I think the problems were fair. Perhaps easier than usual but still plenty of traps for those who aren't paying attention 😄 In my prime I think I would have score well in this IMO. Unfortunately I didn't find the time to seriously attempt these problems. Maybe next year!
For problem 6, note that a common misconception of the proof is that P and Q are the two points of intersection. They are not! They are only points on the radical axis.
Having tried and solved Problem 6, it is really a nice problem. My method was very similar, except for the finish which I used phantom. Dealing with the (many) configuration issues were very frustrating though.
I fully agree that Problem 5 is a beautiful problem! Interesting to see a solution that goes from bottom row to top row while keeping track of a sum of powers of 2, where my solution goes from the top row to the bottom row while keeping track of a simple sum of the max number of red circles on a path. Circle i in row k has a path to the root with a(k,i) red circles. Let S(k) = sum_i a(k,i) and M(k) = max_i a(k,i). Set L(k) = [log_2(k)] + 1, and T(k) = L(1) + L(2) + ...+ L(k-1) + k. Now prove by induction on k that: - S(k) >= T(k) - M(k) >= L(k) The first follows from the observation that S(k) >= S(k-1) + M(k-1) + 1 >= T(k-1) + L(k-1) + 1 = T(k) (when going one row down, extend the path from the max circle in both directions, and the path from each of the other nodes in the remaining direction). The second follows from the observation/calculation below that, if k = 2^n, then S(k) >= T(2^n) = 2^n * n + 1, which means by the pigeonhole principle that there is i with a(k,i) >= n+1 = L(k). Let's see. T(2^0) = T(1) = 1 = 2^0 * 1 + 1. If T(2^(n-1)) = 2^(n-1) * (n-1) + 1, then T(2^n) = T(2^(n-1)) + 2^(n-1) * n + 2^(n-1) = 2^(n-1) * (n-1 + n + 1) + 1 = 2^(n-1) * (2n) + 1 = 2^n * n + 1. So in row 2^n the sum S grows exactly high enough to force M(2^n) to be one higher compared to row 2^n - 1. My method works, but yours is probably a bit simpler.
It turns out that the power of 2 approach works equally well with my trickle-down method. Instead of S(k) and M(k) as above, let S(k) = sum_i 2^(-a(k,i)) and M(k) = 2^(-max_i a(k,i)). With induction we show that S(k) + M(k) =1. Indeed S(1) = 1/2 and M(1) = 1/2, so S(1) + M(1)
For Problem 5, You can try to prove by induction that for each 2^k, the maximum number of circles a path passes through is exactly k + 1, then all such k+1-path needs to cover all possible starting points in the 2^k + 1 row meaning the max. number would increase by 1 in each 2^k + 1
But you cannot assume that the top part of the triangle is "sharp". Maybe there is a path with "too many" circles in the top part, but then all red circles in the bottom part of the triangle might be unreachable. So to get to the bottom you have to skip the circles in the top part and dive into the cluster of circles in the bottom part. My solution to problem 5 is to assign to each circle the max number of red circles on the path from the root to that circle. The sum of these numbers in a row grows by at least 1 plus the max number of red circles in a path to the previous row. sum max row 1 1 1 row 2 3=1+1+1 2 row 3 6=3+1+2 2 row 4 9=6+1+2 Since the sum is 9 for row 4 (having 4 circles), there must be at least one circle in row 4 which has a path to the root with 3 red circles (because 4x2 = 8 < 9). So we continue: row 4 9 3 row 5 13=9+1+3 3 row 6 17=13+1+3 3 row 7 21=17+1+3 3 row 8 25=21+1+3 Since the sum is 25 for row 8 (having 8 circles), there must be at least one circle in row 8 which has a path to the root with 4 red circles (because 8x3 = 24 < 25). And so on.
@@ronald3836 If it is not sharp in that situation the claim is still true because just take that path with > f(n-1) would have the num of red circle only in that path to be ≥ f(n)
@@skylerplayz3713 but now your induction hypothesis seems to change (or maybe not because I don't know what you have in mind exactly). Did you write it out in full? I have seen 3 proofs now: 1. The proof of this video, sum of powers of two and letting it bubble up. 2. Keep track of a simple sum and max for a row and letting that trickle down. (The one I came up with, and it is probably the same as the row-by-row construction of a binary tree mentioned in another comment.) 3. Using a theorem on the length of the longest chain in a poset in terms of partitionings into antichains.
