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.
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.
Was thinking something along the same lines (the crystal lattice structure and formation of crystals). Could be important in several fields of science (though I'm sure it's already been discovered and is being used in industry). Also makes me think of Graphene, borophene and Physical Vapor Disposition.
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.
@@sdspivey a hexagram is not the same as a hexagon. A quick google search reveals that it is "A hollow six-pointed star formed by overlapping two equalateral triangle" (en.m.wiktionary.org/wiki/hexagram)
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.
In all seriousness, I can't write German for shit, but in all honesty, a German video would be great. I mean, I can read it and understand German... So listening to math, and in the language of math, German. As is KORREKT.
Dr. No, as he slowly lowers James Bond into a vat of boiling oil: “I’ll release you if you can answer this simple question: How many perfect pentagons can be drawn in a 5D 3x3x3x3x3 lattice of dots?” Bond: “I got nothing.”
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!
Cliff Pickover has tweeted that fact in a nice diagram recently (on July 15th), search for 'Memory aid'. I can't give the link because YT won't let me.
@@nikitakipriyanov7260 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.
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
Just so I am clear... n = the number of dots per side, not the "length". For example, in the original diagram, what Mathologer showed, it could be called a 4 x 4 grid if thinking of lengths, but it's actually a 5 x 5 grid thinking of dots. So, the formula you gave with n^2 (n^2 - 1) / 12 has to be using dots, right? Note: I've not yet checked for derived this for myself.
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.
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.
@@nanamacapagal8342 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.
@@TheReligiousAtheists 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 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!
“But of course close doesn’t win the game in carnivals or mathematics” Analysts: “Allow us to introduce ourselves” ???: “Amateurs” Analysts: “What did you say?!” Numerical analyst: “AMATEURS”
Same. My thought process was that it seemed like a way to construct sqrt(3) as the hypotenuse of a right triangle with integer sides, which is impossible. I haven't gotten it to work so I feel like this thought process is flawed, but it was my first intuition.
update: I am actually right, if there was an equilateral triangle, you could double it and the height will be a line between two lattice points, which will then be sqrt(3) times the original side length of the triangle. Sadly, you cannot get a factor of sqrt(3) by taking lengths between lattice points, hence a contradiction.
If a number is rational, that means there is something you can multiply it by to get an integer. If all numbers in a set are rational, there will be some number you can multiply them all by to get a corresponding set of integers. The square grid in two dimensions is effectively the points in the plane whose co-ordinates are integers (if we make the smallest distance between two points equal to 1). One of the orthogonal distances in an equilateral triangle is irrational, so it can never fit exactly onto a square grid.
NotaWalrus i had the same train of thought, at school at somepoint I probably thought about it, whilst trying to construct an equilateral triangle on a grid
@@Mathologer Sure, just checking us. There goes my hypothesis that you started writing 120 degrees, then realized you'd basically already written it since cos 120 = -cos 180 - 120 = -cos 60, but left a half-written entry which got merged with the entry for 180 degrees.
Some of the resulting forms are very beautiful, independent of their mathematical origins, some fractal reminiscent, some just joyful. Thank you for a beautiful afternoon's half hour.
StarchyPancakes Yea but does this take into account squares tilted at 45 degrees potentially being double counted? EDIT: Read my main comment (not this reply
So far, I've got SUM(i = 0 -> n-1) { (n - i)^2 } for the non-slanted squares. Haven't added the rest yet, but maybe it's not the right way of thinking about it. The non-slanted squares are like a special case of the slanted squares.
@@danielc1112 I got SUM(i=0 -> n-1) { i * (n-i)^2 } because each tilted square can be thought of as having the same dimensions as the non titled square that encloses it. And for each non titled square of side length L there are L-1 tilted squares that have the same dimensions, hence the multiplication of (n-i)^2 by i.
@@strawberryanimations1035Might as well start the sum at i=1 since the term at i=0 equals 0, yeah? I actually got a formula of SUM(i=1 -> n-1) { (n - i) * i^2 } though it's the same thing as yours effectively, due to symmetry.
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.
@@srsr7258 Yes, you are right. I meant rational not integer. If a*cos α equals to an integer number , then 'a' must be a rational number, since value of cos α is rational only for limited number of angles. Otherwise it is multiplication of Q*Q' or Q'*Q', which must result in Z. This is only possible if a = cos α = sqrt (Z).
@@alexandergoomenuk9930 I did no assumption on the triangle's side -- it could be any (integer, rational, irrational). Also, multiplication of two irrational numbers could result in any number - natural, integer, rational, or irrational.
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!
