Three unsolved problems in geometry

The Toeplitz conjecture and perfect cuboid problem are among easy-to-understand geometry problems that remain unsolved.
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[@GlorifiedTruth]

I have successfully inscribed squares in 140 million different Jordan curves. I figure that's about half way to infinity, so I'm well on my way to proving the conjecture.

[@stanpope6087]

No, no, no! You aren't even 1% of the way because you haven't done 14,000,000,001 of them yet! :)

[@vhawk1951kl]

Presumably because you have no more profitable endeavour with which you occupy Your associative apparatus.

[@timmy1729]

Large numbers are not closer to infinity than small numbers

[@vhawk1951kl]

@@timmy1729 You having exactly how much direct immediate personal expedience of infinity"? Yeah right, turns out that" infinity" is just another dream.

[@howdy832]

My favorite part about this is all the replies that don't get it

[@nzuckman]

I like the 3rd problem a lot. I like the idea of disproving the perfect cuboid by deriving various restrictions and getting two which contradict each other

[@gedlangosz1127]

I've not heard of any of these before! Thanks - another excellent video.

[@LucenProject]

Ahh! I've been working on the perfect cuboid problem as something I randomly thought up! I didn't know it had a name! Thank you for this! Now I can look into it properly!

[@BigParadox]

Such a nice presentation!

[@arekkrolak6320]

The perfect cuboid problem I saw recently on the list of 5 open number theory problems :)

[@dickyarya8204]

I'm not even a math student yet I'm here lol

[@rylaczero3740]

Interesting. Would love to know more of the unsolved problems in other mathematical areas.

[@alexandertownsend3291]

Here are a bunch of open problems in math. I will add more as I think of thwm. Keep an eye on this comment. This should be enough to get you started. ABC Conjecture Beal's Conjecture Birch and Swinnerton Dyer Conjecture Collatz Conjecture Determining If Physics Can Be Axiomitized Determining Which Numbers Can Be Written As the Sum of Three Cubes. Goldbach's Conjecture Hodge Conjecture Legendre Conjecture Navier Stokes Equations P vs NP Proof or Refutation for the Existence of a Perfect Euler Brick Riemann Hypothesis and/or Generalized Riemann Hypothesis Schanuel's Conjecture Twin Primes Conjecture Yang Mills Mass Gap Problem

[@_unkown8652]

twin prime conjecture, riemman Hypothesis,... Look at the mathematical problems of the millenium, only a single of them has been solved so far!

[@adb012]

Collatz conjecture, also called the 3n+1 conjecture, is my favorite. It is so simple that even an elementary school student can understand it and play with it. It we have tried, and it has held true, for numbers up to 300,000,000,000,000,000,000 (300 quintillions), but not only that it could not be proven either true or false in general, but mathematicians doesn't even know how to approach that challenge, What does it say? Take any natural number (i.e. positive integer) at random. If it is even, half it. If it is odd, multiply it by 3 and add 1 (hence the name). You will always end in a 4, 2, 1, 4, 2, 1... loop. For example, let's start with 3, *3+1 = 10, /2 = 5, *3+1 = 16, /2 = 8, /2 =4, /2 = 2, /2 = 1, *3+1 = 4... Will we always end in the same 4, 2, 1 loop no matter with what number we start?

[@rykehuss3435]

@@_unkown8652 I already solved Riemann hypothesis (its true), but unfortunately my dog ate the papers

[@rykehuss3435]

@@adb012 Obviously its true. No matter the number, you eventually end in the loop, its just simple maths. Give me my 1 million now thanks

[@2manypeople1]

I love the video, but there seems to be a small mistake. Should be 1+2^(n-2) instead of 1+2^(n-1).

[@csababekesi-marton2393]

Yes. Here: en.wikipedia.org/wiki/Happy_ending_problem

[@kasuha]

I spent some time on the Toeplitz conjecture some time ago (inspired by a 3blue1brown video) and I'd swear I have a proof for any curves that aren't fractal surfaces without defined tangent at uncountable number of points. And it should hold even for these too, I just did not find how to prove that. But I'm no mathematician and given how long the problem is known most likely someone else already tried my approach and found some shortcomings I couldn't see. Also, if the conjecture holds for Jordan curves, it holds for any curves containing a loop since such curves always contain a Jordan curve.

