Plan ahead for the next school year. Schedule your class today!

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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.

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0 replies
jwelsh
Jul 1, 2025
0 replies
Integer sequences with a two-step recurrence
Assassino9931   1
N a minute ago by Assassino9931
Source: Bulgaria Autumn Tournament 2009 Grade 12
Fix an integer $m$. Determine the number of sequences $(a_n)_{n\geq 1}$ of integers such that $a_na_{n+2} = n^2 + m$ for all $n\geq 1$.
1 reply
Assassino9931
Today at 1:02 AM
Assassino9931
a minute ago
old product!
teomihai   2
N 3 minutes ago by teomihai
It is posible to prove ,without induction :
$(\frac{1}{2}\frac{3}{4}...\frac{2n-1}{2n})^2\leq{\frac{1}{3n+1}} $ for any positiv integer number $n$.?
2 replies
1 viewing
teomihai
Yesterday at 3:51 PM
teomihai
3 minutes ago
2025 IMO Results
ilikemath247365   5
N 4 minutes ago by cheltstudent
Source: https://www.imo-official.org/year_info.aspx?year=2025
Congrats to China for getting 1st place! Congrats to USA for getting 2nd and congrats to South Korea for getting 3rd!
5 replies
1 viewing
ilikemath247365
Yesterday at 4:48 PM
cheltstudent
4 minutes ago
A feasible refinement of GMA 567
Rhapsodies_pro   2
N 38 minutes ago by Rhapsodies_pro
Source: Own?
Let \(a_1,a_2,\dotsc,a_n\) (\(n>3\)) be non-negative real numbers fulfilling \[\sum_{k=1}^na_k^2+{\left(n^2-3n+1\right)}\prod_{k=1}^na_k\geqslant{\left(n-1\right)}^2\text.\]Prove or disprove: \[\frac1{n-1}\sum_{1\leqslant i<j\leqslant n}{\left(a_i-a_j\right)}^2\geqslant{\left({\left(n^2-2n-1\right)}\sum_{k=1}^na_k-n{\left(n-1\right)}{\left(n-3\right)}\right)}{\left(n-\sum_{k=1}^na_k\right)}\textnormal.\]
2 replies
2 viewing
Rhapsodies_pro
Jul 21, 2025
Rhapsodies_pro
38 minutes ago
No more topics!
Balanced grids
BR1F1SZ   2
N Jul 17, 2025 by Tamam
Source: 2025 Francophone MO Juniors/Seniors P2
Let $n \geqslant 2$ be an integer. We consider a square grid of size $2n \times 2n$ divided into $4n^2$ unit squares. The grid is called balanced if:
[list]
[*]Each cell contains a number equal to $-1$, $0$ or $1$.
[*]The absolute value of the sum of the numbers in the grid does not exceed $4n$.
[/list]
Determine, as a function of $n$, the smallest integer $k \geqslant 1$ such that any balanced grid always contains an $n \times n$ square whose absolute sum of the $n^2$ cells is less than or equal to $k$.
2 replies
BR1F1SZ
May 10, 2025
Tamam
Jul 17, 2025
Balanced grids
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G H BBookmark kLocked kLocked NReply
Source: 2025 Francophone MO Juniors/Seniors P2
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BR1F1SZ
593 posts
#1
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Let $n \geqslant 2$ be an integer. We consider a square grid of size $2n \times 2n$ divided into $4n^2$ unit squares. The grid is called balanced if:
  • Each cell contains a number equal to $-1$, $0$ or $1$.
  • The absolute value of the sum of the numbers in the grid does not exceed $4n$.
Determine, as a function of $n$, the smallest integer $k \geqslant 1$ such that any balanced grid always contains an $n \times n$ square whose absolute sum of the $n^2$ cells is less than or equal to $k$.
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alfonsoramires
9 posts
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bumppppp
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Tamam
29 posts
#3
Y by
Lemme solve :yoda:

The answer is $k=n$

$1$) $k\geq n$

Just the first and the $n+1$ th row be $1$'s and other squares $0$'s. Every $n*n$ square has the sum $n$.

$2$) $k=n$ actually works

Assume otherwise(We just assume every squares sums absolute value is greater than $n$.). Let $S_i$ be the $n$ times $n$ square that has its most left highest square as the $i$ th square in the first row. WLOG let $S_1$ have a positive sum. Let $S_i$ be positive. Let $T_i$ and $T_{i+1}$ be the sums of $S_i$ and $S_{i+1}$ respectively. We can see that $\lvert T_{i+1}-T_i \rvert \leq 2n$ because the two squares differ by only two columns($2n$ numbers.). If $T_{i+1}$ was negative it could be at most $-n-1$.And since $T_i \geq n+1$ the inequality $\lvert T_{i+1}-T_i \rvert \leq 2n$ wouldn't hold if $T_{i+1}$ was negative. So all $T_i$ are positive. So the left upmost $n*n$ and the right up most $n*n$ has a postive sum. We can use the same argument for shifting rows instead of columns and get that the left down most $n*n$ is has a positive. And we can use the same argument to get that the right downmost $nxn$ has a positive sum. If we let sum the sums of the left/right upmost $n*n$ s and left/right downmost $n*n$s be $X$ , $X\geq 4(n+1)$ but this contradicts the second condition of the grid.Hence we are done $\square$
This post has been edited 6 times. Last edited by Tamam, Jul 17, 2025, 11:25 PM
Reason: dilly dilly dillin
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