@@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
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
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.
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?
@@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.
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.
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
@@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.
@@megachonker4173Nah, it's not a theoretical vs practical kind of thing. It's just that Maths is too vast a field. Olympiad preparations are done separately by students and it involves only a few areas Geometry, Combinatorics, Number Theory, Functional Equations, Inequalities, Graph Theory and Algebra. You can consider this like the Olympics. Everyone plays those sports but only few trained professionals can participate, if you compete with them or try to break their records, there's no doubt you will fail. They have acquired a special training and only a few select represent their countries.
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.
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😢😢
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.
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.
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
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.
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
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.
@@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!
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.
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.
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
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?
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.
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.
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.
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.
@@arnavdeep8396 unfortunately due to economical issues and dad's work stress I didn't make it. However I participated in and had been selected 4 times to compete Nationally from the state where I resided and have done so with pride. I didn't win medals, but I won valuable experience. And with that experience, it helped me grow and be better, after an admission exam for the major I wanted to study for, everything was easy, and so did the semester compared to my classmates. As of now, currently my major is Mechanical Engineering in the field of Management, and have been called from multiple areas for job opportunities due to my resilience and efforts. If anyone else reads this, just want to let them know that, even if you failed at a goal, the effort on trying is worth more than not to, you gain experience and lessons from failure rather than winning. Even if the opportunity was lost, there are many more opportunities that can be better or the same, as long as you know to keep moving forward. Thank you all for the support! 😄
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.
I tried this type of test in high school, and it led me to a catastrophic self-esteem collapse that ended with the complete derailment of my life plans and a case of morbid depression.
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.
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.
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.
My intuition is telling me to create the convex Hull of the outermost points. Make some argument that no matter what line and point point we start with these will always be touched. Remove these Hull points and create the new convex Hull. Keep applying the same argument until we are left with 1 or two points. If we have 2 points draw a line through them and use one of them. If 1 point choose any line going through it as the starting point.
I think the answer would be proving that this concept of switching pivots is actually the same as pivoting around a single point. You could deduce this since the line keeps rotating and is infinitely long. So you sweep 2pi and you got all points on the plane. Further thinking says that there should at least be points at both sections that L divides. Otherwise points can cover up other points.
I think how to choose a point P from S is just make a 2 point first and then connect them with a stright line for infinite distance after that make another point beside the line and then connect the 3 point together and just repeat it so we can pick a point p everywhere (correct me if i am wrong because i just a junior higschool students)
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.
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
Imagine if we take three points in a shape of equilateral triangle with size length of 1 unit and have a 4th point in the center... Place those triangles on a plane with the distance between the center points of > 3 units( basically in such a way that the triangle boundaries DO NOT crosses each other) ... in such a way that no point is collinear... Now a line can do the windmill action only if the line passes through inside all those triangles... But with our placement it's should not be possible... Because you can pass the line only through two triangles and the third triangle will always receive the line on its outer point instead of the center one which will result the line spinning around the triangle and missing the center point
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.
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.
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.
I am a mechanical engineer I scored A grade in Calculas 1, 2 and 3 and let me tell you i can't even process this question even after seeing the animation.
The answer is choosing point PE S and a line l passing through F such that the resulting windmill process uses each point of S as a pivot infinitely many
This is one of those problems where you cannot expect anyone with just pen and paper to come up with a proof in a day. Exam questions should be questions where there is a method of progressing to the answer using steps accessible to the entrants.
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.
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
I can solve it, the answer is: since it continuously rotates clockwise and never changes direction or stops, all points would be in contact with the line infinitely as much times as the line rotates.
I wouldn't be able to compete even in the paralympic version of this
I wouldn't be allowed admission to the parking lot
@@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.
This made me lose my coffee, bravo 🤣
bro relax 😭🤣
@@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
Please double it and give it to the next person
Literally please give the line to the next dot💀💀💀
I laughed harder than I probably should at this comment
Alternate sequence because there's no other point to prevent S or s corespondebt
Show. the formula
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
You know it's hard when a math problem doesn't have a single number in it
Pure mathematics ☠️
It's the true Mathematics, not the school Mathematics. Also, it's not "hard", it's just how Mathematics really is.
math is about logic and proving stuff. Numbers are lame and boring
@@sigmamaleaffirmationhypnob7340 if you really think numbers are boring then study ramanujan
Its hard even with numbers
POV math question requires more English language comprehension than the English one.
😂😂😂
Fr💀
lmaoo so trueee
I was looking for this comment. Thank you.
Ya and that is true for any standard math questions
I gave up at "the"
Mathematicians talk in this formal way that takes some time to get used to.