[@absalomdraconis]

The number of mathematicians in the world is not infinite, and thus they have not yet had time to touch on all possibilities.

[@aadarshshah1444]

i slept soo well after this video. Thanks

[@joshuasparber156]

It seems to me that the Jordan curve problem might need a proof similar to that of the Poincare Conjecture. I would believe that whether a square can be constructed from a curve, any curve, would depend on whether the curvature on distinct points was such that the tangents or the limit the tangent is going to would be perpendicular to each other. This could be put into mathematical language. For instance, look at the instances in which this would work, the hypocycloid. The limits of the tangents at the cusps are such that a square can be constructed from the points. But flatten one of the points a little and this would no longer be the case. This is a problem in which there are so many cases it might be better off to work it out intuitively rather than with a computer. This would also be the same as saying that there are no two lines in which a Cartesian product domain can be constructed. One would think that there would be cases in which this could not be done. So possibly, not every curve that can be constructed would have 2 lines within which such a Cartesian product could be constructed.

[@black_jack_meghav]

Thanks sir , maybe I'll try to solve some of them

[@kashnigahbaruda]

The perfect cuboid was very surprising not to have been solved or found!

[@DOTvCROSS]

First problem could be used on day 1 of "Erdos-someoneelse" spork-in-the-road theory. Paul and I were talking about the slide @1:53. He even went as far as to demonstrate to me all published work (2024) on this problem. If asked for my input: "Paul, you seem like one person drawing a single point with a single pencil" Akin to (P,dOf,pencil) =P+dOf^(p/frame). Spice it up a little, 2 people or 2 pencils or 1 person 2 pencil 3 dOF, ......(time passes as I explain every permutation of his problem).....

[@robert-skibelo]

Great material, nice graphics too. It's so refreshing to hear a calm, clear presentation, as opposed to all those frenzied, shouty Americans elsewhere on KZbin. And to hear Z pronounced properly. ;-) You've misspelt Toeplitz in at least one place and the vowel of the first syllable should be more like that of "berth", not "go".

[@francogonz]

I think I actually have a counterexample to Toeplitz Conjecture. Im working with a professor from my university on that

[@williamnathanael412]

Wow! Is it concave, non-smooth, or both?

[@poketoscoparentesesloparen7648]

It’s been proved if it’s conVEX and smooth it’s possible, he said the exact opposite

[@francogonz]

@@williamnathanael412 all i can say is that im trying to prove that my shape is continue, besides that, i got this 🥺

[@francogonz]

@@poketoscoparentesesloparen7648 yes! Im argentinian, so i have some troubles typing in English but translate this please: El contraejemplo en el que estoy trabajando es una curva cerrada simple. No interesa que sea convexa o cóncava. Surge a partir de un lema que ya demostré (con herramientas muy básicas porque estoy en primer año de matemáticas 😂) pero que básicamente dice cuáles son todos los cuadrados inscriptos posibles en mi figura propuesta. Eso está verificado por mi profesor, pero no puedo publicarlo porque sería muy obvia la solución después de que el lema sea conocido. Luego, el contraejemplo consiste en eliminar estos cuadrados inscriptos, sin generar otros nuevos. Quizás necesite más ayuda en demostrar que mi figura sea continua... Pero es una figura cerrada simple! 🛐 (Pero no suave)

[@poketoscoparentesesloparen7648]

@@francogonz I actually dont remember what i meant with my last comment. I think i was talking to William but I don’t remember what problem I had with his comment so I’m sorry if i said things that don’t make sense… If you actually find that counterexemple congratulations, I hope you find it 😊 I’m also not English so I think I must have misunderstood what was written or something

[@danishnazir7366]

Appreciated 👍

[@serdarbozdag1707]

Nice!

[@DumAndSmart]

I’m working on the third problem I figured out what the last two digits are for each number when you’ve squared it with just a calculator well I haven’t figured them out but figured three possibilities sides 64,36 diagonal 61,89 Sides 44,56 diagonal 81,69 Sides 04,96 diagonal 21,29 Both one side and the spatial diagonal must end in 25 and one of the diagonals( the 44,56/04,96 diagonal) must end in two zeros

[@danbromberg5909]

Interesting topic, thank you! Speaking of seemingly easy problems in geometry, how about a discussion on the Steiner-Lehmus Theorem?

