable point C you could reach from two points A,B would have a rational distance from the center of the line AB? Because the normalized length of AB will always mean that the length of center(AB)->C will be a factor of sqrt(5) times that length.
There are many ways to prove these results, some of them just as pretty as the ones I present in this video. I don't think that proving the irrationality of root 5 from scratch is any "faster" that what I show here though :)
@Mathologer hey, thanks for replying so fast, I'm honored! I like the video and the visual proofs :) one day I'll be asking for your help to make a video about my own work, if you have any love for differential geometry at least ;)
Both Mathologer and 3b1b are great and I love their visual representations!
About triangles and grid:
1. Imagine all points of that grid having integer coordinates
2. Let's imagine that the equilateral triangle fitting that grid exists and the first triangle's vertice having coordinates of (0, 0).
3. The second triangle vertex would have coordinates (a * cos α; a * sin α) where a is the length of the triangle's side and α -- the angle of between horizontal line on that grid and triangle's side between first two vertices.
4. Assuming the triangle fits the grid, the equations a * cos α and a * sin α are both integers.
5. The third vertice would have coordinates (a * cos (α + π/3); a * sin (α+π/3)). Its X coordinate is a * cos (α + π/3) = a * cos α * cos π/3 - a * sin α * sin π/3 = 1/2 * a * cos α - √3/2 * a * sin α.
6. Since a * cos α and a * sin α are integers, 1/2 is rational and √3/2 is irrational, the third vertice's X coordinate is irrational.
7. Our assumption is incorrect and such triangle does not exist.
@Пётр Зайцев I agree, it's true. But the proof holds for integer coordinates only, right? What about non-integer grid?
@Alexander Goomenuk, my proof would also work for the equilateral triangle with rational coordinates.
Just replace all occurrences of an "integer" term by the "rational" (across all the proof text), and the statement 6 would still apply.
@Alexander Goomenuk, or, if you mean that the distance between two neighbour grid points could be non-integer, we assume that distance as 1. If not 1, it could be reduced (i.e., construct a new coordinates system over the existing one).
@Пётр Зайцев yes, of course. That's correct as well. Thanks for the explanation!
Got inspired by an orchard problem: https://www.youtube.com/watch?v=p-xa-3V5KO8. I too start with one point (A) situated at the origin of a coordinate system. Then the segmentline to the next point (B) with lengt IABI equals N times a unit vector. Point B will have coordinate (0,N). The last point (C) will then have coordinate (N/2, N/2*sqr3). The line with point A and C as endpoints will have rise sqr3. But if point C was on a gridpoint it would have a rational rise.
Very interesting !
Let's try this :
Assuming there's really a Planck length in the Universe and you work with real world coordinates, you couldn't keep shrinking the polygons forever without them converging to a single point. Would that mean there's no such thing as a square grid in the Universe or that triangles are not a thing?
@fede Be careful what you wish for. https://www.wolframphysics.org/technical-introduction/
Wolfram index of Notable Universes https://www.wolframphysics.org/universes/
fede Mother most likely Planck length tiling is not regular. Mass distorts energy density distribution that should also distort distance measurement.
There certainly is a quantity which is called the Planck length; the question is whether or not space is discrete with minimal length the Planck length.
@drdca With all those quasars and supernovae, the Universe isn't quite discreet in deed ;-)
chtoffy made me worry for a moment that I used the wrong homophone, haha
Turning property can be thought of intuitively if you rotate the entire 2D plane 90 degrees clockwise. Each time you do this you are essentially rotating the line counterclockwise. You can do these turns 4 times, each with 90 degrees before ending up at the original shape. This also explains why the turning property doesn’t hold in 3D, you can’t rotate the figure the same way with with the grid staying the same.
14:53 after that point I expected a 4D geometrical animation and explanation. Also, I like how people and even the smartest mathematicians are calling everything "irrational" when it comes to the decimal points.
Among other things, your videos prove the beauty and elegance of mathematics.
Fantastic lecture. Btw no equilateral triangle actually follows quite quickly from picks theorem & area =(1/2)s^2sin(60)
The issue I have with this in my mind is if we were to imagine an infinitely dense grid you'd have points at all points, a solid opaque infinitely large square, of course an equilateral triangle can fit inside this shape.
Perhaps infinite density breaks something, perhaps I am simply missing something.
I loved these proofs - its a nice idea to have in mind! I wonder if there are other fun ways to prove this - maybe by using the fact that the square grid is bipartite or by looking at homomorphisms of Coxeter Groups!
Edit: runways to fun ways
Very nice video. Excellent work, as always.
Proving that there are no equilateral triangles in 2D square grids doesn't seem hard. Basically, if there's an equilateral triangle, you can split it in half to get a 30-60-90 triangle whose sides are d, 2d and d*sqrt(3). A little bit of similar triangles shows that this leads to sqrt(3) being a rational number, which is false, hence no such equilateral triangle exists. But this proof doesn't generalize to other polygons or other dimensions.
I remember asking myself the exact same question (which regular polygons can be embedded into the grid) long time ago, but I didn't find a proof. Really nice to finally see a proof, and what a beutiful one!
