Simple yet 5000 years missed ?
2024 ж. 23 Ақп.
200 701 Рет қаралды
Good news! You really can still discover new beautiful maths without being a PhD mathematician.
Stumbled across this one while working on the magic squares video. Another curious discovery by recreational mathematician Lee Sallows. A simple and beautiful and curious fact about triangles that, it appears, was first discovered only 10 years ago. Really quite amazing that this one got overlooked, considering the millennia old history of triangles.
Wiki page dedicated to Lee Sallows
en.wikipedia.org/wiki/Lee_Sal...
His personal homepage
www.leesallows.com
The relevant subpage
tinyurl.com/y6tzsbjt
t-shirt: www.teepublic.com/t-shirt/300...
music: Campagna - Adventure of a Lifetime
Enjoy!
Burkard
Sometimes you wonder how mathematicians come up with things... Other times, you wonder how mathematicians don't come up with things...
Most modern "science" is full of mathematical nonsense. Like the shape of the Earth. Maths doesn't match reality.
The best math comes from Side Projects. The things you think about when you should be doing something else.
They are "discovering" the things.
how you develop 'wonder' thing? I am not wondering at anything
@@robertveith6383 Not only. They also develop things.
This is a great length of a Mathologer video, nothing wrong with this! Thanks
This short format appears to get the thumbs up from many regulars. Nice :)
@@Mathologer Yes, long, short, medium--all good, and a viewer would be foolish to complain about a short video if the alternative is no video at all. Whatever length suits your schedule and the video's content best. Of course it's not ideal to increase the quantity (minutes of videos produced per year, say) at the cost of quality, but I've never noticed that in a Mathologer video, so not an issue in this case!
@@Mathologer Same here! I don't always have time to sit through a lot of the longer ones unless they're a direct interest of mine, but I'll click on these shorter ones any day of the week.
@@Mathologerisn't there aflaw I. Your reasoning at5:42..why are the areas of the pthertwo triangles 1/3 the total area ? Just because their heights are one third that total length you stay didn't show that their nases are equal to the base of the blue triangle...see what I mean? Hope you can respond. Thanks for sharing.
@@leif1075if you can see the area of the blue triangle is exactly 1/3 of the original triangle, you can use the same reasoning for the red and green triangles. In the latter two, you’ll be using the outer red triangle side and the outer green triangle side, NOT the blue triangle side. Each coloured triangle area is 1/3 total area of the original triangle. They are the same area in all three views (blue, red, green), despite the different bases and heights.
I love the second "simpler" proof. It is intuitive and I can even explain it to members of the family who are not true maths lovers.
The duality relationship between the triangle and its folded form is simply beautiful. As a triangle lover, I absolutely love this video. I cannot believe this was not known.
Please look at the answers i put in the comment of "i l put username later" to link this duallity with the usual midpoint duallity (involving hexagons ABCA'B'C' such that (XY)//(X'Y') \forall X e Y \in {A,B,C} ) This make me wonder if there is not a way to combine opposite triangles and this new duallity in the space of positive triplet that satisfy triangular inequality and that sums is 1 (exept for 000) . The orientation will matter in order to get opposite triangles, we would like to be able to do addition (such that the addition of two opposite gives the emptytriangle (0,0,0)) and a multipplication such that a triangle multiplide with its inverse (defined by the new duality) or maybe the opposite of its inverse gives an equilateral triangle, in such a way that we get a nice structure, why not a field (we will probably get an isomorphism of a well konwns field, I'm thinking about quaternioons because it is the only 3 D field I know and maybe the only one possible, I really don t know much about it lol) Really to many suppositions here so I ll have to stop here not to be ridiculous, but it might be interesting to search something. Note that it is easy to get a tripplet of homogenous coordonates that satisfy triangular inequality (and that are decided equal up to scalar multiplication ) from a triplet of 3 real numbers up to scalar multiplication : take the angles that are obtain with a triangle that vertices are (a,0,0), (0,b,0) and (0,0,c) in the 3-d space. (indeed we get all triangle that angle are all less then pi/2 , which are also triangles s.t. the mesure of angles satisfy the triangular inequality, isnt it nice?^^) I will do a litlle homemade video to talk about this, and I ll give the link^^ Thank you Mathologer for this video and every single other long or short one❤❤❤
I think I'd honestly prefer the first proof, but I was too busy shouting at the screen about the second proof to enjoy it.