EDIT: this is for the first part of the video, where the squares are in 2D space. I did this when he said to try and come up with an explicit formula for it. I got an explicit formula for the number of squares in a n x n square. For those who don't want to read through how I got it, the formula is (n^4 - n^2) / 12, and it's factored version is n^2 * (n+1) * (n-1) / 12. Now, onto the solution. Before I do, though, I just want to say I will refer to a n x n square not as being a square with side length n, but rather a square with n dots. This is because the sums will be in terms of the side length of a square, since that's just how I derived the formula. Also, "side length" will refer to the number of dots on the side of a square. Lastly, since youtube doesn't allow for LaTeX rendering, I'll have to refer to the sum in a different way. Here I'll use the following syntax, "sum(k=1,n)(EXPR)", where "EXPR" is the expression that the sum is taken over First, I realized that each tilted square has an untilted bounding square around it(in other words, each tilted square is contained within an untilted one). This means that for me to include the number of tilted squares, I need to multiply the number of untilted sqaures of a given size(which we already know how to calculate) by the number of tilted squares inside of it(plus one to include the untilted square itself), then sum the terms up. So, how many tilted squares can be put in an untilted n x n square? Well, n-2. You just need to choose a point on the side of the square that isn't one of the corners. Since there are n points on the side, and since there are two endpoints, the number of points to choose from is n-2. But we need to add 1 to this to include the untilted bounding square itself. If you had trouble understanding why the number of squares is n-1, then I recommend trying it yourself. Make an n x n square, and inscribe as many squares as you can in it, systematically of course. Second, I came up with an expression that would go in my sum. So, where k is the number of times we reduce the side length of the square, the number of squares per value of k is k^2 * (n-k), where k goes from 1 to n. Although this doesn't make much sense, a simple change of variables from k -> n-k+1 yields (n-k+1)^2 * (k-1). Here, k is now the side length of the square. The (n-k+1)^2 is the number of untilted squares, and the (k-1) is the number of tilted squares you can make within the k x k square. Despite the second expression making more sense, the first one it much easier to deal with, so I used that one. Now, I manipulated the sum as follows, sum(k=1,n)(k^2 * (n-k)) = n * sum(k=1,n)(k^2) - sum(k=1,n)(k^3) (distribute k^2 over (n-k) and split the sums) = n * (n^3 / 3 + n^2 / 2 + n / 6) - (n^4 / 4 + n^3 / 2 + n^2 / 4) (convert sum of squares / cubes to their explicit versions) = (n^4 / 3 - n^4 / 4) + (n^3 / 2 - n^3 / 2) + (n^2 / 6 - n^2 / 4) (distribute and group by power) = n^4 / 12 - n^2 / 12 = (n^4 - n^2) / 12 = n^2 * (n^2 - 1) / 12 = n^2 * (n + 1) * (n - 1) / 12 Nice, what a great compact formula for computing how many squares you can put in an n x n square.
a faster proof, on a square grid the area is given by: Area = B/2 + I - 1 but an equilateral triangle area is: Area = L²sqrt(3)/4 where L is a square root of something, given by Pythagoras So, Area must be a rational number by the first formula, but an irrational number by the second formula, proof by contradiction
Pick's Theorem! Very nice idea. You don't actually need the full Pick's theorem, only the fact that the area of any polygon with vertices on the grid points must have rational area, which is pretty easy to show.
Cool! There's one step missing in you proof though. You still need to make sure there isn't some way of having an equilateral triangle with side length that is equal to 2*k*√3 for some integer k. Because then the area of the triangle is 2*k*√3*√3/4=3*k/2 which is still rational (The factor of 2 is needed to ensure that the area is a multiple of 1/2, which we know it must be from picks theorem). The proof still works though, because 2*k*√3=√(4*k^2*3), and 4*k^2*3 can never be written as the sum of 2 squares because it's prime factorization has 3^(odd power). And any number with a prime of the form 4k+3 to an odd power in its prime factorization can't be written as the sum of 2 squares. This comes from the looking at the gaussian primes in the complex plane. 3blie1brown has an excellent video on the topic: ruclips.net/video/NaL_Cb42WyY/видео.html
@@jaapsch2 If a polygon has all its vertices on a grid, then it can be decomposed into a collection of rectangles and right-angled triangles with their bases and heights lying along the grid lines, and thus having integer lengths and rational areas.
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.
For a good 3 mins there I was making equilateral triangles using the 3d grid printed on your shirt (Amazing how well all of those cubes were lined up, kudos to the artist for representing like 5 different things in one brilliant shirt) anyways you were talking about how there is no equilateral triangle while im creating them and wondering WTF, then you went on to the 3d grid and explained to me exactly what my mind was doing and why it works. Awesome vid.
Well it's not going to happen. It's an idea that upon closer inspection is not helpful. You're already watching the English version so what's the point in making a video for you in German, excluding most other people? It would only make minor sense if there were an English video with the same content as well, and this would just be duplication of effort. From the looks of these videos, it looks like quite a bit of effort. Note that the only gain is to allow you and a few like you to watch a video that _you would have watched in English just the same._
The German philosopher Immanuel Kant would agree with you. He thought that all mathematics was not known on the grounds of experience nor was it derived from a definition. The only other option is something similar to what you give: simply “sensing” mathematics (I’m oversimplifying a bit). Gottlob Frege, a German mathematician and one of the inventors/discoverers of formal logic, criticized this idea and tried to prove that all mathematics could be derived from definitions. I think I tend to agree with Frege
Hum. I wouldn't say that. Do we "sense" axioms or do we invent them ? Can we say that higher dimensions exists, that complex numbers exists ? I know that complex numbers can be used to represent 2D points but I don't hold that to be the same thing as them "existing". And if they don't exist then they were invented by mathematicians.
Some of the things I like about this video - Nice music at end - Beautiful mind-blown visual animations (as always) - You made this video! ** Mathologerisation (>
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.
