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
A diophantine equation of the form ax^4+by^2+c=dy^2
AtakanCICEK   0
a few seconds ago
Source: I think I have seen this question somewhere in this site before but couldnt locate (5-6 years ago)
Find all (x,y) pairs of integers such that $$(2x^2+5)^2=20y^2+29$$holds.
\[
\left\{\, (\pm1,\,\pm1),\; (\pm3,\,\pm5),\; (\pm7,\,\pm23) \,\right\}
\]Note: I have faced similar type of equation before $$2a^4-2a^2=b^2-1$$which leads to $$(a^2)^2+(a^2-1)^2=b^2$$and got a solution from pythagorian parametrization but I dont have a clue about this one.
0 replies
AtakanCICEK
a few seconds ago
0 replies
JBMO type Combinatorics
Sadigly   2
N 6 minutes ago by Just1
Source: Azerbaijan Junior MO 2025 P3
Alice and Bob take turns taking balloons from a box containing infinitely many balloons. In the first turn, Alice takes $k_1$ amount of balloons, where $\gcd(30;k_1)\neq1$. Then, on his first turn, Bob takes $k_2$ amount of ballons where $k_1<k_2<2k_1$. After first turn, Alice and Bob alternately takes as many balloons as his/her partner has. Is it possible for Bob to take $k_2$ amount of balloons at first, such that after a finite amount of turns, one of them have a number of balloons that is a multiple of $2025^{2025}$?
2 replies
Sadigly
May 9, 2025
Just1
6 minutes ago
AOPS MO Introduce
MathMaxGreat   90
N 6 minutes ago by CrazyInMath
$AOPS MO$

Problems: post it as a private message to me or @jerryZYang, please post it in $LATEX$ and have answers

6 Problems for two rounds, easier than $IMO$

If you want to do the problems or be interested, reply ’+1’
Want to post a problem reply’+2’ and message me
Want to be in the problem selection committee, reply’+3’
90 replies
MathMaxGreat
Jul 12, 2025
CrazyInMath
6 minutes ago
IMO 2025 P2
sarjinius   35
N 9 minutes ago by jrpartty
Source: 2025 IMO P2
Let $\Omega$ and $\Gamma$ be circles with centres $M$ and $N$, respectively, such that the radius of $\Omega$ is less than the radius of $\Gamma$. Suppose $\Omega$ and $\Gamma$ intersect at two distinct points $A$ and $B$. Line $MN$ intersects $\Omega$ at $C$ and $\Gamma$ at $D$, so that $C, M, N, D$ lie on $MN$ in that order. Let $P$ be the circumcentre of triangle $ACD$. Line $AP$ meets $\Omega$ again at $E\neq A$ and meets $\Gamma$ again at $F\neq A$. Let $H$ be the orthocentre of triangle $PMN$.

Prove that the line through $H$ parallel to $AP$ is tangent to the circumcircle of triangle $BEF$.
35 replies
+4 w
sarjinius
Today at 3:38 AM
jrpartty
9 minutes ago
Inequality
Martin.s   5
N Yesterday at 1:11 PM by solyaris


For \( n = 2, 3, \dots \), the following inequalities hold:

\[
-\frac{1}{3} \leq \frac{\sin(n\theta)}{n \sin \theta} \leq \frac{\sqrt{6}}{9}
\quad \text{for } \frac{\pi}{n} \leq \theta \leq \pi - \frac{\pi}{n},
\]
and

\[
-\frac{1}{3} \leq \frac{\sin(n\theta)}{n \sin \theta} \leq \frac{1}{5}
\quad \text{for } \frac{\pi}{n} \leq \theta \leq \frac{\pi}{2}.
\]
5 replies
Martin.s
Jun 23, 2025
solyaris
Yesterday at 1:11 PM
Quadratic surface and its tangent plane
RainbowNeos   0
Yesterday at 12:18 PM
Given a n*n symmetric real matrix $A$ with full rank. Suppose that $A$ has at least two positive eigenvalues and at least one negative eigenvalue. Show that for all $x\in\mathbb{R}^{n*1}$ such that $x^T A x=1$, there exists $y\neq x$ such that $y^T A x = y^T A y = 1$.
0 replies
RainbowNeos
Yesterday at 12:18 PM
0 replies
positive derivative at local max
tobiSALT   1
N Yesterday at 12:15 PM by Mathzeus1024
Source: CIMA Math Olympiad 2023 P3
Let $f : [0, 1] \to \mathbb{R}$ be a function with continuous derivative such that $f(0) = 0$ and $f(1) = 1$. Show that there exists a real number $t$ such that $f'(t) > 0$ and $f(t) > f(s)$ for all $s$ such that $0 \le s < t$.
1 reply
tobiSALT
Nov 18, 2024
Mathzeus1024
Yesterday at 12:15 PM
Putnam 2016 B1
Kent Merryfield   22
N Yesterday at 11:56 AM by SomeonecoolLovesMaths
Let $x_0,x_1,x_2,\dots$ be the sequence such that $x_0=1$ and for $n\ge 0,$
\[x_{n+1}=\ln(e^{x_n}-x_n)\](as usual, the function $\ln$ is the natural logarithm). Show that the infinite series
\[x_0+x_1+x_2+\cdots\]converges and find its sum.
22 replies
Kent Merryfield
Dec 4, 2016
SomeonecoolLovesMaths
Yesterday at 11:56 AM
Generalization of 2014 Brazilian Revenge Problem
zqy648   0
Yesterday at 9:30 AM
Source: 2023 June 谜之竞赛-7
Let \( k \geq 2 \) be an integer, and let \( f, g \) be non-constant integer-coefficient polynomials with positive leading coefficients. A positive integer \( n \) is called good if $f(n) \mid k^{g(n)} - 1.$

