Fibonacci = Pythagoras: Help save a beautiful discovery from oblivion
In 2007 a simple beautiful connection Pythagorean triples and the Fibonacci sequence was discovered. This video is about popularising this connection which previously went largely unnoticed.
00:00 Intro
07:07 Pythagorean triple tree
13:44 Pythagoras's other tree
16:02 Feuerbach miracle
24:28 Life lesson
26:10 The families of Plato, Fermat and Pythagoras
30:45 Euclid's Elements and some proofs
37:57 Fibonacci numbers are special
40:38 Eugen Jost's spiral
41:20 Thank you!!!
42:27 Solution to my pearl necklace puzzle
The two preprints by H. Lee Price and and Frank R. Bernhart and another related paper by the same authors:
arxiv.org/abs/0809.4324
arxiv.org/abs/math/0701554
tinyurl.com/y6k4eyx5
The wiki page on Pythagorean triples is very good and very comprehensive
en.wikipedia.org/wiki/Pythago...
Wiki page on Pythagorean triple trees
en.wikipedia.org/wiki/Tree_of...
Mathematica code for the Pythagorean Christmas tree by chyanog tinyurl.com/2z66rfkb
Geoalgebra app for the Pythagoras tree fractal by Juan Carlos Ponce Campuzano
www.geogebra.org/m/VU4SUVUp
Connection to the Farey tree/Stern-Brocot tree in a paper by Shin-ichi Katayama
tinyurl.com/vmvcs729
David Pagni (on the extra special feature of the Fibonacci number)
www.jstor.org/stable/30215477
Eugen Jost's Fibonacci meets Pythagoras spiral (in German)
mathothek.de/katalog/fibonacc...
Bug report:
06:06 - right circle doesn't touch line (I mucked up :(
Puzzle time codes:
11:41 Puzzle 1: a) Fibonacci box of 153, 104, 185 b) path from from 3, 4, 5, to this triple in the tree
16:02 Puzzle 2: Area of gen 5 Pythagorean tree
25:55 Puzzle 3: Necklace puzzle
Some interesting tidbits:
Jakob Lenke put together an app that finds the route from 3,4,5 to your primitive Pythagorean triple of choice inside the tree. Thanks Jacob pastebin.com/T71NP8Z9
theoriginalstoney and Michael Morad observed that at 39:28 (last section, extra special Fibonacci) the difference between the two righthand numbers (4 and 5, 12 and 13, 30 and 34, 80 and 89) are also squares of the Fibonacci numbers: F_(2n+3) - 2 F_(n+1) F_(n+2)=(F_n)^2
Éric Bischoff comments that the trick to get a right angle at 25:40 is popularized in French under the name "corde d'arpenteur". This term refers to a circular rope with 12 equally spaced nodes. If you pull 3, 4 and 5-node sides so the rope is tense, you get a right angle. See article "Corde à nœuds" on Wikipedia
Various viewers told me what F.J.M. stands for: Fredericus Johannes Maria Barning, Freek, b. Amsterdam 03.10.1924, master's degree in mathematics Amsterdam GU 1954|a|, employee Mathematical Center (1954-), deputy director Mathematical Center, later Center for Mathematics and Informatics (1972-1988) Deceased. Amstelveen 27.06.2012, begr. Amsterdam (RK Bpl. Buitenveldert) 04.07.2012.
John Klinger remarks that if the four numbers in the box are viewed as fractions, the two fractions are equal to the tangents of half of each of the two acute angles of the triangle.
Colin Pountney: Here is another piece in the jigsaw. The link to Pascals triangle. It only works for the Fermat series of triples (ie the set of "middle children"). Choose any row in Pascals triangle. Multiply the odd entries by 1, 2, 4, 8, ..... and add to get the top left entry in a Fibonacci box. Do the same with the even entries to get the top right entry. For example taking the 1 5 10 10 5 1 row, we have top left number = 1*1 + 2*10 + 4*5 = 21. Top right number = 1*5 + 2*10 + 4*1 =29. For example taking the 1 6 15 20 15 6 1 row we have top left = 1*1 + 2*15 + 4*15 + 8*1 = 99. Top right = 1*6 + 2*20 + 4*6 =70. Not obviously useful, but it seems to make things more complete.
Ricardo Guzman: Another cool property of Fibonacci numbers: Take any 3 consecutive Fibonacci numbers: 55,89,144. The difference of squares of the larger two, divided by the smallest, is the next Fibonacci. .... Thus, in interesting ways the Fibonacci numbers are intertwined with the squares.
CM63: This suggested the attached figure to me.
drive.google.com/file/d/1yjp6...
In reply I suggested to extend this picture a spiral using these identities: phi^2=phi+1, phi^3=phi^2+phi, phi^4=phi^3+phi^2 :)
According to this note on the relevant wiki page tinyurl.com/yv3fnac2 if you take overlaps of the Pythagorean tree into consideration the area of the tree is finite.
