Відео

I used a Python library called Manim CE, there are lots of ytb videos that explain how to install it and how to code using manim. It's hard at first, but I got use to it fairly quickly.

I thought about n-sections, but not about increasing the number of dimensions ... 🤔 In 3d, would this mean to cut the solid angles of the polyhedra ?

​@@MatheFysyk Not sure actually. I haven't thought it through. It's not obvious to me that it does generalise in the way I had in mind. There might be something in the logic that breaks immediately. In the proof, you reference the dihedral symmetry group. (You say D6 in the video but I think it should be D3 which I didn't notice first time around. So that might be an error. Double check that.). Before considering exactly what notion to consider at vertices or edges, I'd start by thinking about a candidate invariant shape and its corresponding symmetry group. A group that might play an analogous role to the dihedral group in the proof in the video. Take for example a regular tetrahedron with its corresponding symmetry group in 3 dimensions. One could look it up or try to tease it out without too much issues. A hypothesis might be that the outer shape is an irregular tetrahedron. (Maybe one needs to be more clever. ) Now if the proof is to work in a similar way, you'd want to find a set of transformations preferably something like rotations centered at the vertices that play a similar role as 1/3 of the vertex angles in this proof. Solid angles probably won't be the way to go since they don't really represent rotations. Rather, you'll probably want to have a set of multiple rotations centered at each vertex (point), acting along different planes (or put another way, rotating around different axes). Another option might be to rotate around the edges (lines). Maybe one needs both notions in combination. I'd have to think about it. My guess is that we'd want some the rotations to be some integer fraction (1/3) of the angles available in the shape. Now there are two tasks to solve if it should work. The first would be to combine these fractional vertex and or edge rotations and check that they leave the candidate shape invariant. The second part would be to interpret the rotations in some nice way that highlights the analogy to the 2d case with 1/3 angles. Those are my thoughts on it off the cuff. Reflections in the faces of the shape would be another type of transformation that would be needed for a similar proof. If something like this turns out to work against all odds, which it may or may not, one could try higher dimensions. A pessimistic guess is that if it doesn't fail in 3 dimensions it might at least fail eventually. But maybe not. What makes it interesting to me is that I don't see immediately if it would work or fail. It could be a unique theorem for 2d. Or it could be a theorem that easily generalises arbitrarily. Or it could work for sufficiently low dimensions due to some specific group properties. Or it could work in some dimensions and not others, for other interesting reasons (Since you used complex numbers in the proof, maybe one could run with that and generalise using quaternions etc. If you need some special algebraic structure that might only work in very special dimensions for example.). Some more experienced mathematician might see clear answers here immediately. I don't currently. I would have to think about whether the later steps where you get equations and solve them, would work with this set of ideas. If you do take any time to consider this or see any obvious problems with the idea please let me know =)? If I have time, maybe I'll see if I can make any progress in making it work or rejecting the possibility. Do you have any other thoughts? ps. My first comment sounds a bit overly critical about using complex numbers. They are a great tool, and your usage of them makes the proof very clear. I want to stress that I think the video is terrific!

@@HyperFocusMarshmallow There are 2 ways to denote the symetry group of the eq triangle, D_3 or D_6 (D_n or D_2n in some french books). But you're right, it makes more sense to denote it as D_n, so I'll change my notations in the future, as everyone seems to work with D_n. What makes the proof work is more the affine group over C than the dihedral group, that was kind of a little digression when talking about eq triangles and their algebraic characterization a+bj+cj²=0. Maybe there is something to dig with generalizations of this theorem, but I can't see something simillar with Morley's theorem in more than 3 dimensions. Maybe 4 when working with quaternions as you mentionned it ? We could consider the affine group of R^3, but we can't multiply vectors in R^3, so we can't really do the computations that are done using the affine group of C ... I don't think this particular proof can be generalized in greater dimensions.