Prove that there exists a positive real constant \( c \) such that for any integer \( n \geq 3 \), the number of good integers among \( 1, 2, \cdots, n \) does not exceed \(\dfrac{cn \ln \ln n}{\ln n}.\)

Created by Mucong Sun, Tianjin Experimental Binhai School
0 replies
zqy648
Yesterday at 9:30 AM
0 replies
Function x_1!...x_n!=m!
zqy648   0
Yesterday at 9:26 AM
Source: 2023 May 谜之竞赛-6
Let \( m, n \) be positive integers, and \( x_1, x_2, \cdots, x_n \) be integers greater than 1 satisfying
\[x_1! \cdot x_2! \cdots x_n! = m!.\]Prove that: (1) \( n \leq 1 + \log_2 m \); (2) There exists a positive real constant \( c \) such that for sufficiently large \( m \),
\[\max\{x_1, x_2, \cdots, x_n\} > m - c \cdot \frac{\ln m}{\ln \ln m}.\]Created by Mucong Sun, Tianjin Experimental Binhai School
0 replies
zqy648
Yesterday at 9:26 AM
0 replies
a sequence of a polynomial
truongphatt2668   3
N Yesterday at 3:28 AM by truongphatt2668
Let a sequence of polynomial defined by: $P_0(x) = x$ and $P_{n+1}(x) = -2xP_n(x) + P'_n(x), \forall n \in \mathbb{N}$.
Find: $P_{2017}(0)$
3 replies
truongphatt2668
Sunday at 2:22 PM
truongphatt2668
Yesterday at 3:28 AM
Minimum value
Martin.s   5
N Yesterday at 2:52 AM by aaravdodhia
What is the minimum value of
$$
\frac{|a + b + c + d| \left( |a - b| |b - c| |c - d| + |b - a| |c - a| |d - a| \right)}{|a - b| |b - c| |c - d| |d - a|}
$$over all triples $a, b, c, d$ of distinct real numbers such that
$a^2 + b^2 + c^2 + d^2 = 3(ab + bc + cd + da).$

5 replies
Martin.s
Oct 17, 2024
aaravdodhia
Yesterday at 2:52 AM
Aproximate ln(2) using perfect numbers
YLG_123   7
N Yesterday at 12:04 AM by vincentwant
Source: Brazilian Mathematical Olympiad 2024, Level U, Problem 1
A positive integer \(n\) is called perfect if the sum of its positive divisors \(\sigma(n)\) is twice \(n\), that is, \(\sigma(n) = 2n\). For example, \(6\) is a perfect number since the sum of its positive divisors is \(1 + 2 + 3 + 6 = 12\), which is twice \(6\). Prove that if \(n\) is a positive perfect integer, then:
\[
\sum_{p|n} \frac{1}{p + 1} < \ln 2 < \sum_{p|n} \frac{1}{p - 1}
\]where the sums are taken over all prime divisors \(p\) of \(n\).
7 replies
YLG_123
Oct 12, 2024
vincentwant
Yesterday at 12:04 AM
Putnam 2003 B3
btilm305   35
N Sunday at 2:50 PM by SomeonecoolLovesMaths
Show that for each positive integer n, \[n!=\prod_{i=1}^n \; \text{lcm} \; \{1, 2, \ldots, \left\lfloor\frac{n}{i} \right\rfloor\}\](Here lcm denotes the least common multiple, and $\lfloor x\rfloor$ denotes the greatest integer $\le x$.)
35 replies
btilm305
Jun 23, 2011
SomeonecoolLovesMaths
Sunday at 2:50 PM
Finding a subsquare from the main square
goodar2006   2
N May 30, 2025 by quantam13
Source: Iran 3rd round 2012-Special Lesson exam-Part 2-P4
Prove that if $n$ is large enough, in every $n\times n$ square that a natural number is written on each one of its cells, one can find a subsquare from the main square such that the sum of the numbers is this subsquare is divisible by $1391$.
2 replies
goodar2006
Sep 15, 2012
quantam13
May 30, 2025
Finding a subsquare from the main square
G H J
Source: Iran 3rd round 2012-Special Lesson exam-Part 2-P4
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goodar2006
1347 posts
#1 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Prove that if $n$ is large enough, in every $n\times n$ square that a natural number is written on each one of its cells, one can find a subsquare from the main square such that the sum of the numbers is this subsquare is divisible by $1391$.
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ocha
955 posts
#2 • 6 Y
Y by goodar2006, Adventure10, Mango247, and 3 other users
This is also Gallai's theorem (see here, page 12).

Let $A = \{a_{i,j}\}_{1\le i,j \le n}$, be the matrix corresponding to entries in the square.
Define $B = \{b_{i,j}\}_{1\le i,j \le n}$, where $b_{i,j} = \sum_{k=1}^i\sum_{\ell=1}^j a_{k,\ell}$.

Then the sum of the values of $A$ over a sub-square, $(i+1,i+k) \times (j+1,j+k)$, equals the sum $b_{i+k,j+k} + b_{i,j} - b_{i,j+k} - b_{i+k,j}$.

Let the entries of $B \mod 1391$ be a colouring of the $n\times n$ lattice. If $n$ is sufficiently large, then by Gallai's theorem there exists $4$ monochromatic lattice points that are vertices of a square. These correspond to four entries of $B$ such that $b_{i+k,j+k} + b_{i,j} - b_{i,j+k} - b_{i+k,j} \equiv 0 \mod 1391$. This is exactly what we wanted to prove.
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quantam13
147 posts
#3
Y by
@above
Good solution
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