A beautiful international math olympiad problem

[@jeremydarcangeli7093]

I wouldn't be able to compete even in the paralympic version of this

[@SephresX]

I wouldn't be allowed admission to the parking lot

[@gredangeo]

@@SephresX I don't even know if it's even legal for me to watch this video. It's above my pay grade to understand this.

[@zacharyhall2012]

This made me lose my coffee, bravo 🤣

[@airbornesnow4541]

bro relax 😭🤣

[@spicywasab]

​​@@gredangeoas a non-native english speaker who just watched "The Night Agent" on Netflix, I'm sorry to ask this dumb question but is it a ref to the series ? *it's 32 past 2am please don't mind the stupidity of this question

[@KaramAlayan]

Please double it and give it to the next person

[@SXLSkyya]

Literally please give the line to the next dot💀💀💀

[@lianifire]

I laughed harder than I probably should at this comment

[@marcomabr]

Alternate sequence because there's no other point to prevent S or s corespondebt

[@marcomabr]

Show. the formula

[@KaramAlayan]

So i am talking from experience the comment is really from experience LMAOOO at the time i was 14 If we put a × (multiply sign) between the second and third digits of the number 2020 meaning like this 20 × 20 = 400 a complete square and the result was a complete square how many numbers between 2010 and 2099 have the same property ? This was a question that i got in the 2021 Syrian math olympiad its so hard i didnt get it right good luck enjoy And the answers : 1 or 2 or 3 or 4 or 5

[@funkyschnitzel]

I did quite a lot of math in my undergrad degree, and consider myself fairly competent. If this question had been posed to me without the video we got here, I wouldn't even know what it was asking, let alone how to answer it.

[@abhyudaykrishna9714]

that's a bit much, but i agree with the spirit of this lmao. im both disappointed and relieved that my undergrad (3.5 years of which are over already) has almost nothing to do with math?

[@lgooch]

You need to have some Olympiad math experience to even attempt these

[@andersnaugle4105]

My friend got 22nd in their country on this test in 8th grade

[@TheLudwigWan]

There is a big difference between school math and olympiad math.

[@Zorisura]

​@@TheLudwigWanI've joined a math olympiad before, and big is an absolute understatement. Even at the beginner stages of the events, you can already see how olympiads involves ALL the lessons you've studied one part at a time in school into one problem. It's genuinely insane, most of the people in these contests have a hard time comprehending it unless you're from one of those schools which have diehard trainings for their students. Math is the language of the universe, and these people are linguists.

[@prasannabantu284]

Mitochondria is the power house of the cell.

[@mikebar42]

😂

[@idlepopz7136]

a²=b²+c²

[@ayshaaktherjannat7624]

😂

[@filipultr4435]

Mitochondria are the power house of the cell

[@PrinceMeNb1]

Midi-chlorians are cells and the power house of the force

[@gbear1005]

I gave up at "the"

[@waarschijn]

Mathematicians talk in this formal way that takes some time to get used to.

[@KewalTailor]

Same, lol

[@frozenpeak1524]

I was like ‘where does the problem say the’ then my slow ass finally finally got it when the video looped

[@zarrowthehorse]

LOL

[@keorinkoluexl6365]

"BREAK TIME!!"

[@tiffanytan2183]

Without the animation, I wouldn’t even understand the question. No, I still don’t know how to approach the question anyway.

[@spelcheak]

I still don’t know what’s being asked

[@mcgaming1172]

​@@spelcheakI don't even know whats going on lmao

[@AnmolArpit-iy2pk]

I think its a sequence and series and a progression based question but my brain is also fried

[@wabalaka1565]

You know it's hard when a math problem doesn't have a single number in it

[@Jee8655]

Pure mathematics ☠️

[@megachonker4173]

It's the true Mathematics, not the school Mathematics. Also, it's not "hard", it's just how Mathematics really is.

