Fibonacci = Pythagoras: Help save a beautiful discovery from oblivion

[@johnklinger2868]

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.

[@Mathologer]

Glad you mentioned that one :)

[@TillerSeeker]

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)

[@thephilosophyofhorror]

Despite serious competition, Mathologer remains the greatest math channel imo ^^ Thanks for another awesome video!

[@Mathologer]

Glad that you think so :)

[@Filipnalepa]

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.

[@PhilBagels]

I'm not sure I agree, but I'd definitely put it in the top 3.

[@thedominion6643]

Top three of NBA all-time best players are completely interchangeable. Same with math KZbin.

[@DendrocnideMoroides]

@@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

[@aeschynanthus_sp]

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.

[@Mathologer]

My father was a civil engineer :)

[@anteshell]

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.

[@FLScrabbler]

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...

[@frankharr9466]

@@Mathologer Hey! So was my uncle!

[@frankharr9466]

@@FLScrabbler And then you discover that it's square at three feet off the floor and at no other height.

[@royalninja2823]

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.

[@Mathologer]

That's it :)

[@yanniking7350]

Well thought man

[@vevericac3294]

@@yanniking7350 its not that hard

[@BabaBabelOm]

Didn’t think to iron out such a simple problem myself, but for those in tow you did good work. 🤟🏻

[@PC_Simo]

I got the same answer, through the same reasoning 😃👍🏻.

[@hydra147147]

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).

[@danielhmorgan]

degenerate as it seems, does it help to think of 2,1,1,0 as 1.618,1, 0.618,0.382 ?

[@Mathologer]

That's it :)

[@haniamritdas4725]

@@danielhmorgan There's the rub. Answering it from this perspective leads to analysis which apparently leads to centuries of excited confusion lol

[@landsgevaer]

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.

[@misterlau5246]

@@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

[@johnmeyers1344]

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.

[@Mathologer]

That's great :)

[@johndoe-rq1pu]

YOU discovered them?? Thank you for your contributions to mathematics!

[@markdoldon8852]

7 y

[@PC_Simo]

@@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.

[@contrawise]

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!

[@TigruArdavi]

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. 🙏

[@Mathologer]

Glad you like the videos so much :)

[@danielhmorgan]

I was moved by this, almost to tears. what a great treatment of a rich subject. Thank you, thnkyu, thku, thx...

[@jakoblenke3012]

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)

[@JaceMillerr]

In general, area of tree of depth x is x.

[@kevinmartin7760]

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.

[@margue27]

Is there a formula, which takes the overlapping regions into account, and calculates the real visible area?

[@jakoblenke3012]

@@margue27 I imagine it to be complicated, but I’ll work on it. For n=6 the area is 6-1/16

[@DeclanMBrennan]

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. 🙂

[@jonadabtheunsightly]

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.

[@DeclanMBrennan]

@@jonadabtheunsightly 🤣

[@alexanderstohr4198]

@@jonadabtheunsightly - dig out the totally volatile fine-structure-number and you will advance a good bit.

[@jesuschristislord4538]

I'm pretty sure 345 is connected to π. How sure? 110%

[@PC_Simo]

@@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). 😅

[@sixhundredandfive7123]

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.

[@richardfredlund8846]

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'

[@richardfredlund8846]

@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.

[@wyattstevens8574]

​@@richardfredlund8846Maybe Numberphile's "Journey to 3264" would help- that (not the conjectured 6^5) is the maximum amount of tangents for 5 conics.

[@richardfredlund8846]

@@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.

[@richardfredlund8846]

@@wyattstevens8574 it's approx min 23 of Tangent conics and tangent quadrics | Differential Geometry 5 | NJ Wildberger (vidoe on youtube)

[@richardfredlund8846]

@@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 ).

[@axisjayy7625]

The beauty of the interconnectedness of mathematics

[@denelson83]

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 primitive 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.

[@iveharzing]

Every single video of yours has so many interesting mathematical connections! I always get excited while watching them!

[@BrandonWillWin]

I’m stunned. What a sublime concept, especially the animations that produce the cool little fractal trees

[@wgoode97]

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!