[@rosiefay7283]

The Steiner-Lehmus Theorem is known to be true, and there are proofs of it. Why is this theorem particularly worthy of discussion?

[@KaliFissure]

Surface(cos(u/2)cos(v/2),cos(u/2)sin (v/2),sin(u)/2) 0>u>4π 0>v>2π Klein or not Klein? I developed this shape as part of a physics model and I'm curious about your opinion of it. "Shirley's Surface".

[@tokajileo5928]

are there closed formulas for any khi(n) where n is positive odd integer (and not=1) and khi is the riemann zeta. ?

[@cbunix23]

How does five dots in a straight line fit into the first problem?

[@pathanson1377]

OK try this one - take an ellipse with semi axes a and 1. Inscribe a circle in one quadrant of the ellipse. What's its radius? I suspect that it can't be solved (quartic equations etc) although I have a good approximation.

[@venkybabu8140]

If you take three waves orthogonal with one as a multiple of some integer you may be close to some number. Written as sin cos tan cosec sec cot.

[@zanti4132]

I'm curious if there are any 4D Euler bricks. Expessed algebraically, this would mean four positive integers exist such that the sum of the squares of any two of them is the square of a positive integer. It would also mean that selecting any three of the integers would form an Euler brick. With there being an infinite number of Euler bricks, it seems likely that 4D Euler bricks exist, but I am betting the integers are quite large.

[@artsmith1347]

1:40 Typo on the screen. "1 + 2^(n-2)" is spoken, but "1 + 2^(n-1)" is displayed. The wiki page says the conjecture is for "1 + 2^(n-2)". A table of the values would have been interesting to see: n 1 + 2^(n - 2) 4 5 5 9 6 17

[@RJSRdg]

By my reckoning, the third problem may have solutions which satisfy h^6 - h^5 - h^4-h^2 - h + 1 = 0 where h is the square of the cube root of the shortest side. Now I just need to find a way of solving that equation (or not!).

[@RJSRdg]

And no, that equation doesn't have an integer solution. There may be other ways of solving the problem, but that isn't one of them....

[@seanh1463]

For the first problem, any three points should be non-collinear! This is really obvious cuz you can't get any shape on a straight line, but to be more rigorous, we should mention that!

[@MarkWusinich]

That was fun!

[@IntegralKing]

I love how these problems are so easy to describe, but immensely difficult in their execution

[@ekrao9047]

If we want to construct a cuboid taking absolute length ie planks length whose fraction is not possible then we can not construct its diagonal as it comes in fraction of absolute length

[@bornbad6738]

The answers are towards you out side of the image and forward out of perception into the image.

[@moer6025]

If someone were to come up with a solution, where would they go to show their work?

[@arhylle]

Cambridge, Harvard, etc.

[@someone1059]

publish it in annals of mathematics or send it to american or european mathematical society.

[@joakimsiljelind118]

Why is it imposible for random dots to form a line?

[@DanielWalvin]

It sounds like that last conjecture should be that there *isn't* a perfect cuboid, if

[@enantiodromia]

But relatively speaking, "exhausting" all numbers up to only 5*10^11 is next to nothing, given the infinite supply of greater numbers.

[@warrendewson5395]

when i was in school my geometry teacher would have us solve it all the way out. what i mean is she would tell us to find 3 times the square root of 3 which is approximately 5.2

[@scarletevans4474]

3:28 Is it "convex AND smooth" or it should actually be "convex OR smooth"? I.e. these are two different cases for which the conjecture is proved, right? Not BOTH of these restrictions being required at the same time.

[@laserstandard3027]

Isnt a triangle also a jordan curve? How would you draw a square in this shape?

[@lgooch]

It’s very easy to

[@sxkjknjw2]

2:52 isn't it supposed to be true? Since topologically, a circle and a Jordan curves are the same, so if you can inscribe a square in a circle, can't you do it in a Jordan curves?