[@sigmamaleaffirmationhypnob7340]

math is about logic and proving stuff. Numbers are lame and boring

[@Search957]

​@@sigmamaleaffirmationhypnob7340 if you really think numbers are boring then study ramanujan

[@mamtagrover3748]

My dumbass brain just thought of drawing the answer Blowed up damn

[@paranoyd1]

Why not? It’s still math

[@mamtagrover3748]

@@paranoyd1 how

[@liitutereuiui4687]

​@@paranoyd1You have to prove ot works with any finite amount of dots in any order. The picture is just for visualisation and was not included in the excersize

[@capsey_]

​@@paranoyd1 I mean, if you can prove for general case using drawing then yes, it's valid answer

[@paranoyd1]

@@liitutereuiui4687 it’s harder to prove with drawings but technically it’s possible. Aren’t words and numbers drawings too?

[@Stepbrohelp]

I have a bachelors degree in pure mathematics, and I have absolutely no idea where to even begin with a question like this. The fact that actual children are able to solve these kinds of questions just blows my mind.

[@ohwow512]

Not sure if I'm correct, but i think this problem leans more so onto logic rather than advanced mathematics, which is why some people who are well-versed in pure maths fail to solve this

[@megachonker4173]

Basicly, theorical and pratical knowledge.

[@zchelmerjoashgamboa7366]

​@@ohwow512 So much of high level math is just purely elaborate trains of logic, so yes, it's that in a way. The actual solution will blow your mind though.

[@nampita]

He lost me at " Let S..."

[@dotand0]

He lost me at " Problem 2. "

[@Jee8655]

He lost me at "the international Olympiad"

[@mcgaming1172]

He lost me at "the"

[@gauravkoirala8399]

He lost me at "th"

[@Maths_3.1415]

He lost me at "t"

[@imaginaryunit123]

My brain melted

[@Shabudana]

Me after hearing the questions: "Good question "

[@Beanskiiii]

Humans that are really good at math are so cool

[@n4n4sh1_]

Mi buddy the alien:

[@AnshuSingh-et1ms]

​@@n4n4sh1_😂😂

[@PsyloAlpha]

POV math question requires more English language comprehension than the English one.

[@shameem8743]

😂😂😂

[@moonmartex]

If you don't understand anything don't worry, the IMO is the final BOSS of math olympics and even a person with a decent math training would still struggle with this.

[@youbigtubership]

I want to ask Alexa this and see if it melts down or explodes.

[@SATsolver]

What happened? 😂

[@youbigtubership]

@@SATsolver It started spinning and singing 'Windmills of Your Mind."

[@SATsolver]

@@youbigtubership ha ha ha 🤣🤣😅

[@joshbull6467]

Holy shit balls. Where would you even start with a problem like that!

[@K4vin60]

Watch the video it’s really cool

[@dinojello]

Drawing some scenarios on the paper for cases with a small number of points I guess. If you figure out something even if it’s not complete solution you can get some points.

[@geordiepunchingahorse423]

What makes this problem special is that unlike most hard IMO problems (2023 P6, 2021 P2, 2020 P6) is that this problem has a relatively simple solution. The solution is posted on their channel but it revolves around the idea that you draw a line pasing through one of the points and divide the plane into 2 parts. Notice that the number of points on each side does not change (other than a brief moment when the line passes through 2 points). These kinds of questions with relatively simple ideas but little solves are dubbed “windmill effect” problems because of this problem. They usually have simple solutions that not many people will think of.