[@jhrhew]

Truly amazing. Finding wonderful hidden connections among the known things.

[@S.G.Wallner]

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.

[@robertunderwood1011]

What is the Dedekind cut? How does it relate to this sequence of irreducible fractions

[@PC_Simo]

I know the feeling 😅.

[@jollyroger9286]

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!

[@amoswittenbergsmusings]

And Farey sequence!

[@MrJepreis]

Wow Mathologer and Standupmath videos on the same day! great work as always! many thanks and respects Mathologer!

[@lennyvoget8725]

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 :)

[@Mathologer]

Correct :)

[@cstiger4]

the factors of 153 and 104 made it rather easy!

[@nilofido411]

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.

[@bot24032]

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

[@user-yc3fw6vq5n]

Wow

[@Geenimetsuri]

Elegant!

[@stevennunn]

Brilliant video, never stop making these, all the best, Steve!

[@AnonimityAssured]

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.

[@Mathologer]

Glad you are having fun :)

[@rogerdudra178]

Greetings from the BIG SKY. Even when I get old I find I enjoy a puzzle.

[@sinecurve9999]

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).

[@bscutajar]

So a 1^2+0^2=1^2 triple, which corresponds to the first two terms of the Fibonacci sequence 0,1

[@calvincrady]

@@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.

[@alexdemoura9972]

Fiboghoras and Pythanacci, my favorite duo

[@bentationfunkiloglio]

Best content on The Tube (and elsewhere)!

[@Mathologer]

Glad that you think so :)

[@crigsbe]

This is a MUST VIDEO for all professors in mathematics ! ❤ Mathologer makes my retirement very colorful. Thank you very much.

[@insertcreativenamehere492]

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.

[@Mathologer]

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 :)

[@insertcreativenamehere492]

@@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.

[@mrwess1927]

@@insertcreativenamehere492 would there be more or less overlap if it was translated into 3 dimensions?

[@misterlau5246]

Very well explained, with those graphics, it's a question of watching this carefully. Bravo signore!

[@jimmy685]

Wow, so this sequence stands as another proof of infinite pythagorean triples! (As the Fibonacci sequence is, itself, infinite) Neat!

[@valentinmihairusu6643]

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!

[@Mathologer]

That's great. My mission is accomplished as far as you are concerned then :)

[@brahmanandamperuri5721]

Hello sir , Namaste, I am big fan of your teaching.

[@robhillen8007]

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.

[@Nusret15220]

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.

[@Mathologer]

I am living the dream :)

[@Nusret15220]

@@Mathologer Yeah, and I wanna live it too. *Wait for me sir, I'm on the way.*

[@phiarchitect]

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.

[@Mathologer]

I am just such a thoughtful guy ... :)

[@phiarchitect]

@@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.

[@seanm7445]

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.

[@briourbi1058]

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

[@seanm7445]

@@briourbi1058 Thanks!

[@antosandras]

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.

[@reekrev]

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.

[@Mathologer]

Thank you very much for sharing this with me. Anything else you know about him?

[@willemvandebeek]

My condolences and thank you for sharing this, I had been looking for this information myself and failed to find it.

[@R4ndomNMBRS]

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 :)

[@willemvandebeek]

@@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.

[@y.herstein2199]

@@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.

[@bens4446]

You have done a great service in raising this from obscurity.

[@jusinocasino11]

This is the coolest most beautiful math video I've ever seen. Thank you

[@nejcvranic3421]

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.

[@thorntontarr2894]

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.

[@vladimir_woland]

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.

[@beautifulsmall]

dimensionless pythagoran triples. Its connections like this that feel like foundations rather than amusements. Stunning.

[@cauchym9883]

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?

[@KanaalJo]

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?

[@geon79]

11:47 The box is 9 - 4 -13 -17. To reach it from the bottom (1-1-2-3), go right (1-2-3-5), then right (1-3-4-7), then right again (1-4-5-9) then left (9-4-13-17). In summary, R-R-R-L. I'm in awe! Thank you for sharing all this wonderful math.