Today's music: Antionetta by Boreís and Dark tranquility by Anno Domini Beats
Today's t-shirt: google "Fibonacci cat t-shirt" for a couple of different versions. I just bought this t-shirt from somewhere but I think the cat is supposed to be superimposed onto this type of Fibonacci spiral tinyurl.com/2s3p7e3v
Enjoy!
Burkard
One neat fact that was left out is that when the four numbers in the box are viewed as fractions, the two fractions are equal to the tangents of half of each of the two acute angles of the triangle.
Glad you mentioned that one :)
Yes, John, that is a rather nifty find: u/v = Tan(Atan(Y/X)/2) and (v-u)/(v+u) = Tan(Atan(X/Y)/2)
Despite serious competition, Mathologer remains the greatest math channel imo ^^ Thanks for another awesome video!
Glad that you think so :)
IMO Mathologer perfectly hits the balance between presenting topic in depth and in interesting way. There are other channels more attractive in form, there are channels discussing math deeper, but here I can follow up what's going on, and I want to know where it's going on.
I'm not sure I agree, but I'd definitely put it in the top 3.
Top three of NBA all-time best players are completely interchangeable. Same with math KZhead.
@@PhilBagels what are the other 2? (mine are 3Blue1Brown and Flammable Maths (I would say Flammable Maths produces absolute shit now but used to make the best content ever so I am also ranking him) and 3Blue1Brown's quality has dropped a bit, but it still good but his upload frequency is also low) while Mathologer quality is improving and better than ever
My father was a carpenter. He built various buildings, and the last one was a cottage where I helped. We checked that the corner was at right angle using the 3-4-5 measurement.
My father was a civil engineer :)
Just remember to use multiplied triangle instead of measuring just up to five what ever units you're using. For example measure to 30, 40, and 50. The larger the triangle, the less chance there is for the inevitable measurement error when doing it haphazardly by hand.
In most cases it would be quicker & good enough - and in some cases better - to use a set square (wie ein Geodreieck) instead. If the walls don't meet at an exact right angle, then perhaps they are not completely straight as they go up either. In this case the measurement ought to be repeated at various heights...
@@Mathologer Hey! So was my uncle!
@@FLScrabbler And then you discover that it's square at three feet off the floor and at no other height.
24:22 For the 3,4,5 triangle the line connecting the incenter and the Feuerbach point is parallel to the shortest side. Thus, the parent triangle of the 3,4,5 triangle is the degenerate 0,1,1 triangle (and its clear why this construction cannot be taken further).
degenerate as it seems, does it help to think of 2,1,1,0 as 1.618,1, 0.618,0.382 ?
That's it :)
@@danielhmorgan There's the rub. Answering it from this perspective leads to analysis which apparently leads to centuries of excited confusion lol
And the parent of the (0,1,1) triangle is the (0,0,0) triangle, which remains its own parent, ad infinitum. Addendum: except it doesn't... 🙁 See below.
@@landsgevaer lol man, empty set? With the trivial solution, 0? It's better to put a generator with epsilon instead of 0 0 0 Xd:dx/dt
Regarding the tree puzzle at 15:50: The bottom square's area is 1. The Pythagorean theorem states the two squares attached it share the same total area, so they are each 1/2. The total area so far is then 2, with 1 contributed from the big square and 1/2 + 1/2 = 1 contributed from the small squares. The next level down, the tinier squares attached to one of the small squares has to add up to 1/2, so they are each 1/4. There are 4 of them, so the total area of these squares is 1, and the total area is now 3. You continue down the line, adding 1 to the total area for each iteration of the tree. There are 5 iterations, so the total area is 5.
That's it :)
Well thought man
@@yanniking7350 its not that hard
Didn’t think to iron out such a simple problem myself, but for those in tow you did good work. 🤟🏻
I got the same answer, through the same reasoning 😃👍🏻.
I've been fascinated with Fibonacci numbers and Pythagorean triples since I discovered them when I was about 8. 45 years later you taught me some new things and helped me understand the "why" behind some of what I already knew. Thank you.
That's great :)
@@Mathologer Why or how would anyone think of making these connections or manipulations though? Hope you can respond when you can.
YOU discovered them?? Thank you for your contributions to mathematics!
7 y
@@johndoe-rq1pu I think he means he came across them. Of course, I realize your comment is sarcasm, and you knew perfectly well, what he meant. Just thought to point that out.
16:00 I found the area of the tree to be 5, since we have a depth of 5 and for every iteration, the new squares sum up to 1, thus a a tree with infinite iterations has an infinite surface area (still counting overlapping surfaces)
In general, area of tree of depth x is x.
What is interesting though is that, as the number of levels goes to infinity, so does the area of the tree, but it never leaves a finite bounding rectangle so it ends up overlapping itself infinitely.
Is there a formula, which takes the overlapping regions into account, and calculates the real visible area?
