วีดีโอ

This guy is 16 years old? kudos!!!

Before watching the video here is how i would handle it: prove by induction. For n=3, there is a pair with the shortest distance, they watch each other and the third is not watched. For higher n: there is again a pair with shortest distance that must watch each other. If one of them is watched by another, we can use pidgin hole principle to see that there must be one left unwatched. If the pair is not watched by another, then we can disregard the pair and refer to the n-2 case by induction.

1=2😊

4:51 Here's my proof: There are only 50 even and 50 odd numbers in the set, so there are an even number and an odd number in the subset, and their sum is odd For me (an autist), it's easier, although it uses a fact that some people may not know (the oddity of a sum)

This problem looks like it generalizes to some sort of upper bound for information loss in KNN clustering, neat 😂.

my reasonning: i have an algo to find the one loop: take the longest edge if its a pair, you can remove it without altering the rest else, break loop end: the origin of the longest edge is not seen by anyone

you could do it in two cases, though none are satisfying. could have L,W and D = 0 or Either L,W or D = ∞ assuming none of L,W or D = 0

Strange wording, it was presented as astronomers on different planets for me. It's a middle school level problem.

why are they always criminals

What will be if criminals make equal sided polygon?

Before I finish the video, I'll share my solution: for each node, draw a circle that just reaches the closest node. Let's consider the biggest circle. If there is another circle of the same radius, this means that they're watching each other. Let's remove them from our consideration and consider the next biggest circle (there is gonna be at least one since the number is odd). We found the biggest circle for which there is no circle of equal radius. Now, since each node's circle includes just one other node, on the edge (none inside), this means that for the node with the biggest circle there is no such other circle that it includes the current node, which means that there is no other node for which the current one is the closest, which means we've found a criminal who isn't watched by anyone.

Currently at 7:08 and there is one more constraint on k. If k = 1, Alice can choose 1 or 2 for that one number, and Bob’s task is impossible. So 1 < k < 593

I dumbed it down to: As there's an exact number of "watchers" and "watched", any time someone is being watched more than once, there is someone not being watched at all. There will always be one shortest line, where the 2 criminals will be watching eachother, and that would either have to be far enough away from the remaining criminals to be unwatched by any other person, OR have a third (or more) person spectating, thus forcing there to be one person watched twice, therefore one person unwatched. If they are disconnected from the rest of the group, then the remaining group will also have one smallest line where 2 are watching eachother, and you come across the same conundrum where this group would also be seperate or watched by a third party. No matter how high N is, you can either keep breaking off pairs until you get down to only 3 people, where the last person has to be unwatched, OR you will have already found someone being watched by 2+ people, which will inevitably mean someone somewhere down the line is unwatched.

Is there a general equation for the line cutting the quadrilateral into a triangle and a tangential quadrilateral?

The way the problem is presented confused me at first since it didn’t say any k of the set just exactly k. Maybe it’s a standard that these problems are worst case scenarios but I assumed it was alice working to maximize k, in which case she would choose the smaller numbers over the larger.

At LEAST one person will be alive. It is also possible for multiple to be alive.

Who watches the watcher?

If every criminal is watched, every criminal is part of a cycle. At least one has length greater than 2, a contradiction. (this was before watching)

a pair of sum b, and a pairs of sum 2024.

To my own surprise I actually solved this one. So, what I did is in order for Bob to ALWAYS be able to add numbers to the be the same is Alice, his total of his selection must amount to greater than or equal to Alice's. In order for Alice to maximize her total using k numbers, she chooses the k largest numbers in that set. So basically, the maximum value for k arises when the sum of the last k numbers in a set of successive integers from 1 to n+k is less than the sum of the first n integers. We can set up an inequality where n(n+1)/2>((n+k)(n+k+1)-n(n+1))/2 and using our knowledge that k is equal to 2023-n, solve for n and round up using the quadratic formula. By doing this we get n is equal to 1431 and therefore k is 592 since k = 2023-n. With this you can generalize for any set of successive integers. Edit: Just saw you covered this solution after skipping straight to the answer and that I did this for no reason :(

Is it just me or is this the wrong answer. If k=1 and Alice colours just the number one then Bob can’t solve it. So the largest k is should be 0. This is a weird edge case but I can’t see why this shouldn’t be a solution. However I really enjoyed the video and the problem

0:15 excuse me sir, Hungary is not a Balkan country (on paper)

uhhhh i don't get it, why we're taking pairs of numbers and why k=51... i feel bad :C

Wouldnt you just sum all of the numbers and divide by 2 to find the maximum sum? 0.5(2023)^2 +0.5(2023) = 2047276 Work backwards to find how many numbers are needed to surpass 1023638 -----> bob needs 1431/2023 numbers, so K is 592. It is OBVIOUS there will always exist a number you can leave out to obtain the required sum. I think that is what students wrote but the graders had a stick up their ass

i love this video

Not sure what it is yet, but can't be more than 592. Otherwise Alice would pick the highest numbers and Bob would not be able to get a sum of equal size.

k=1011 alice colours 1-1011 red (1011) bob colours 1012-2022 blue (1011) it says some so there can be uncoloured numbers so 2023 remains as the only uncoloured one 1011=1011 proof: if k>1011 then alice colours 1-1012 red (1012) bob colours 1013-2023 blue (1011) 1012>1011

This does beg the further question, "which quadrilaterals can be divided into fewer kites?" For 1, it is obvious; the kites themselves. For 2 through 5, I do not know. For 6 and more, every quadrilateral can.

Edit: failed at reading comprehension, see replies Counter example: Alice chooses the first 1428 integers, they will sum up to 1020306. then if Bob chooses the remaining integers they add up to 1026970. However Bob can remove 1664, 1665, 1667, and 1668 from his choices, making his sum 1020306 - same as Alice's. So k=592 is wrong

“you got a solution in a way we didn’t want so you lose points” is such bs 😭

Thermonuclear warhead as the weapon of choice. Checkmate athiests

Why are they criminals? Why do they watch each other?

Obviously. Because the distance between people is different for everyone, this has to be. For there to be no one left, there was to be an odd numbered amount of people in a group, equidistant to each other, and they cannot pick someone that someone else picked. But since there can be no groups like this, one person will always be left standing

What do you use to make your videos, its very similar to veritasiums style

Fun fact: The person who solved the problem entirely (got 10/10) is Romanian and he also got a max score in the IMO that year. Being Romanian too, I see this guy as the Romanian Jesus of Mathematics, you can show him any problem and he will solve it for. Anyway, completely coincidentally, the author of the problem is also Romanian, and the original text used "n natural number" instead of "2023", but they changed it before the contest as it probably would have been an overkill.

At 20:23 you say we have 2 edge cases with bi=1 or bi=2, but doesn't bi/2-2 imply another 2 edge cases in bi=3 and bi=4 ?

5:00 I solved this a different way: if there are more than 50 numbers in the subset, we know that they can't all be odd / can't all be even (since there are only 50 odds and 50 evens in the original set) - thus, there is at least one odd number and one even number, which we can add to get an odd number.

Polyamath posted, hence I stop everything I am doing

I’ve no problems in math in any placements: schools, universities, but I always been struggled at the olympiads. This questions wants some impossible thinking during short time, which requires that you knows these types of questions and know how they are solving, otherwise you’re just a genius, who can solve any sort of problems in few hours

Hi, I’m that guy who made a bunch of problems in the math problem channel in your discord server. Good vid

Legend of the Problem 4

I'll figure it out one day