[@OmnipotentNoodle]

Omg exactly like the space-filling atoms problem at the finale of Assassination Classroom :O

[@henry1227]

2021 p2 is disgusting lol I remember Luke (Robitaille) explaining a combinatorial solution back at MOP and it had to be split into like 3 sessions

[@GeeBeeOhThree]

I wasn't quite there but I did know that having it so no more that 2 points fall Colinear means there cannot be a point that will *never* be hit (because the colinear would take priority), and given the nature of the space every point has to be culinary woth at least 1 other point. It's not proof, but it's a decently sound set of data to assume the statement is correct. Given the highest math I have mastered is Differential Equations, I'm happy with that

[@eeeeeeeeeeeeeeeeeeeeeeeeee9594]

Windmill Effect 💀💀💀

[@StupidusMaximusTheFirst]

The number of points on each side can change, don't let that illustration fool you. Since no 3 points from the set can form a line, any subset of 2 points you pick to draw a line is guaranteed to have at least one which does not define another possible line. It doesn't really go indefinitely, you can only use a certain pivot to draw as many distinct lines as your set allows, which is gonna be -1 to the total number of points. So you cannot use a pivot indefinitely to draw distinct lines using a finite set of points?

[@chomumama5978]

Makes me think of the pen spinning in hand trick that bored students do in math class. Tips of fingers as points and the pen as the line, only the pen doesn't go exactly on the finger but its a good analogy for this problem.

[@mrbenwong86]

I did this once in 1991, 2 days, 4.5 hours each, shit hard.

[@NameSpot06]

Damn how did you even reach the level to attempt that test thats ao damn hard I failed in my 1st regoinal olympiad exam (lagged by 10 marks) in atronomy😢😢

[@taizu55]

Where's that random indian guy in comments claiming he already did this question in 4th grade? 😂

[@tishanchamoditha9033]

😂

[@Maths_3.1415]

JEE Advance maths is the toughest 😂

[@emperor8716]

LMAOO ikr 😂

[@sakinurrahman3890]

Loool

[@TristanVandeloock-ue6us]

"could you repeat that?" I've already told you it 6 times, Jimmy.

[@wjudee]

i think this problem is actually really simple if you have good familiarity with invariants, because you can give an orientation to the line and it’s easy to find out that the number of points on each side stays the same

[@J3NKA]

You could still have a set S containing some points creating a convex closed bound with no points outside that bound, and choose the line to be tangent to that bound. Then none of the points inside that bound will be reached even though there are the same amount of points at either side of the line at any time. You have to choose P such that it doesn't create a convex closed bound with no points outside of that bound. Edit: I guess you can say that you choose P and l such that you always have atleast one point on each side.

[@wjudee]

@@J3NKA i solved it this way: let’s assume for simplicity that there’s an odd number of points. we say that “bright points”are those on the right and “dark points” are those on the left. if we chose a starting line in such a way that there are as many dark points as bright ones, then when the line pivots by 180 degrees it will be on the same point as the beginning (because of the invariant thing) but dark points and bright points will be swapped. since to change color one point must be hit, all points are hit and it loops. if there is an even amount of points than you can assign a colour to the pivot and it works similarly, but there’s a slight catch which is easy to figure out (so i encourage you to try). it’s a really beautiful problem!

[@user-en7dx1qp3k]

It is a simple solution but a few of the finer details make it hard to find and prove. For instance, the intended solution involves giving the line an orientation and then placing it in a starting position for which it is "in the middle" (with the same number of points on either side and the line passing through only one point). It's not actually clear that this is something that you can do, and, even though it's not hard to prove you can, omitting this part would potentially lead to a 1-point deduction. The main reason this problem is hard (aside from it being combinatorial geo lol), however, is similar to that of IMO 2014/5, in that while the intended solution is short and simple, there really is only one correct solution despite there being so many seemingly reasonable directions to go with. Nothing else works, which is rare for the IMO.

[@donspaceye9730]

There is one possible solution. If we can draw a graph satisfy the following 3 conditions, then this statement holds: 1. There should be an incoming edge and an outgoing edge for every node p. This outgoing edge must turn clockwise relative to the incoming edge of node p. The inward angle of this edge must be the smallest among all possible edges of p to its neighbors. 2. This graph must be cyclic, which means that the graph must contain a path that forms a loop. This path should be unique, which means that for every node in this graph, there should be exactly one incoming edge and one outgoing edge. 3. This graph must contain every single node in set S. if for every set S, we could draw a graph that holds for the above 3 condition, than the statement in the question prompt holds.

