welcome to another Mathologer video.
today I'll animate some absolutely gorgeous visual shrink proofs for you.
really nice super accessible and at the same time
deep mathematics plus at the end I'll prove some really cool irrational
trigonometry for you. something you may have always felt in your bones but never
captured in words. but what's that? you've never heard of a shrink proof? well I've
actually shown you some in previous videos but don't worry all will be
explained. ready for a fun journey down a very deep rabbit hole? ok here we go.
first to motivate things and to get you hooked here's a little puzzle: take a
look at the grid of dots over there. how many squares can you make with all four
corners on the dots? well plenty of you will be familiar with puzzles like this. of
course it's easy to spot lots of squares but unless you're very systematic it's
really easy to overlook a few and you're bound to get it wrong. shall we count the
squares together? systematically? ok there is this one 4 times 4 square. then there
are the 3 times 3 squares. how many of those are there. well 1 2 3 4
ok, next are the two by twos. how many are there 1 2 3 4 5 6 7 8 9. and finally
there the 1 by 1s. how many? it's not hard to guess right 1 4 9 so what's next
yep that's the squares, so a good guess is that 16 is next and of course that's
true. and so the total number of squares is 1 plus 4 plus 9 plus 16 which is 30.
right? wrong :) and this comes as a bit of a surprise when you first see it. what did
we miss? well there are lots of tilted squares in the square, like this one or
that one. now first challenge for you: complete our
count. systematically of course. how many squares are there altogether
including the tilted ones. and for the super keen among you: can you find a formula
for the number of squares in an n x n grid?
as always leave your answers in the comments. ok time for the next puzzle:
how many equilateral triangles are there in this grid? even the puzzle nuts among you
may not have seen that one before? so how how many equilateral triangles can you spot
in the grid? that's a tough one isn't it? not obvious at all. now how can we
systematically look for such triangles? well we can just scale and move the red
triangles so that two of its vertices end up on dots like this. ok and then we just
have to check whether the third vertex is also on a dot. and .. no not in this case.
let's just do it as a couple of times to build some intuition. so there... doesn't
work. that one? nope. that one? not even close.
what if we start tilting like this? nope. time for a prediction? now by now a
lot of you will have probably begun to suspect that there are no equilateral
triangles at all in this grid. and you are right no equilateral triangles in
there. so then what if we start with a larger grid of dots? a pretty natural
question to ask right? and being mathematicians let us go all the way and ask: are
there equilateral triangles in the infinite square grid. so if the grid goes
on forever in all directions are there any equilateral triangles? you can
probably guess but the first mathematician to ask this question to prove
the answer was the French mathematician Eduard Lucas in 1875. you may have come
across his name before. the important relatives of the Fibonacci numbers, the
lucas numbers are named after him. there 2 plus 1 is 3, 3 plus 4 is 7, 4 plus 7 is
11, etc this sequence grows just like the Fibonacci numbers except the
sequence starts with 2 and 1 rather than two 1s.
Edoard Lucas is also the inventor of the Tower of Hanoi puzzle. you know that one? and he
was the first to publish a description of the well-known game dots and boxes.
I played that when I was young. anyway enough gameplay back to our triangle
hunt. yep back to the hunt. and first one important and easy observation. we can
find triangles with vertices on the grid points that are really really really
close to equilateral triangles. why is that? well let's suppose that our grid
has tiny spacing of, doesn't really matter, one millimeter. okay now take a
huge equilateral triangle with let's say 1 kilometer sides. ok now toss the
triangle however you like onto the grid. it's it's like a carnival game and
you're trying to have all three vertices of the triangle hit the dots. will you
succeed? of course not. no one ever wins at carnival games. but you will get very
very close. well perfect for getting you to pay for another toss. right? so because
of the tiny grid spacing each vertex will be no more than one millimeter away
from some grid point. so we can just deform our huge equilateral triangle
just a tiny little bit by moving each vertex to its closest grid point, like
this. ok this new triangle will be pretty much indistinguishable from an exact
equilateral triangle right? you can't tell the difference even with this much
smaller set up. and in this way, by making the beginning equilateral triangle
larger and larger, we find triangles with vertices on grid points that are as
close as we like to being equilateral. but of course close doesn't win the game
in carnival games or mathematics. the fact that these ultra close
approximations exist does not prove that they are exact equilateral triangles
hiding in the square grid. and in fact there aren't. Edward Luca proved the
surprising fact that there aren't any equilateral triangles there. again that
happened about 150 years ago. Lucas's proof over there
was very pretty but algebraic. what I want to show you instead is a visual proof
the Mathologerisation of a really amazing proof by the mathematician Joel
Hamkins that he posted on his blog in 2016. if you've got mathematics in your
blood then I guarantee this stuff will make your day. okay
to set up the proof we'll give two simple properties of square grids. for the first
property connect any two of the grid points with a segment and color the
endpoints of a segment blue and green. there now turn the segment 90 degrees
around the blue end and the green end will lend on another grid point. easy to
see right. same thing if you turn around 180 degrees or 270. that's our turning
property. here's how we can use this property. let's say somebody tells us
that the two dots are part of a square grid then we can use the turning
property to discover other points in this grid there that's another point of
a grid and that's another one and another one. got it?
