Welcome to another Mathologer video. You are all familiar with the diagram over
there, right? Yes, of course, it’s the diagram powering
the Mathologer logo :) Yeees, but that’s not really it, right? I am sure for the majority of people watching
this video this diagram will scream "Pythagoras". If this is not the case, ask for your money
back from your school and definitely keep on watching for a, hopefully, life changing
experience. And even for those of you who’ve treasured
this twisted square diagram since primary school keep watching. Today will be about all the secrets of this
iconic diagram that not even most professional mathematicians will be familiars with. But first a bit of revision to get us all
on the same page. A squared plus B squared equals C squared. You all know the drill. Take any right-angled triangle with short
sides A and B and hypotenuse C. Then the area of the square on the hypotenuse, C squared,
is equal to the combined area of the squares on the two shorter sides, A squared plus B
squared. A stunning visual proof of this fundamental
mathematical truth runs like this. Take four copies of the triangle and combine
them into a large square like this. Okay, this large square is partly covered
by the four triangles. The area that is not covered is what? Well, C squared of course. Now move one of the little triangles a bit. How large is the blue area now? Well, still C squared of course. How about now? Still C squared. And this will be true no matter how we position
the triangles inside the blue square as long as they don’t overlap. Right? Okay, ready for a big AHA moment :) Move the
triangles one more time. No overlaps and so the area is still C squared. However, this area is also A squared plus
B squared. And so A squared plus B squared equals C squared. What an amazingly simple and beautiful proof
and what a killer punchline, don’t you think! Alright, now, how about this... The area of the large square is equal to the
area of the little square plus the area of the four triangles Okay, the area of the large
square is A plus B squared. The area of the little square is C squared. And the area of one of the right-angled triangles
is 1/2 base times height, 1/2 A times B. Now simply go on algebra autopilot. Fantastic again. Maybe not quite as fantastic as the first
proof but still fantastic. Just out of interest, can you let me know
in the comments which of these two twisted square proofs you were already familiar with
and when and where you first came across them. Will be interesting. Of course, most kids never get to see either
proof, which in my book is a crime which should entitle them to get their money back from
their schools. As well, most school textbooks that feature
a proof only mention the second less spectacular algebraic one. Go figure. Anyway, no matter how pretty these proofs
are, that’s really all there is to this iconic diagram. Right? Wrong! There is more, lots and lots more. There is Fermat’s four square theorem, the
TRIthagorean theorem, the bug chasing vortex, and much more. Ready to be amazed. Here we go. The twisted square diagram has been around
for a couple thousand years and features in the earliest documented proof of Pythagoras
theorem which is part of an ancient Chinese manuscript on astronomy and mathematics. Well, Pythagoras neither discovered “his”
theorem nor was he the first to prove it and so to have this theorem named after him is
really a bit of a scandal in mathematics. Anyway, let’s have a look inside the manuscript. What you find there is pages upon pages of
this. Lots and lots of Chinese characters. However, eventually we come across a page
featuring our diagram. Well, that’s definitely our diagram. But there is a bit more to this drawing, isn’t
there? There, a second very different twisted square
diagram made up of the same four right angled triangles. In fact, this second diagram can also be used
to prove Pythagoras theorem in two different ways, one algebraic and one one based on shuffling
triangles. Just looking at areas again we have: Four
times the area of the triangle plus the area of the little square is equal to the area
of the large square. Expressing this identity in terms of A, B
and C, and some algebra auto-pilot spits out A squared plus B squared equals C squared
in pretty much exactly the same way as before. On the other hand, the C squared area of the
large square stays unchanged when we move some of the triangles. And so C squared is equal to A squared plus
B squared. Also a very pretty proof and in a way even
slicker than the first triangle shuffle proof since we only have to shuffle two of the triangles
instead of all four. The Chinese drawing also features a grid. What’s that grid for? Well, it’s clear that no matter what right-angled
triangle you start with, everything we said so far will stay true. Of course, this is very important. Otherwise none of our proofs would actually
count as a general proof of Pythagoras theorem. On the other hand, the grid shows that in
our example, we are dealing with a super famous right-angled triangle. Can you see it? Yep, it’s a 3-4-5 right-angled triangle.