[@donspaceye9730]

For every set S, there must exist a convex hull (a polygon hull that bounds every single node in set S). If this convex hull contains every node in set S, then we are done. If there are nodes inside the convex hull, we simply create a smaller convex hull inside. We can make these two convex hull a single graph component by erasing two edges and add two new connecting edges that connect these two hulls together. We can repeat this process recursively by connecting this component with other inner convex hulls, until we reach the innermost nodes. This proof is done.

[@Powerdeadmau5]

Imma be real honest, I tried going for this by getting in the OMM (Olimpiada Mexicana de Matemáticas), which is the Olympiad in Mexico. Since the only way to go to the IMO here is by getting Silver or Gold in OMM. I tried getting selected, but failed the 3rd stage inside the OMM. So I couldn't go for this year's OMM, which I believe it was in Durango. I've met someone who got Bronze or Silver in IMO, which was in Japan last year. He's a cool guy and even took a photo with him. Currently I've only got 1 more chance to enter the OMM, just so I can try and enter the IMO.

[@heyshubh9851]

I missed the part, "where that's my problem".

[@funcompilator1202]

thats why everyone can't be awesome at math matematicians don't need a problem to be practical, their target is to solve it to show themselves they can it isn't bad, just a reminder

[@skeltek7487]

Easy to understand problem. If the pivot and line chosen randomly, some points may never become the pivot as the line just circles around them. There is a lot of ways to go on about this. The least elegant solution is probably one of the most simplistic and uninteresting... To start with any three points and do an induction, where any new point added either doesn't break the validity of the statement or the initial pivot may be redefined to keep the validity of the statement true.

[@johnjoe9976]

Is the answer 'Australia'?

[@youbigtubership]

Close. It was New Zealand.

[@I_am_Alxui]

Nah it was Papua New Guinea

[@arham_shah]

Nah bro, it was mitochondria is the power house of the cell

[@rishisaluja2008]

Broo u all are sooo wrong !! It was willy wonka and the chocolate factory!

[@mistersingh6470]

Umm, actually it is Punjab

[@SpartanForces117]

„Mom? The math puzzles are getting scary again!“

[@NayeemMohammad-bq9gx]

I am in my PhD Program in CS. Even if I go through all the maths from grade 6 till college, I still wouldn't come up with something like this question. Solving it is out of question.

[@adhirathpatil]

BTW making a good problem is quite a lot harder than solving a good problem

[@r35ct12]

solving it is difficult on its own, however i got lost at the explanation to the problem already

[@pulkitjain8135]

My computer science degree degraded my maths skills

[@ohwow512]

Just let a simulation run for a sufficiently long amount of time and just conclude it at that lol

[@ThatOneAmpharos]

Id rather kms than solve it. Edit: on a second thought, let me cook

[@2nd-place]

When I was a young man, I used to think I was pretty smart. I didn’t know enough to know how dumb I was.

[@ujwalj7001]

I am going to write the basic level for selection for this exam from India next month. It is called IOQM -Indian Olympiad Qualifier In Math Why when i am just a 10th grader? Bc my coaching centre told me to. (Pay attention to the fact that i mentioned 'Coaching centre' and not School or Tution Centre) The syllabus for the exam is based on class 12th topics nd even some advanced ones. Still students from 8th to 12th write them. This is the 2nd time writing the exam nd i just write without any prep as the prep would cost me a great fraction of my time nd won't make much difference to the result. Last time i got 7 out of 100 nd i still don't know how my 2 intuitive guesses, 1 from sets nd 1 from permutations, came right. I got 1 mark from another permutation qn which is actually funny. I finished the full exam in 2hr nd around 30 min. so 30 min remaining nd i drew a full 8 by 8 triangle with each mini triangles nd counted the no. of triangles manually nd proceeded to get 1 mark for that qn. I hope u understood that i finished the exam this fast as i didn't know any answers. But that result motivated me greatly bc i only went 3 marks short of the cut off - 10 marks It would have qualified me for ROQM - Regional Olympiad Qualifier in Math - nd lost a spot for another student who at least made some effort for the exam. The qns nd the sylubus is actually absurd from the POV of a physics enthusiast who learn math out of necessity but appreciates it's beauty so much that sometimes it makes me think math is my field rather than physics. I've seen this vid before but thought i should share my experience when it again got recommended to me.