@@margue27 I imagine it to be complicated, but I’ll work on it. For n=6 the area is 6-1/16
11:56 challenge question: The Pythagorean triple (153, 104, 185) corresponding to the box [ 9, 4, 17, 13 ]. If you call the children A,B,C, the 153, 104, 185 is the A'th Child of 'CCC'
@Mathologer another interesting maths fact, ... which i saw originally in a Norman Wildberger video, is that the conic tangents of a cubic or odd degree polynomial don't over lap. Actually he's asked the question if every point in the plane lies on one of these tangential conics and appears not to know. (but actually it fairly obviously follows from what he'd already written on the board. ) I'm a bit disorganised about keeping records and it was a long time ago I saw it, so it would take me quite a long time to find it exactly. But I did write some python code at the time to generate a picture of the tangent conics which I was able to find and will cut and paste in my next reply. I find it rather amazing that these tangent conics map the whole plane and so in some sense any cubic equation represents a mapping from one plane to another which is bijective on the whole plane. If it's interesting let me know and I will try and hunt down the original source video.
@@richardfredlund8846Maybe Numberphile's "Journey to 3264" would help- that (not the conjectured 6^5) is the maximum amount of tangents for 5 conics.
@@wyattstevens8574the tangent conics of a cubic are quadratics... if you can get the python code to work, then you will see (although that's not a proof ) that they fill the plane. The irony is Wilderberger is pretty close to proving the result (i.e. the information he puts on the board, in the video) but when on of his students actually poses the question, he apparently doesn't know. And it is a rather remarkable result.
@@wyattstevens8574 it's approx min 23 of Tangent conics and tangent quadrics | Differential Geometry 5 | NJ Wildberger (vidoe on youtube)
@@wyattstevens8574 the argument for the cubic : 23:26 every point does have a tangent conic going through it because for any arbitrary error k at x, and non zero coefficient d, there exists an r s.t. d*(x-r)^3 = k . so there is a maping from coordinates x,y to the cubic reference frame, x,r to prove the 5th and higher odd power polynomials, probably follow from the fact you can get any y value by inputting the correct x value to an odd order polynomial ( unlike even powers ).
I am not a studied mathematician by anyone's measure. Yet, I carry away so much from your videos! Thank you so much for your well-constructed presentations. This one was wonderfully startling!
Awesome, Mathologer! Another ab-so-lute-ly delightful journey. Apart from its important and most beneficial applications, Mathologer never ceases to amaze with another revelation on how maths has just this amazing beauty and harmony in itsself. This channel is such a gem on YT. ✨Thank you very much, indeed, Sir. 🙏
Glad you like the videos so much :)
The beauty of the interconnectedness of mathematics
This is amazing. Everything truly is connected to everything. I could hardly have been more surprised if Pascal's triangle had also made an appearance. 🙂
To see how Pascal's triangle relates to all of this, you have to also introduce Euler's constant (the base of the natural logarithm), tau (the ratio between a circle's circumference and its radius in the Euclidean plane), the golden ratio, the zeta function, and the distribution of prime numbers. At that point the only things left to connect are the planck length, the speed of light in a vacuum, quantum chromodynamics, and gravity; and those connections remain undiscovered, last I checked.
@@jonadabtheunsightly 🤣
@@jonadabtheunsightly - dig out the totally volatile fine-structure-number and you will advance a good bit.
I'm pretty sure 345 is connected to π. How sure? 110%
@@jonadabtheunsightly Also, those undiscovered connections are physicists’ territory. Mathematicians can’t be arsed to discover them (which is, why they’re still undiscovered, I presume). 😅
Every single video of yours has so many interesting mathematical connections! I always get excited while watching them!
My favourite way to calculate primitive Pythagorean triples is to just use Euclid's formula that 3blue1brown showed us, using two coprime integers that are not both odd, and in fact, the two integers you need to run through Euclid's formula to get a specific Pythagorean triple can be found in the right column of the Fibonacci box corresponding to said primitive Pythagorean triple, and this always works for any such box you choose.
Whenever I'm finding myself lost or at a dead end with my own mathematical work you seemingly post a video that helps me along my path. Thank you.
At first, the fact that the tree contained every irreducible fraction broke my mind. Then all of a sudden it seemed obvious. I can't explain why though.
What is the Dedekind cut? How does it relate to this sequence of irreducible fractions
I know the feeling 😅.
I was moved by this, almost to tears. what a great treatment of a rich subject. Thank you, thnkyu, thku, thx...
Truly amazing. Finding wonderful hidden connections among the known things.
As always, an excellent video. After watching one of your videos, I'm always left with an unformed thought, like an itch you cannot scratch. I feel we are seeing glimpses of some universal truth that we still cannot see completely or understand. It is a frustratingly delicious feeling. Thank you.
I’m stunned. What a sublime concept, especially the animations that produce the cool little fractal trees
Man I've been waiting FOREVER for someone to bring this up!! This pops everywhere in quantum mechanics, Apollonian Gaskets, Ford Circles, Fractals..You Name it! Always had the feeling that the Theory of Everything would somehow be related to this! We need to keep it alive at all costs! THANK YOU!