great! now for our second property which I'll call the shift property. start again
with two points in the grid. so same guys. grab the segment and shift the segment
so that the Blue Point ends up at another point of the grid like this. okay
then the Green Point will also wind up at a point of the grid. now let's do it
one more time. okay shift over. again the green point coincides with a point of the
grid. pretty obvious right. just like for the turning property we can
use the shift property to capture points of a mystery square grid. as an example
say we know that those three points over there belong to a square grid. then
here's how we can find another point of the grid. right just grab it
and shift like that and we found another point. all clear?
piece of cake :) okay let's get to work. let me show you why there are no equilateral
triangles in the square grid. let's assume what we see over there is an
equilateral triangle with all three vertices in some square grid. now.
important, we have no idea whatsoever what that grid might be, its spacing or
how it might be rotated. all we know about the grid is that the three
triangle vertices are grid points. but using our turning and shift properties
we can capture some of more points and, and this will be important for later, so
remember it. okay note that for the moment we'll only use the shift property.
okay time to begin our hunt. well that's kind of cool isn't it this
means that if a square grid contains an equilateral triangle then it also
contains a regular hexagon. okay let's keep on going and pin down some more
points of the grid. remember, so far we've only used a shift property but now the
turning property kicks in all we'll need are 90-degree turns. and here we go. this
means that if a square grid contains a regular hexagon then it also contains a
smaller regular hexagon and at this point are your mathematical spider senses
tingling? no? well keep watching. Got it now? we've created an incredibly
shrinking nest of hexagons with all the vertices on our square grid. but of
course that's completely impossible. Think about it no matter what the
spacing of the grid is eventually the hexagons will be too small to reach from
grid point to grid point. right? there. no chance. so where does this leave us?
okay so mathematicians may do some astonishing things but we aren't still
not allowed to do the impossible. we can't really make that shrinking nest
of hexagons. but those hexagons all came straight
from the assumption that our square grid contains an equilateral triangle. and so
that assumption just cannot have been true, the grid could not have possibly
contained an equilateral triangle. and that's the proof. a beautiful proof by
contradiction. absolutely marvelous isn't it? of course
along the way we also showed that there are no regular hexagons in the square
grid. so what about the other regular polygons? Pentagons and heptagons and
octagons, and so on? well exactly the same argument allows us to show that all the
remaining regular polygons also cannot occur in a square grid. all we get are
squares. for example here are the shrinking regular pentagon's that do the
trick. okay to summarize: square grids do not
contain any regular polygons, apart from squares of course. end of story?
nope just the end of the chapter. time for another puzzle. we'll return to our
incredibly shrinking polygons in the plane soon, but first let's do some
cheating let's add a dimension and see what that gives us. that's a three by
three by three cubical grid. I've drawn in the struts since without them the
grid is pretty confusing. pretty damn confusing right? can't see the 3d at all.
anyway time for more puzzles. you know what I'm going to ask right? yep how many
squares are in this three by three by three grid? and then how many equilateral
triangles? regular pentagons? regular hexagons etc? I'll leave figuring out the
number of squares as a challenge for you. in the comments right. what about the
other regular polygons? anything in here or in the infinite cubical grid? hmm
well doesn't look very triangly, does it? but have a closer look. yep those
three corners of one of the cubic cells are the vertices of a perfect
equilateral triangle. why? well because the three edges are diagonals of faces
of that cubical cell and of course all these diagonals have the same lengths.
but since there's an equilateral triangle in this grid, wouldn't the shift
property then give us a regular hexagon just like in 2d? yep and here it is. huh
remember what happened last time. so won't the turning property now allow us
to create a smaller hexagon in the grid and then a smaller one and so forth.
mm-hmm does that mean the equilateral triangle that really is there really
isn't there? have we exploded the mathematical universe? well let's try.