2:15 I thought of an arbitrary right angle triangle with integer lengths a and b, the hypotenuse being the base of a square that may fit several times in the grid, and its right angle aligned with any right angle of the smallest square that contains the grid. for the square to fit in an n times n grid, we have 0
Great video! When you brought up how the only rational trig results are fom fractions of 3,4, or 6 of the full circle I had a grin on my face from then onwards, your "mathematical spidey sense" phrase was very fitting
my answer for the nxn points in a square is: sum of (i*i*(n-i)) from i=1 to n-1 so for 5x5: 1*4+4*3+9*2+16*1 wolframalpha says that can be simplified to (1/12)*(n-1)*(n^2)*(n+1)
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.
Just discovered RUclips and loving these videos....I’ve got a university degree majoring in Mathematics but I’ve been working as an engineer for the last 15 years - bringing back good memories.
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
Well, my thoughts are : there are n² squares 1x1 in a square nxn there are (n-1)² squares 2x2 in a square nxn and so on... there are (n+1-k)² squares kxk in a square nxn didn't prove it but it was intuitive (on paper i guess) so the sum of powers defintively has something to do with counting squares but the "dividing by 2" technique only seems to work with this 4x4 square
@@farissaadat4437 I have no idea. I was just trying to figure out a way to produce the answer in my head, without having to count all of the boxes. I'm an engineering student and I'm not great at math or I would be in a physics program. All I know is that it works, though I have no proof to offer. I was hoping that someone else could explain it to me. As Teumi said, I can intuit this process but I don't know what it means, really.
@@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).
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!
After all this time running this channel I still find it impossible to predict what people will like. Believe it or not but with this one I was pretty convinced that relatively few people would end up watching it :)
i spent way too much time doodling in class on my square grid paper to construct a 60deg angle using just the grid and a straight edge. Though fruitless, it was a fun exercise
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.
@@zuthalsoraniz6764 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... commons.m.wikimedia.org/wiki/File:Bundesarchiv_Bild_183-33006-0004,_Bauer_beim_Melken.jpg
At 6:13 you say that we get triangles arbitrarily close to being equilateral in the square grid. This is clearly true if you are measuring closeness by differences of the angles from 60 degrees. Is it true though if we take different measures of how "close" a triangle is to being equilateral? An example of such a measure of closeness could be the distance from the triangles centriod and circumcenter (or distance between two such triangle centres).
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.
This presentation remind me a experience. At high school my math is bad, at exam my teacher give one question, how much summary number from 1 until 1000, i slove this one with draw a diagram from 1-1000 like stair case , then divide that draw to be a large triangle and many minor triangle, and calculate that's area. And next day, she call me and said to me how dumb i'm, she said i'm not even use right formula and get right answer, said that i'm cheated and give me 0 score i just laugh at thats time. But my economic teacher pass by on righ time and right place like superman, hearing her bit anger he said to me "what have you done?". She explain to him and give my paper to him, then suddenly he said "ok... i get it, leave it to me, you can back now." I saved by economy teacher at math problem, truly i cant forget that. 😂😂😂
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.
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.
I knew that there were no equilateral triangles due to the fact that equilateral triangles always have a multiple of root 3 as their perpendicular height.
@@Mathologer If there are two points that are connected, then the halfway point of a double-scale version of the triangle is also connected, which means a connection exists between two points which are a multiple of sqrt(3). (And, you can't have multiples of sqrt(3) because 3 is not the sum of two squares.)
The square of the length of a segment between two points on the grid is an integer. The area of an equilateral triangle is √3/4 (which is irrational) times the square of the length of a side. But the area of a polygon with all vertices on 2D grid points is an integer multiple of 1/2.
@@Mathologer Corners of tilted figures need to align with dots of the original grid - because of how you construct them. If you leave only those points and "untilt" the figure - those points will form the untilted, again square grid (why? - because of what you have said at the beginning about the tilted squares in the square grid!) with larger spacing between the dots. It means that tilting the figures does not change that much - maybe only the scale of the grid. And the scale does not matter here. Tada! Great animations though, it helps so much! P.S.: The patreon "My Son..." is your son or the patreon's nick? I'm just curious.
Really loved this one! Got my own half-baked proof based on the necessarily irrational heights of integer-sided equilaterals trying to grease my gears as I type :D
Here in the Netherlands they do learn about the double angle formulas (e.g. cos(2\alpha)). Might be interesting to include the higher multiples as well!
I've come back to this one a few times and I think it is one of the prettiest visual proofs on the channel. Second only to the triangle-of-triangles pythagoras proof.
3:12 I thought about this problem when I was in middle-school. I was trying to draw equaliteral triangels using dots on my notebook and but no matter how much close I get, how much points I use there was tiny bit missing. Then I realized a perfect triangels height is square root of 3 times half of its floor. Square root of 3 is irrational so I will never get there. I was really, I mean REALLY dissapointed. (Also there are no perfect hexagons or octagons or pentagons in the grid.) (I am not sure about 12-sided perfect polygon, I will be pleased if someone posts me a proof of that one about whether or not you can do it.) (Okay I watched the video nevermind.)
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?
Since we have (at least) three spatial dimensions, triangles and squares are equally compatible with the Universe having a “resolution”. Non-plane-tiling regular polygons, however, would not be able to exist in the physical Universe.
Is there any reason to assume that if there's a grid, it should be regular? Could be non periodic, or even amorphous. I would like my universal grid in a beautiful Penrose tiling, please.
@@frechjo Be careful what you wish for. www.wolframphysics.org/technical-introduction/ Wolfram index of Notable Universes www.wolframphysics.org/universes/
Iam loving your very entertaining and interessting videos ❤ Your love in mathematics can always be seen in every single topic, you are presenting. Thank you 😌 Ich freue mich schon auf dein Video auf deutsch 😊😊😊
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.