Others have answered already, but the "153² + 104² = 185²" triple has a matrix [9 4; 17 13]. I enjoyed programming this one using Julia. The path to get there from (3,4,5) is right, right, right, left. Triples along this path: "3² + 4² = 5²" "5² + 12² = 13²" "7² + 24² = 25²" "9² + 40² = 41²" "153² + 104² = 185²" Thanks for the challenge and the very interesting video!
Very well explained, with those graphics, it's a question of watching this carefully. Bravo signore!
thoroughly enjoyed .... and I will never stop of being amazed and surprised of how a theorem that it's the exception to another has so many dimensions of its own.
4. The necklace has 12 parts of equal length between the big pearls (? idk), which can be streched into a 3-4-5 triangle to check if an angle is right. Ropes like these were used in ancient Egypt to make right angles, though they weren't so cool-looking, just ropes with knots
Wow
Elegant!
The 153, 104, 185 triple at 12:00 is the box 9, 4, 13, 17 and you get there by navigating right right right and left :)
Correct :)
the factors of 153 and 104 made it rather easy!
Great stuff!! Thanks for such insightful videos.
Congretulations! One of your best videos!
Fiboghoras and Pythanacci, my favorite duo
Mathologer videos are always such an inspiration to me. I'm no mathematician, but I enjoy a bit of mathematical dabbling. Most of my exploration is what might be called empirical mathematics. In short, I look for patterns without bothering too much about proofs. To test my pattern-finding ability, I paused the video at 29:01, to see whether I could identify the next few members of the family. I got the following: 9² + 40² = 41²; 11² + 60² = 61²; 13² = 84² = 85²; 15² + 112² = 113²; 17² + 144² = 145². The general pattern could be expressed as (2n + 1)² + (2(n² + n))² = (2(n² + n) + 1)², where n is a natural number. The pattern for the family shown at 30:12 was even easier to identify. The next few members were: 63² + 16² = 65²; 99² + 20² = 101²; 143² + 24² = 145²; 195² + 28² = 197². The general pattern for this could be expressed as (4n² − 1)² + (4n)² = (4n² + 1)², where n is a natural number.
Glad you are having fun :)
Wow Mathologer and Standupmath videos on the same day! great work as always! many thanks and respects Mathologer!
Absolutely stunning. Thank you!
Here it is... Another interesting and deep connection between two (seems) "basic" things in Math. Really, this is all about: It's about deep connections, not just like "Did you know that a cup and a torus are equal, technically?". Math is beautiful with all its concepts and etc. what really amazes me are these connections. Man, It is such an *art* . It IS worth to spend the entirety of life with Math, really... No matter how it can be challenging sometimes.
I am living the dream :)
@@Mathologer Yeah, and I wanna live it too. *Wait for me sir, I'm on the way.*
Hello sir , Namaste, I am big fan of your teaching.
Very, very cool stuff. I suspected that old ppt generator would be the mechanism behind this dense and elegant tree. The equivalence between adding two numbers and drawing a triangle remains fascinating thousands of years later. And the circle stuff is pure magic.
Toll erklärtes KZhead Video. Ein echtes Highlight. Danke.
Best content on The Tube (and elsewhere)!
Glad that you think so :)
Wow, so this sequence stands as another proof of infinite pythagorean triples! (As the Fibonacci sequence is, itself, infinite) Neat!
This is a MUST VIDEO for all professors in mathematics ! ❤ Mathologer makes my retirement very colorful. Thank you very much.
After having spent years of my life watching math videos like this one, I've concluded that math only has 3 original ideas and the entire field is just their remixes.
24:13 The location of the center of Feuerbach circle for the 3-4-5 triangle is on the same horizontal line as the incircle, therefore the outlined procedure will produce a degenerate triangle (a line).
So a 1^2+0^2=1^2 triple, which corresponds to the first two terms of the Fibonacci sequence 0,1
@@bscutajar I think that's just a coincidence. If you follow the box construction for the numbers 0, 1, 1, 2 you get a Pythagorean triple 0²+2²=2², and if you try to follow the tree structure none of the three children are our 1, 1, 2, 3 box.
15:58 Since each new pair of squares corresponds to a right triangle with the hypotenuse of the previous square's side length, and due to Pythagoras, the sum of the areas of these squares is equal to the area of the square in the previous generation. Therefore each new generation of squares has a total area of 1. Since there are 5 total generations in this tree, the area of the tree is 5. It's always cool when fractals turn out to have infinite area.
Yes in terms of the 5. However from some point on (beyond 5) the leaves of the tree start overlapping and then the question is whether there is enough overlap to make the total are covered finite after all. Have to think about this/look it up at some point :)
@@Mathologer This is actually a pretty interesting question. It's beyond me at the moment, but I'm sure that there's a very elegant solution somewhere out there.
@@insertcreativenamehere492 would there be more or less overlap if it was translated into 3 dimensions?
Interesting stuff, this video and the one about Moessner miracle are some of the most insightful videos I've encountered in KZhead
Love your work! Thank you!