3 squared plus 4 squared equals 5 squared. 3,4,5 the prototype Pythagorean triple, a
triple of integers that satisfies the equation A squared + B squared equals C squared. Since we’ve all been familiar with the 3,4,5
triple since kindergarten, we are taking the fact that such special triples exist for granted. That the existence of such integer triples is in fact a little bit of
a miracle is illustrated by Fermat’s last theorem which says that
if you change the exponent to 3 , 4 , or any other larger integer the equation will not
have any positive integer solutions. But of course, the wonderful 3,4,5 Pythagorean
triple does exist. And you can get more of these wonderful integer
triples by simply changing the number of lines in the grid. There, doubling the number of gridlines give
this 6 squared plus 8 squared equals 10 squared. And tripling gives this But there are also
other essentially different Pythagorean triples and many of you will be familiar with this
particular one. 5, 12, 13, 5 squared + 12 squared equals 13
squared. Anyway, there are infinitely many essentially
different Pythagorean triples and let me just mention that we know how to construct all
of them. And of course we can use any of these triples
to make up one of these diagrams. But let’s stick with 3,4,5. Alright, let’s have another close look. How many squares are there? Well there is this one Then that one And one
more The area of the little square is ..... 1 squared the area of this green square is 5
squared and what about the largest one? Well 3+4 is 7 and so the area is 7 squared. Cool Now, what’s 25 - 1 25-1 that’s 24
of course. What about 49 minus 25? You got it? 49 minus 25 is also 24 :) Interesting, both
differences are the same. Is that a coincidence? What do you think? Well, no, that we get the same difference
is always true and really easy to see. Have a look. What’s the difference between blue and green? Well that’s just the area of four copies
of our right-angled triangle. And how about the difference between the green
and the orange. Same thing. Very pretty isn’t it? So what this means is that within the square
numbers we can find three equally spaced squares and that every Pythagorean triple translates
into three such special squares. Okay so that’s what we get for 3,4,5. And here is what we get if we double the number
of grid lines. In maths an arithmetic progression is a sequence
of numbers in which consecutive terms differ by a fixed amount like 2, 4, 6, 8, 10, etc.
with common difference 2 or 3, 6, 9, 12, etc. with common difference 3. And so another way of expressing this equal
difference business at hand is to say that there are arithmetic progressions of length
3 within the squares. This fact was already known to the great 17th
century mathematician Pierre de Fermat whose super famous last theorem I mentioned just
a moment ago. And this fact prompted Fermat to ask a very
natural question. Are there any longer arithmetic progressions
hiding in the squares? Right, here we’ve got 4 plus 96 is 100 plus
96 is 196. What if we add another 96? Do we get another square? That would then be an arithmetic progression
of length 4. Well, no, because 196 plus 96 is 292 and 292
is not a square. But in general, are there arithmetic progressions
of 4 squares, or 5 squares? How long can such progressions be? What do you think? Well, this is a tricky one. Let me just try to confuse you a little bit
more. In 2004 the mathematical superstars Ben Green and Terry Tao proved that
that there are arithmetic progressions of arbitrary length among the
primes. Insane right? Here is one of length 5 that you’ve been
staring at all your life without realising it. There 5 plus 6 is 11 plus 6 is 17 plus 6 is
23 plus 6 is 29, an arithmetic progression of 5 primes with common difference 6. Anyway, how long can arithmetic progressions
of squares get? Fermat figured out that in fact 3 is the surprising
answer. You cannot do any better than the mini sequences
that our diagram produces from Pythagorean triples. Allegedly Fermat also had a proof, but nobody
knows what that proof looked like exactly, a situation similar to that with his famous
last theorem where he also claimed to have a proof but none was found in his writings. Here is a quote from the well-known Australian
number theorist Alf van der Poorten. I looked this up in reference [3] and found
at page 54 the unhelpful footnote “Fermat could show by descent that one cannot have four squares in arithmetic
progression . . . . Gerry Myerson has pointed me to a reference but
the argument there seems utterly soulless and I remain searching for a decent descent
argument warranting report to you.” I decided recently that such a proof was most
readily found on a (previously) blank page of my notebook. And here is what he ended up being written
in his notebook. This is the nicest and most natural proof
of Fermat’s four square theorem that I am aware of and it may very well be similar to