As someone who is considering a career in mathematics, may i ask your occupation?
Mehmed : stochastic processes , statistics and mathematical biology. I work in the tech industry now.
@@arthurmaruyama5331 That sounds very interesting. Thank you for your answer. You are living my dream.
Statistically speaking a man walking with his dog represent a three legs being. (this involves as well the knowlege of statistics as biology)
@@adammarkiewicz3375 this requires (2+4)/2 = 3
You're the Bob Ross of mathematics.
Glad you think so :)
But without the "fro".
happy little polygons :p
Reverse fro
Fro of size zero?
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.
In AI, minimizing the Shannon entropy is analogous to optimal sphere packing in a rectangular object.
Was thinking something along the same lines (the crystal lattice structure and formation of crystals). Could be important in several fields of science (though I'm sure it's already been discovered and is being used in industry). Also makes me think of Graphene, borophene and Physical Vapor Disposition.
Well, after that one, I (probably irrationally) suppose mathematicians tend to avoid too much sun exposure cos tan is a sin.
This is the funniest comment I've seem for months :)
@@Mathologer Pin it!
@@Lavamar yeeaah, *PIIIN IT!!! PIIIN IT!!! PIIIN IT!!!*
@@proto3139 For all values of x, or do you need to be careful about division by zero?
Arghh
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.
Yes its so awesome were did you got it from Mathologer?
It isn't a hexagram, it is a dodecagon, it has 12 sides. Although I would accept calling it an equi- augmented hexagon.
@@sdspivey a hexagram is not the same as a hexagon. A quick google search reveals that it is "A hollow six-pointed star formed by overlapping two equalateral triangle" (en.m.wiktionary.org/wiki/hexagram)
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.
Had to pause so many times just to truly appreciate the beauty of this visual proof. So much to reflect on.
On the rational main-angles in a goniometer.
"At some point I'll do a whole Mathologer Video in German. Promised."
What's a main-angle? I know Haupt und Winkel, but not the combination.
@@pierreabbat6157 It's literally "head-angle" or the same angle found along in an even n-sheet.
Großartig. Burkard Polster in der Kinematografische Meisterwerk, "Mathologer: Jetzt auf Deutsch, kein Englisch".
In all seriousness, I can't write German for shit, but in all honesty, a German video would be great. I mean, I can read it and understand German... So listening to math, and in the language of math, German. As is KORREKT.
There seems to be a little Australian accent in your German 😉
10:50 this is what bond villains see before they die
Dr. No, as he slowly lowers James Bond into a vat of boiling oil: “I’ll release you if you can answer this simple question: How many perfect pentagons can be drawn in a 5D 3x3x3x3x3 lattice of dots?”
Bond: “I got nothing.”
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!
My physics teacher tought us that so we would finally be able to remember those.
This is how they should write them in books for schools.
@@D-Bar I meant √¼, √¾, etc. but I can't type 0/4, 2/4, or 4/4 that way!
Cliff Pickover has tweeted that fact in a nice diagram recently (on July 15th), search for 'Memory aid'. I can't give the link because YT won't let me.
I was taught to sketch an equilateral triangle (side = 2) and divide it in half. You can then see ratios for 60deg and 30deg.
It’s kinda nice for a quick memory trick, but doesn’t really help build intuition. Drawing triangles are much better
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.
I was just thinking of commenting this, lol
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?
@@nikitakipriyanov7260 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.
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
????????????
Just so I am clear... n = the number of dots per side, not the "length". For example, in the original diagram, what Mathologer showed, it could be called a 4 x 4 grid if thinking of lengths, but it's actually a 5 x 5 grid thinking of dots. So, the formula you gave with n^2 (n^2 - 1) / 12 has to be using dots, right? Note: I've not yet checked for derived this for myself.
I can't take my eyes off of your epic Escher rubic cube shirt.
Yes, the star inside (maybe trivial, but like it anyway)
Yeah, I kept missing what he was saying cause my eyes kept getting stuck in a loop on the cube design.
@@invisibledave, infinity is everywhere...
The perspective is false also mathematically w r ooooo n g
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.
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.
@@nanamacapagal8342 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?
@@TheReligiousAtheists 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 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!
“But of course close doesn’t win the game in carnivals or mathematics”
Analysts: “Allow us to introduce ourselves”
???: “Amateurs”
Analysts: “What did you say?!”
Numerical analyst: “AMATEURS”
lol.... nice one
Somehow, my first reaction to the equilateral triangle version was "Huh? Surely there can't be any?"
Same. My thought process was that it seemed like a way to construct sqrt(3) as the hypotenuse of a right triangle with integer sides, which is impossible. I haven't gotten it to work so I feel like this thought process is flawed, but it was my first intuition.
update: I am actually right, if there was an equilateral triangle, you could double it and the height will be a line between two lattice points, which will then be sqrt(3) times the original side length of the triangle. Sadly, you cannot get a factor of sqrt(3) by taking lengths between lattice points, hence a contradiction.
If a number is rational, that means there is something you can multiply it by to get an integer. If all numbers in a set are rational, there will be some number you can multiply them all by to get a corresponding set of integers.
The square grid in two dimensions is effectively the points in the plane whose co-ordinates are integers (if we make the smallest distance between two points equal to 1).