[@Ken-no5ip]

Im letting this play when I fall asleep so I can have mathematical revelations in my dreams

[@bradforddavis6497]

The people making math curricula seriously need to tune in to this channel.

[@martinmontero3517]

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.

[@Honey_B_River]

what does sqrt mean????

[@martinmontero3517]

@@Honey_B_River it means square root

[@whydontiknowthat]

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.

[@seanm7445]

@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!

[@bscutajar]

You get a 1^2+0^2=1^2 triple.

[@Pedritox0953]

The videos where calculus, geometry combines are awesome

[@dl9ar245]

Toll erklärtes KZbin Video. Ein echtes Highlight. Danke.

[@gcewing]

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.

[@ffggddss]

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

[@ffggddss]

Drat! Looks like I lost my "heart" by editing in some extra explanation. :-(

[@Mathologer]

@@ffggddss We cannot have that. Just gave you two hearts back :)

[@ffggddss]

@@Mathologer Beaucoup thanx for all 3 hearts! And for churning out such marvelous videos!

[@ffggddss]

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.

[@jacksonstarky8288]

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.

[@Mathologer]

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 :)

[@misterlau5246]

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

[@jkid1134]

Where did the first baby rabbit come from? 🤔

[@JohnDoe-ti2np]

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.

[@jacksonstarky8288]

@@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. 🙂

[@sidds2004]

This is the best channel on Mathematics

[@KyleDB150]

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

[@maxweinstein1537]

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)

[@HatterTobias]

Interesting stuff, this video and the one about Moessner miracle are some of the most insightful videos I've encountered in KZbin

[@ruferd]

I'm only halfway through the video but so far I've learned a new geometric fact every 30 seconds. This is absolutely insane.

[@karlslavinjak5904]

Congretulations! One of your best videos!

[@misterlau5246]

Master! You have to work a lot to make these awesome animations and it's a demoivre set

[@martw3240]

ps ,I am a tiler, to get around non square rooms, you tile off the centre lines. one wall may be square, to another , and yet the other walls may not agree with your initial mark out. Working off centre lines, you fade out the difference towards the edges

[@Mathologer]

Interesting :)

[@seanm7445]

Here’s my answer for @12:00 9,4,13,17 And you go Right, right, right, left

[@Mathologer]

Correct :)

[@alexdemoura9972]

He was a member of the first Dutch Math Group to attend a course of a brand new revolutionary computer language ALGOL, in Darmstadt, Germany, 1955.

[@zaprod]

This is the most compelling proof I've ever seen. It's truly miraculous as you say.

[@jkid1134]

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.

[@oliverpackham6278]

Holy cow, the pythagoras fibonacci thing works for other variations too! Look! If you start the fibonacci sequence with 3 and 4 and check the 29, 18, 11, 7 part it makes 203^2 + 396^2 = 445^2 which checks out.

[@charliearmour1628]

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.

[@xXstealseekerXx]

I started to play around with fibonacci sequence after you demonstrated how to bend it. I wanted to make a zig-zag pattern and see it if it takes me anywhere. And I got something like this: 1 1 2 3 5 8 13 21 34 55 89 144 ... If the zig-zag pattern is split down the middle you get two sequences: 1 2 5 13 34 89 ... and 1 3 8 21 55 144... First sub-sequence is sequence of odd terms of fibonacci (F1 F3 F5 ...) and second one is sequence of even terms (F2 F4 F6 ...). Now if you look closely you can see that to compute n-th term in odd terms of fibonacci sub-sequence you can double the n-1 th term and add all the previous terms: 1 + 2*2 = 5 1 + 2 + 5*2 = 13 1 + 2 + 5 + 13*2 = 34 1 + 2 + 5 + 13 + 34*2 = 89 Similarly you can do this with even terms: 1 + 3*2 = 7 1 + 3 + 8*2 = 20 1 + 3 + 8 + 21*2 = 54 1 + 3 + 8 + 21 + 55*2 = 143 and if you add 1 you get the right terms =) I am not sure if this is a known fact so I hope someone here in comment section can tell me.

