What does this prove? Some of the most gorgeous visual 'shrink' proofs ever invented

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:
There is also a whole chapter about all this and much more related maths in his new book

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:

Scherrer's and Hadwinger's articles:
Scherrer, Willy, Die Einlagerung eines regulären Vielecks in ein Gitter, Elemente der Mathematik 1 (1946), 97-98.
Hadwiger, Hugo Über die rationalen Hauptwinkel der Goniometrie, Elemente der Mathematik 1 (1946), 98-100.

Another, nice paper on rational (and algebraic) cosines

Here is a solution to the first puzzle (one way to find the general formula):

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.
Remarkable :(
b) This transition to the good stuff I clearly did not think through properly.

Enjoy!

Burkard

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14. Sep. 2021: Thank you very much Michael Didenko for your Russian subtitles.

[arthurmaruyama]

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.

[Scrogan]

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.

[cassied]

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.

[alexanderli]

You're the Bob Ross of mathematics.

[abcrtzyn]

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.

[ryanjude]

Had to pause so many times just to truly appreciate the beauty of this visual proof. So much to reflect on.

[miruten]

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

[davidrosa]

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<=a, 0<b<n, a+b<n or a<n-b. For a given square, you may see how many times it fits on the grid by counting how many horizontal traslations of 1 unit (distance between 2 adjacent horizontal/vertical dots of the grid) you may move it before it's no longer contained inside the grid; it's easy to see that the square fits tha last number squared. This turns out to be (n-a-b)^2. I then expressed the solution to the puzzle as 2 nested summations. Then, assuming that all the possible values of a and b that satisfy the inequalities, yield all squares that fit on the grid without repetition, we can express the solution as 2 nested sums: I then proceeded to painfully and carefully algebra-pilot this guy (using the identities for sums of powers 1, 2 and 3 of consecutive integers) until it gave me the beautiful solution n^2(n^2-1)/12. The road I traveled was VERY windy, a simpler proof is very welcome. I might come back to try and get one now that I know the final expression I'm trying to get, it's suprisingly simple and n^2 can be replaced by the total number of dots of the grid, that may be hint. Oh, and the solution for the 5 times 5 grid is 5^2(5^2-1)/12=50, which I think is correct; I hope I didn't mess up anything, haha.

[Tehom]

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.

[jeskomatthes]

Well, after that one, I (probably irrationally) suppose mathematicians tend to avoid too much sun exposure cos tan is a sin.

[PeterZaitcev]

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.

[PC_Simo]

2:24 There are 1+4+9+16+9 = 39 squares, altogether, in the 5*5 -grid (measured in dots), and Σ(k²; k = [0, n]) + (n-2)², in an n*n dot grid.

[zygoloid]

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!

[rafaelhenrique-hpbo]

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

[HaoSunUW]

Fantastic lecture. Btw no equilateral triangle actually follows quite quickly from picks theorem & area =(1/2)s^2sin(60)

[MusicThatILike]

General nxn grid is: n^2 + SUM(i = 1 -> n-1) { 2 * (i)^2 }, so it's palindromic

[eliyasne]

Among other things, your videos prove the beauty and elegance of mathematics.

[freshtauwaka]

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)

[xyz.ijk.]

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.

[aksela]

The regular polygons you can fit inside a 3D grid are also the regular polygons you can use to tile a surface. Coincidence?