what Fermat came up with a couple hundred years ago. I won’t go into the nitty gritty details
here. Let me just say that this proof features the
kind of infinite descent argument that we’ve already encountered in several Mathologer
videos. At its core is a well-known construction of
all Pythagorean triples and so it appears that this proof may have had its origin in
the little Pythagorean triple construction that I just talked about. Pythagoras theorem has been around for thousands
of years and so you would think that everything worth discovering about Pythagoras theorem
was discovered a long time ago. Far from it! What I’ll talk about now was only discovered
around twenty years ago. Okay, so instead of twisting squares we now
twist equilateral triangles. Let’s try to mimic as much as possible what
we did before. So, instead of a right-angled triangle we
start with a triangle that features a 60 degree angle. Make three copies. Arrange them into a large equilateral triangle
Next we have to shuffle the triangles around. After a bit of fiddling it turns out that
for this to produce something interesting we first have to get rid of one of the red
triangles. There, go away ... Now shuffle. Aha, that looks very promising. As before, side by side our two diagrams amount
to an equation. Cancel two triangles each on the left and
right side of this equation. So what’s left? Well, we’ve got one equilateral triangle
each for the three sides A, B and C of our original triangle. Let’s name the areas of these triangles
like this. Using these names our identity looks like
this. Fantastic. A Trithagorean theorem for 60-degree triangles
:) Note that in the real Pythagoras the A and B areas add up exactly to the C area. That almost also works in this Trithagorean
theorem. Right? A area plus B area equals C area PLUS there
is a correction term, a really nice correction term, the area of the original triangle T
:) For the next surprise, let’s make the 60 degree angle into a right angle but keep
the equilateral triangles on the sides. Experts, what am I about to say? :) Well, we can make the identity on top work
by simply removing the T :) Really? Yes, Pythagoras not only works for squares
but also for equilateral triangles. A area plus B area equals C area. In fact, even better, Pythagoras works for
three suitably scaled copies of any shape glued to the sides of a right angled triangle. Semicircles, Capital letter M for Mathologer,
anything whatsoever. I already talked about this at length in a
separate video on Pythagoras theorem a couple of years ago. Also check that on out. Anyway, back to equilateral triangles. 60
is 90 MINUS 30. On the other hand, 90 PLUS 30 is 120. And it turns out that just like 60 degree
triangles, 120 degree triangles also have a nice Trithagorean theorem. The only difference between these two Trithagorean
theorems is that the plus correction term turns into a minus correction term. Also nice, isn’t it? Anyway, back to 60 degree triangles. Well, our Trithagorean theorem is very cute. However, it is definitely not as applicable
as Pythagoras theorem. Right, if you give me the length of any two
sides of a right-angled triangle, Pythagoras immediately tells me what the third length
is. Not so with our Trithagorean identity. However, there is also a nice A squared plus
B squared equals C squares Pythagoras counterpart for 60 degree triangles that is just as applicable
as its 90 degree original. Here it is. A squared plus B squared is equal to C squared
plus A times B. Supernice. And also very easy to see why this is true
if you know the cosine rule a generalisation of Pythagoras to general triangles taught
as part of trigonometry. Here is another nice way to conjure up this
identity which also shows that it is simply another way of writing our original Trithagorean
identity. Let’s call the area of the equilateral triangle
with sides of length 1 F There F is that red area. Then the 60 degree triangle with sides A and
B bounding the 60 degree angle has area ABF. Bit of a challenge. Can you see at a glance why this is the case? If you can let us know in the comments. Anyway, this means T is equal to A times B
times F. What’s the area of the A triangle. Well this is an equilateral triangle and so
all its sides are equal to A. And so its area is A times A times F, A^2F. And similarly with the B and C equilateral
triangles. Now just divide by F and we are done. Slick, hmm? What about Trithagorean triples, counterparts
to 3,4,5 and Co.? Are there 60 degree triangles with integer
sides? What do you think? You should be able to answer that one :) Are
there 60 degree triangles with integer sides? Of course, there are ! :) Any equilateral
triangle with integer length sides will work. 1,1,1, 2,2,2 and so on. Obvious. What about other 60 degree triangles? Not so obvious but, yes, there are such triangles. In fact our example over there is such a triangle.