One of the orthogonal distances in an equilateral triangle is irrational, so it can never fit exactly onto a square grid.
NotaWalrus i had the same train of thought, at school at somepoint I probably thought about it, whilst trying to construct an equilateral triangle on a grid
@@bluerizlagirl Jup, that was my first intuition as well and I canceled my studies after just a year.
22:21 cos 120 = -1? wtf?
Just checking whether people pay attention :)
Mathologer Nice “save” from a small mistake in the video.
@@Mathologer Sure, just checking us. There goes my hypothesis that you started writing 120 degrees, then realized you'd basically already written it since cos 120 = -cos 180 - 120 = -cos 60, but left a half-written entry which got merged with the entry for 180 degrees.
That's new angle system, where 120=π
@@alapandas6398 then the "nice" angles would be 20,30,40 and not 30, 45, 60
Some of the resulting forms are very beautiful, independent of their mathematical origins, some fractal reminiscent, some just joyful. Thank you for a beautiful afternoon's half hour.
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!
General nxn grid is: n^2 + SUM(i = 1 -> n-1) { 2 * (i)^2 }, so it's palindromic
StarchyPancakes Yea but does this take into account squares tilted at 45 degrees potentially being double counted?
EDIT: Read my main comment (not this reply
Not sure about it (still haven’t proved it) but the 2 gives the correction of s(n)
So far, I've got SUM(i = 0 -> n-1) { (n - i)^2 } for the non-slanted squares. Haven't added the rest yet, but maybe it's not the right way of thinking about it. The non-slanted squares are like a special case of the slanted squares.
@@danielc1112 I got SUM(i=0 -> n-1) { i * (n-i)^2 } because each tilted square can be thought of as having the same dimensions as the non titled square that encloses it. And for each non titled square of side length L there are L-1 tilted squares that have the same dimensions, hence the multiplication of (n-i)^2 by i.
@@strawberryanimations1035Might as well start the sum at i=1 since the term at i=0 equals 0, yeah? I actually got a formula of SUM(i=1 -> n-1) { (n - i) * i^2 } though it's the same thing as yours effectively, due to symmetry.
So excited when the notification popped out!
Cool Profile Picture like it!! 👌👍
I think Mathologer notifications are the only notifications I get excited about. Every time!
I once used the shrinking pentagon to write a new chapter of the Zahlenteufel from Hans Magnus Enzensberger. Maybe it is somewhere on the net.
Was totally ambushed by the contradiction at the end. Very elegant! Great video. Thanks
I deeply deeply enjoyed the videos on this channel. Watched three in a row and cannot stop. Thank you so very much :)
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.
Why do you assume that a (length of triangle's side) is integer?
@@alexandergoomenuk9930 He didn't - he assumed a * cos α and a * sin α are integers, which are the horizontal and vertical grid separations
@@srsr7258 Yes, you are right. I meant rational not integer. If a*cos α equals to an integer number , then 'a' must be a rational number, since value of cos α is rational only for limited number of angles. Otherwise it is multiplication of Q*Q' or Q'*Q', which must result in Z. This is only possible if a = cos α = sqrt (Z).
@@alexandergoomenuk9930 I did no assumption on the triangle's side -- it could be any (integer, rational, irrational).
Also, multiplication of two irrational numbers could result in any number - natural, integer, rational, or irrational.
Furthermore, this proof also works for rational number while the proving presented in the video does not.
Some carnival guy right now: *scribbling furiously*
Hrm?
Pi He mentioned at an early point in the video that the triangle-fitting thing seemed like one of those impossible carnival games.
@@NStripleseven ok
Some carnival guy right now: "You just spoiled my business!"
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!
Among other things, your videos prove the beauty and elegance of mathematics.
is this a proof by the infinite descent? Nice. Also, 16:42 looks like a rope bridge.... perspective is crazy.
At first I saw an elevator shaft looking up from the bottom, but now I can only see the rope bridge
EDIT: this is for the first part of the video, where the squares are in 2D space. I did this when he said to try and come up with an explicit formula for it.
I got an explicit formula for the number of squares in a n x n square. For those who don't want to read through how I got it, the formula is (n^4 - n^2) / 12, and it's factored version is n^2 * (n+1) * (n-1) / 12.
Now, onto the solution.
Before I do, though, I just want to say I will refer to a n x n square not as being a square with side length n, but rather a square with n dots. This is because the sums will be in terms of the side length of a square, since that's just how I derived the formula. Also, "side length" will refer to the number of dots on the side of a square. Lastly, since youtube doesn't allow for LaTeX rendering, I'll have to refer to the sum in a different way. Here I'll use the following syntax, "sum(k=1,n)(EXPR)", where "EXPR" is the expression that the sum is taken over
First, I realized that each tilted square has an untilted bounding square around it(in other words, each tilted square is contained within an untilted one). This means that for me to include the number of tilted squares, I need to multiply the number of untilted sqaures of a given size(which we already know how to calculate) by the number of tilted squares inside of it(plus one to include the untilted square itself), then sum the terms up. So, how many tilted squares can be put in an untilted n x n square? Well, n-2. You just need to choose a point on the side of the square that isn't one of the corners. Since there are n points on the side, and since there are two endpoints, the number of points to choose from is n-2. But we need to add 1 to this to include the untilted bounding square itself. If you had trouble understanding why the number of squares is n-1, then I recommend trying it yourself. Make an n x n square, and inscribe as many squares as you can in it, systematically of course.