@@MatheFysyk Good about the dihedral group. Many notations in maths sometimes, I just wanted to alert you about it if it was an error. =) You are probably right about generalisations failing. I'm totally new to this theorem so you're quite apparently way more familiar with the details after making the video. I made the first comment without knowing whether it would work and my last comment was just me starting to think about it on the fly without trying to figure out any details. So thanks for even indulging me. =) But trying to write comments under a theorem video is also a great way to realise that I didn't fully appreciate some parts of the proof. Well I wouldn't expect to from memory and I watched the video two weeks ago so maybe I can be excused. Indeed the Affine group over C plays a big role. We can't multiply vectors in R^3, but the affine group over R^3 is as the name suggests a group and it does act on R^3 which has an addition. So one should be able to set up quite similar constructions. It's just that one has to keep the group and the objects it acts on distinct. So there would be some work but I don't think there is any big obstruction in expressing the geometric conditions or the group and it's action. My guess is that rather the problem preventing the generalisation will come down to that somewhere in the logic the number of parameters required to specify two structures won't match up. The number of parameters going in to the problem is fixed. And those degres of freedom must be enough to construct a polynomial (or rather several) with certain fancy properties. And if that fails then that's it. Well maybe that's a trivial way to put it. Finally. I skimmed some papers without trying to understand anything in depth but as far as I understand it, it seems to fail because the most natural generalisations of trisecting angles define too many independent equations to get the right type of invariant object. There might still be some interesting constructions in the same spirit, but I think I'm ready to leave it for now. Thanks for the interaction! =)

@@HyperFocusMarshmallow I made the video nearly 3 months ago, so there are little details that I can't really remember without doing carefully the proof again 😅 Yes, you can "multiply" transformations of the affine group, but what does work in the proof is that the composition of elements of the affine group over C relies on the possibility to multiply and invert complex numbers, so that you are left with a null product of terms which contains a+bj+cj², after some computations. But maybe there is a way to use the affine group over R^n to show something similar. Another thing is being aware of what we're trying to show. If not, it might be difficult to know how to use the previous affine groups 😅 Sorry for the late answer on your first comment, I haven't received a notification, that explains the 2 weeks gap before my answer.

Le meilleur, ça reste quand même de faire un dessin 😎

I finished just in time for the SoME, so I used the same music as last year to save some time. I'll change it for the next video. Do you have a recommandation of a free music I can use that would be better ?

If you have studied the real affine functions (x maps to ax+b), then yes, you are familiar with the elements of the affine group over IR. That's just a technical reformulation.

Intéressant 🤔 Je m'étais posé la question il y a 2 ans, et je n'avais rien trouvé. Si on souhaite pouvoir calquer la méthode pour des n-sections, il faudrait exprimer les points au centre comme des points fixes de certaines transformations affines, et trouver un élément d'ordre 4 à partir de ces transformations, qui serait donc analogue à i ...

@@MatheFysyk ...quitte à ne plus se placer dans le triangle, mais dans d'autre polygones. Je passe un peu du coq à l' ane : la (ou plutôt une) généralisation (de l'enoncé) aux polyèdres est tentante mais on dealerait sans doute avec les quaternions si on veut tenter de generaliser (la demo) ce qui est moins ergonomique^^ Je reviens sur le fait de prendre un angle quelconque central au lieu du secteur du milieu dans le cas des trisecteurs, car il y a peut-être un résultat de majoration intéressant : le fait que sur geogebra le cas n=4 et n=8 donne des triangles quasi equilateraux provient sans doute de la proximité des 2x4-secteurs - (resp. 2x8-secteur ) avec les 3-secteurs. Ce qui m'interpelle est que la proximité en entrée relatives au sections d'angle (1/2 et 1/4 proche de 1/3) donne une proximité avec le triangle equilateral qui à l'air bcp plus forte (ceci ne veut pas dire grand-chose d'une part parce que je n'ai pas défini proximité et d'autre part parcequ'il suffit de peu de contrainte en plus des inégalités triangulaires pour faire qu'un triangle "ressemble" à un triangle equilateral) Mais tout de même , la proximité semble résister à toute déformation du triangle initial : dans mon expérience "manuelle" geogebra , on peut triturer et aplatir tous les autres triangles, et notre triangle quasi équilatéral ne bouge vraiment pas beacoup !!! - ce qui serait évidement le cas de la situation de Morley (cas des 3-sections) où le triangle reste equilateral, mais on pourrait penser que si on triture le triangle de base assez on peut s'elogner de l'equilateralité... mais non! D'où l'idée d'une part d'une convergence rapide ( quand le rapport tend vers 1/3) de la fonction [ rapport d'angle angle commun des secteurs centraux avec angle sommet] vers [ différence du plus gros angle et du plus petit] et d'autre part de l'existence d'une borne sup assez basse pour cette fonction sur un intervalle raisonnable (et pourquoi pas sur tout l'intervalle ]0,\pi/3[, y compris au voisinage de 0 et de 1!) D'ailleurs on peut se demander si le triangle central est toujours acutangle (aucun angle >pi/2) même en prenant pour tout sommet un secteur central dont le rapport de sa mesure est un meme khi pour les trois sommets avec khi aussi proche de 1 qu'on veut (je viens de penser aussi à khi aussi proche de 0, là je n'ai aucune idée de ce qui se passe, remarque pour le reste non plus🤣🤣🤣) Merci pour ta réponse et encore bravo et merci pour tes vidéos!! Je viens d'en visionner une autre !! 👍👍👍

