imo 2025 (1 Viewer)

tywebb

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for the first time since 1988 the imo will be held in australia, so it will only be the second time it is held in australia

website is at https://imo2025.au

contest days are July 15 and 16

2 other websites which are of interest are https://www.imo-official.org which will have results and https://artofproblemsolving.com/community/c6t1881404f6_imo_2025 which probably will have some discussions of unofficial solutions

shortly after the 2025 imo the 2024 sl will go up at https://www.imo-official.org/problems.aspx

but we'll have to wait a whole year for the 2025 sl

australian team is

• William Cheah, Scotch College VIC
• William Liu, The King’s School NSW
• Xiangyue Nan, Fintona Girls' School VIC
• Jayden Pan, Shore School NSW
• Justin Tran, Sydney Grammar School NSW
• Zihui Zhang, Clayfield College QLD

imo-2025-australia.jpg
 

tywebb

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it is interesting that the russian team is named as follows

1. Dmitry Grishko, Moscow
2. Ilya Zamotorin, Saint Petersburg
3. Ivan Chasovskikh, Moscow Region
4. Vasily Patrushev, Vladivostok
5. Artem Sadykov, Chelyabinsk
6. Ivan Kokarev, Chelyabinsk.

due to war in ukraine the last few years russians were allowed quietly to compete as individuals with unassigned country (rather than ban them) and it took a fair bit of sleuthing to ascertain russian contestants.

whether they will be recognised as "russian team" in official results remains to be seen, but up-front naming them from the outset is a change. this article says russia is still suspended: https://www.theguardian.com/world/2...s-compete-at-the-international-maths-olympiad

maybe they know by now the sleuthing would go on anyway so may as well name them now.
 
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Socialism

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I believe the sports and IMO type of competitions should be kept out of political debates.
i get that there's a lot of debate about israel vs palestine but like,,, who is saying russia should be allowed to invade ukraine 😭 there is no political debate just universally condemned violations of international law... like there is no debate. 😭 no-one is going out onto the street saying russia should be allowed to annex ukraine 😭 ⁉
 

coolcat6778

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kid on the left is the epitome of
for the first time since 1988 the imo will be held in australia, so it will only be the second time it is held in australia

website is at https://imo2025.au

contest days are July 15 and 16

2 other websites which are of interest are https://www.imo-official.org which will have results and https://artofproblemsolving.com/community/c6t1881404f6_imo_2025 which probably will have some discussions of unofficial solutions

shortly after the 2025 imo the 2024 sl will go up at https://www.imo-official.org/problems.aspx

but we'll have to wait a whole year for the 2025 sl

australian team is

• William Cheah, Scotch College VIC
• William Liu, The King’s School NSW
• Xiangyue Nan, Fintona Girls' School VIC
• Jayden Pan, Shore School NSW
• Justin Tran, Sydney Grammar School NSW
• Zihui Zhang, Clayfield College QLD

View attachment 48177
🤓... like are you a fucking 9 year old child? have some decency in your own hair style
 

coolcat6778

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the guardian article mentions a letter with 714 signatories to suspend israel from the 2025 imo, just as they had done for russia for the last few years

letter attached
i'm sorry, what year were you on the Australian International Math Olympiad team?
How the fuck are these people in high school doing the olympiad, like none of this is even remotely in the syllabus. These retards don't even get state ranks in math for the hsc
 

tywebb

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I believe the sports and IMO type of competitions should be kept out of political debates.
On the other hand , it is a clear double standard to ban Russia but not Israel.
yeah many in the maths community say maths must stay out of politics

however when they suspended russia, they already became political

so there’s no turning back.

on suspending israel, well iran behaved badly too so suspend them too?

how about the us? some legal commentators are saying the us bombing iran was a breach of international law, like this: https://www.smh.com.au/world/middle...s-may-be-his-worst-crime-20250622-p5m9ea.html

and so suspend the us too?

it’s a slippery slope. where does it stop? suspend anyone involved in any way in a war, you end up suspending everyone, including australia.

anyway the imo board have responded to the letter by saying they divert the decision to the imo jury which meets in july. so decision on israel hasn’t been made yet, but may come later, so it will be interesting to see what the decision is going to be.
 
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tywebb

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there is another way to deal with it too, maybe not popular, but still may get some support.

lift the suspension on russia and don’t do suspensions anymore, not on russia, not on israel or anyone

it’s not putin or netanyahu who are coming, they are just maths students
 

killer queen

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I keep seeing this thread and reading the title as "in my opinion 2025" 😭 😭 and think it's someone ranting about this year

man this sounds fun though I wish I was good enough for it / had the opportunity to try out for this kind of thing
 

Tony Stark

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there is another way to deal with it too, maybe not popular, but still may get some support.

lift the suspension on russia and don’t do suspensions anymore, not on russia, not on israel or anyone

it’s not putin or netanyahu who are coming, they are just maths students
Well this is pretty much the only way to keep math, probably the most universal language that we have, out of politics. Probably they don't want to get bad publicity or be crapped on by the media, or there are some people in their with really heavy beliefs about the Russia v. Ukraine war. I think with the middle-east, people experience 'compassion fatigue', where there has been so much conflict that it doesn't really impact people who only know about it through the media; it's less personal to them. But because of the global involvement in condemning Russia, people feel like it is the right thing to do.