Hahahah very meta 😂
9:48 apparently he could hear us from the past
Meta I meant @czertify’s comment some how remind me of Pierre de Fermat’s comment, I am not sure if it’s intentional to subliminal but it is just somehow made it even funnier for me
Meta I meant @czertify’s comment some how remind me of Pierre de Fermat’s comment, I am not sure if it’s intentional to subliminal but it is just somehow made it even funnier for me
You had me going at the beginning. Because of the particular choice of original triangle, you briefly had me wondering whether the "folded" triangle might be (geometrically similar to) the mirror image of the original. But no, not in the general case.
only in exactly one case, the equilateral, the triangle is the same as it's folded counterpart.
@@yonaoisme You're wrong! You will also get the similar triangle if centroid coincides with one of the Humpty points (projection of orthocenter onto the median). This is because medians to sides will be in ratio sqrt(3):2. In case of equilateral triangle orthocenter already coincedes with centroid, so it's a simple case
@@notEphim Yay.
What do you nwan..isn't this video a bit unclear to everyone??
I like both proofs. They scratch different intuitional itches. 😁 I've always found it satisfying to arrive at the same place by different mathematical routes. I think it helps cement the ideas, and also expands intuition.
Same here. I've been obsessing about identifying the "second best proofs" for theorems for a long time :)
The dot proof is more emotionally satisfying. :)
As a color blind person, I didn't like the dot proof as much. I got the idea, but it wasn't as visual for me
@@kilianvounckx9904 Haha. I’m colorblind as well (red/green). I couldn’t tell which dots were which, mostly.
Another gem from Mathologer. It's because of Mathematicians like you out there, Maths is still beautiful and elegant.
Maths would still be elegant and beautiful without him (or anybody for that matter), but he certainly does an excellent job of helping a broader audience appreciate it!
@@jrbrown1989my thought exactly! He's brilliant at communicating things in such a way that a broad audience can see their beauty 😊
Very nice theorem! This folding process gives some sort of "sides-medians duality": -The sides of the folded triangle are each 2/3 the length of the corresponding medians of the original triangle; -The _medians_ of the folded triangle are each 1/2 the length of the _sides_ of the original triangle. This proves the 1-time-folded triangle is in general not similar to the original one, but the 2-times-folded one is similar to the original one, with a lengths ratio of (2/3)*(1/2)=1/3.
so this is a kind of duality between two different triangles, neat
Yes, I was wondering as I watched the video if the folded triangle would be called the “dual” of the original (as you have wjth polyhedra) or has some other name. And, are there other interesting properties of the dual as relate to the original?
Well duel is supposed to mean something very different. You put points at the middle of line segments then draw a line between any 2 points that once shared a point.
The term "dual" is very general and is used all over mathematics. It means an operation that yields the same type of object (a triangle in this case) and brings you back to the original if applied again. So no, what you describe is not "THE dual", but just "SOME dual".
@@pauselab5569 there are many many kinds of dual in loads of situation. A duality is typically some pair of objects that you can swap between by exchanging some property. And I think this triangle situation can work with that. Duals are typically great because, by proving something about one object, you automatically get an equivalent proof for the other, and sometimes it's very easy to get a proof for one but you care about the other.
Thought the same! I wonder what to do with it, what properties are preserved by this type of duality, whether it is the same as some other type of duality, and whether there are analogues of this in higher dimensions!