[@Mathologer]

Glad you are having fun there. Yes, all of this is well-known to mathematicians. Still, a good idea and so keep doing these sort of experiments and eventually you'll find something new :)

[@Muhahahahaz]

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

[@SaturnCanuck]

Thanks for this. Always love your videos on a lazy Sunday afternoon

[@jsmunroe]

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.

[@mlmimichaellucasmontereyin6765]

Wow!!! Bravo! Once again, you guys rock the "world" of maths to the core. For exampkle, I finally see the way to use geometry + graphic programmiong to find the exact location of each primal positive number ('prime') n (=p) in the sequence n + 1 of N => positive infinity. For example, since all p = 6n +/-1 and there are only primals, coprimes, and pseudo-primal composites at 6n +/-1 then, in any decan of N at magnitude/cardinality M/C, we can check for primality by using the pythagorean-fibonacci geometry (PTG) rule. In other words, by progressing along the number line of N+ (or R+), we can eliminate multiples of n & p, yet also check for primality at 6n +/-1 by using the PTG Rule. Voila! We find no mystery of primal numeric logic or locations of noncomopsites p, and no mysterious patterns of p (determined by the symmetries and regularities of the preceding composites n). Clearly, this verifies my 2017 insight (& mapping). The noncomposites p are gaps in the sequences of composites n, due to the result of dyadic arithmetic continuation of n + 1. This also confirms the intrinsic interdependence of geometry and "numbers" as expressions of geometric-numeric logic, enabled by the natural metalogical principles of being (the cosmos, or life). QED. For more extensive consideration andf/or discussion, see my preprints (at ResearchGate .net). Thanks & best of luck etc. ~ M

[@wyattstevens8574]

My approach to (153, 104, 185) : First, I knew how to solve a PPT for m and n (m > n >= 1) based on the eventually revealed Euclid formula: odd leg+hypotenuse= 2m² => 2m²=153+185=338 => m²=169 => m=13 Hypotenuse-odd leg= 2n², but I already know m=13: 104=2*13n=26n N= 104/26=4 Seeds= (13, 4) Box has 9/17 and 4/13 (this is just a compact way to say the matrix) Inradius=36 I(B)= 68 I(A)= 117 I(C)= 221 I brute-forced the tree given all 4 upper bounds, and terminated each box that finally had one of the values too big. I ended up finding RRRL as the sequence.

[@binathiessen4920]

the area of the tree at 16:00 is infinite assuming that overlaps are excluded. The first square has sides of length 1 and area 1. The next square has sides of length 1/sqrt(2) and area 1/2, and there are 2 of them, so the area of the next step is also 1. Each step the area of each square is divided by 2, and the quantity of triangles is multiplied by 2, these effects cancel out, and so the area of the tree is equal to the number of steps in the sequence, and so the area grows without a finite limit.

[@PC_Simo]

15:55 The total area of the tree is 5. Each smaller square has half the area of the previous square; giving the formula: 1 + (2 * 1/2) + (4 * 1/4) + (8 * 1/8) + (16 * 1/16) = 1 + 1 + 1 + 1 + 1 = 5 * 1 = 5. 🟩🌳

[@sapegrillansa9531]

Glad i found this, i'm going to explore everything on this channel! Looks fantastic! thanks :-)

[@orchestrain88keys]

Great stuff!! Thanks for such insightful videos.

[@matthewschellenberg8969]

This was one of the best videos yet!

[@Mathologer]

Glad that you think so :)

[@sankarsana]

Excellent! Math and math videos are a treat. Thanks Mathologer.

[@Mathologer]

You're very welcome!

[@Muhahahahaz]

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)

[@Drachenbauer]

16:00 Area of this tree : 5 there are symmetrical triangles (half squares) between the colorful squares, so it´s pretty easy to see, because all squares of one growth step have the same size. Each square-size (growth-step) sums up to 1. And there are 5 Square sizes displayed so far. But the area thing also works the same for asymmetrical trees, like the one that looked like perfect to support a swing on it´s bigger side or that cristmas tree that also looked like a fern leaf. It´s just harder to see because squares of the same growth step wary in size. You can make that easyer by using different couors for each growth step as shown (but it´s a bit broken for the cristmas tree here...)

