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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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0 replies
jwelsh
Jul 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Number theory
MathsII-enjoy   2
N 2 minutes ago by MathsII-enjoy
Find all positive intergers $m$ that satisfies: $$\sigma(m^2+8)+\phi(m^2+8)=2m^2+24$$
2 replies
MathsII-enjoy
Jul 19, 2025
MathsII-enjoy
2 minutes ago
Inspired by old results
sqing   0
29 minutes ago
Source: Own
Let $ a,b> 0, a^2+b^2+ab=3 .$ Prove that
$$(a+b)^2(\frac {a} {b^2+1}+\frac {b} {a^2+1})\geq 4$$Let $ a,b> 0, a+b+2ab=4.$ Prove that
$$(a+b)^2(\frac {a} {b^2+1}+\frac {b} {a^2+1})\geq 4$$$$(a+b)^2(\frac {a+1} {b^2+1}+\frac {b+1} {a^2+1})\geq 8$$Let $ a,b> 0, a+b+ab=3 .$ Prove that
$$(a+b)^2(\frac {a+1} {b^2+1}+\frac {b+1} {a^2+1})\geq 8$$
0 replies
1 viewing
sqing
29 minutes ago
0 replies
Numbers on cards (again!)
popcorn1   84
N an hour ago by numbertheory97
Source: IMO 2021 P1
Let $n \geqslant 100$ be an integer. Ivan writes the numbers $n, n+1, \ldots, 2 n$ each on different cards. He then shuffles these $n+1$ cards, and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square.
84 replies
popcorn1
Jul 20, 2021
numbertheory97
an hour ago
Find the minimum value
sqing   1
N an hour ago by sqing
Source: 2025 China Mathematical Olympiad Hope Alliance Summer Camp
Let $ a,b,c> 0, abc=\frac{1}{1024} .$ Find the minimum value of $ a^2+2b^2 +4c^2+\frac{2ac}{a+2c}.$
1 reply
sqing
an hour ago
sqing
an hour ago
The refinement of GMA 567
mihaig   4
N an hour ago by mihaig
Source: Own
Let $a_1,\ldots, a_{n}\geq0~~(n\geq4)$ be real numbers such that
$$\sum_{i=1}^{n}{a_i^2}+(n^2-3n+1)\prod_{i=1}^{n}{a_i}\geq(n-1)^2.$$Prove
$$\left(\sum_{i=1}^{n}{a_i}\right)^2+\frac{2n-1}{(n-1)^3}\cdot\sum_{1\leq i<j\leq n}{\left(a_i-a_j\right)^2}\geq n^2.$$
4 replies
mihaig
Yesterday at 11:22 AM
mihaig
an hour ago
Integer-Valued FE comes again
lminsl   216
N 2 hours ago by TwentyIQ
Source: IMO 2019 Problem 1
Let $\mathbb{Z}$ be the set of integers. Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that, for all integers $a$ and $b$, $$f(2a)+2f(b)=f(f(a+b)).$$Proposed by Liam Baker, South Africa
216 replies
lminsl
Jul 16, 2019
TwentyIQ
2 hours ago
Peru IMO TST 2022
diegoca1   0
2 hours ago
Source: Peru IMO TST 2022 D2 P1
Let $N$ be a positive integer. Determine all positive integers $n$ that satisfy the following condition:

For any list $d_1, d_2, \ldots, d_k$ of divisors of $n$ (not necessarily distinct) such that
\[
\frac{1}{d_1} + \frac{1}{d_2} + \cdots + \frac{1}{d_k} > N,
\]there exists a subset of the fractions $\frac{1}{d_1}, \frac{1}{d_2}, \ldots, \frac{1}{d_k}$ whose sum is exactly $N$.
0 replies
diegoca1
2 hours ago
0 replies
A nice property of triangle with incircle (I)
TUAN2k8   0
2 hours ago
Source: own
Let \( ABC \) be a non-isosceles triangle with incircle (\( I \)). Denote by \( D, E, F \) the points where (\( I \)) touches \( BC, CA, AB \), respectively. The A-excircle of \( ABC \) is tangent to \( BC \) at \( G \). The lines \( IB \) and \( IC \) meet \( AG \) at \( M \) and \( N \), respectively.

a) Prove that the circumcircles of triangles \( MBF \), \( NCE \), and \( BIC \) are concurrent at a point.

b) Let \( L \) and \( K \) be the midpoints of \( AG \) and \( BC \), respectively, and let \( J \) be the orthocenter of triangle \( IMN \). Show that the points \( L, K, J \) are collinear.
0 replies
TUAN2k8
2 hours ago
0 replies
Inequality
SunnyEvan   3
N 2 hours ago by SunnyEvan
Find the smallest positive real number \( k \) such that the following inequality holds:
\[
x^k y^k z^k (x^2 + y^2 + z^2) \leq 3
\]for all positive real numbers \( x, y, z \) satisfying the condition \( x + y + z = 3 \)
Click to reveal hidden text