3,8,7 works :) Check for yourself that this triple really satisfies that equation: 3 squared
+ 8 squared = 7 squared + 3 times 8. If I ever change the Mathologer diagram from
a twisted square to a twisted equilateral diagram I’ll base it on this special 3,8,7
triple of integers. Something like this. What do you think, time for a change of logo? Anyway, the official name for our Trithagorean
integer triples is Eisenstein triples named after the 19th century genius mathematician
Gotthold Eisenstein. Just like with Pythagorean triples there are
infinitely many essentially different Eisenstein triples. Before I move on, here is something cute. Overlaying these two triangles immediately
gives a second Eisenstein triple. Right, the sides of this blue 60 degree triangle
are 7, 8 and 8-3 = 5 There a new Eisenstein triple 5,8,7. Nice, hmm. Here is something else that you’ll like. Superimposing these two triangles gives a
red 120 degree triangle with integer sides These sides have length 7, 3 and 8-3 = 5 Also
to get the ABC identiy for 120-degree triangles just change the plus in front of the AB to
a minus. Also super pretty. Anyway, do these triangular twists to the
original Pythagorean theorem get your seal of approval? I sure hope so :) Okay so we have this pair
of diagrams for equilateral triangles. and this very similar pair for the squares We
can make these two pictures look even more similar by relocating these two triangles. Now a bit of a facelift. Very pretty, don’t you think. Anyway, while preparing for this Mathologer
video I discovered a third type of diagram that nicely complements this pair. What does THIS diagram show. Well, ... the white triangles are all the
same. There are 12 of them on the left and 6 on
the right. So let’s cancel 6 on both sides. Also beautiful. The Hexagorean theorem :) Ready for another
nice AHA moment? Okay, divide by 6. What’s a regular hexagon divided by 6? Let’s see :) There the Hexagorean theorem
for 120 degree triangles divided by 6 is the Trithagorean theorem for 120-degree triangles. You saw it first on Mathologer :) As I already
mentioned all the twisted triangle shuffle stuff only seems to have been around for about
20 years. Over there is a painting of amateur mathematician
Wayne Roberts created in 2003, which is one of the possible origins of the nice triangle
shuffle proof of our Trithagorean theorem. This painting was also the inspiration for
the way I prettified the diagrams in this chapter. Okay back to the original twisted square diagram. Have a look at this. What you’ve just witnessed is yet another
really weird proof of Pythagoras based on the twisted square diagram. Right, it’s another proof? The top two squares have the same area as
the bottom square because it is possible to chop the top squares into ten pieces that
can be reassembled exactly into the large square at the bottom. Really wonderful and unexpected, isn’t it,
with an added bonus miracle being the overall shape at the top and bottom dissections which
mirror the two diagrams in our original proof. Of course, with all these visual proofs it
is very important to chase around the angles and lengths in these diagrams to make sure
that looks are not deceiving, that things really fit together exactly as suggested by our pictures for all possible
choices of right-angled triangles. Anyway, I checked, trust me :) And if you
don’t trust me check for yourself :) Here is another miraculous chop up proof based
on the twisted square diagram. There are a number of interesting properties
of right-angled triangles that appear to be very mysterious at first sight. But once you remember the twisted squares
it all becomes crystal clear in a flash. Here are three great examples. Okay, take any right-angled triangle and attach
the square on the hypotenuse. Next, draw the line that cuts the right angle
in two. Now, no matter what right triangle you started
with this line will always pass through the center of the square. If you encounter this property in the wild
it will most likely come as quite a surprise. But then you remember your twisted squares. And now everything is clear. Draw in one of the diagonals of the big square. Obviously this diagonal cuts the right angles
in half. And its also clear that it passes through
the center of the square. Tada. Nice. I hope you still experienced a bit of an AHA
moment despite me advertising the twisted squares as the solution in advance. Okay, here is how the second property is usually
presented. Start with a square and insert a right-angled
triangle like this. Make three copies and attach them to the remaining