I gambled on day two and completely failed, found a wrong bound for P5 and tried to prove that (didn't work, wonder why). Pretty frustrating considering how I solved P4 in 15 minutes after the contest (got scared of inequality + prefered combi anyway)
That’s the same that happened to me but my problem was that I started with p6 then last 40 min I tried p5 (overconfidence in combi failure) I got it but didn’t write up a complete proof and didn’t get on it a 7 sadly
I didn't mention it explicitly, it's to prevent AA_1A_2 from being collinear, similarly for BB_1B_2, CC_1C_2. To get some intuition, if we have isoceles, then there is symmetry about the corresponding perp bisector and the _1 and _2 will lie on the perp bisector.
There's nothing wrong with that result, but the situations where a_2023 attains a value in [2023, 3034) do not meet the other restrictions of the problem, e.g. x_1, x_2 ,..., are pairwise distinct. For example, in order for the equality case (a_2023 = 2023) to hold, you need x_1 = x_2 = ... = x_2023.
So i feel like i kind of solved question 4 can anybody tell me what they think? So i guessed (lol) that the minimum value one could make under the square roots would be if x1 x2 x3... Would simply be the integers 1,2,3 ...2023. And so the value under the square root sign would be (2023*2024/2 )*(1 + 1/2 + 1/3 ... +1/2023)). The sum of the reciprocals there can be approximated using integrals i.e ln(2023) which is less than the actual sum. Using more approximations in order to make the arithmetics easier i find that the sum for a_2023 >4700 .... Does this constitute a proof ??
You can read books on your own. There are some free ones, such as an olympiad combinatorics book, by Pranav A. Sriram. Of course, working on problems is a great way to learn. AOPS has lots of problems
Lets make a petition to remove geometry from IMO. It the lest least useful, ugliest, and most confusing subject in IMO. No mater how talented you are, to solve p3 or p6 problems you have to train at least 1000 hours a year for several years with olympiad geometry problems just to get a skill that you are not going to use again when you enter college. That must cause some kind of brain damage.
If you're struggling with them, consider looking for olympiad prep books made in China or Russia. Cause they have like 1000s types of geometry problems, theorems and solutions.
@@michaelblankenau6598 It is unfair for some country like China and USA which probably at least in total 5000 people can get gold; and let some african country come without even solve partial P1. This is unfair to be fair we should give everyone an opportunity!
@@ybc8495 I'm African and I don't understand what do you mean. 6 people in each team is a lot already and this is IMO, the hardest math Olympiad competition in the history, it's better to have the best in the world competing then just letting everyone participate and not do anything, it takes the prestige of the competition, the national math Olympiads, team selection tests wouldn't make sense if we allowed that.
What do you think of the IMO 2023 problems? Were the statistics at the end of the video surprising? 😃
i think its not too hard
@@fakecreeper9645 Good that you found it manageable!
Realmente el problema 5 es muy disfrutable.
I think the problems were fair. Perhaps easier than usual but still plenty of traps for those who aren't paying attention 😄
In my prime I think I would have score well in this IMO. Unfortunately I didn't find the time to seriously attempt these problems. Maybe next year!
😮
Problem 5 is nice, there is a short solution where you go from row to row and build a binary tree
Not really.
For problem 6, note that a common misconception of the proof is that P and Q are the two points of intersection. They are not! They are only points on the radical axis.
What a nice set of problems, I like them very much.
Nice solutions 👌
Fantastic problems and solutions!
Problem 5 reads like a monster editorial proof for an otherwise simple intuition for a codeforces competitive programming problem
That often happens in math when you have to write something down!
Nice job! Thank you :)
Late comment but after trying all problems myself I must admit P6 is one of the most interesting geometry problems I saw in few years.
I'm glad you enjoyed P6!
Great video!
Having tried and solved Problem 6, it is really a nice problem. My method was very similar, except for the finish which I used phantom. Dealing with the (many) configuration issues were very frustrating though.
I fully agree that Problem 5 is a beautiful problem!
Interesting to see a solution that goes from bottom row to top row while keeping track of a sum of powers of 2, where my solution goes from the top row to the bottom row while keeping track of a simple sum of the max number of red circles on a path.