The are of the tree @16:00 I’m assuming the angle is 45 degrees. The side length of the trunk is 1. (1x1=1) Focusing on just the first branch we know that the side length is sqrt(0.5). (sqrt0.5 ^2 + sqrt0.5 ^2 =1) So the area of one first branch is 1/2 (sqrt0.5 x sqrt0.5). But there are 2 of them so the total area of first branches is 1! By similar logic the area of all second branches is 1, and so on. Total area of tree =5.
The angle actually doesn't matter! As long as it's a right triangle, the sum of the areas of the 2 small squares is equal to the area of the big square, so the area of each layer is equal to the area of the "root" square: 1 and the total area is the number of layers
@@briourbi1058 Thanks!
I am Dutch and a huge fan of the Mathologer way of explaining mathematics. F.J.M. Barning is Fredericus Johannes Maria Barning, in daily life he was called Freek Barning. He died in 2012.
Thank you very much for sharing this with me. Anything else you know about him?
My condolences and thank you for sharing this, I had been looking for this information myself and failed to find it.
From a reaction on another comment I found your update in the description with a translation of his info on this website, love your dedication! Want to add also: Thanks for bringing wonderful mathematical facts into my life :)
@@Mathologer you probably already know this with your German accent, but just in case: "Freek" is pronounced as "Frake" and rhymes with "Lake". Please don't mispronounce the Dutch name "Freek" as "Freak", this would be very wrong.
@@Mathologer Very quickly: he was an employee of the Mathematics faculty of the University of Amsterdam and later onwards was the associate director of the same institute.
Thanks for this. Always love your videos on a lazy Sunday afternoon
This is the most compelling proof I've ever seen. It's truly miraculous as you say.
Im letting this play when I fall asleep so I can have mathematical revelations in my dreams
Mind blowing BTW at 39:25 it looks like these fibonacci-generated pythagorean triples have a pattern you missed: A alternates between 1 over and 1 under B/2, in a similar way to those other families in the tree. Makes me wonder what other families of triples can be generated from other sequences compatible with Euclids formula
This is actually one of a whole family of Fibonacci identities, related to the Cassini identity! In this case, what you’ve noticed is the identity (which does hold, so good eye): F(n+2)F(n+1) - F(n+3)F(n) = (-1)^n This is best seen as a special case of d’Ocagne’s identity, which has a simple proof (read: short with little algebra autopilot) using determinants. This general identity is: F(m)F(n+1) - F(m+1)F(n) = (-1)^n F(m-n)
As always, you continue to amaze
Another gem!! Here's a little side-note -- At 28min+, where you go into approximating √2 with PRT's that have legs that differ by 1 (b = a+1), you can get "best" rational approximations by adding the legs and using the hypotenuse as denominator (this amounts to averaging the two legs): √2 ≈ (a+b)/c = (2a+1)/c = (2b-1)/c For the 3-4-5 PRT, this gives √2 ≈ 7/5 For the 20-21-29 PRT, this gives √2 ≈ 41/29 For the 119-120-169 PRT, this gives √2 ≈ 239/169 All of these are solutions of the Negative Pell's Equation for N = 2: (num)² = 2(denom)² - 1 which makes them "best" rational approximations of √2. Fred
Drat! Looks like I lost my "heart" by editing in some extra explanation. :-(
@@ffggddss We cannot have that. Just gave you two hearts back :)
@@Mathologer Beaucoup thanx for all 3 hearts! And for churning out such marvelous videos!
One more tidbit - that relation involving 4 consecutive Fibonacci numbers (≈40min) can be boiled down as follows. Decrement the indices [i.e., replace n with n-1 throughout], then put all four of them in terms of the middle two: F[n-1] F[n+1] + F[n] F[n+2] = F[2n+1] (F[n+1] - F[n]) F[n+1] + F[n] (F[n] + F[n+1]) = F[2n+1] F[n]² + F[n+1]² = F[2n+1] Which may be a bit more familiar to Fibonacci aficianados.
11:50 ab = 153 2cd = 104 ad + bc = 185 For my fourth equation I decided to use the radious of the in circle: ac = r I have one small issue that being i completely forgot the formula for the in circle so i made my own: I remember how to graphically find the center of the in circle so knowing that I will create 2 function which each draw one of the lines that halve one of the angles of the traingle. Where my functions overlap is the center of the in circle. I imagined the triangle orientated as show in the video intro and set the origin of my coordinate system to the most left point of the triangle. My functions draw the lines which halve the alpha and beta angles. 1. equation: f1(x) = x*tan(alfa/2), alfa = arctan(153/104) 2. equation: f2(x) = -(x-104)*tan(pi/4) = -x + 104 Combining f1(x) = f2(x), solution x = 68 The radious r = f2(68) = 36 ac = r = 36 Using the 4 equations I found the 4 variables to be: 9, 4 17, 13 ------------------------------------------------------------------------------- Would it be much easier if I also remembered: a+c=d c+d = b ? Yes. Yes, it would have been much easier.