[@robertthompson5908]

The property of being a square is not a topological property. If you inscribed the square in a circle, and then did the topological transformation that takes the circle to your Jordan curve, it wouldn’t take the square to a square.

[@ganrimmonim]

Whenever I remembered it, the absence (so far) of a perfect cuboid keeps me up at night. 😞

[@jayghann5880]

The circle shape is the happy ending for everything to discover anything humans they want to achieve and I will explain how ... just think you old a pen and blank paper and I try to right any number and soon you pen touched the paper will create a dot like this .... but if you zoom in to the dot you will see the shape of dot is cycle shape and the more you zoom in you will find the same shape..! that’s mean the cycle is the happy ending because you could expand the circle to anything you want for everything

[@theobolt250]

I imagine that there is a n number of mathematisians who think of the Jordan Curve/Toeplitz problem "who cares? There are more worthwhile problems in the world". But there are to many unknown variables to solve that question satisfyably.

[@SquirrelASMR]

Omg guys click in description to hear him sing and play guitar, he's great!

[@1.4142]

We know some properties of the perfect cuboid if it exists: the odd edge must be greater than 2.5 × 10^13 the smallest edge must be greater than 5×10^11 the space diagonal must be greater than 9 × 10^15 Some facts are known about properties that must be satisfied by a primitive perfect cuboid, if one exists, based on modular arithmetic: One edge, two face diagonals and the space diagonal must be odd, one edge and the remaining face diagonal must be divisible by 4, and the remaining edge must be divisible by 16. Two edges must have length divisible by 3 and at least one of those edges must have length divisible by 9. One edge must have length divisible by 5. One edge must have length divisible by 7. One edge must have length divisible by 11. One edge must have length divisible by 19. One edge or space diagonal must be divisible by 13. One edge, face diagonal or space diagonal must be divisible by 17. One edge, face diagonal or space diagonal must be divisible by 29. One edge, face diagonal or space diagonal must be divisible by 37. In addition: The space diagonal is neither a prime power nor a product of two primes The space diagonal can only contain prime divisors ≡ 1(mod 4). If a perfect cuboid exists and {\displaystyle a,b,c}a,b,c are its edges, {\displaystyle d,e,f}{\displaystyle d,e,f} - the corresponding face diagonals and the space diagonal {\displaystyle g}g, then The triangle with the side lengths {\displaystyle (d^{2},e^{2},f^{2})}{\displaystyle (d^{2},e^{2},f^{2})} is a Heronian triangle an area {\displaystyle abcg}{\displaystyle abcg} with rational angle bisectors.[10] The acute triangle with the side lengths {\displaystyle (af,be,cd)}{\displaystyle (af,be,cd)}, the obtuse triangles with the side lengths {\displaystyle (bf,ae,gd),(ad,cf,ge),(ce,bd,gf)}{\displaystyle (bf,ae,gd),(ad,cf,ge),(ce,bd,gf)} are Heronian triangles an equal area {\displaystyle {\frac {abcg}{2}}\in \mathbb {N} }{\displaystyle {\frac {abcg}{2}}\in \mathbb {N} }. Thanks wikipedia

[@lcarliner]

What about not being able to trisect a triangle using just a strait edge and compass?

[@Monkeyslinger123]

It's been proven that there are angles that can't be trisected with just a straight edge and compass.

[@MikeRosoftJH]

Speaking about curves, here's a solved problem: Is it true that every continuous, non-intersecting closed curve divides the plane into two parts, the interior and the exterior? (This seems obviously true, but it was only answered in the affirmative 135 years ago.)

[@TranquilSeaOfMath]

Jordan Curve Theorem

[@mouryapaladugu3414]

The second one is so simple. I wonder why it's so difficult to prove.

[@discovermaths]

Sometimes easy-to-state problems are incredibly hard to solve. Think of Fermat's Last Theorem.