Second, I came up with an expression that would go in my sum. So, where k is the number of times we reduce the side length of the square, the number of squares per value of k is k^2 * (n-k), where k goes from 1 to n. Although this doesn't make much sense, a simple change of variables from k -> n-k+1 yields (n-k+1)^2 * (k-1). Here, k is now the side length of the square. The (n-k+1)^2 is the number of untilted squares, and the (k-1) is the number of tilted squares you can make within the k x k square. Despite the second expression making more sense, the first one it much easier to deal with, so I used that one.
Now, I manipulated the sum as follows,
sum(k=1,n)(k^2 * (n-k))
= n * sum(k=1,n)(k^2) - sum(k=1,n)(k^3) (distribute k^2 over (n-k) and split the sums)
= n * (n^3 / 3 + n^2 / 2 + n / 6) - (n^4 / 4 + n^3 / 2 + n^2 / 4) (convert sum of squares / cubes to their explicit versions)
= (n^4 / 3 - n^4 / 4) + (n^3 / 2 - n^3 / 2) + (n^2 / 6 - n^2 / 4) (distribute and group by power)
= n^4 / 12 - n^2 / 12
= (n^4 - n^2) / 12
= n^2 * (n^2 - 1) / 12
= n^2 * (n + 1) * (n - 1) / 12
Nice, what a great compact formula for computing how many squares you can put in an n x n square.
So happy to see you here today! Yay content!!😊
I got so excited about the final proof ! Thank you very much Mathologer.
This channel is great like its viewers. We always get excited when notification popped out.
a faster proof, on a square grid the area is given by:
Area = B/2 + I - 1
but an equilateral triangle area is:
Area = L²sqrt(3)/4
where L is a square root of something, given by Pythagoras
So, Area must be a rational number by the first formula, but an irrational number by the second formula, proof by contradiction
Pick's Theorem! Very nice idea. You don't actually need the full Pick's theorem, only the fact that the area of any polygon with vertices on the grid points must have rational area, which is pretty easy to show.
that was my immediate thought as well
Cool! There's one step missing in you proof though. You still need to make sure there isn't some way of having an equilateral triangle with side length that is equal to 2*k*√3 for some integer k. Because then the area of the triangle is 2*k*√3*√3/4=3*k/2 which is still rational (The factor of 2 is needed to ensure that the area is a multiple of 1/2, which we know it must be from picks theorem).
The proof still works though, because 2*k*√3=√(4*k^2*3), and 4*k^2*3 can never be written as the sum of 2 squares because it's prime factorization has 3^(odd power). And any number with a prime of the form 4k+3 to an odd power in its prime factorization can't be written as the sum of 2 squares. This comes from the looking at the gaussian primes in the complex plane. 3blie1brown has an excellent video on the topic: ruclips.net/video/NaL_Cb42WyY/видео.html
same idea!
@@jaapsch2 If a polygon has all its vertices on a grid, then it can be decomposed into a collection of rectangles and right-angled triangles with their bases and heights lying along the grid lines, and thus having integer lengths and rational areas.
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.
For a good 3 mins there I was making equilateral triangles using the 3d grid printed on your shirt (Amazing how well all of those cubes were lined up, kudos to the artist for representing like 5 different things in one brilliant shirt) anyways you were talking about how there is no equilateral triangle while im creating them and wondering WTF, then you went on to the 3d grid and explained to me exactly what my mind was doing and why it works. Awesome vid.
Ich würde wirklich gerne ein Mathologer Video auf deutsch sehen. Greetings to Australia!
Great Video!
Na ja! Aber wie lange werden wir warten?
Well it's not going to happen. It's an idea that upon closer inspection is not helpful. You're already watching the English version so what's the point in making a video for you in German, excluding most other people? It would only make minor sense if there were an English video with the same content as well, and this would just be duplication of effort. From the looks of these videos, it looks like quite a bit of effort. Note that the only gain is to allow you and a few like you to watch a video that _you would have watched in English just the same._
@@xbzq pls dawg, don't kill our hope, dude
Who are you again mate? greetings from australia.
@@xbzq I'm not so sure about that; he promised it at 20:58
Mathologer often uses the expression “mathematical spidey sense”. He is right. Math is not invented or discovered,; it is sensed.
I sense a disturbance in the Matrix ;)
The German philosopher Immanuel Kant would agree with you. He thought that all mathematics was not known on the grounds of experience nor was it derived from a definition. The only other option is something similar to what you give: simply “sensing” mathematics (I’m oversimplifying a bit). Gottlob Frege, a German mathematician and one of the inventors/discoverers of formal logic, criticized this idea and tried to prove that all mathematics could be derived from definitions. I think I tend to agree with Frege
@@bwhit7919 When you say you agree with Frege, do you mean you were convinced by his arguments?
phxcppdvlazi I agree with Frege, at least partially. I think that all math is derived
Hum. I wouldn't say that. Do we "sense" axioms or do we invent them ? Can we say that higher dimensions exists, that complex numbers exists ? I know that complex numbers can be used to represent 2D points but I don't hold that to be the same thing as them "existing". And if they don't exist then they were invented by mathematicians.
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.
Some of the things I like about this video
- Nice music at end
- Beautiful mind-blown visual animations (as always)
- You made this video!