ha it doesn't work, the vertices of the new smaller regular hexagons are not
grid points. the mathematical universe is safe. in fact, while the 3d grid
definitely has the shifting property or a shift property it doesn't have the
turning property. but now what about all those other polygons? remember all our
proofs by contradiction in the 2d square grid used the turning property to create
the shrinking polygons. that doesn't work in 3d. so maybe all of those other
polygons are also hiding in the cubical grid. that'd be really cool and it's not ...
true. the devil comes to spoil our fun in the
form of a pentagram. have a look at this. well that's pretty cool an Eiffel Tower
vanishing inwards rather than upwards. so what just happened?
well we constructed a sequence of shrinking pentagons just using the
shifting property which, remember, also works in 3d. and so this proves that
regular pentagons also don't exist in the 3d grid. the same sort of star
argument also works for lots of the other regular n-gons. for example, here's
a shrinking diagram proving that there are no regular heptagons in the grid
and here's the one for the regular 9-gon. however this really cool argument
doesn't work for all n-gons. of course it better not works for hexagons and since
they really are there. but it also doesn't work for octagons see if you can
figure out what's goes wrong in these cases. in fact the star argument fails
for any regular n-gon where n is a power of two. but all is not lost. there's a
third shrinking argument again just using the shifting property that takes
care of all regular n-gons beyond the hexagons. this argument actually dates
back to 1946 and a paper by the Swiss mathematician Willy Scherrer. I only found
this one after quite a bit of digging. now here's an animation of Willy
Scherrer's shrinking argument. Enjoy Isn't that absolutely beautiful? as a
pretty easy challenge for you figure out why Scherrer's shrinking argument fails for
the other regular polygons for hexagons and below. great stuff isn't it? let's
summarize what we've got so far. so the 2d square grid only contains one type of
regular polygons the squares. all right. the 3d cubical grid contains three types
of regular polygons: equilateral triangles, squares and regular hexagons.
not sure whether you're into this stuff but what about higher dimensions? nope
nothing new happens there since these higher dimensional hypercube grids also
have the shift property everything that we said for the 3d cubical grid remains
true. so they contain equilateral triangles and squares and regular
hexagons but that's it. and, finally, end of story?
nope, one more chapter to go. we'll answer a question that was staring at us all
through school but that probably none of us ever asked. ready to go deeper? okay last chapter. ready for some trick?
no, wait, come back. there is no flagpoles or ladders against the wall. you really
like this one I promise. this is Willy Scherrer's 1946 paper on shrinking
polygons. it's very short and believe it or not it doesn't contain a single
diagram. amazing isn't it? just looking at this you would never suspect it there is
a nice visual proof hiding there. mathematicians can be really weird at
times. and it gets weirder Scherrer says that the only reason why he published
his proof is because another Swiss mathematician his student Hugo Hadwiger
found a really good application for the shrinking argument. and what was
Hadwiger's application? let's keep scrolling. there in the same
journal right after Scherrer's paper we find Hadwigers paper entitled Über die rationalen Hauptwinkel der Goniometrie All clear? what's that? you
don't speak German? that's okay I forgive you :) Irgendwann mach ich auch mal ein ganzes Mathologer Video auf deutsch. versprochen. Anyway, German are not there's plenty here that should be very
familiar. up top there's fractions of 2 pi and down below
I'm sure you'll all recognize the angles in that list 0 30 45 60 90 120 etc. these
are the nice angles that we all know from school, the angles that pop up in
equilateral triangles and squares and regular hexagons the sort of regular
polygons we've been going on about. the niceness of these angles shows up in
other ways as well. most obviously these angles are all rational, they are simple
fractions of a full circle. but of course just like there are infinitely many
rational numbers they are infinitely many rational angles. but what makes
those particular rational angles especially nice is the trigonometry.
those are the angles for which getting the exact values of trigonometric ratios
sine and cosine and tan is really straightforward. and there's something
extra special about those trig value that you would have always known but
perhaps never focused upon. for each of those rational angles above at least one
of the trigonometric ratios is also rational. all rational numbers there on
the right. looks very familiar doesn't it? and now comes Hadwiger's insight. if we
look at Scherrer's shrinking polygons in just the right way Scherrer's argument
proves that the angles up there are the only rational angles with at least one
of the trigonometric ratios also a rational number. so really just school
stuff and something you probably never knew. definitely it was new to me.