sides. Now what’s the special property? Pretty obvious, right? All right angles are on a line! :) Very pretty. Again, very hard to see why this should be
the case, unless you remember our twisted squares. One more. There are hundreds of different proofs of
Pythagoras theorem. One of them is by the 20th American president
James Garfield. It runs like this. Start with a right-angled triangle. Make a copy. And complete the resulting figure to a trapezium
Now, as usual, calculate the area of this trapezium in two different ways. First using the formula for the trapezium
and then by adding up the area of the three triangles. Then some algebra autopilot, and out comes
A squared plus B squared equals C squared. All, by now, a complete no-brainer. But doesn’t this trapezium remind you of
something? Yes, of course. It’s really just half our favourite diagram
And, really, on close inspection president Garfield’s proof is exactly our original
algebraic proof “divided by 2” and so nothing really new. But, of course, I can see why it is great
to be able to say to kids that being good at maths and becoming an American presidents
can go together. And so for propaganda purposes let’s also
say in the future that an American president came up with a new proof :) Having fun so far? Right, let’s start with this special case
of our twisted square diagram with, the little blue square twisted 45 degrees. Let’s squash the diagram a bit. Okay, relocating the triangles as usual is
still possible. Let’s say the sides of the blue rectangle
are root A and root B. Why root A and not just A? Don’t worry about it. Whatever the lengths are in the diagram, we
can certainly write them as square roots of some numbers. Now the area of one of the two rectangles
is root A times B. Okay, now what about the diagram on the right? Using Pythagoras we see that one side of the
blue rhombus is root A+B Okay, the blue area on the left is 2 times root AB. which is equal
to the area of the blue rhombus on the right. but this area is definitely not larger than
that of the square with sides of the same length. Right? The areas of both the rhombus and the square
are base times height. Both shapes have the same base but the height
of the square is larger and so its area is also larger. What’s the area of the square? Easy! Now this is a very famous inequality between
the two most useful types of averaging two positive numbers, the geometric mean on the
left and the arithmetic mean on the right. So what we just proved is that the geometric
mean of two positive numbers is always less or equal to their arithmetic mean. we just started with. First on the left. Then, reshuffled, on the right we get this
tilted parallelogram. Now, if we do exactly the same as we just
did with a different sort of labelling, we get another super-famous inequality, the so-called
Cauchy-Schwarz inequality. You need a bit more maths background to be
able to appreciate that one and so I’ll just do an animation of the relevant bits
at the end of this video for those of you in the know and interested. But I also would like to show you one more
surprising completely different application that everybody watching
will be able to appreciate. To set it up we modify things a bit more. First we swap the two diagrams. And now we adjust so what all the sides of
the parallelogram are equal to 1 And now we adjust so what all the sides of the parallelogram
are equal to 1 Then on the right things change like this. Okay, now pencil in some of the angles. Then the lengths of the sides of our triangles
will be the sines and cosines of these angles, like this. Also, we can find the two angles again inside
the blue parallelogram. Okay, in term of areas of the blue regions
what have we got? Almost there. The area of the parallelogram is equal to
base times height. Base is 1. And height? Well that’s just the sine of alpha plus
beta, right? And so the area base times height is this. Very nice, and so in total we get the addition
formula for sine. How nice is that :) Did you see that one coming? :) Let’s finish off with something totally
different. Start with a square and place some bugs on
the corners of this square. Let’s say that at some point in time each
bug will start running towards the bug it is facing right now and that all four bugs
will always be running at the same constant speed towards each other. With a setup like this what are some natural
questions to ask? Well, how about this: Will the bugs ever catch
up to each other? If they do how long will their journey be? What are the equations of their journey? Which way will they be facing when they meet? How do things change if we use different starting