** Mathologerisation (>
Fantastic lecture. Btw no equilateral triangle actually follows quite quickly from picks theorem & area =(1/2)s^2sin(60)
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.
2:15 I thought of an arbitrary right angle triangle with integer lengths a and b, the hypotenuse being the base of a square that may fit several times in the grid, and its right angle aligned with any right angle of the smallest square that contains the grid. for the square to fit in an n times n grid, we have 0
This was a very good one! I had not thought about why only those angles are convenient in those terms before, and what an interesting way to prove it!
Nice :)
Nice
69
420
Both Mathologer and 3b1b are great and I love their visual representations!
Great video! When you brought up how the only rational trig results are fom fractions of 3,4, or 6 of the full circle I had a grin on my face from then onwards, your "mathematical spidey sense" phrase was very fitting
Wows! What an outstanding result!!
20:48 Never realised, that he has an non-native-english accent, until I hear him speaking German.
Grüße aus Leverkusen, der Heimat des Aspirin.
13:57 how many reg triangles and hexagons?
minecrafters: YES
Thank you so much! You always brings us great and surprising stuff. I love your channel.
Awesome video, already waiting for the next one! 😄
Can you do one on why the platonic solids fit so nicely inside each other?
Yes, would be nice to explain how any two Platonic solids are related :)
my answer for the nxn points in a square is: sum of (i*i*(n-i)) from i=1 to n-1
so for 5x5: 1*4+4*3+9*2+16*1
wolframalpha says that can be simplified to (1/12)*(n-1)*(n^2)*(n+1)
I get the same, n²(n²-1)/12
Nice, you reminded me that you don't have to consider "parallel and no paralel" cases separetly
but for n=1 you get 0
@@guyarbel2387 I consider that n=1 is just one point, n=2 is 4 points, n=3 is 9 points and so on.
@@guyarbel2387 yes that's true.
I've been waiting for it. Thank you
Believe it or not, but I am already well into the next video :)
I love it man, small small things in a gorgeously animated way.
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 think all proof papers would be better if instead of QED they ended with "ta-dah!"
lol, that's so true!
LOL. I will use that in my math lessons from now on.
How about "Told ya so"
Or, bada bing bada boom, like Grant does
I always learn something new. I love the fact, you use diagrams stuff like that instead of complicated algebra equations.
Just discovered RUclips and loving these videos....I’ve got a university degree majoring in Mathematics but I’ve been working as an engineer for the last 15 years - bringing back good memories.
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
Is that not just a coincidence? What does the sum of powers of two have to do with counting squares?
Well, my thoughts are :
there are n² squares 1x1 in a square nxn
there are (n-1)² squares 2x2 in a square nxn
and so on... there are (n+1-k)² squares kxk in a square nxn
didn't prove it but it was intuitive (on paper i guess)
so the sum of powers defintively has something to do with counting squares but the "dividing by 2" technique only seems to work with this 4x4 square
@@farissaadat4437
I have no idea. I was just trying to figure out a way to produce the answer in my head, without having to count all of the boxes. I'm an engineering student and I'm not great at math or I would be in a physics program. All I know is that it works, though I have no proof to offer. I was hoping that someone else could explain it to me. As Teumi said, I can intuit this process but I don't know what it means, really.
@@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).
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!
This might be your best yet!
After all this time running this channel I still find it impossible to predict what people will like. Believe it or not but with this one I was pretty convinced that relatively few people would end up watching it :)
Your animation skills are superb. Thanks so much!
Ich freue mich schon auf das Mathologer-Video auf Deutsch!
Schöne Grüße aus Marburg.
Grüße aus Leverkusen. Wusste nicht, dass sich Deutsche überhaupt seine Videos anschauen.
Ach noch jemand aus Marburg, witzig :D
So viele Marburger hier :)
Music, "Chris Haugen - Fresh Fallen Snow" (I hear it on a lot of videos, love it)
Well spotted.
@@Mathologer Warum ist das nicht in der Beschreibung angegeben? So habe ich Glück gehabt, daß Paul Miner es hier erwähnt hat.
Thank you for mentioning.
Nice video, as always.
i spent way too much time doodling in class on my square grid paper to construct a 60deg angle using just the grid and a straight edge. Though fruitless, it was a fun exercise
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.
@@zuthalsoraniz6764 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...
commons.m.wikimedia.org/wiki/File:Bundesarchiv_Bild_183-33006-0004,_Bauer_beim_Melken.jpg
At 6:13 you say that we get triangles arbitrarily close to being equilateral in the square grid. This is clearly true if you are measuring closeness by differences of the angles from 60 degrees. Is it true though if we take different measures of how "close" a triangle is to being equilateral? An example of such a measure of closeness could be the distance from the triangles centriod and circumcenter (or distance between two such triangle centres).
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.
I have waited long for a Math video to come out
This presentation remind me a experience. At high school my math is bad, at exam my teacher give one question, how much summary number from 1 until 1000, i slove this one with draw a diagram from 1-1000 like stair case , then divide that draw to be a large triangle and many minor triangle, and calculate that's area.
And next day, she call me and said to me how dumb i'm, she said i'm not even use right formula and get right answer, said that i'm cheated and give me 0 score i just laugh at thats time.
But my economic teacher pass by on righ time and right place like superman, hearing her bit anger he said to me "what have you done?".
She explain to him and give my paper to him, then suddenly he said "ok... i get it, leave it to me, you can back now."