[@newwaveinfantry8362]

Typically if it's simple to state but seemingly impossible to prove, it's because it's a statement about an infinite set of objects, that is way too general for the current mathematical tools proven from current axiomatic models for us to rigorously generalize. Sometimes it's really easy to generalize an intuition in your head, but difficult to mathematically and formally generalize over the existing proofs for different subcases of the problem, that for whatever reason use different tools with no discernable pattern. This is why it was so hard to prove Fermat's last theorem, the poincare conjecture and why it's so hard to prove that every even number is a sum of two primes, or the twin prime conjecture, or the Collatz conjecture, or the Rieman hypothesis. All it would take to disprove 3 out of 4 of these is a simple example, yet nobody believes there are any. We know they are true for the most part and for most numbers, but we don't have a tool that allows us to generalize and induct over the examples we have and rigorously argue for those conjectures being true everywhere.

[@SquirrelASMR]

@@newwaveinfantry8362 wow that's super interesting! How similar those problems might be. Imagine someone finds a nice way to handle all those proofs with some new tools for working with infinite problems?

[@aguspueyrredon9997]

If I understand correctly the second inscribed square inside a curve problem, I propose this to solve s big chunk of cases: you start with a little square inside of the curve and you make it grow progressively (like Paint eraser in Windows) untill it touches one side, then you let it rearrange itself untill it touches another side. If things goes right, it naturally touches all four sides. In case it doesn't, you rotate it and it continues the growth process untill the goal is done. If it doesn't work, you start in a different place inside said curve. This strategy works in convex curves and in order for it to work in concave ones, you should slightly adapt it: let one corner of the square to be already located in some point of the curve then you let it grow untill another corner reaches another point in the curve and repeat untill it works. As you can see, this is not mathematically rigorous, rather a computer solution that can work as an algorithm in a simulation platform

[@mrcombustiblelemon2902]

2:54 "The most obvious example being a perfect circle" Me: "erm... How about a perfect square?"

[@randydewees7338]

Stumbling block 👏

[@vhawk1951kl]

Here's an idea: stop anyone you like at random and ask them if they have a "problem" With adjoining up random dots. However here in the tiny rural village of Little Bridlington on the March we speak of little else.

[@CatPerson136]

Now I'm wondering is it possible to have an unsolvable equation in mathematics

[@DAkshar]

I think no

[@Maniclout]

what do you mean by unsolvable?

[@osbinrock]

I think you mean that something like e^x-4x.....etc.. =0 Right? Like something large and big with one incognite... right??

[@camerontankersley3184]

It's possible

[@osbinrock]

@@camerontankersley3184 Nop Its impossible bro XD

[@mihalykundean4675]

It feels like I am listening to a creepy pasta, but these are just fantastic matematical problems that makes the world beautiful and not scarry.😅

[@azzteke]

Mathematical

[@colinmccarthy7921]

I have studied Mathematics at Degree Level.I know there are many area’s of Mathematics.ie. Calculus,Statistics,Probability,Etc.

[@robertgoldman8064]

On the perfect cuboid , could it be a number not greater than 1.

[@Asixoid]

Check out my attempt to prove the inscribed square problem: kzbin.info/www/bejne/d3nFkImKeNaAjJo

[@khconlineeducation]

👍👍

[@BrijeshEducare]

Please send pdf of these problems in detail. And thank you very much for this video

[@ShadNex]

No

[@carlosdelossantos5115]

And my homework consists of three geometry problems

[@SlimThrull]

Can't you construct a perfect cuboid using known Pythagorean triples? It seems this would be fairly easy to do. Though, I doubt I'm the first person to think of this so I suspect it doesn't work as nicely as I'd like.

[@CharlesPanigeo]

If it was that easy to do, mathematicians would have solved the problem by now. If that is indeed a way to construct a solution, i imagine the solution is so large as to be impossible to compute with current tech.

[@RJSRdg]

It may be possible, but you get an equation involving powers of 6, which may not be solvable! But yes, in principle. You start with a length of the shortest side, which is the shortest side in three Pythagorean triples. The smallest number which is the shortest side in three Pythagorean triples is 27: 27-36-45 27-120-123 27-364-365 However the middle sides also need to form a Pythagoran triple, which they do not do in this case. Similar combinations exist for all odd cubes > 8 and their multiples, so it may be that there is a correct combination out there somewhere.

[@leefisher6366]

For the first problem, I would love to know how you are defining 'random', or 'generally positioned'. Could not five points that all lie on one straight line be plotted randomly; in the same way that an unbiased die could roll a hundred sixes simultaneously? Randomness can include such things; and if we don't allow it to, we aren't accepting the true nature of the beast.