Circle i in row k has a path to the root with a(k,i) red circles.
Let S(k) = sum_i a(k,i) and M(k) = max_i a(k,i).
Set L(k) = [log_2(k)] + 1, and T(k) = L(1) + L(2) + ...+ L(k-1) + k.
Now prove by induction on k that:
- S(k) >= T(k)
- M(k) >= L(k)
The first follows from the observation that S(k) >= S(k-1) + M(k-1) + 1 >= T(k-1) + L(k-1) + 1 = T(k) (when going one row down, extend the path from the max circle in both directions, and the path from each of the other nodes in the remaining direction).
The second follows from the observation/calculation below that, if k = 2^n, then S(k) >= T(2^n) = 2^n * n + 1, which means by the pigeonhole principle that there is i with a(k,i) >= n+1 = L(k).
Let's see.
T(2^0) = T(1) = 1 = 2^0 * 1 + 1.
If T(2^(n-1)) = 2^(n-1) * (n-1) + 1, then
T(2^n) = T(2^(n-1)) + 2^(n-1) * n + 2^(n-1) = 2^(n-1) * (n-1 + n + 1) + 1 = 2^(n-1) * (2n) + 1 = 2^n * n + 1.
So in row 2^n the sum S grows exactly high enough to force M(2^n) to be one higher compared to row 2^n - 1.
My method works, but yours is probably a bit simpler.
It turns out that the power of 2 approach works equally well with my trickle-down method.
Instead of S(k) and M(k) as above, let S(k) = sum_i 2^(-a(k,i)) and M(k) = 2^(-max_i a(k,i)).
With induction we show that S(k) + M(k) =1.
Indeed S(1) = 1/2 and M(1) = 1/2, so S(1) + M(1)
I really appreciate problem 4 it's quite good
Amazing P6.
For Problem 5, You can try to prove by induction that for each 2^k, the maximum number of circles a path passes through is exactly k + 1, then all such k+1-path needs to cover all possible starting points in the 2^k + 1 row meaning the max. number would increase by 1 in each 2^k + 1
But you cannot assume that the top part of the triangle is "sharp". Maybe there is a path with "too many" circles in the top part, but then all red circles in the bottom part of the triangle might be unreachable. So to get to the bottom you have to skip the circles in the top part and dive into the cluster of circles in the bottom part.
My solution to problem 5 is to assign to each circle the max number of red circles on the path from the root to that circle. The sum of these numbers in a row grows by at least 1 plus the max number of red circles in a path to the previous row.
sum max
row 1 1 1
row 2 3=1+1+1 2
row 3 6=3+1+2 2
row 4 9=6+1+2
Since the sum is 9 for row 4 (having 4 circles), there must be at least one circle in row 4 which has a path to the root with 3 red circles (because 4x2 = 8 < 9).
So we continue:
row 4 9 3
row 5 13=9+1+3 3
row 6 17=13+1+3 3
row 7 21=17+1+3 3
row 8 25=21+1+3
Since the sum is 25 for row 8 (having 8 circles), there must be at least one circle in row 8 which has a path to the root with 4 red circles (because 8x3 = 24 < 25).
And so on.
@@ronald3836 If it is not sharp in that situation the claim is still true because just take that path with > f(n-1) would have the num of red circle only in that path to be ≥ f(n)
@@skylerplayz3713 but now your induction hypothesis seems to change (or maybe not because I don't know what you have in mind exactly). Did you write it out in full?
I have seen 3 proofs now:
1. The proof of this video, sum of powers of two and letting it bubble up.
2. Keep track of a simple sum and max for a row and letting that trickle down. (The one I came up with, and it is probably the same as the row-by-row construction of a binary tree mentioned in another comment.)
3. Using a theorem on the length of the longest chain in a poset in terms of partitionings into antichains.
I gambled on day two and completely failed, found a wrong bound for P5 and tried to prove that (didn't work, wonder why).
Pretty frustrating considering how I solved P4 in 15 minutes after the contest (got scared of inequality + prefered combi anyway)
Oh no, that is sad to hear :(
you participated in IMO?? what country were you
That’s the same that happened to me but my problem was that I started with p6 then last 40 min I tried p5 (overconfidence in combi failure) I got it but didn’t write up a complete proof and didn’t get on it a 7 sadly
Explained with perfection. Even people like myself (little olympiad knowledge) can follow through the solutions easily.(P6 was still a monster though)
Thank you! The idea is to make the solution understandable and motivable :)
This is like playing a puzzle game on the hardest difficulty ever
By the way, interesting as a hardcore academic/video game otaku
Need more help on p5
I've lost the track a bit, where in #6 you have used that A1B1C1 is scalene?