[@ricardolichtler3195]

Mathologer is the very best channel ever

[@AdrianBoyko]

iSeeker says, on StackExchange: “For those interested in the origin of spinors, one might take a look at a couple of papers by Jerzy Kocik, in which he argues that Euclid’s parametrisation of Pythagorean triples is (by a couple of millenia!) "the earliest appearance of the concept of spinors”

[@Mathologer]

That's an interesting observation :)

[@tree_relics]

This is an incredible topology of harmonic relations!!

[@BartvandenDonk]

The nice thing about this is that it doesn't change if you're using decimal, hexadecimal or which other number system you choose. It is always true. The magic with circles and triangles is very intriguing.

[@Mathologer]

In general, mathematical equations and identities hold true irrespective of your choice of representing numbers :)

[@briourbi1058]

38:45 if you take the sum of the numbers on a row ( for exemple 3+4+5) you get the number in the middle of the row above it: 3+4+5=12 ; 5+12+13 = 30 ; 16+30+34 = 80

[@Scrolte6174]

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.

[@jursamaj]

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.

[@e11e7en]

Madness. I’m absolutely blown away

[@05degrees]

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.

[@admarquis]

About the parametrization of the unit circle using Pythagorean triples scaled down to have a hypothenuse of unit length: Take the first box, about the 3-4-5 or 4-3-5 triangle. The fractions 1/2 and 1/3 are the Y-axis intercept (the parameter t of the parametrization) of the line between the point (-1,0) and the points (3/5, 4/5) or the (4/5, 3/5) respectively that exist on the unit circle. Each box is for one triangle, representing the two possibilities of which side lies on x axis. This is a very rich subject! This rational, pure math parametrization is explained beautifully by Norman J. Wildberger on his KZbin channel. Also there's another subject worthy of Mathloger exploration& perspective perhaps one day, the Miller Protractor and the core circle, x²+y²=x as key between projective (2 variants, an eighth of a turn of rotation away of each other, basically the x=y asymptote of the hyperbola of the red projective quadratic norm x²-y² rotated on the coordinate axis to yield what Norman calls the the green geometry based on the projective quadratic norm 2xy, (x²+y²)²=(x²-y²)²+(2xy)² is the among founding identities of chromogeometry) and the Euclidean, x²+y² -based blue geometry, it is marvelous. 😁 Thanks again for the great video I have been waiting for a long time! 😁🙏🙋

[@didierblasco8116]

What an amazing contribution !

[@martw3240]

the run towards a straight line of pythagorus triples is easy to see in either direction, 345 is the clue. A balanced rise over run would suggest an integer relationship of 1/root2 ,which is clearly untrue

[@bestcreations4703]