I did a bit of trigonometry to express the six angles with the coloured dots in terms of the angles of the given triangle. Here's what I figured out. (I'm sure this is known to the triangle experts.) With the usual notation, let's call A, B, C the points of the triangle, a, b, c the edges and alf, bet, gam the angles. The median through A divides alf into the angles alf_b and alf_c (to the side of the edges b and c respectively). Similarly, bet=bet_c+bet_a and gam=gam_a+gam_b. With this, one gets: cot alf_b = 2 cot alf + cot gam, cot alf_c = 2 cot alf + cot bet and two similar pairs of equations. (The proof uses the law of sines and the addition formula for cot.) Btw., it can be checked that cot(alf_b+alf_c)=cot(alf). Now the folded triangle has angles alf_F = bet_a + gam_a, bet_F = gam_b + alf_b, gam_F = alf_c + bet_c, and one obtains cot alf_F=(-cot alf + 2 cot bet + 2 cot gam)/3 and two similar expressions for cot bet_F and cot gam_F, i.e. a linear relation between the cotangents of the angles! So, if one forms a 3-vector from the cotangents of the angles, then the folding operation from the video is the multiplication of this vector with the 3×3-matrix M which has -1/3 on the diagonal and +2/3 in all other entries. This matrix satisfies M^2=1, reflecting the fact that folding twice reproduces the triangle up to size.
α β γ δ ε ζ η θ ι κ λ μ ν ξ π ρ ς σ τ υ φ χ ψ ω ϐ ϑ ϒ ϕ ϖ Ϛ Ϟ Ϡ ϰ ϱ ϲ There is some greek letters for you. You can copy and paste them to tidy up the text if you want. :)
One of the best math channels out there. Your glee is contagious!
My first thought with seeing this was a way of defining a Dual of a triangle (up to scaling), following up with some theorems saying "A triangle has property X iff its dual has property Y". Time to explore.
beautiful! thanks!! medians of median-triangle give back the scale-down sides -- i also re-discovered this in middle school, when i tried to compute the formula for median lengths using pythagoras and area formula, and noticed that it was a reversible formula, in the sense that you can apply the same formula to get back the sides, if the median lengths are known, with a scaling factor. now seeing the animation today looks very beautiful.
And I thought train spotters were strange. Now I'm aware there are triangle spotters, too.
@qnbitsWhat about ch**trail spotters then? 😮
Thanks! Lovely reminder why I love elegant mathematics like this.
We were meditating over these principles in 1988, when I was in 7th grade and we demonstrated tons of problems around that. How was this discovered only 10 years ago?
Exactly! I am almost 50 years old, and I can distinctly remember playing with these 'folds' in the EXACT same way as shown in the video when I was around 10 years old with my own arts and craft projects at home. In fact, I might still have it laying around somewhere in some boxes at the attic. I suspect when he says _'discovered'_ he actually means either *A)* officially described in some math paper, and/or *B)* a proof was found. Which are VERY different things than _'known/discovered'_ .
@@CookieTube I'd be interested in seeing those folds!
It was probably only published by someone 10 years ago. People playing around with triangles ABSOLUTELY have discovered this many times over history.
@@LeoStaleylet's raise a glass to all those awesome people throughout history that loved the beauty of form and function. May everyone that wants to share in curiosity and wonder.
Some things are so simple that no one who stumbles upon it would have the nerve to publish it.
Beautiful theorems. Elegant presentation. Bravo!
Kudos to the discoverer of this. A very organized mind.
Check out some of his other inventions by following the links in the description :)
Cool! I wonder how many mathematicians/geometrists have realised this in the past but either assumed it was already widely known or thought that it was trivial, so never bothered to publish it.
Probably quite a few, but possibly/probably never visualised in the way I showed in the video :)
Agreed - it follows easily from the centroid being 1/3 along each median, as demonstrated in this video.
Well, if you put a lid on a pot with boiling water, it is pushed up and clacks sometimes. Probably people noticed that quite some thousand years ago. But not seeing it as trivial and going from there to a steam engine took quite a while. The next step towards many inventions was hidden in plain sight and regarded as trivial, until someone took a really close look and pointed out that it's anything but. So I have the deepest respect for people who have an eye for those things and look behind seemingly trivial things.
What a nice story from the 2D world. Thanks for finding time to share it with us.
Wonderful series of presentations. Very clear, very entertaining and educational. Thank you!
This is my favorite mathologer video in a while. Quite easy to digest, and beautiful.
Glad you liked it!