Humans are, after all, only human.
 

Eagle Mum

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How the fuck are these people in high school doing the olympiad, like none of this is even remotely in the syllabus. These retards don't even get state ranks in math for the hsc
There are people who love discovering the wonder and elegance of maths. Maths underpins almost everything in the universe. I haven’t formally studied maths since 4U and AMO at high school, but I find that having a good maths foundation is useful for many aspects of my work.

Some of my contemporaries who scored very well in 4U maths struggle now with Yr 9 maths, so I suspect they did a lot of rote learning of curriculum content, whereas the problem solving exercises with Olympiad questions encourages you to look at problems from a different angle (excuse the pun) to appreciate the universality of maths.
 

killer queen

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q6 is giving pigeonhole principle, no idea how you're supposed to do it but I wanna take a stab anyways
my brain says 4046 but no way it's that many
 

coolcat6778

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q6 is giving pigeonhole principle, no idea how you're supposed to do it but I wanna take a stab anyways
my brain says 4046 but no way it's that many
We are given a 2025×20252025 \times 2025 grid of unit squares. Matilda must place rectangular tiles (aligned with the grid lines, and no overlapping) so that:


  • Each unit square is covered by at most one tile, and
  • Each row and each column has exactly one unit square that is not covered by any tile.

We are to find the minimum number of tiles needed to satisfy this condition.


Step 1: Understand the requirements

Let’s denote:


  • n=2025n = 2025
  • There are nn rows and nn columns.

The requirement is that:


  • Each row has exactly one uncovered square.
  • Each column has exactly one uncovered square.

So overall, the number of uncovered unit squares is:

n+n−xn + n - x
where xx is the number of overlaps between the uncovered squares in rows and columns (i.e., squares that are simultaneously the uncovered square in both a row and a column). The maximum possible overlaps is when the row and column uncovered squares align on the diagonal — so xx is at most nn.


But since each row and column has exactly one uncovered square, the total number of uncovered squares is:

n (from rows)=number of distinct uncovered squaresn\ (\text{from rows}) = \text{number of distinct uncovered squares}
So we must have:


  • Exactly one uncovered square per row → nn uncovered squares total.
  • Each of these uncovered squares must also be the only uncovered square in its column.

Thus, these uncovered squares must lie on a permutation of columns across the rows: i.e., in row ii, the uncovered square is in column π(i)\pi(i) for some permutation π\pi of {1,2,...,n}\{1, 2, ..., n\}.


So the uncovered squares are at positions:

(i,π(i)) for i=1,2,…,n(i, \pi(i)) \text{ for } i = 1, 2, \dots, n
Step 2: All other n2−nn^2 - n unit squares must be covered

There are n2=20252n^2 = 2025^2 total squares.


Since there are n=2025n = 2025 uncovered squares (one in each row and column), the number of covered squares is:

n2−n=2025⋅2024n^2 - n = 2025 \cdot 2024
We are to cover these with rectangular tiles, such that:


  • No tile overlaps with another.
  • Each tile covers at least one square.
  • The goal is to minimize the number of tiles.

Step 3: Strategy — cover as many squares per tile as possible

To minimize the number of tiles, each tile should cover as many unit squares as possible.


One way to do this is to use large rectangular tiles that avoid the nn uncovered squares.


Note:


  • Each row and column has exactly one square that must be excluded from tiling.
  • This means, in row ii, we can tile the remaining n−1n - 1 squares (i.e., all but π(i)\pi(i)).
  • Similarly, in each column, the uncovered square is unique.

So here's the key idea:

Step 4: Cover each row except the one missing square

For each row ii, the uncovered square is in column π(i)\pi(i). That means, we need to cover the other n−1n - 1 squares in row ii.


But the challenge is to avoid overlapping tiles across rows, so we must arrange the tiles cleverly to not cover the same square twice.


Suppose we define the tiles to be vertical, each covering a segment in a column (i.e., tiles of size k×1k \times 1), and design the tiling to ensure no overlapping and full coverage of all but the diagonal (i,π(i))(i, \pi(i)) squares.


This still might use many tiles.


Let’s now consider the optimal known construction:


Step 5: Use permutation rectangles – Optimal Tiling Construction

Let’s again note:


  • The uncovered squares are at (i,π(i))(i, \pi(i)), for some permutation π\pi of {1,2,...,n}\{1, 2, ..., n\}.
  • So each row has all squares covered except column π(i)\pi(i).
  • Each column has all squares covered except row π−1(j)\pi^{-1}(j), where j=π(i)j = \pi(i).