Conjecture : pour chaque sommet S du triangle on considère un secteur angulaire ayant même bissectrice que l'angle du triangle associé et tel que le rapport entre sa mesure et celle de l'angle est la même pour les trois sommets. Chaque secteur angulaire a un côté gauche et un côté droit pour observateur placé au sommet regardant vers l'intérieur du triangle. Pour chaque sommet, chaque coté droit intercte le côté gauche de celui qui est à droite en un point, ça fait trois point en tout qui forment le triangle central. Conjecture : la mesure du plus petit angle intermédiaire du triangle central est plus grande que pi/6 , celle de l'angle intermédiaire supérieure à pi/4, et celle du plus grand inférieur à pi/2. . Je l'ai vérifié pour de nombreux cas où le rapport commun des angles est une fraction décimale de base 2, pour *chacun de ces cas* on peut presque considérer ces expériences manuelles comme des preuves par informatique (en invoquant la continuité) bien sur il serai exagéré de conclure que c est une quasi preuve pour tous les décimaux mais disons que ça rend la conjecture assez vraisemblable. Il serait très rigolo (mzis sans doute trop beau^^) que si on considère des fractions triadiques on obtienne des triangles equilateraux parfaits, il faudrait tester le cas de 1/9 pour deja avoir un cassage ou un espoir dans ce sens.... mais c'est plus difficile à expérimenter à la main pour des raison de constructibilité ! - à moins de trouver une astuce pour contourner l'impossibilité de trissecter... je vais y réfléchir tranquillement. A ce propos : pour le cas Morley (rapport 1/3) as-tu trouvé une astuce de construction (genre tripler les deux premiers angles et obtenir la trissection du troisième à l'aide du triangle equilateral central) ou as-tu utilisé des approximations numériques ?

@@savonliquide7677 Oulala, beaucoup de choses dans ce com, faut que je prenne le temps de tout lire en détail, ça m'a l'air intriguant.

@@savonliquide7677 Il faudrait qu'on en parle autre part (sur discord par ex), ou un autre endroit plus adapté, les com UA-cam ne permettent pas d'exposer de long raisonnements comme ça ... Mais après, il faut se dire que même pour des n-sections avec n > 3, les intersections des n-sectrices sont toujours les points fixes de g_ig_j où g_i, g_j sont les rotations d'angle 2/n * les angles en les sommets centrées en ces sommets. Et peut-être que les calculs algébriques qui suivent "survivent bien" même lorsque n > 3 ... il faudrait faire les calculs.

That's right, but they could have tried to use a compas and a ruler with graduations, which makes the angle trisection possible ... Moreover, the method to do such trisections was discovered by Archimedes roughly to periods contemporary with Euclid.

It's too much honor, thanks :) I hope you'll enjoy the next video as much as this one.

​@MatheFysyk Do all or most of those specialists really seek fame? Aren't they mostly just cueuous..and how can I be like them? I could never admit I'm not a math genius as great as Ramanujan or Connes or anyone else...Thanks for sharing..how do they not get bored or frustrated reading math papers though??

​@MatheFysyk what in earth is a non flat triangle..a triangle in 3 dimensions?..don't you think k most ppl will wonder what that term means? Or what all that jargon trimesters means? Even mathematicians?