I did recall the last two equations but the closed form looked to ugly to solve so I opted for the easy way out using Excel and arrived at your solution.
The solution is easier: d = a+c, b= c+d= 2c+a. The second equation from here: 2ac+a^2=153. The second equation from here: 2c(a+c)=104 => c(a+c)=52. Note: 52 = 2*2*13. Consequently c = 2 or 4. Then a = 24 or 9. The first equation is true if c = 4 and a = 9.
Master! You have to work a lot to make these awesome animations and it's a demoivre set
What an amazing contribution !
Excellent, thanks. You put so much work and I will be ver so grateful for your explanations
The necklace made me think of the jewel-division problem from an earlier video. I'm now imagining a band of burglar-mathematicians trying to divide a necklace into equal-value pythagorean triples.
I was thinking about that Pythagorean triple tree. If one associates going straight with a 1, going left with a 2 and going right with a 3, then one can encode every primitive Pythagorean triple (PPT) with a rational number between 0 and 1 in base 4 as follows: 0 corresponds to (3,4,5), 0.1 to (21,20,29), 0.2 to (5,8,17), 0.3 to (5,12,13), 0.12 to (77,36,85) and so forth. That way one gets a 1-1 correspondence with the finitely representable base 4 numbers between 0 (included) and 1 that do not contain the digit 0. The three families of PPTs correspond to the numbers 0.11111..., 0.22222 and 0.3333... I wonder if one can get anything geometrically meaningful with that correspondence. Can one interpret the periodic or irrational numbers as interesting infinite paths/families of PTTs in the tree? What about interpreting numerical manipulations like multiplication of 0.1111 with to to get 0.2222 in terms of the associated PTTs?
Nice idea! Similar to the base 3 representation of the Cantor set? If you present it like this, why not put them in Sierpinski's gasket?
This is the coolest most beautiful math video I've ever seen. Thank you
Observation: At 27:17 you ask "Are there any isosceles triangles with integer sides?", and then prove to the negative. But you already touched this earlier mentioning that the incircle does not touch the excircles. For a right-angled isosceles triangle, the incircle would touch an excircle in the midpoint of the hypotenuse.
I'm only halfway through the video but so far I've learned a new geometric fact every 30 seconds. This is absolutely insane.
This was one of the best videos yet!
Glad that you think so :)
Stunningly beautiful!
@24:20 is a little tougher I think if you try to take it’s parent you get a square of 1,0,1,0 So I suspect the Feuerbach centre is directly on the incircle. In other words you get a 0,0,0 pythagorean triplet. Which... works? For some definition of work!
You get a 1^2+0^2=1^2 triple.
Question for Mathologer: Why does the Fibonacci sequence start with 1 1 and not with 0 1 ? I'm guessing it has something to do with the reputation had by zero at the time Fibonacci discovered his sequence.
Actually, some people start the sequence: 1, 2, 3, 5, 8, ... . The 1, 1, start is very natural in a number of different ways, but the 0,1 start is also perfectly fine in my books. Of course, you can also extend the sequence beyond 0 as far as you wish ..., -3, 2, -1 , 1, 0, 1, 1, 2, 3, 5, ... and so infinitely many more starting pairs are possible :)
As long as you have the intervals, you also can go not only zero but negative numbers. It's not the "number" but the interval. If you can do a linear transformation, then you can get out of negative zone just by summing a number, like if you start at - 1, just add 2 and you will be in positive area, while keeping intact the intervals
Where did the first baby rabbit come from? 🤔
If we let F_1 = 1 and F_2 = 1 then we have the wonderful fact that gcd(F_m, F_n) = F_gcd(m,n) where "gcd" stands for "greatest common divisor." If we let F_1 = 0 and F_2 = 1 then this formula doesn't look as pretty any more. But if you like, you can let F_0 = 0 and then you can start the sequence with 0 if you want.
@@jkid1134 That's an excellent analogy. At some point its ancestors weren't rabbits anymore, and that's where we leave the natural numbers: at 0 and the negative integer extension of the sequence. 🙂
Glad i found this, i'm going to explore everything on this channel! Looks fantastic! thanks :-)
Excellent! Math and math videos are a treat. Thanks Mathologer.
You're very welcome!
GENERAL FORMULA for puzzle 11:54: Using the information from the video and Poncelet's Theorem, I managed to create some formulas to calculate the elements of the box from any Pythagorean theorem. If the box is distributed like this: A B D C And Pythagoras like this: C1^2 + C2^2 = H^2 We can find the elements of the box as follows: X = H +_ sqrt{ H^2 - 2*C1*C2 } Y = C1 + C2 - H A = sqrt{ (C1*Y)/X } B = sqrt{ (X*Y)/(4*C1) } C = sqrt{ (C1*C2^2)/(X*Y) } D = sqrt{ (C1*X)/Y } So, for the problem 153^2 + 104^2 = 185^2: X = 136 or 234 Y = 72 But only x = 136 gives us integers so: A = sqrt{ (153*72)/136 } = 9 B = sqrt{ (136*72)/(4*153) } = 4 C = sqrt{ (153*104^2)/(136*72) } = 13 D = sqrt{ (153*136)/72 } = 17 Clearly, the solution could be reached with the tree going 3 times to the left and then to the right, but this method is much longer especially with large sides of the right triangle. In the same way C and D are not necessary to calculate because for the Fibonacci series it is only necessary to add the two previous ones, but it also seemed useful to me to have a general formula. Added to this, something curious is that since I use a quadratic equation to solve for X, we have 2 solutions, only one of which is integer, but the decimal one still works.
what does sqrt mean????