14:40 That's the first thing I thought of. The silhouette of a cube viewed at an angle is a hexagon. So, if you were to squash down a cube from opposite angles, you'd get a hexagon and equilateral triangles. That'd kinda be cheating but it's probably considered true in a non-euclidean way.
Truly amazing, as always. Thank you so much for making these videos, I always get so happy when I see a new one in my subscription feed!
The 3D grid does still have the turning property. In 2D, you need 1 angle. In 3D, you need 2 (or 1 angle about an axis). The reason it appears to fail is that you are choosing an arbitrary axis, or equivalently not turning the appropriate amounts for the 2 angles.
It also has more options.
You can turn it 90 degrees about any of the 3 principle axes.
Mr Mathologer, do you know any visual proof of Pick's theorem?
Using that, it's easy to prove that no equilateral triangle fits into a square grid:
Let's suppose that one of these triangles exists. By Pick's theorem, the area of any polygon with its vertices on grid points must be n/2 for some integer n. (For the equilateral triangle, this is also easy to see by inscribing the triangle into a rectangle and then subtracting three right triangles with integer legs.)
But if we name "s" the side of the equilateral triangle, then its area is (√3/4)s². As s² is integer (because it's the Euclidean distance between two grid points), and using the fact we previously saw, then (√3/4)s²=n/2, meaning that √3 is rational.
As always, thanks for your amazing videos!
I love how the ever shrinking polygons is a great analogy to their irrational coordinates.
12:58 At this point I thought "Why not just use triangles to do the argument?" so I've tried to do it, but then realised that the triangles, when you apply the rotation, actually grow in size rather than shrink, so the infinite descent argument won't work here.
21:00 I guess I'll prepare for that video more.
I cannot express how much I like this channel... the equilateral triangle proof was amazing and reveals some really important aspects of general problem solving. Genius!
Is there also a pattern to which angles give algebraic numbers when trigonometric functions are applied to them?
I always learn something new. I love the fact, you use diagrams stuff like that instead of complicated algebra equations.
Knowing something 'in your bones' but then seeing a visualisation of a mathematical proof by contradiction just makes me so happy. Thank you for this.
One question came to my mind for the 3D square grid: cubes can be found within that grid. But how about the other 3D n-gons? Tetrahedron, Octahedron, Dodecahedron and Icosahedron? Can that be answered (or is it just trivial)?
There are also tetrahedrons and octahedrons in the 3d grid. If you take every second vertex of a cube, those vertices are the vertices of a tetrahedron. The 6 closest vertices to a fixed vertex of the grid are the vertices of an octahedron. Icosahedra and Dodecahedra are not contained in the grid.
Amazing video, as allways. Thank you so much for sharing those incredible proofs.
The first time I stumbled with that equilateral triangle problem was in my childhood: basicaly, I wanted to draw a pixelated equilateral triangle in paint, for some reason.
When I got a little older and had some contact with mathematics, I managed to formulate the problem in a more rigorous way (some way how its formulated in the video, basicaly), and, eventualy, managed to prove the impossibility using sines and cossines.
Pretty cool to see such a beautiful alternative proof for that.
Needless to say, the proof shown in the second part of the video was also amazing.
I feel like there'd be a connection between calculating an infinitely repeating decimal and constructing an infinitely shrinking polygon. I wonder if that's related to which rational numbers generate those vs. which rational numbers generate repeating decimals in certain bases (i.e. base 60).
13:51 I count 54.
Each 2D plane has 6: 4 small squares, 1 large square, and 1 diagonal square with corners at the midpoints of the edges. There are 3 of these planes in each orientation, and 3 × 3 × 6 = 54.
Intuitively, I felt like there were also 6 more, each with two corners at the centres of opposite faces of the large cube, two corners at the midpoints of two of the edges that link those faces, and each side of the square being the long diagonal of a grid cube. It turns out these aren't actually squares, since one diagonal is √2 times longer than the other.
Aw man, I totally counted those as well and got 72🤦♂️ I can't believe I forgot the diagonals were longer than the edges 😅
I counted 60. Ther are 6 diagonal i n 3d.
@Werni Nah, the diagonal edges are longer than the orthogonal edges, so they are rectangles, I also got 54.
Interesting. The square grid in 2D has some simple symmetry, described by a D(4) group (it has 8 elements). Since we know we're in the plane, we knew how rotate and that gives us a "rotating" property. An infinite lattice also has a translational symmetry and that gives us a "translatin" property.
In the 3D, the same grid corresponds to a well known point symmetry group, isomorfic to S(4)×C(2) group, with 48 elements. This is the most symmetric crystallographic lattice, m3m, for example, NaCl, the well known cooking salt crystal has that symmetry. But the sad fact we don't know around which axis to rotate takes away all perfect symmetry of this.
To me I think about rotation symmetry of 3 folds and 4 folds being different.
Excellent video, as always. You’re a treasure, thank you!