[@user-ef8kc4rv7n]

I assume we reject cases with zero probability?

[@poketoscoparentesesloparen7648]

@@user-ef8kc4rv7n All cases have zero probability. Even for choosing one point, if you choose again the probability that it is the same point is 0

[@good_brake]

General position in this case means that no set of 2 points are at exactly the same position, and no set of 3 points are colinear. The generalization for n dimensional space it a little bit more involved, but the general pattern holds, no k+2 points lying in any k dimensional flat subspace for all k

[@good_brake]

@@user-ef8kc4rv7n Close, the set of configurations not in general position is measure zero, so a randomly selected set of points (assuming some reasonable continuous distribution we're pulling from) will almost surely be in general position. The qualifier "general position" discards a measure zero set, but not every conceivable measure zero set, since any configuration considered by itself is measure zero.

[@evanknowles4780]

In the original problems statement the 5 points have to be in general position, which means no more points than required to define a proper Affine subspace actually sit in one. In the case of the plane this means no 3 points lie in a line and no 2 points coincide.

[@mystychief]

bricks with a, b, c integer and the spaceal diagonal integer do exist: 3, 4, 12 gives D=13; and 8, 9, 12 gives D=17 (a*a+b*b+c*c=D*D).

[@tudormilea7306]

won't you have a diagonal equal to 4^2 + 12^2 !€ Z?

[@kabivose]

But the 4,12 face doesn't have an integer diagonal

[@bornbad6738]

It's like ask8ng melting cheese on a chocolate bar , is how fast per second.

[@Ggdivhjkjl]

Maths is so romantic.

[@user-uv9bi3xz1o]

😂😂 never been good at math but today the math Gods visited me n told me to find an unsolved math problems n solve it. Keep in mind that I got a solid D- in maths. But I'm here. I can't get the idea out of my head. The answer will be 83641 or 83640.(some things) I might be off my meds

[@alantaylor2694]

Am I being dumb or misunderstanding something. If the jordan curve is a rectangle then you can't draw a perfect square on it, right? You even show a rectangle in the video as a valid Jordan curve. I'm confused. Edit: Re looked at it. I'm just dumb. (Should I delete all this and just put 'Interesting video?'...nah)

[@goodtoGoNow1956]

toeplitz must be true but I don't have the skills to prove it. But it has to do with closing unclosed arcs I am sure.

[@jaopredoramires]

wasn't Teoplitz solved?

[@Mrbobinge]

Travelling (potholes)salesman dilemma. Modify the software to optimize maximum four progressive destinations' new direction as mandatory 90 degrees to the previous one.

[@swedishpsychopath8795]

Consider Toeplitz conjecture solved (by me): I literally spent 5 seconds to find a Jordan curve that doesn't allow for a square to be inscribed. Just draw a triangle with convex sides so that the sides bends towads the "center" of the triangle. Problem solved! Even if you make the "corners" rounded (to make it "smooth" there won't be any squares). So where is my Nobel prize in mathematics?

[@FreeFireFull]

The edges of the square are allowed to stick outside the shape, it's just the corners that have to lie on the curve. Can't you just have one of the sides be "parallel" to the triangle's side, and lay out the rest as appropriate?

[@yuripalchik8636]

Nobel prize in math does NOT exist.

[@swedishpsychopath8795]

@@FreeFireFull Yes - the edges may stick outside the shape - but the vertices ("corners") must be on the curve. But it is just impossible to draw a square inside my Jordan curve, since you always lack the four'th corner :) Just try it on paper yourself. The trick is that the edges MUST bend towards the center with a partial clotoidal slope.

[@bobh6728]

I’m still trying to draw a triangle with convex sides that bend towards the center (I assume you mean interior) of the triangle. My problem is having a triangle with curved sides. If we are talking about a “triangle” like shape with curved sides, then for it to be convex the sides would have to bend away from the center. Please clarify. We will keep your prize in escrow until clarification is received. lol

[@swedishpsychopath8795]

@@bobh6728 Lol! Not to worry - I'll draw an example and make a link to it. I'll let you know (in a day or two). Could you at least send a small advance of the Nobel math price amount? The easiest (and safest) way to transfer the money is to put it in an envelope and place it under the seagulls nest on top of the Monolith statue in Frognerparken in Oslo.