@@leif1075 a nonflat triangle is a triangle on a space with curvature. Think if you drew three geodesics (or equators) on a sphere. That’s still a triangle but because it’s in a non flat space its interior angles add to greater than 180. as for your other comments, sounds like you’re a young person in need of a reality check. None of us are ramanujan. Almost all of us aren’t “geniuses”. We do mathematics because it’s fun and challenging. And frankly, most mathematical progress is made by the collective power of us “dummies”. There are way more important things in life than needing to feel like the smartest person in the room; you’ll grow out of it.

@SPVLaboratories why do you think or assume we are not geniuss or I am not?? You have no evidence I'm not and presented no proof. Wasn't Ramanujan just a "dummy" too then? He was just smart and worked hard then like others? What if it's too depressing and discouraging and I don't want to be a dummy or ordinary and I need to be and feel like I'm a genius and that it's true? And it's not necessarily childish or immature to want or need that.

I'm sorry, I can't see what you're trying to tell me, could you make a drawing of this situation ? The triangle is made of the same thread as the thread that spins around the spool ? The center point is the center point of the triangle ? Why do you think that's physics and not math ?

@@MatheFysyk It is physics and math. But it is more physics. No, the spool was the one that the three threads we spun on when the threads were sliding into a small turning drum that was twisting them like a helix. The triangle was made out ot three different out points that the threads were coming out of. And once in a while my grandma would change the position of like little hooks so a different pattern of thread would be weaved. I noticed that the central point of a cog was unchanged despite movements of the sides of the triangle. That weaving contraption resembled something form 19th century and everything was made out of wood except metal hooks and joints. Drawing this is beyond the scope of this YT entry and because of the fact I was 8 years old and simply do not remember details. I am 67 now, so if even I drew that I would not be able to reproduce it from my memory exactly. I only remember that very phenomena and not the device in detail my granda used which she had to pedal with one of her feet. I know one thing, though. Since I was a kid I was fascinated by such occurences and that is why I picked up physics in college and University which I never finished because of my serious illness a while ago back in my life. But it is not forbidden to be an amateur lover of physics. And a question for you: what causes the speed of light?

@@user-ky5dy5hl4d Indeed, the center of a triangle is unchanged under the action of many symetries. And for your question, I'd say that the speed of light (or rather a limiting speed) appears naturally when you consider special relativity fundamental postulates and derive the correct way to change between different frame of references. Maybe I'll talk about fundamental physics in future videos, but I don't want to make something that already exists and is done 1000x better by other youtubers.

@@MatheFysyk I metioned only the speed of light. And in my question of ''what causes the speed of light'' has nothing to do with the speed which here is not the issue; the speed can be 1000 miles an hour. What I mean is that if I push a cart on wheels I will give it some speed. So, where did the speed of the cart come from? It came form mu pushing it. So, the same is for a photon which leaves the Sun, for example. What gave that photon a ''push'' to travel at any speed for this matter?

@@user-ky5dy5hl4d Photons have no mass, but they have momentum and energy. No one pushes the photons so they accelerate to reach the speed of light. That's like asking why time is passing or what causes time to pass, or why do we use the word "blue" to talk about blue things, I'd say it's like that and there is no answer. Maybe there is a complicated explanation in quantum field theory, but I don't know about it.

Thanks ! I bought a new microphone, so the quality of my voice should be much better in the next video 🙃

I hope you'll also like the next videos 😊

Thanks 🙃 The next video I'm preparing will also be about geometry, but with less algebra 😅 I think I'll make it in french, but I'll provide subtiles in english.

@@MatheFysykI’m terrible at making these sorts of videos but would be happy to do an English voice over track for you if you’d like a version with an Australian accent!

@@MatheFysyk I speak French as well, so no problem 😄

Thanks for your comment ! (Sorry, I'm replying a bit late 😅) Feel free to use my code as you wish.

Thanks a lot for your comment ! Yes, I wanted also to talk a bit about the history and the background of this theorem, I think I'll keep doing this in the future :)

@@MatheFysyk I'm sure there's audience for this kind of content. Good luck finding it (hope UA-cam figures out quickly who those people are).

Thanks for your comment ! It's a lot of work, but I'm proud of the result, and even if I'm not selected for the competition, it was fun to make this video 🤗