Find the smallest positive real number \( k \) such that the following inequality holds:
\[
x^k y^k z^k (x^2 + y^2 + z^2) \leq xy+yz+zx
\]for all positive real numbers \( x, y, z \) satisfying the condition \( x + y + z = 3 \)
Click to reveal hidden text
3 replies
SunnyEvan
Yesterday at 11:19 AM
SunnyEvan
2 hours ago
Perpendicularity in Two Tangent Circles
steven_zhang123   1
N 3 hours ago by aaravdodhia
Source: 2025 Hope League Test 2 P3
Circle \(O_1\) and circle \(O_2\) are externally tangent at point \(T\). From a point \(X\) on circle \(O_2\), a tangent is drawn intersecting circle \(O_1\) at points \(A\) and \(B\). The line \(XT\) is extended to intersect circle \(O_1\) at point \(S\). A point \(C\) is taken on the arc \(TS\) of circle \(O_1\). The line \(SC\) is extended to intersect the angle bisector of \(\angle BAC\) at point \(I\). The circle passing through points \(A, T, X\) and the circle passing through points \(C, T, I\) intersect at another point \(E\). Prove that \(EO_2 \perp XI\).
Proposed by Luo Haoyu
1 reply
steven_zhang123
Yesterday at 8:24 AM
aaravdodhia
3 hours ago
number theory
Hoapham235   5
N 3 hours ago by Jjesus
Let $x >  y$ be positive integer such that \[ \text{LCM}(x+2, y+2)+\text{LCM}(x, y)=2\text{LCM}(x+1, y+1).\]Prove that $x$ is divisible by $y$.
5 replies
Hoapham235
Wednesday at 4:51 AM
Jjesus
3 hours ago
Peru IMO TST 2022
diegoca1   0
3 hours ago
Source: Peru IMO TST 2022 D1 P4
Let $\Omega$ be the circumcircle of triangle $ABC$, with $\angle BAC > 90^\circ $ and $ AB > AC $. The tangents to $\Omega$ at points $B$ and $C$ intersect at $D$. The tangent to $\Omega$ at point $A$ intersects line $BC$ at $E$. The line through $D$ parallel to $AE$ intersects line $BC$ at $F$. The circumference $\Gamma$ with diameter $EF$ intersects line $AB$ at points $P$ and $Q$, and line $AC$ at points $X$ and $Y$.
Prove that one of the angles $\angle AEB$, $\angle PEQ$, $\angle XEY$ is equal to the sum of the other two.
0 replies
diegoca1
3 hours ago
0 replies
centroid lies outside of triangle (not clickbait)
Scilyse   2
N 3 hours ago by EthanWYX2009
Source: 数之谜 January (CHN TST Mock) Problem 5
Let $P$ be a convex polygon with centroid $G$, and let $\mathcal P$ be the set of vertices of $P$. Let $\mathcal X$ be the set of triangles with vertices all in $\mathcal P$. We sort the elements $\triangle ABC$ of $\mathcal X$ into the following three types:
[list]
[*] (Type 1) $G$ lies in the strict interior of $\triangle ABC$; let $\mathcal A$ be the set of triangles of this type.
[*] (Type 2) $G$ lies in the strict exterior of $\triangle ABC$; let $\mathcal B$ be the set of triangles of this type.
[*] (Type 3) $G$ lies on the boundary of $\triangle ABC$.
[/list]
For any triangle $T$, denote by $S_T$ the area of $T$. Prove that \[\sum_{T \in \mathcal A} S_T \geq \sum_{T \in \mathcal B} S_T.\]
2 replies
Scilyse
Jan 26, 2025
EthanWYX2009
3 hours ago
Peru IMO TST 2022
diegoca1   0
3 hours ago
Source: Peru IMO TST 2022 D1 P3
Consider an $n$-gon (a polygon with $n$ sides) with $n \geq 3$. Distinct non-negative integers are assigned to each side and diagonal of the n-gon in such a way that, for any triangle whose vertices are vertices of the n-gon, the numbers assigned to its three sides form an arithmetic progression. Determine the maximum value of $n$ for which this is possible
0 replies
diegoca1
3 hours ago
0 replies
f(x + 13/42) + f(x) = f(x + 1/6) + f(x + 1/7)
orl   3
N Jul 18, 2010 by BaronShadeNight
Source: IMO Shortlist 1996, A7
Let $ f$ be a function from the set of real numbers $ \mathbb{R}$ into itself such for all $ x \in \mathbb{R},$ we have $ |f(x)| \leq 1$ and

\[ f \left( x + \frac{13}{42} \right) + f(x) = f \left( x + \frac{1}{6} \right) + f \left( x + \frac{1}{7} \right).\]