I saved by economy teacher at math problem, truly i cant forget that. 😂😂😂
Small Gauss solved this from its head. 1+1000=1001, 2+999=1001, ... 500+501=1001; hence 500*1001=500500.
11:01 this moment is meme-potential. had me laugh out loud at his expressions
math is beautiful and so is your channel
words cannot express how much I love this channel
...express it in grid form ;p
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 😅
@Werni Nah, the diagonal edges are longer than the orthogonal edges, so they are rectangles, I also got 54.
Imagine having to use 360° because the formulas look weird with 2pi. If only there was a better constant to represent full circles...
Someone came up with "eta" which is equal to pi/2. This gives 4eta for 360 degrees, which makes sense because it's four right angles together.
*cough* *cough* Tau *cough* *cough*
Imagine common folk caring about scaling factors so strongly
@strontiumXnitrate Mmmmmnnnn! Donuts!
You can cherrypick whatever you want. Tau works with angles, pi works with area.
Burkard Polster's presentations are superb! Interesting and engaging--with no sacrifice of rigour.
Long days back I'm seen your updated video welcome 🙏👍
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.
I knew that there were no equilateral triangles due to the fact that equilateral triangles always have a multiple of root 3 as their perpendicular height.
Tilted equilateral triangles?
but how would you have verified that thee were no equilateral triangles that are slanted also in such diagam ?
@@Mathologer If there are two points that are connected, then the halfway point of a double-scale version of the triangle is also connected, which means a connection exists between two points which are a multiple of sqrt(3). (And, you can't have multiples of sqrt(3) because 3 is not the sum of two squares.)
The square of the length of a segment between two points on the grid is an integer. The area of an equilateral triangle is √3/4 (which is irrational) times the square of the length of a side. But the area of a polygon with all vertices on 2D grid points is an integer multiple of 1/2.
@@Mathologer Corners of tilted figures need to align with dots of the original grid - because of how you construct them. If you leave only those points and "untilt" the figure - those points will form the untilted, again square grid (why? - because of what you have said at the beginning about the tilted squares in the square grid!) with larger spacing between the dots. It means that tilting the figures does not change that much - maybe only the scale of the grid. And the scale does not matter here. Tada! Great animations though, it helps so much! P.S.: The patreon "My Son..." is your son or the patreon's nick? I'm just curious.
That was refreshing. Very good job. Subscribed :)
Really loved this one! Got my own half-baked proof based on the necessarily irrational heights of integer-sided equilaterals trying to grease my gears as I type :D
25:15 "This is really just high school stuff"
Too bad I didn't go to school in Germany
I did, but we didn't learn it either
Here in the Netherlands they do learn about the double angle formulas (e.g. cos(2\alpha)). Might be interesting to include the higher multiples as well!
14:53 Fun fact: The entrance to the Museum of Math in New York City is a glass cube, with this hexagon drawn.
Museum of Math? In New York City? How in the world did i not know this existed???!!!
and...
i want a poster of the shrinking pentagon version on the 2D grid. That is beautiful.
I liked seeing those familiar angles from trig, but I loved the picture of the spiral formed by the shrinking octagons!
I've come back to this one a few times and I think it is one of the prettiest visual proofs on the channel. Second only to the triangle-of-triangles pythagoras proof.
3:12 I thought about this problem when I was in middle-school. I was trying to draw equaliteral triangels using dots on my notebook and but no matter how much close I get, how much points I use there was tiny bit missing. Then I realized a perfect triangels height is square root of 3 times half of its floor. Square root of 3 is irrational so I will never get there. I was really, I mean REALLY dissapointed.
(Also there are no perfect hexagons or octagons or pentagons in the grid.) (I am not sure about 12-sided perfect polygon, I will be pleased if someone posts me a proof of that one about whether or not you can do it.)
(Okay I watched the video nevermind.)
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?
It means math isn't about the universe, even if it turns out to be incredibly useful in it.
Since we have (at least) three spatial dimensions, triangles and squares are equally compatible with the Universe having a “resolution”. Non-plane-tiling regular polygons, however, would not be able to exist in the physical Universe.
Is there any reason to assume that if there's a grid, it should be regular? Could be non periodic, or even amorphous. I would like my universal grid in a beautiful Penrose tiling, please.
Oh, it gets worse, way worse. writings.stephenwolfram.com/2020/04/finally-we-may-have-a-path-to-the-fundamental-theory-of-physics-and-its-beautiful/
@@frechjo Be careful what you wish for. www.wolframphysics.org/technical-introduction/
Wolfram index of Notable Universes www.wolframphysics.org/universes/
This is so cool. Thanks for making these videos
Iam loving your very entertaining and interessting videos ❤
Your love in mathematics can always be seen in every single topic, you are presenting. Thank you 😌
Ich freue mich schon auf dein Video auf deutsch 😊😊😊
Shrink proof (adj): when you've been through ten different psychiatrists and you're still depressed
...when you shrink wrap a bowl of liquid or food and the content still spills out ;p
11:24 Looks like an origami Flower Tower. 😀
Great stuff!!! love it. well done
Love your videos!!
4:07 ... Ron Swanson traveled back in time? This is the proof!
21:15 shows the shrinking algorithm in real life. 🙂
What a beautiful video, love it
It's 2:30AM in Tehran and I'm going to bed. Ooooh look a Mathologer video. No sleep tonight