[@bald_agent_smith]

So what’s the problem? 😅

[@VersedYT]

To prove that the “windmill” will function infinitely many times due to the switching of point S to point P.

[@lauriesteveescoses590]

The last sentence of the problem. Show we can choose a point in the set of points such that every point in the set will become a pivot at some point

[@varunittigi9288]

I wonder what patrick would answer 😂

[@LightningStrike691]

Haven’t seen the video yet, but this reminds me of the group action used on elliptic curves.

[@Sumpydumpert]

Great video just watched one on reapeating symmetry

[@pgc6290]

Its good that the problems get increasingly tough since then only your knowledge and intelligence is tested and not how good you can solve the paper//ie test.

[@Courett]

"Let s be" I know where this is going

[@user-ot8it3bn4n]

What was that question again? 😢

[@fieryscorpion]

Show that we can choose a point P in S and a line l going through P such that the resulting windmill uses each point of S as a pivot infinitely many times.

[@derianvandalsen]

​@@fieryscorpion so it should be a closed loop?

[@khazixlol]

​@@derianvandalsenyou dont draw it, is supposed to be some mathemagician spell shit idk what they call it, equilibrium or something

[@yasmngeorge3802]

Writing the problem must be just as hard as solving it. I’d start by plotting what I can then going over the question again

[@elmoe01]

PASS! Next question please.

[@kongolandwalker]

I do not remember exactly, but I think at school (in an olimpiad preparation class) we had a problem, where we had to prove that some special line across a pointcloud exists, and the intended proof was the same windmill process.

[@bobing1752]

Okay I think I found a solution in about 10 minutes but I'm not sure how to prove it: The first observation was that if you put a point in a triangle and start on an edge of the triangle with a line not crossing it, then the windmill will never actually touch the inside point. So here's my dumb engineer theory which seems to work: Take your subset of points and link the outermost points together to create a polygon with only the outermost points AND NO OBTUSE ANGLES for the pilygon. All the remainint points should be inside this polygon. Repeat the process for the remaining points so that you have "matriochka polygons", until there are either: 1 point remaining, then just pick it as your starting point and the direction of the line so that it hits a point of the innermost polygon as the very first 2 points remaining, then pick a point and a line so the very first point point the line will hit is the other one No point remaing, then pick a point on the innermost polygon such that it will another point of this polygon next I have tested my solution and it seems to work. I'd be very interested in knowing whether people think it doesn't work (if you can provide an example I'd be happy), or else if people can actually prove it rigurously. I can intuite that it should work because starting from the innermost, the line will necessarily cross all the points in its way, but I don't think it's enough to be consideres proof.

[@necoluga7950]

i kind of thought of a way to approach it a little bit. it probably take a good amount of research into set theory (i am noob i baraly can use like the logical operator things to form a statement) so how i thought of it is that cense any kina slope in a polar form of like xcos(θ)-ysin(θ) = ycos(θ)+xsin(θ) would have to have some angle that the slope intersects with the point. and then i kind of relised that the parameter that you couldn't have 3 co-plainer points made it so there weren't any weird edge cases. so eventually it would rotate into a position where the points would intersect with the slope given that it continuously rotated clockwise. so yeh idk its kind of bad and ill spoken but i that's how i would approach it. it maybe isn't right idk.