Same, lol
I was like ‘where does the problem say the’ then my slow ass finally finally got it when the video looped
LOL
"BREAK TIME!!"
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.
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?
You need to have some Olympiad math experience to even attempt these
My friend got 22nd in their country on this test in 8th grade
There is a big difference between school math and olympiad math.
@@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.
Without the animation, I wouldn’t even understand the question.
No, I still don’t know how to approach the question anyway.
I still don’t know what’s being asked
@@spelcheakI don't even know whats going on lmao
I think its a sequence and series and a progression based question but my brain is also fried
Shows the power of visualisation, mathematics wouldn’t be anywhere without it
@@spelcheakProve the line touches every point infinite times as it spins
Mitochondria is the power house of the cell.
😂
a²=b²+c²
😂
Mitochondria are the power house of the cell
Midi-chlorians are cells and the power house of the force
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.
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
Basicly, theorical and pratical knowledge.
@@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.
@@megachonker4173Nah, it's not a theoretical vs practical kind of thing. It's just that Maths is too vast a field. Olympiad preparations are done separately by students and it involves only a few areas Geometry, Combinatorics, Number Theory, Functional Equations, Inequalities, Graph Theory and Algebra.
You can consider this like the Olympics. Everyone plays those sports but only few trained professionals can participate, if you compete with them or try to break their records, there's no doubt you will fail. They have acquired a special training and only a few select represent their countries.
the children are asian so yea math
My brain melted
He lost me at " Let S..."
He lost me at " Problem 2. "
He lost me at "the international Olympiad"
He lost me at "the"
He lost me at "th"
He lost me at "t"
Me after hearing the questions:
"Good question "
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.
I did this once in 1991, 2 days, 4.5 hours each, shit hard.
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😢😢
I want to ask Alexa this and see if it melts down or explodes.
What happened? 😂
@@SATsolver It started spinning and singing 'Windmills of Your Mind."
@@youbigtubership ha ha ha 🤣🤣😅
Humans that are really good at math are so cool
Mi buddy the alien:
@@n4n4sh1_😂😂
We have an alien in our midst.
maths question without a single number in it goes crazy😭😭
Most of them dont
Holy shit balls. Where would you even start with a problem like that!
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.
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.
"could you repeat that?"
I've already told you it 6 times, Jimmy.
instructions unclear, built a functioning windmill
That is actually one of the most interesting maths problems i ever saw
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
I missed the part, "where that's my problem".
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
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.
My computer science degree degraded my maths skills
Just let a simulation run for a sufficiently long amount of time and just conclude it at that lol
Where's that random indian guy in comments claiming he already did this question in 4th grade? 😂
😂
JEE Advance maths is the toughest 😂
LMAOO ikr 😂
Loool
@@Maths_3.1415 Yeah because they’re 1.5 billion of them
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
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.
@@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!
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.
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.
Omg exactly like the space-filling atoms problem at the finale of Assassination Classroom :O
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
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
Windmill Effect 💀💀💀
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?
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.
Dunning Kruger effect
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.
BTW making a good problem is quite a lot harder than solving a good problem
Man this is not maths, it’s straight up depression
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.
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.
YOU DONT NEED TO EXPLAIN IT.
YOU ARE THE ONLY ONE WITH BRAIN HERE.@@donspaceye9730
Preach
„Mom? The math puzzles are getting scary again!“
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.
I am commenting here so that I can know if u made it or not!
Bro did u make it?
@@arnavdeep8396 unfortunately due to economical issues and dad's work stress I didn't make it. However I participated in and had been selected 4 times to compete Nationally from the state where I resided and have done so with pride. I didn't win medals, but I won valuable experience.
And with that experience, it helped me grow and be better, after an admission exam for the major I wanted to study for, everything was easy, and so did the semester compared to my classmates.
As of now, currently my major is Mechanical Engineering in the field of Management, and have been called from multiple areas for job opportunities due to my resilience and efforts.
If anyone else reads this, just want to let them know that, even if you failed at a goal, the effort on trying is worth more than not to, you gain experience and lessons from failure rather than winning. Even if the opportunity was lost, there are many more opportunities that can be better or the same, as long as you know to keep moving forward.
Thank you all for the support! 😄
"Let s be"
I know where this is going
solving it is difficult on its own, however i got lost at the explanation to the problem already
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.
Is the answer 'Australia'?
Close. It was New Zealand.
Nah it was Papua New Guinea
Nah bro, it was mitochondria is the power house of the cell
Broo u all are sooo wrong !! It was willy wonka and the chocolate factory!
Umm, actually it is Punjab
Never gone higher than state lvl in maths Olympiad 💀
I tried this type of test in high school, and it led me to a catastrophic self-esteem collapse that ended with the complete derailment of my life plans and a case of morbid depression.