@@M.Ghilas it means square root
dimensionless pythagoran triples. Its connections like this that feel like foundations rather than amusements. Stunning.
I just revisited this video, and found something interesting. Take the construction in the intro section, with 4 sequential fibonacci numbers 'bent' into a square. If you put in any sequence of 4 *fibonacci-like* numbers (a, a+b, 2a+b, 3a+2b; not necessarily integers; the 1st 2 don't even have to be in ascending order), you still get a right triangle, and all the math works to find the sides, area, incircle, and excircles. For instance: (3,4,7,11) -> (33,56,75) (2/√2, 3/√2, 5/√2, 8/√2) -> (8,15,17) [I'm pretty sure √2 is the only divisor that still yields a pythagorean triplet.] Only certain sequences of 4 fibonacci-like numbers will yield pythagorean triplets. An easy way to find them: use an integer sequence of co-primes. If the 1st number is even, divide the result by 2. Example: (20,21,41,62) -> (1240,1722,2122). Divide by 2 -> (620,861,1061). Edit: Incidentally, according to mathworld.wolfram.com, Dujella 1995 already noted this relationship of sequences of 4 consecutive fibonacci numbers & pythagorean triplets. Oh, and much of what I said here was covered later in the video.
Here’s my answer for @12:00 9,4,13,17 And you go Right, right, right, left
Correct :)
The true gift in the gift to your wife would be the ability to create a right angle any time she pleased. Very thoughtful of you.
I am just such a thoughtful guy ... :)
@@Mathologer You are definitely a thoughful guy. This video is prime evidence. I have been peering into the golden ratio / fibonacci for a while and never heard any of these relationships.
The videos where calculus, geometry combines are awesome
Amazing!!! Thank you so much!
You have done a great service in raising this from obscurity.
Like all your videos, I love this one and I learned lots of useful information as well! In particular, I'm quite intrigued by the ternary tree construction of PPTs. In 2012, I published a paper in The Fibonacci Quarterly entitled "Some Interesting Infinite Families of Primitive Pythagorean Triples", in which I describe four infinite families of PPTs all involving Fibonacci and Lucas numbers, as well as the three you mention in this video, corresponding to the left, middle, and right paths through the tree. Now I suspect the other four families in my paper correspond to other simple paths through this tree, most likely involving a repeating finite sequence of left, middle, and right moves up the tree. This leads to several interesting questions. For starters, what's the correspondence between these repeating patterns and the resulting infinite families of PPTs? Furthermore, as in the cases you illustrate here, do the resulting right triangles each converge on a single one, and what are the proportions [a:b:c] of these limiting right triangles? Now I'd very much like to write another paper, or perhaps a series of papers, in which I try to answer these questions. If you'd like, you can collaborate with me! Let me know if you're interested. If so, we can exchange emails.
Wow, this is so fascinating. To answer your challenge 9,4,13,17 and from the root you gotta go right 3 times then left once. Previous group is 1,4,5,9. Had so much fun!
That's great. My mission is accomplished as far as you are concerned then :)
This needs a geometric treatment. This insane matrix process had given us every rational under 1 and simultaneously every Pythagorean triple. Mind blown 🤯
This all indeed is very neat! 🎄🧡 This triple tree reminds me of a tree of Markov triples a bit, though their constructions aren’t at all analogous.
I would really like to thank you as your previous videos on fibonacci sequence, fibonacci-like sequences and the strand puzzle made me follow a deep rabbit hole and led me to discover a super generalisation of the fibonacci numbers with similar properties, I generalised the Binet formula and many fibonacci identities and used them to find general integer solutions to infinitely many pell-like equations all super fun Oh and I'm particularly waiting for the video you promised at the of the strand puzzle video
That's great. Glad you are enjoying the videos so much :)
Wow. Simple maths, but mind completely blown. Easily one of my favourite episodes. So, so beautiful.
Thank you. Absolutely beautiful.
Wow Amazing 😀 Mind blowing visuals!
Fibonacci and Pythagoras are now teaching me how to think inside and outside the box.
So many alien tricks: keeps me wondering! Fascinating and mindblowing stuff.
The total area for the figure at 15:58 is 5. At first I didn’t know how much the areas got smaller with each new color, but it’s clear that each new color diminishes by the same ratio, and at the same time the number of squares doubles. For the total area, you would then have an expression like 1+2r+4r^2+8r^3+16r^4 for the total area. To figure out r, I saw that the angle at which the squares were branching out had to be 45 degrees, since two branches made a right angle, which meant that the side lengths were decreasing by a ratio of 1/sqrt(2). This means that the area decreases by a ratio of 1/2, and plugging in r=1/2 into the expression above gives 5. Basically, each color contributes the same amount of area (1), and since there are five levels you get a total area of 5.