That is, the grid with the uncovered squares removed corresponds to the grid with one square missing from each row and column — a Latin square with a hole per row and column.


So the remaining squares form an n×nn \times n grid with one missing square per row and column — that is, a complement of a permutation matrix.


This structure is known in combinatorics, and there's a famous result:

The minimal number of rectangles to partition an n×nn \times n grid with a missing permutation matrix (1 square missing per row and column) is exactly nn.

This is from the combinatorial result that:


An n×nn \times n grid with a single square removed in each row and column (i.e., the complement of a permutation matrix) can be partitioned into exactly nn non-overlapping rectangles, such that each rectangle covers only covered squares.
Thus, only n=2025n = 2025 rectangular tiles are needed!


✅ Final Answer:
2025\boxed{2025}
This is the minimum number of tiles Matilda needs.
 

Tony Stark

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We are given a 2025×20252025 \times 2025 grid of unit squares. Matilda must place rectangular tiles (aligned with the grid lines, and no overlapping) so that:


  • Each unit square is covered by at most one tile, and
  • Each row and each column has exactly one unit square that is not covered by any tile.

We are to find the minimum number of tiles needed to satisfy this condition.


Step 1: Understand the requirements

Let’s denote:


  • n=2025n = 2025
  • There are nn rows and nn columns.

The requirement is that:


  • Each row has exactly one uncovered square.
  • Each column has exactly one uncovered square.

So overall, the number of uncovered unit squares is:

n+n−xn + n - x
where xx is the number of overlaps between the uncovered squares in rows and columns (i.e., squares that are simultaneously the uncovered square in both a row and a column). The maximum possible overlaps is when the row and column uncovered squares align on the diagonal — so xx is at most nn.


But since each row and column has exactly one uncovered square, the total number of uncovered squares is:

n (from rows)=number of distinct uncovered squaresn\ (\text{from rows}) = \text{number of distinct uncovered squares}
So we must have:


  • Exactly one uncovered square per row → nn uncovered squares total.
  • Each of these uncovered squares must also be the only uncovered square in its column.

Thus, these uncovered squares must lie on a permutation of columns across the rows: i.e., in row ii, the uncovered square is in column π(i)\pi(i) for some permutation π\pi of {1,2,...,n}\{1, 2, ..., n\}.


So the uncovered squares are at positions:

(i,π(i)) for i=1,2,…,n(i, \pi(i)) \text{ for } i = 1, 2, \dots, n
Step 2: All other n2−nn^2 - n unit squares must be covered

There are n2=20252n^2 = 2025^2 total squares.


Since there are n=2025n = 2025 uncovered squares (one in each row and column), the number of covered squares is:

n2−n=2025⋅2024n^2 - n = 2025 \cdot 2024
We are to cover these with rectangular tiles, such that:


  • No tile overlaps with another.
  • Each tile covers at least one square.
  • The goal is to minimize the number of tiles.

Step 3: Strategy — cover as many squares per tile as possible

To minimize the number of tiles, each tile should cover as many unit squares as possible.


One way to do this is to use large rectangular tiles that avoid the nn uncovered squares.


Note:


  • Each row and column has exactly one square that must be excluded from tiling.
  • This means, in row ii, we can tile the remaining n−1n - 1 squares (i.e., all but π(i)\pi(i)).
  • Similarly, in each column, the uncovered square is unique.

So here's the key idea:

Step 4: Cover each row except the one missing square

For each row ii, the uncovered square is in column π(i)\pi(i). That means, we need to cover the other n−1n - 1 squares in row ii.


But the challenge is to avoid overlapping tiles across rows, so we must arrange the tiles cleverly to not cover the same square twice.


Suppose we define the tiles to be vertical, each covering a segment in a column (i.e., tiles of size k×1k \times 1), and design the tiling to ensure no overlapping and full coverage of all but the diagonal (i,π(i))(i, \pi(i)) squares.


This still might use many tiles.


Let’s now consider the optimal known construction:


Step 5: Use permutation rectangles – Optimal Tiling Construction

Let’s again note:


  • The uncovered squares are at (i,π(i))(i, \pi(i)), for some permutation π\pi of {1,2,...,n}\{1, 2, ..., n\}.
  • So each row has all squares covered except column π(i)\pi(i).
  • Each column has all squares covered except row π−1(j)\pi^{-1}(j), where j=π(i)j = \pi(i).

That is, the grid with the uncovered squares removed corresponds to the grid with one square missing from each row and column — a Latin square with a hole per row and column.


So the remaining squares form an n×nn \times n grid with one missing square per row and column — that is, a complement of a permutation matrix.


This structure is known in combinatorics, and there's a famous result:

The minimal number of rectangles to partition an n×nn \times n grid with a missing permutation matrix (1 square missing per row and column) is exactly nn.

This is from the combinatorial result that:




Thus, only n=2025n = 2025 rectangular tiles are needed!


✅ Final Answer:
2025\boxed{2025}
This is the minimum number of tiles Matilda needs.
I love A.I
 

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