[@sapnupuas1475]

There are finite amount of sets and all points are colinear, a line that intersects two points is unique thus if you rotate the line it is bound to hit another one point, you can do this indefinitely like intersecting the same points again but the total amount of points is still finite

[@dorelbaz8939]

We can solve this problem by way of negation. This way, we'll get at least one point P that will never be used as a pivot by the windmill and this point P (examining one such point will suffice) will necessarily have to violate the condition for it to achieve this status. i.e - a contradiction with the condition set by the problem. Thus solving the problem.

[@patricktoo]

instructions unclear, built a functioning windmill

[@gurgigir]

I think honestly the most complicated part of this is explaining this in writing

[@brucemillercanada]

it will be a nice cs prject too , like try to code the dots and line and you can even sort them in groups and extract the data for 2 dots , 3 dots , random spawn , organized spawn ....

[@entertainer5189]

I think the answer is that maybe since there are N of points and N of pivots the process much cycle through each point like if there is windmill rotating clockwise so its basically going 360 degree thats all we know about that so during this every points of S may encountered the line L once a rotation and all this go through indefinitely because there are only N number of points so its often revisited infinitely untill the windmill is running so Means Every single point of S is a pivot untill the windmill stops moving? Well not know coz not so good in maths just got interest in this question so!

[@Phlegethon]

I thought I was smart….

[@Alexander-mw1ek]

I don’t know how to show this pedantically but its pretty obvious that every point will be encountered since the line will eventually rotate 360° and given the exception that no 3 points are collinear.

[@laughingbrick4824]

You quit the question as soon as there is more english in math than numbers.

[@a.v.y8331]

without knowing too much math and having not watched the video, I imagine there are two proofs that the solution has to fulfill. 1) this process produces a regular pattern that will return to point p with line l having the same gradient as the origin 2) the line must pass through each point at least once over iterations of the process (this would prove that at some point, each point will be used, and from 1, each point will therefore be infinitely passed through over infinite iterations) I am assuming this problem has something to do with graph theory but I have no idea

[@Whoiskimb]

I can’t even comprehend the question with the animation… Let alone the written words…

[@Eps398]

What a beautiful problem 💕

[@jedagelijksebraintraining]

I was there as a guide for Cyprus.

[@AryanKumar-vo1ic]

Sir, could there be more of geometry based videos? It is such a delightful branch of mathematical sciences, I m sure you would find something enjoying to teach us about. Thank you.

[@harshgada4325]

Without the visual representation, i wouldnt even be able to understand the question... Let along figuring out thr answer.

[@PieletPi]

Solving it is very difficult, but imagine how hard it was to come up with it.

[@voltairespuppet]

I think i came up with one that wont visit them all if starting at one of them. 4 points. Triangle and a point at the centre; start at any of the outer points. Generalisation for infinite points on an outer circle and one or more points on another circle inside. Start on any outer point and it will just rotate around the outer circle. Start on the inner circle and it will glitch step from the inner to the outer for each turn.

[@likusbikus5892]

my head hurts listening to this 😩

[@n.m.2624]

So its a set, if we see the question carefully its asking us to prove all of them, Basically a set is of 3 types, reflexive, symmetric and transitive... We have to find a general Formula and prove that no exceptions are there in it, it proves the questions point and thus it is the answer, this is probably the outline of the Process

[@samuil7365]

Guess and check

[@pressfinchat]

I can’t even understand what the question wants me to do

[@zekmaro]

Imagine if they gave animation with problem description

[@AmanSingh-nw7lw]

What a beautiful problem

[@KotakMeister]

I joined the dots...still haven't figured out what picture it made.