When we go to a 2-dimensional grid to a 3-dimensional one, my first reaction was thinking about 2-dimensional sections. Any polygon on our 3-dimensional grid will be, in particular, inside one of these 2-d sections. What "allows" us to create triangles and hexagons on out 3-d grid is the fact that not all 2-d sections are square grids, in fact "tilted" sections will never be square grids. This is also why the turning property fails in these 3-d grids: had we restricted ourselves to polygons in vertical and horizontal sections, it would obviously work.
On n-dimensional grids, for n>3, I also found it useful to think about sections. I believe that for any polygon in our n-dimensional grid, we can find a "horizontal" 3-dimensional section fully containing it (maybe requiring a change of coordinates). Then the same argument we used for 3-d grids will apply.
I'm confused, what if I take a 5d hypercube, if I pick a random point won't the 5 points connected to it form a regular pentagon?
Edit: Aha, these points are not even on the same plane.
The 5 points connected to it should form whatever is the 4D equivalent of a regular tetrahedron, just like the three points connected to one of a 3-cube's corners form an equilateral triangle, and the four points connected to one of a 4-cube's corners from a regular tetrahedron.
@Zuthal Soraniz the regular tetrahedron plays two 'roles' in 3 dimensions: simplex (the simplest regular polytope) and demicube (what happens when you remove every second vertex from a hypercube and connect the rest).
simplexes and demicubes are generally not the same thing:
2-simplex: regular triangle | 2-demicube: line segment or digon
3-simplex and 3-demicube: regular tetrahedron
4-simplex: regular pentachoron aka 5-cell | 4-demicube: regular 16-cell (which happens to also be the dual of the 4-cube)
the demicubes of 5 dimensions and higher arent even regular anymore
polytopes are weird and awesome. :)
The sides may have the same length, but it will be a very crinkly pentagon!
Triangles are the only shape that is always perfectly flat. This is why milking stools traditionally have three legs: all three will always all be touching the floor, even if it is uneven (unless crazily so). Perfectly smooth floors in cow sheds are a modern thing 😁
I at first also thought your argument worked. But my best way to describe why it fails is the following: from the set of 5 points you described, ANY pair of two has the same distance to each other. However, in a flat regular polygon, each point has a shorter distance to its 2 neighbours than to any other points. That's also why it works with the 3D grid and triangles. With your construction you get a set of 3 points. Pick any one of these and you just get the 2 neighbours in the polygon, but no additional points that would need to be further away.
@bluerizlagirl traditionally these stools often had only 1 leg. This would allow the milker to tilt the seat to the most comfortable position...
https://commons.m.wikimedia.org/wiki/File:Bundesarchiv_Bild_183-33006-0004,_Bauer_beim_Melken.jpg
incredibly simple and beautiful proof using contradiction and geometry. I love proofs that use symmetry.
Is there a video on infinity or infinitesimals? I keep getting myself stuck in endless loops of thinking because of the properties of infinity. I often use 10 to the power of negative infinity in my attempts to understand infinity. For example I explored the theoretical possibility that 3 (0.33333 + infinitesimal/3 ) = 1. However that doesn’t work because nothing is smaller than an infinitesimal and therefore it cannot be divided by three. Is there a way to get a better understanding of infinity and infinitesimals without making my brain implode?
Arthur Maruyama - 2020-07-25
I have made a career of mathematics, but these videos make me feel that childhood joy of mathematics all over. Thank you so much for making these.
Coenraad Loubser - 2020-08-15
It makes me feel the extreme childhood frustration at all the missing blanks that aren't filled in, and compels me to do mathematics and fill them in myself... when I have better things to do, like eating chocolate and sleeping.
Paul the Audacious Bradford - 2020-09-13
Adam Markiewicz
Male or female dog?
Adam Markiewicz - 2020-09-15
@Paul the Audacious Bradford Hmmm... I haven't realized that until now, but man walking with female dog also makes hermaphrodite!
Rock Brentwood - 2020-10-30
Well, I have an interesting problem for you (and you might be able to link it to your line of work): find a geometry for the background space for the narrator's T-shirt which makes all the depicted right angles and straight lines actually so, as depicted. In the answer, it is allowed for the space to be curved, and "straight line" to be interpreted as geodesic.
Now ... a related problem. Look up the flight distances for New York, Miami, Chicago and Houston and treat the flight paths as straight lines. Determine which geometry they fit in. Is it 2 dimensional or 3? Is it even Euclidean?
Do the same, this time assuming all the flight paths have equal constant curvature (with D being the associated diameter of curvature). If the answer to the previous question ("Euclidean, even?") is no, then (a) what's the largest value of D in which the answer flips? and (b) prove that, for flight paths with the curvature associated with that diameter, the cities all fit in a plane.
Repeat the above this time adding Los Angeles as city #5 and then Seattle as city #6 to the list. What geometries do the first 5 cities' distances fit in - as a function of the diameter D? Is there any D in which the cities all fit in 3D Euclidean space (or even 2D)? And is it unique? If so, then do all 6 cities fit in 3D for this diameter? (If not, then prove that the 6th city must diverge as you allow the diameter to approach the unique value determined by the first 5 cities.)
Paul the Audacious Bradford - 2020-10-30
@Rock Brentwood
Oh. OK.