obviously this is an absolute gem but unfortunately Hadwiger write-up is a
bit of a mess. and so for you and for posterity let me present you with a
Mathologerisation of Hadwigers proof. I'll focus on the cosine part of the
proof, the proof that the only rational angles with rational cosines are the
ones over there together with these obvious companions. right the cosines of
all these simple angles are either 0 1 minus 1 1/2 or - 1/2. and the claim
is that the cosines of all other rational angles are irrational numbers.
that's what we want to prove. ok so just as a reminder a rational anger is an
angle of this form. here p over q is a non-negative fraction less than 1 in
lowest terms so the integers p and q have no common factor and p is less than
q. so the fractions in our short list over there are just the rational
fractions with q equal to 2 3 4 & 6. so what we want to show is that all cosines
of rational fractions with q equal to 5 7 8 9 etc are irrational. 5 7 8 9 etc. notice that these are exactly
the numbers corresponding to polygons for which the shift property gives us
shrinking arguments. not a coincidence :) okay let me show you how the proof works
using a specific example. let me show you that the cosine of the angle 3/8 of 360
is irrational. as you can probably guess since the proof is based upon our
shrinking argument this will also be a proof by contradiction. so let's assume
that the cosine of our angle is a rational number. now let's look at
multiples of our angle 2 times alpha, 3 times alpha, 4 times alpha, 5 times alpha and so on. it turns out, and this is really just high school stuff,
that all the cosines of the multiples of our angle alpha can be written as
integer polynomials of cos alpha. above I've listed the first few of these
polynomials. let's have a look at the very first one. there that one. cos of 2
alpha equals minus 1 plus 2 times cos squared alpha. well since we are assuming
that cos alpha is a rational number r that means cos of 2 alpha is also a
rational number right? if r is a rational number then minus 1 plus 2 r squared is
also a rational number. in fact, in exactly the same way the other cosine
polynomials show that all the cosines of these multiples of alpha are rational.
We will use this fact in a moment. again the the specific angle we want to focus on is
3/8 of 360. ok what's next? well if we want to apply our shrinking argument we
better bring in a regular polygon to shrink. the denominator of our angle is 8
and so we start with this regular octagon centered at the origin of the XY
plane and with outer radius 1. here's our angle. here's 2 times this angle and 3
times okay and one more 4 times and so on. because the numerator 3 and the
denominator 8 are relatively prime we hit all corners of the octagon. okay
remember the cosines of all the multiples of our angle are rational.
because we chose radius 1 this is the same as saying that the x-coordinates of
all the vertices of our octagon are rational. now let's make everything
integers. right? rational numbers are good but integers are better. suppose all
those fractions there have common denominator s. okay then scaling this
picture by s gives an octagon all of whose vertices have integer
x-coordinates. let's just do this. okay so we scale up and it makes all those
rational x-coordinates into integer x- coordinates. all good? and now all is ready
to use shifts to perform the shrinking construction just as before. so let's
just do this again quickly. you get it? the x-coordinates of the
original octagon are all integers but because we're using shifts that means
all the x-coordinates of the second smaller octagon must also be integers.
all those guys are also integers. and you can see where we're going with this
right? the exact same will be true for all the subsequent smaller octagons. so
all the infinitely many colored dots that we come across this way stand for
integers. and of course that's impossible since the spacing between the dots
shrinks to zero they can't possibly all be integers. and that's our contradiction.
tada!! so we've used our shrinking regular
octagons to prove that the cosine of the angle 3/8 of 360 is irrational and
similarly we can use shrinking regular q-gons
to prove that the cosines of all these general rational angles are irrational.
and the sine and tan part of the proof are also easy. isn't that beautiful.
okay leave a note in the comments and let me know what you think of all this.
well and that's all from me for today. until next time.
What about squares in equilateral triangle or hexagon grids? Should work the same way.
Filling in the gaps:
I guess one way to justify the "shifting property" and "turning property" of each respective lattice would be to represent each point as a complex integer a+ib (a and b in Z): a rotation by 90 degrees would be multiplying by "i", which results in -b+ia which also belongs to the lattice (also integers), and a translation is just adding another complex integer.
Probably a similar representation works for a hexagonal lattice, but with complex rationals in Q(sqrt(3)).
web archive links to the blog posts (the original website seems down to me): integer lattice, hexagonal lattice.
Two nice proofs of the impossibility of the equilateral on a square grid on math stackexchange here: one proof is a clever one-liner that uses Pick's theorem, and the other is a clever mix of geometry, algebra, and number theory.
And finally, can't access the linked jstor article "Triangles with Vertices on Lattice Points", wondering if the author has made it available somewhere else (couldn't find it so far.)