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.
Solving it is very difficult, but imagine how hard it was to come up with it.
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.
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.
There is no problem to solve. It works just beautifully...
PASS! Next question please.
My intuition is telling me to create the convex Hull of the outermost points. Make some argument that no matter what line and point point we start with these will always be touched. Remove these Hull points and create the new convex Hull. Keep applying the same argument until we are left with 1 or two points. If we have 2 points draw a line through them and use one of them. If 1 point choose any line going through it as the starting point.
Haven’t seen the video yet, but this reminds me of the group action used on elliptic curves.
I can’t even comprehend the question with the animation… Let alone the written words…
My absolute respect to whoever crafted that problem🙌
Without the visual representation, i wouldnt even be able to understand the question... Let along figuring out thr answer.
The cell and his house have a beautiful mitochondria behind their power wall
I think the answer would be proving that this concept of switching pivots is actually the same as pivoting around a single point. You could deduce this since the line keeps rotating and is infinitely long. So you sweep 2pi and you got all points on the plane.
Further thinking says that there should at least be points at both sections that L divides. Otherwise points can cover up other points.
I don’t even understand what the question is asking😅It is even harder to comprehend than the monologues of Sir Humphrey in Yes Minister 😂
I think how to choose a point P from S is just make a 2 point first and then connect them with a stright line for infinite distance after that make another point beside the line and then connect the 3 point together and just repeat it so we can pick a point p everywhere (correct me if i am wrong because i just a junior higschool students)
What was that question again? 😢
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.
@@fieryscorpion so it should be a closed loop?
@@derianvandalsenyou dont draw it, is supposed to be some mathemagician spell shit idk what they call it, equilibrium or something
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
I think honestly the most complicated part of this is explaining this in writing
so we are just going to ignore the fact that only six people got the problem no 6 right
Imagine if we take three points in a shape of equilateral triangle with size length of 1 unit and have a 4th point in the center... Place those triangles on a plane with the distance between the center points of > 3 units( basically in such a way that the triangle boundaries DO NOT crosses each other) ... in such a way that no point is collinear... Now a line can do the windmill action only if the line passes through inside all those triangles... But with our placement it's should not be possible... Because you can pass the line only through two triangles and the third triangle will always receive the line on its outer point instead of the center one which will result the line spinning around the triangle and missing the center point
This is the problem you encounter on a test that your teacher said is “Not challenging”
I am filled with sorrow and disappointment towards the self whenever I am reminded of my intellectual inferiority by such enigmas.
Wh-wher-where’s the numbers?!?
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.
Even my most optimized algorithm would scream TLE on this.
the 22 people who got a perfect score really are patient geniuses
I didnt even realize there was a question on the problem
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.
I thought I was smart….
I joined the dots...still haven't figured out what picture it made.
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.
Great video just watched one on reapeating symmetry
(a+b)^2 = a^2+2ab+b^2
FYI, there are guys that's not either student or scientist like me but enjoys your videos)
I am a mechanical engineer I scored A grade in Calculas 1, 2 and 3 and let me tell you i can't even process this question even after seeing the animation.
Put 'em in a circle and start with a line not passing through the inner part of the circle.
The answer is choosing point PE S and a line l passing through F such that the resulting windmill process uses each point of S as a pivot infinitely many
This is one of those problems where you cannot expect anyone with just pen and paper to come up with a proof in a day. Exam questions should be questions where there is a method of progressing to the answer using steps accessible to the entrants.
It’s the international math Olympiad those people are very smart
They are ✨ premium ✨ human
I wonder what patrick would answer 😂
I can’t even understand what the question wants me to do
Show that it's possible to pick a line and point for any set of points such that the line touches every point infinitely many times while windmilling.
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
I cannot comprehemd math without visualizing things in a physical reality, hell I still even struggle at elementary physics.
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.
"only 6 questions? phew, i can do that easily in 10 minutes!"
Dude that rotation to the bottom right totally has 3 collinear points.
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
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
Put a point at each corner of a square
day 2s last q casually being even more difficult
Thanks to this short I watched that video
I was there as a guide for Cyprus.
This seems quite similar to the gift wrapping algorithm used in 2D physics engines to compute the convex "shell" of a polygon.
THIS ONE IS SO FUN
It's much harder to understand the question than solving it
You quit the question as soon as there is more english in math than numbers.
I wouldn’t even be able to remember the whole question 😭🤣
Him : P2 is very hard (only 22 students completed it)
Meanwhile p6 with only 6 students :🗿
I can solve it, the answer is: since it continuously rotates clockwise and never changes direction or stops, all points would be in contact with the line infinitely as much times as the line rotates.