configurations/numbers of bugs, etc.? Lots of mathematical fun to be had here :) Some
of these problems and their very surprising answers were made popular
by the famous mathematics populariser Martin Gardner in his column in the July 1965 issue of the Scientific American. Let me show you a nice twisted square way
to get a feel for what is going on here and for how to come up with answers to some of
our questions. First ask the bugs to forget about Martin
Gardner’s instructions. Instead, ask them to run along 1/5th of the
sides of the square and then stop, like this. Now ask them to turn towards each other like
this. Whoa, twisted squares and we’ve arrived
back at the starting configuration just with a slightly smaller rotated square. Now let’s ask the bugs to repeat everything
they’ve done so far over and over. So, cover 1/5th Turn towards each other. Cover 1/5th. Turn towards each other. And so on. Pretty spectacular. Two simple observations. 1. Because of the symmetry of everything that
is happening here the bugs will form a square at all times. 2. From the way we repeat the same overall movement
over and over it follows that from one square to the next we are rotating the orientation
of squares by the same angle and we scale the size down by the same factor. Conclusions from this. Since we scale down by a fixed factor the
size of the squares will decrease but will do so infinitely often. Hence there are infinitely many squares and
their sizes shrink to 0. And, since the orientations of consecutive
squares differ by a fixed angle and there are infinitely many squares, on their journey
ideal bugs will circle the center infinitely often. That’s a bit unexpected, isn’t it? Now, does this mean that a bug’s paths is
infinitely long? Actually, no, although these paths wind infinitely
often around the center, they are of finite length. Let’s figure out how long one of these paths
is. If the side of the square is 1 unit long,
a bug walks 1/5th of 1 and 1/5th of 1 is 0.2 The remaining distance is 0.8. We find the side-length of the smaller square
with Pythagoras. It’s approximately 0.825. Since the outer side length 1 shrinks to 0.825
in the second square, we conclude that 0.825 is the factor by which the squares shrink. This allows us to figure out the lengths of
all the segments of one of the bug’s path. Right? Since the first segment is 0.2 long, the second
segment is 0.2 times 0.825 long To get the length of the next segment we just have to
multiply by 0.825 again And so on This means that the total length of the path is this. The infinite series in the brackets is a geometric
series. Since you are still here chances are you know
the formula for the sum of a geometric series by heart. It’s just 1 over 1 minus the scale factor. With this the overall length of a path pans
out to be this: This is approximately 1.14 which is a little bit longer than one of the
sides of the square. Interesting. If we reduce the fraction of the sides covered
by the bugs from 0.2 to 0.1 the numbers change like this. If we reduce the fraction of the sides covered
by the bugs from 0.2 to 0.1 the numbers change like this. Aha, the length of the path is even closer
to 1. In fact, if we let the fraction of the sides
covered by the bugs go to 0, the picture will turn into the picture that corresponds to
the original Martin Gardner problem. And the length of the paths of the bugs in
the original problem can be seen to be exactly 1. How nice an answer it that :) Bye the way,
this was the picture on the cover of Martin Gardner’s Scientific American article. Again, the bugs paths are exactly as long
as the sides of the square. It’s a pretty surprising answer, yes, but
in hindsight is there maybe a simpler way to see why the bugs path should be exactly
as long as their original distance apart. Again leave your thoughts in the comments. I could report on many more ingenious results
as far as bugs chasing bugs are concerned, but let’s call it a day for today. If you are interested in more neat details
about any of the topics I touched upon today, check out the links in the description of
the video. One last nice insight. Early on we encountered 2 types of twisted
square diagrams. This one and that one There is actually a
third one. In this third diagram the triangles overlap
Here then is your final puzzle. Let’s say A=1/2 and B=1 as in this example
, what is the area of the square in the middle? Once again, leave your hopefully surprising
answers in the comments. Okay and that’s it from me for today. As promised, I’ll leave you with an animation
of the proof for the Cauchy-Schwarz inequality in the case of two numbers A and B.