That was very very awesome, thank you ;^> Also, your (ten minutes in) guess at how some of us would prove the result was exactly spot on; once you told me the result, that's how I had worked out it was right.
Glad you enjoyed it!
Loved the dotted one.... really gave a good overview after the flipping. Thanks for sharing.
What a joy! Just amazing! Thank you!
Amazing quality kept for another Mathologer video. Thank you so much for spreading glorious mathematics ideas to mathe-maniacs like us, Mathologer!
Both proofs have their merits, but I prefer the second one just a tiny bit. On the one hand, when I follow a proof I like to be sure that I haven't missed some tricky step that might undercut the whole proof and that was easier with the first proof by just following the angle dots, but on the other hand the second one is quite short which is a big advantage.
Another beautiful video. These geometric proofs are really something for the soul. Thank you.
Thank you very much!!! I wish you long life!❤
Cool shirt as always doc!
Triangles come in pairs that you turn into each other by folding them inside out, lovely! And thinking about the vertex angles: assigning identical angles the same color. an Isosceles triangle (RR, GB, GB) turns into a different isosceles triangle (RG, RG, BB) but only an equilateral triangle actually turns back into itself? edit. Whoops, i got too ahead of myself and wrote this, right before you explained the isosceles
It's amazing! Thank you so much!
Fantastic animation! Really helps to understand.
1. I LOVE the mini Mathologer videos. Just enough to not overstretch my attention span. More of those, please. Bitte schön! 2. Straight up going-for-the-kill proof FTW.
i like the dot proof because i wouldn't have thought of it, and it's very beautiful.
I ❤ this stuff & highly appreciate the cleverness w/o numbers. Definately going to use this for tutoring, thank you for sharing.
That's great :)
So beautiful! This is why I LOVE Euclidean geometry and ratios more than numbers. My eyes are moist.....
I loved the dotted angles proof!
Very nice geometric proofs. It might be also interesting to look at it a bit more algebraically. Let's say a',b',c' are the big parts (as in 2/3) of the respective medians. From Steiner theorem (or law of cosines) we have a'^2=-a^2/6+b^2/3+c^2/3 and similar equalities hold for b', c'. So if we represent the triangle by the vector (a^2,b^2,c^2), "folding" is just a matrix multiplication (a'^2,b'^2,c'^2)=M*(a^2,b^2,c^2), where M=[[-1/6,1/3,1/3],[1/3,-1/6,1/3],[1/3,1/3,-1/6]]. Since M^2=I/4 (where I is the identity matrix), folding twice means making the squares of sides 4x smaller, i.e. scaling the triangle down by a factor of 2.
Nice! So the squares of the sides behave better than the sides themselves. That's sort of a dual version of what I wrote about (the cotangents of) the angles an hour ago.
Beautiful and brilliant
I always enjoy your videos and look forward to them coming out, so let's hope you get some respite at work ;)
Yes, just really insane at the moment. Hardly any free time :(
The 2:1 median split is taught in school... And quite early at that.
Used to be taught in school. These days at least in Australia hardly any nice geometry is covered in school anymore :(
@@Mathologer really? But that seems like a basic property, not something I'd classify as "nice".
This is why we should never stop playing with all the subjects.. there's a lot of beauty still hidden! Thank you for the video :)
Wow! Wonderful discovery! Thanks a lot for the video!!! I like the second proof more, due to its simplicity. But the first proof is also inspiring!
Great video Mathologer, I thought that the proof using right angles was slightly more concice and elegent but definately aprieciated the the longer and more thourough proof about the angles it was good to see both of those proofs done as they explained slightly diferent parts of this concept to me at least and that is why I whatch your channel so much, You are among the best math content creators that I have found on KZhead to date that truly seems interested in teaching a wide audience with wide rangeing capacities.
Something I like underpinning the two proofs is the duality of representations of triangles implicit in them. Three sides, or three angles and area. That one proof exists demands that the other one should be there as well. Nice.
Duality is a great way of explaining things that should be used to replace straight single logical proof in certain more complex subjects, such as economics and mechanics. (Axiom: Mankind seeks to satisfy his needs with the least effort:yet mankind''s ambition to meet these needs is unending (Henry George). Action and reaction are equal and opposite (Isaac Newton).