[@BlackHattie]

You can make career with 1/137

[@frankfahrenheit9537]

Seit wann kann Edmund Stoiber so gut Englisch?

[@fawziekefli2273]

I'm pretty sure the problem of the _Happy Ending_ has been solved. 😉😎

[@allwaizeright9705]

They never got around to solving it because they got married..

[@TymexComputing]

"Are there are perfect cuboids?" - I know i dont know! :)

[@rainerausdemspring894]

The perfect cuboid problem has nothing to do with geometry. This is number theory.

[@totenkopf1479]

I'm too dumb to understand this so I'm in the right place. Oh ye of high intelligence please answer my question! Does Tenet make sense?? Thank you

[@kumarsapkota559]

I can solved this problem 💪

[@tokajileo5928]

if you think deeper, you realize all problems are geometric , like the famous Riemann hypotesis. Fermat tehorem was also related to elliptic curves --->geometry

[@nabinnyc]

i think "the happy ever after" problem would be better lol. happy ending. smh. ok. nevermind. let's math.

[@hndfhngtool-assistedspeedr2128]

It saddens me to know that paxpayer money is wasted on those mathematicians who are searching for answers that bring absolutely no improvement to our lives. Think of all the expensive supercomputers that mathematicians use to bruteforce their searches for pointless answers.

[@rainerausdemspring3584]

The perfekt cuboid problem has little to do with geometry. This is number theory.

[@nickallbritton3796]

It's both actually. Kind of obvious isn't it?

[@ujjwalbhattarai8670]

Under rote h=b²+p² Solve pythagoras therom. Problems solve

[@willemkoning2142]

0:04 « problems in mass ». You mean like being overweight or something?

[@mikethered123]

I would imagine that AI will be solving at least one of these within just a few years. :)

[@jerrywbrice]

Of course, one day the AI will surpass even human genius in cleverness. Then I don't think people will be worried about tricky geometry problems very much.

[@Temari_Virus]

@@jerrywbrice High dimensional geometry may prove useful. For example if you had an n+m dimensional space, with n being the number of inputs and m being the number of outputs, you could assign n axes the values of the inputs and m axes the corresponding values of the outputs. A good understanding of this geometry could help you trick the AI, like say for example create a function that transforms an image such that a human can still recognise it, but it would seem like something completely different to the AI (the only way to completely eliminate these "blind spots" would be to have enough parameters to account for every single possible input, which just isn't feasible, good AI can only make these spots harder to find, but they still exist)

[@rajarani6517]

Actually there is no perfect cubiod

[@nickallbritton3796]

Proof?

[@Skargardsmatte]

I have one and it is wonderful, but it is too large to fit in this comment section :-)

[@bornbad6738]

Stop false questions, an idea ,you can't draw a multi demantional image on a 2 demantional paper in a 4 demantional reality. But a introduction into quantum reality.

[@ottolehikoinen6193]

Perfect cuboids do not exist, at confidence level = large.

[@malakaibarber]

I made my curve infinitely long through time and a square can't be made? How can you understand what a graph is for if you don't understand what to compare. A graph is always a multiplication of two or more things after the root. Other wise its just a straight line right? Well if your mixing two or more things doesn't it make sense that the outcome will always be some format of a multiplication? Why is that baffling people I really don't understand what isn't supposed to make sense. Whats wrong with the result being simply..it will always formulate an operator command. Why waste energy proving a computer can multiply when we already know it can?

[@karigucio]

wowowow, what

[@Jkauppa]

bs

[@Jkauppa]

next

[@Jkauppa]

if you dont give a context, your problem is air babbling

[@Jkauppa]

if you dont have a reason, even something, outside the stated problem, then it has no use

[@Jkauppa]

do you think people will die for your trash theories, better have some non-dead direction "why" you are doing something

[@Jkauppa]

boring

[@ophello]

“Maths” should sound as wrong to you as “Englishes” or “biologys” or “chemistrys.” Can you brits please just say “math” like a sensible person?

[@SA38178]

Thumb down. Check formulas before publishing the video.