[@gomtipatel1717]

Man this is not maths, it’s straight up depression

[@beinganangeltreon]

depending on whats allowed, the line is at least as long as the smallest interval length between points,

[@beinganangeltreon]

and it is along with having a minima, is also utilizing the math like the math of a sprirograph, because the line is said to rotate, it only traces an overlap 2D area that is nonfinitely wide and tall, but has an overlap region with the line rotation of the next, or other (2 dot system) point. It might be that the utilization of the word "line", can refer to the solvers utilization of a finite length line, or a nonfinite length line. A finite length line would have a length an infinitesimal longer on both distal parts of the line than the distance between the very longest distance between the different two or greater points as an upper boundary of an inequality math statement. with a finite length line, then one possibly provable statement is that the number of points combinatorial all have finite length windmill lines, and from being finite can be characterized as having an overlap region, where the continuing rotation produces an amount of overlaps that is identical at amount to the geometry or topology of a duocircle venn diagram produced amount, like an integer amount of overlaps. its possible that trying to prove an exception causes proof from contradiction, or possibly other geometry or topology proof components known to the solver, like angle relations, could contribute to a proof. actually, I'm clueless,

[@olsa4427]

Thanks to this short I watched that video

[@_Thomas.Shelby_]

It's indeed beautiful, just like Blobfish

[@aarshjain4527]

Never gone higher than state lvl in maths Olympiad 💀

[@JuperLuperBruh]

You should do a video on 2013 IMO Q2, it's not as hard as this but I regardless still think it's a really beautiful problem

[@ayzaanashhad7063]

(a+b)^2 = a^2+2ab+b^2

[@P0leP0le]

Put 'em in a circle and start with a line not passing through the inner part of the circle.

[@ScientObject40]

I am filled with sorrow and disappointment towards the self whenever I am reminded of my intellectual inferiority by such enigmas.

[@cahdoge]

So you have to proove, that "doing the windmill" eventually touches evry point in an arbitrary set with no three points on the same line? Intuitively, it has to be true, sicne the rotating line is describing a plane and evry point is on the same plane.

[@R3VISION3]

This seems quite similar to the gift wrapping algorithm used in 2D physics engines to compute the convex "shell" of a polygon.

[@xanderortega4359]

I cannot comprehemd math without visualizing things in a physical reality, hell I still even struggle at elementary physics.

[@angrytedtalks]

Presumably each pivot rotation would eventually encounter another point from which to transfer pivot status. Clearly in a two dimensional plain there is no reason to suppose the rotation wouldn't reach other points. Mathematically, it is as easy as Pi.

[@robinwang6399]

Haven’t given any time to think about it but I feel like this have to do with complex numbers. Going to come back later.

[@UsuarioDEAD]

It would be much simple for me if i could draw the problem like this in my mind as i read the statement..

[@thebluefriend]

Even my most optimized algorithm would scream TLE on this.

[@user-oj4kg1yi1t]

If I read the question without imagery shown there’s no way in hell I would even try this one 💀

[@freddy353]

The problem with most math problems isn’t how difficult they are to resolve but the way in which they’re presented. The verbiage used seems to be intended to confused rather than guide.

[@6-4crusader55]

I’m just about done with calculus now. I genuinely don’t know if I would be allowed to watch the math Olympics on tv

[@JamesAlbertcohen]

I know I’m dumb these videos making it clear I’m more dumb than I think

[@user-ns8br8ex8q]

Wow you must be a genius to solve it

[@Daily_Wits]

how do you make animation? i love your animations?

[@PhysicsProtagonist]

I gave a simpler version of this problem to my 6th grade sister, and she was able to solve it

[@CatBahptista]

Yeah I went to the comments pretty much straight away

[@KK-yv8qr]

I had a brain fart.

[@latoyadowns8934]

Ngl I feel like I can brute force this. I might go insane, but I will get it eventually

[@lawscriteria]

the 22 people who got a perfect score really are patient geniuses

[@daviddickey9832]

I could do this in a programming language but haven't a clue how to do this with mathematical formality.

[@HelloWorld-ts7xb]

I didn't even understand the question