I prefer the dots proof. If you do the right triangle again do you get the right Triangle? Yes I think you do. Love these videos on a nice Sunday afternoon. Love the shirt btw
I wouldn't say it was missed, but rather everyone who noticed it never bothered to write a paper on it. It's all part of the beautiful symmetry of mathematics in nature.
My thoughts too. I'd be suprised if the pythagoreans, ancient Indians and who knows before didn't know of this.
Exactly!
Beautiful, just one word! It must have been such a joy to discover this for the first time.
u lied
Excellent graphics presentation.
@Vihart will love this episode :)
I've always been weak in Geometry, and I won't pretend to have a ready sense of the correctness of what you're presenting even with the help of your excellent graphics. But, even so, I can tell that you really do the graphics well, and I'm a bit envious of those for whom the graphic presentation is intuitive.
Faszinierend, wie immer, danke! 👏 Die beiden Varianten sind schön, jede auf ihre eigene Weise. Im allgemeinen, ich finde es nur gut, wenn es mehrere unterschiedliche Lösungen gibt.
Wow !!! Always the best of best. I am mind blown.
A motivational video for discovering mathematics.
go awey
The first theorem ["Triangles formed by the centroid and the vertices have area 1/3 of that of the whole triangle"] can also be proven easily, though slightly less elementarily, this way: 1) ratios of lengths and ratios of areas are affine invariants; 2) every triangle is affine to an equilateral triangle; 3) the thesis of the theorem is obviously true for equilateral triangles.
Wow, this is beautiful!
Thank you for the video.
Another wonderful aesthetic video, with amazing background music, came to rekindle our mathematical passion! And of course, as a Greek and a lover of geometry, I love both proofs, but I prefer the final "at a glance" proof, because beauty, lies in simplicity :) And as always, thank you for the video!
Ah, the infamous "Moebius C. Escher" Monopoly Edition! Love it!
Yes, that one :)
My favorite channel!
Glad you think so :)
I want that t-shirt in the shop!!
Not my design, but check out the link in the description of this video :)
This is amazing in its simplicity and apparent obviousness... how it went so long unspotted (before the dot proof, of course... see what I did there?) is a testament to how the human brain works (or, often, doesn't work). My undergraduate degree is in cognitive science, and I never went to graduate school due to a combination of the effects of late-diagnosed neurodivergence (autism diagnosed in 2005, five years after graduating; ADHD diagnosed earlier this month) and shifting gears from academia to tech work the year after I graduated (which, again, was probably the ADHD talking). Mathematics, especially geometry and number theory, have always been fascinating to me, and your videos and those of 3Blue1Brown have done more to maintain that fascination than anything else. Thank you.
Beautiful!
Very nice indeed. I prefer the second proof, nicely showing the scaling factor as well.
Brilliant!
How beautiful!!
Fun! I have memories of folding construction paper triangles in half and noticing SOME of this…so I like the second proof. It’s more intuitive. But the colored angle proof was gorgeous!
I love this one. I will teach it to my kid ❤
Both proofs are very beautiful each in their own way and using color and animation is so much more elegant and easy to understand than filling the screen with Greek letters. I wonder does the original theorem generalize to 3 dimensions (and greater) where the "folding" takes place with three pyramids instead of two triangles? And the lines dropped from each apex would cut each other in the ratio 1 to 3.
Truly Beautiful! I think I liked the last proof the best, probably since I already knew the 2:1 ratio of intersecting medians.
Best KZhead Channel ever, this man deserves more than a million subscribers... I really love your stuff despite being a medical student...
Both proofs have their own path of application. A great discovery!!