Prove that $ f$ is a periodic function (that is, there exists a non-zero real number $ c$ such $ f(x+c) = f(x)$ for all $ x \in \mathbb{R}$).
3 replies
orl
Aug 9, 2008
BaronShadeNight
Jul 18, 2010
f(x + 13/42) + f(x) = f(x + 1/6) + f(x + 1/7)
G H J
Source: IMO Shortlist 1996, A7
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orl
3647 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ f$ be a function from the set of real numbers $ \mathbb{R}$ into itself such for all $ x \in \mathbb{R},$ we have $ |f(x)| \leq 1$ and

\[ f \left( x + \frac{13}{42} \right) + f(x) = f \left( x + \frac{1}{6} \right) + f \left( x + \frac{1}{7} \right).\]

Prove that $ f$ is a periodic function (that is, there exists a non-zero real number $ c$ such $ f(x+c) = f(x)$ for all $ x \in \mathbb{R}$).
Z K Y
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BaBaK Ghalebi
1182 posts
#2 • 5 Y
Y by noway, Adventure10, Mango247, Jakjjdm, and 1 other user
" wrote:
let $ a=\frac 16,b=\frac 17$ so $ a+b=\frac{13}{42}$ and:

$ f(x+a+b)+f(x)=f(x+a)+f(x+b)$

now in this equation,if we plug in $ x+a,x+2a,\ldots ,x+5a$ instead of $ x$,and then sum the equations up,we get that:

$ f(x+1+b)+f(x)=f(x+1)+f(x+b)$

now in this equation,plug in $ x+b,x+2b,\ldots ,x+6b$ instead of $ x$,and then sum them up to get that:

$ f(x+2)-f(x+1)=f(x+1)-f(x)=c$

hence for every natural number $ n$,we have:

$ f(x+n)-f(x)=nc$

now note that if $ c\neq 0$,then $ f(x+n)$ would tend to $ \infty$ as $ n$ gets bigger,so according to $ |f(x)|\leq 1$ we must have $ c=0$,so $ f(x+1)-f(x)=0$.i.e. $ f(x+1)=f(x)$ for all $ x\in\mathbb{R}$.

QED
Z K Y
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Zhero
2043 posts
#3 • 4 Y
Y by huricane, Adventure10, Mango247, Jakjjdm
I claim that for all real $a$, $f(a+1) = f(a)$. If we define $f_a(x) = f(x-a)$, we see that $f_a$ trivially satisfies the functional equation of $f$, and that $f_a(0) = f_a(1)$ would imply that $f(a) = f(a+1)$.

Let $c_i = f_a\left(\frac{i}{42}\right)$ for all $i \geq 0$. We will show that $c_{n+42} = c_n$ for all integers $n$. $\{c_i\}$ satisfies $|c_n| \leq 1$ and $c_{n+13} = c_{n+7} + c_{n+6} - c_n$ for all integers $n$. Its characteristic polynomial is $x^{13} - x^7 - x^6 + 1 = (x^7 - 1)(x^6 - 1)$, which has 1 as a double root and the 6th and 7th roots of unity not equal to 1 as single roots. Let $r_1, r_2, \ldots, r_{11}$ be the 6th and 7th roots of unity not equal to 1, in some order. $c_n$ must satisfy $c_n = an + a_01^n  + a_1 r_1^n + a_2 r_2^n + \cdots + a_{11} r_{11}^n$, where $a, a_0, a_1, \cdots, a_{11}$ are some complex numbers. If $a \neq 0$, then we can pick some integer $n$ with $|an| > 1 + |a_1| + |a_2| + \cdots + |a_{11}|$, whence
\begin{align*}
|c_n|
&= |an + a_1 r_1^n + a_2 r_2^n + \cdots + a_{11} r_{11}^n| \\
&\geq |an| - |a_1 r_1^n + a_2 r_2^n + \cdots + a_{11} r_{11}^n| \\
&\geq |an| - (|a_1| + |a_2| + \cdots + |a_{11}|) \\
&> 1,
\end{align*}
a contradiction. Since $r_i^{42} = r_i$, $c_{n+42} = c_n$. In particular, $c_{42} = c_0$, so $f_a(1) = f_a(0)$, i.e., $f(a+1) = f(a)$, as desired.
This post has been edited 2 times. Last edited by Zhero, Jul 4, 2011, 2:23 AM
Z K Y
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BaronShadeNight
211 posts
#4 • 2 Y
Y by Adventure10, Mango247
IMO shortlist 1996
Z K Y
N Quick Reply
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