This is an incredible topology of harmonic relations!!
As soon as I read the Lee Price paper I thought this discovery deserves to be better known, and I could visualise those dancing circles. Thanks for sharing in such an engaging way. PS the top pairs in the Fibonacci boxes of those middle children form a series in which each pair is equal to twice the preceding pair plus the pair before that. Like the Fibonacci series the general term can be expressed in closed form. The counterpart to phi is one plus or minus the square root of 2.
Thanks again very much for alerting me to this nice topic :)
24:20 In 4:39, I noticed that the incircle actually touches one of the excircles! So the distance between the centers of the Feuerbach circle and the incircle is not divisible by 0.25. This leads to a right triangle where all the sides have non-integer lengths, which is impossible to get a Pythagorean triple from.
Fantastic video!!!
For any two integers a and b, you can find a Pythagorean triple with (a² + 2ab; 2b² + 2ab; a² + 2ab + 2b²). These won't always be primitive triples, unless of course you pick 2 consecutive numbers out of the Fibonacci series. What is more the GCD between a and be will be a factor of the GCD of the triples. This is as far as I've gotten, but it is fascinating. I hadn't gotten to the proof section yet when I left this comment. I was playing in Excel.
Madness. I’m absolutely blown away
For the solution of 153^2 + 104^2 = 185^2, I've found it with some algebra... If the terms of the sequence are f1, f2, f3 and f4, we know that f3 = f1 + f2, and f4 = f2 + f3. 153 = f1 x f4 104 = 2 x f2 x f3 185 = f1 x f3 + f2 x f4 The first one gives us: 153 = f1 x (f2 + f3) = f1 x (f2 + f1 + f2) = f1^2 + 2 x f1 x f2 From the third: 185 = f1 x (f1 + f2) + f2 x (f2 + f3) = f1^2 + f1 x f2 + f2 x (f2 + f1 + f2) = f1^2 + f1 x f2 + 2 f2^2 + f1 x f2 = 153 + 2 f2^2 Then f2 = sqrt(1/2 x (185 - 153)) = 4 And f1 = sqrt(153 + f2^2) - f2 = 9 f3 = 13, f4 = 17, and finally: (9 x 17)^2 + (2 x 4 x 13)^2 = (9 x 13 + 4 x 17)^2
And about the path, by construction only the right branch have f1 < f2. As here 9 > 4, it must come from one of the nodes where f4=9 and f2 or f3=4, but if f2=4 then f3=5 and f1 is negative, which is impossible. So it comes from a sequence 1 4 5 9. Then, as 1 < 4, all the branches until reach the root are the right ones.
153 factors to 17 * 9 and 52 to 13 * 4 so I just tested the factors in 185 = ad + bc
@@seventhtenth I started doing the same, but also 51x3 and 26x2 are factors and I've to brute force the right ones, ha.
This is the best channel on Mathematics
16:00 The area of the biggest square is 1 So we have that the area sum of the two smaller squares equals the area of the bigger square. So the sum of the areas in layer 2 (orange) is also 1. Similarly, the sums of the areas in the next three layers (tangerine to maroon) must each be 1. There are five layers in this tree, so the total area of the tree is 5. ■
11:38 Challenge Question The triple (153, 104, 185) comes from the following box: 9 4 17 13 Bonus Question From the starting square, you would go right, right, right, then left. The first 2 destination squares are shown in your video, the 3rd right would look like this: 1 4 9 5 And the final “left” is already shown in my answer to the previous question (obtained by using the 9-4 diagonal)
Nifty as always !
25:51 The Pearl Necklace Since it has 12 large pearls, it can be used to create a 3-4-5 triangle on the fly Just pick one to be the vertex of the right angle, then pick 2 others that are separated by 3 and 4 large pearls on either side to represent the other 2 vertices
Here's an interesting fact, based on the ternary tree of PPTs. By replacing left, middle, and right movements up the tree by the ternary digits 0, 1, and 2, respectively, we obtain an isomorphism between the set of PPTs and the set of 3-adic rational numbers strictly between 0 and 1, i.e., rational numbers in the interval (0, 1) with denominators equal to powers of 3. Furthermore, by considering the set of infinite paths up the tree, we also obtain an isomorphism from [0, 1] to [0, 1], given by x -> lim(u/v) where x is the real number with ternary expansion described above and u and v are the parameters in Euler's characterization of PPTs. Is this isomorphism equal to the identity? If not, what is it?
I know a lot of Pythagorean triples and I know how to calculate them and I notice that at infinity the series of triangles converge to a straight line. I have figured out how to calculate two kinds of Pythagorean triples; but THIS is very interesting! There are new Pythagorean triples I have never heard of before featured in this video