I didn't mention it explicitly, it's to prevent AA_1A_2 from being collinear, similarly for BB_1B_2, CC_1C_2. To get some intuition, if we have isoceles, then there is symmetry about the corresponding perp bisector and the _1 and _2 will lie on the perp bisector.
I competed irl, got 777-770 lmao, P6 is absolutely brutal no one on my team scored anything on it
Congrats! That is darn impressive!!
@@dedekindcuts3589 i went ultra dirty on p5 lmao i got the funny mirsky kill
@@whitemountain4851 OH LOL! That's a good one!
@@whitemountain4851Hi , Can i have a conversation with you?
44:06 the problem was so long, it sounded like it was the next day... now that's dedication
i tried using the cauchy-shwarz inequality for problem 4 , and i got that a_2023 >= 2023 . why ?
There's nothing wrong with that result, but the situations where a_2023 attains a value in [2023, 3034) do not meet the other restrictions of the problem, e.g. x_1, x_2 ,..., are pairwise distinct. For example, in order for the equality case (a_2023 = 2023) to hold, you need x_1 = x_2 = ... = x_2023.
So i feel like i kind of solved question 4 can anybody tell me what they think? So i guessed (lol) that the minimum value one could make under the square roots would be if x1 x2 x3... Would simply be the integers 1,2,3 ...2023. And so the value under the square root sign would be (2023*2024/2 )*(1 + 1/2 + 1/3 ... +1/2023)). The sum of the reciprocals there can be approximated using integrals i.e ln(2023) which is less than the actual sum. Using more approximations in order to make the arithmetics easier i find that the sum for a_2023 >4700 .... Does this constitute a proof ??
my brain has become a mush, yet it is bigger (im still in middle school)
Hi bro, you participated in this year's Alibaba Math Competition?
Yup!
Haha ive wasted time trying pascal on p5, (i didnt watch the video) i think pascal is getting nowhere though, i'll give up in a day
Good first idea to try!
Bro can you please tell me how do i study for imo i dont have money to join a coaching etc.😢😢
You can read books on your own. There are some free ones, such as an olympiad combinatorics book, by Pranav A. Sriram. Of course, working on problems is a great way to learn. AOPS has lots of problems
@@dedekindcuts3589 thank you
@@dedekindcuts3589 thanks so much for free resources :3
what in the genuine shit is this LMAO
😂😂
Kindergarten IMO: In a cage there are goats and chickens. We could see 88 animal heads and 246 animal feet. How many chickens and goats??? LOL
second!!!!
Lets make a petition to remove geometry from IMO. It the lest least useful, ugliest, and most confusing subject in IMO. No mater how talented you are, to solve p3 or p6 problems you have to train at least 1000 hours a year for several years with olympiad geometry problems just to get a skill that you are not going to use again when you enter college. That must cause some kind of brain damage.
If you're struggling with them, consider looking for olympiad prep books made in China or Russia. Cause they have like 1000s types of geometry problems, theorems and solutions.
You're being too dramatic. Geometry is beautiful and it doesn't really matter if we're not going to use it at college.
geometry is sometimes fun when you understand it
why they only allow 6 per country to attend? they should invite everyone, equal chance! it is not fair!
And who is supposed to grade the tens of thousands of papers ?
@@michaelblankenau6598 99.9% will be blank, the rest will be graded by the high school Math chairs.
@@ybc8495If you think High school math chairs have the skills to grade IMO problems you are delusional .
@@michaelblankenau6598 It is unfair for some country like China and USA which probably at least in total 5000 people can get gold; and let some african country come without even solve partial P1. This is unfair to be fair we should give everyone an opportunity!
@@ybc8495 I'm African and I don't understand what do you mean. 6 people in each team is a lot already and this is IMO, the hardest math Olympiad competition in the history, it's better to have the best in the world competing then just letting everyone participate and not do anything, it takes the prestige of the competition, the national math Olympiads, team selection tests wouldn't make sense if we allowed that.