Isn't the double folded triangle exactly the tiling triangle? Both are similar and have 1/9 of the original area
Yes it is. This is also a more intuitive way to see the three parts are of equal area: The tiled and then median-cut original consists of 18 triangles, all of which are exactly half of one of the tiles, so they have equal area. And since each of the parts has six half-tiles they are all of the same area. It seems like you should be able to show the rest like this as well, by colouring in the six different types of triangles. (Three median directions to cut a tile, each cutting a tile in half.)
at the "heart" is all understanding. yet not void of re-flective as most prominently seen in the pairing of things. top two bottom front two back side two side. in a body "whole" is expressed a temple for understanding the "whole" beyond" all material di-mensioning. Gratitude and appreciation for all your beautiful re-flections as seen from glorious realm of math. one beautiful branch for understanding wholly.
Second proof is more simple and easy to spot, first proof is a little more complicated but elegant, in my opinion
Amazing!
finally a mathologer video i was able to keep up with! Thanks
Gotto do more like these :)
I just wanted to mention that I always enjoy your different T-shirts. And today was no exception. The only thing is…. I was so focused on the content (which is fascinating) that I was 8 or 9 minutes into the video before I thought to look at what you had on today. Wonderful video, as always! 😊❤ Best regards from Québec 📐
At last one I could follow and understand! 😀
great video Kudos to Sallows- and you too
11:02 "The proof is elementary." Aha! A true mathematician!
Did you notice that there are exactly five squares inside of the equilateral triangle? The top X is inside of a square. There are three complete squares with the last two being the left corner and left central area of the bottom triangle added to it and the same with the right corner and right side of the central bottom triangle.
This is very cool. I do a lot of tessellations and this is related. I've thought for a long time that there are many unknown simple mathematical theorems and proofs yet to be discovered. Math is a science with infinite possibilities so it seems intuitive to think this.
the first folding example reminded me of a great train -- the first gold is a smaller copy of the original rotated 180d about the height line -- making it the idler gear for the copy after the second folding.
In fact we were taught this in our correspondence math school in the USSR in the 1970s.
this is amazing! i wonder if there are any interesting properties about the triangles you get when folding right triangles
A very nice theorem and observation. Thanks for sharing this with us. The second you suggested folding the two neighboring triangles I paused the video and both got the result and the (second) proof, with the conclusion of the result of a second fold being similar to the original taking only a few extra seconds. Obviously, I prefer the 2nd proof 😊
P.S. It helped that I already knew the 1:2 ratio of the medians Euler point split.
Obviously :)
Very nice video and perfect length for me ;)
You know that when Mathologer posts, it will be a good day
Thank you, it's simple and interesting, people will enjoy in math more.
Glad you think so!
Just one line is crossing my mind watching this "It's a kind of magic, MAJIK!!!" ✨
That was fun, thank you! As an aesthetician, I found the first method to be intricately beautiful, as a novice mathematician I loved the simplicity and directness of the second.
Glad you liked both. I do too :)
@@Mathologer 😊
The Monopoly on the Penrose Triangle is just brilliant.
Fascinating tale! Another aspect of the folding process - perhaps better called an unfolding in this case - is to apply the folding procedure to (three copies of) the whole original triangle: that is, consider each median in turn and fold the corresponding edge about the midpoint - splitting the corresponding vertex and opening the median to give a new triangle with two of the original sides and a new side of twice the length of that median. The three resulting triangles can be put together by matching the surviving original edges to make a bigger version of the standard folded triangle (side-lengths now twice the original medians). The original edges now lie on the medians of the folded triangle (as far as the centroid) ready for the next 'unfolding'.
Pretty sure that I could solve one of the Math exam questions if I used this method. It was so much hassle with counting of right triangles inside the triangle but if I were to turn it inside out it would turn into a one big right triangle.
This is so cool. Numerical simulation on computers of 3-D objects (for display visualization or structural analysis) requires creation of a mesh sometimes made of triangles. Areas of an object that comprise sharper curves/folds need to use a more refined mesh to better capture the shading, forces and deformations with greater resolution. But to save computational cost you don't want to use such a high res mesh everywhere. I'm not an expert in this but it seems like starting out with a coarse triangular mesh and then selectively dividing those triangles using the 1/3 scaling into 9 smaller triangles (or 1/2 into 4?) might be a good way to create an efficient mesh. Hopefully others who know more about automatic mesh generation might like to comment.
Regardless of which proof people prefer, your explanation is, as usual, on point.