For the puzzle at 11:37, I was too tired yesterday to get it, but now that I had time today I gave it another shot. My square was defined as: ┌──┬──┐ I defined the layout of any pythag.. triple as:. │B1│B2│ X^2 + Y^2 = Z^2 . ├──┼──┤ . │B4│B3│ . └──┴──┘ . I started out pretty standard with the main three equations of: B1*B4 = X 2*B2*B3 = Y B1*B3 + B2*B4 = Z I also included these: B1 + B2 = B3 B2 + B3 = B4 To define the innercircle, I used r_i to denote its radius. The equation is then: r_i = B1*B2 I started by creating an equation to find r_i. To do this, I used the slope of the hypotenuse (Y/X) to solve for an angle, divide it by two, and convert it back to a slope. This gives a slope which bisects the angle XZ. The equation is: y=tan(1/2 * arctan(Y/X)) * (x+X) (my origin is placed at the corner XY so x+X places the lines origin at the corner XZ) Using trig identities, namley half angle identity and inverse trig identities, I could simplify the expression: tan(1/2 * arctan(Y/X)) => (sin(arctan(Y/X))) / (1 + cos(arctan(Y/X))) => (((Y/X)/sqrt[1 + (Y/X)^2]) / (1 + (1/sqrt[1 + (Y/X)^2])) => (Y/X) / (sqrt[1 + (Y/X)^2] + 1) => Y / (sqrt[X^2 + Y^2] + X) This means that the equation is: y = (Y / (sqrt[X^2 + Y^2] + X))(x+X) Which is our equation for the first bisector line. Since the angle XY is 90°, it has a bisector with a slope of -1. This results in the equations: y=-x Setting them equal to find the centre of the incircle results in a simplified equation of: x = -XY / (X + Y + sqrt[X^2 + Y^2])) Since X^2 + Y^2 = Z^2, this becomes: x = -XY / (X + Y + Z)) OR: r_i = (XY) / (X+Y+Z) Plugging in the values, we get: r_i = 36 since r_i = B1*B2 = 36, this means B1 = 36/B2. plugging into the equation: B1 + B2 = B3 yields: 36/B2 + B2 = B3 => 36 = B2*B3 - B2^2 using Y = 2*B2*B3 = 104, we know B2*B3 = 52. 36 = 52 - B2^2 => B2^2 = 52-36 => B2 = 4. So, using B2*B3 = 52 we find: B3 = 13 Since B1 + B2 = B3: B1 + 4 = 13 Or: B1 = 9 Since B2 + B3 = B4: 4 + 13 = B4 Or: B4 = 17 To Conclude, my guess as to the numbers in the square are: ┌──┬──┐ │ 9│ 4│ ├──┼──┤ │17│13│ └──┴──┘

[@wyattstevens8574]

You're right! The numbers in the right column are the seeds of Euclid's formula for this triple!

[@sohammukherjee4131]

And the bonus part: on the tree, starting from the first block, go right child, right child, right child, left child

[@Muhahahahaz]

24:04 Parent of 3-4-5 Triangle? Ultimately, things “fail” here because you end up with a horizontal line of length 1/4 between the 2 centers in question. (Details below) Let’s label any given “Fibonacci square” in our tree as follows: A B D C (So that A + B = C, and so on) One property I noticed is that the hypotenuse of the parent triangle is always given by (A - B)^2 + B^2 It helps to know the traditional m and n formula for Pythagorean triples here… (m^2 - n^2, 2mn, m^2 + n^2), for all m > n > 0 Basically, our B and C are exactly n and m from this formula, which is why we take 2BC = 2mn to find one of the legs. But more importantly, the hypotenuse is given by m^2 + n^2 = B^2 + C^2 (One can prove that this formula always works out for our Fibonacci squares, even though our formulas for the hypotenuse and the other leg look like AC + BD and AD… Just use the fact that A + B = C and B + C = D, then rearrange in terms of B and C, which are n and m) Anyway… In the child, B is always either B or C from the parent, and it turns out that |A - B| is always C or B from the parent (respectively). Thus parent hypotenuse is parent B^2 + C^2, which for the child is B^2 + (A - B)^2, or (A - B)^2 + B^2 Now let’s apply this formula to the Fibonacci square for the 3-4-5 triangle: 1 1 3 2 We have (A - B)^2 + B^2 = (1 - 1)^2 + 1^2 = 0^2 + 1^2 = 1 Thus any “parent” of the 3-4-5 triangle would have to have a length 1 hypotenuse. Also, its Fibonacci square would have to look like one of the following: A 1 OR A B OR 1 B 1 C 1 1 D 1 But because A + B = C and B + C = D, this means that either B or C must be zero (whichever isn’t already 1). In other words, we would have a leg of length 2BC = 2*1*0 = 0, so the entire triangle “collapses” into a length 1 hypotenuse/line Thus the incenter and the nine-point center must be either vertically or horizontally aligned, with a separation of exactly 1/4 (keeping in mind the 1/4th scaling of the parent) In the case of your drawing, I believe the 2 centers in question should make a horizontal line with a length of 1/4. (I don’t have a proof that it should specifically be horizontal rather than vertical, but I looked up the coordinates of the various centers for the 3-4-5 triangle to check my work, and that’s what I saw)