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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
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
Grade 10 Inequality
BomboaneMentos316   0
35 minutes ago
Prove that for any x,y,z real and positive the following is true:
(PS: It is written correctly)

\begin{align*}
\frac{x^3}{y^2 + y z + z^2}
\;+\;
\frac{y^3}{2 z^2 + y z x}
\;+\;
\frac{z^3}{x^2 + x y + y^2}
\;\ge\;
\frac{x + y + z}{3}.
\end{align*}
0 replies
BomboaneMentos316
35 minutes ago
0 replies
Prove that <HMA=<GNS
Goutham   11
N an hour ago by Bonime
Source: Serbia NMO 2010 problem 2
In an acute-angled triangle $ABC$, $M$ is the midpoint of side $BC$, and $D, E$ and $F$ the feet of the altitudes from $A, B$ and $C$, respectively. Let $H$ be the orthocenter of $\Delta ABC$, $S$ the midpoint of $AH$, and $G$ the intersection of $FE$ and $AH$. If $N$ is the intersection of the median $AM$ and the circumcircle of $\Delta BCH$, prove that $\angle HMA = \angle GNS$.

Proposed by Marko Djikic
11 replies
Goutham
Mar 11, 2011
Bonime
an hour ago
Azerbaijan NMO 2015
IstekOlympiadTeam   17
N an hour ago by Just1
Source: Azerbaijan NMO 2015
Let $a,b$ and $c$ be positive reals such that $abc=\frac{1}{8}$.Then prove that \[a^2+b^2+c^2+a^2b^2+a^2c^2+b^2c^2\ge\frac{15}{16}\]
17 replies
IstekOlympiadTeam
Oct 21, 2015
Just1
an hour ago
Where did the IMO threads go
KevinYang2.71   11
N an hour ago by a_507_bc
Where did the IMO problem threads go? The problems are publicly released already
11 replies
KevinYang2.71
2 hours ago
a_507_bc
an hour ago
Trivial Fe
Tofa7a._.36   2
N an hour ago by Kempu33334
Find all functions $f : \mathbb R \rightarrow \mathbb R$ such that:
$$f(f(2x+y)) + f(x) = 2x + f(x+y)$$For all real numbers $x,y \in \mathbb{R}$.
2 replies
Tofa7a._.36
2 hours ago
Kempu33334
an hour ago
Tricky FE
Rijul saini   21
N an hour ago by Lunatic_Lunar7986
Source: LMAO 2025 Day 1 Problem 1
Let $\mathbb{R}$ denote the set of all real numbers. Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that
$$f(xy) + f(f(y)) = f((x + 1)f(y))$$for all real numbers $x$, $y$.

Proposed by MV Adhitya and Kanav Talwar
21 replies
Rijul saini
Jun 4, 2025
Lunatic_Lunar7986
an hour ago
Hexagon geometry
Tofa7a._.36   0
an hour ago
Let $ABC$ be a triangle such that $\angle{ABC}=3\angle{ACB}$. In the circumcircle of this triangle, let $D,E$ and $F$ be points such that: $(AD)\parallel (BC)$, $(DE)\parallel (CA)$ and $(EF) \parallel (AB)$.
Let $J$ be the intersection of lines $DF$ and $AC$ and let $\omega$ be the circle that passes through $J$ and is tangent to the line $BD$ at $D$.
Finally, let $L$ be the intersection point of $\omega$ and $(ABC)$.
Prove that points $E,J$ and $L$ are collinear.
0 replies
Tofa7a._.36
an hour ago
0 replies
nt with primes and gcd
Tofa7a._.36   1
N 2 hours ago by speedcode
Find all positive integers $a,b$ such that $a^2+ab+b^2=p^k$ and $ab=m^2$ for some prime $p$ and positive integers $m,k$.
1 reply
Tofa7a._.36
Today at 9:39 AM
speedcode
2 hours ago
Six variables (2)
Nguyenhuyen_AG   2
N 2 hours ago by arqady
Let $a, \, b, \,c, \, x, \, y, \, z$ be six positive real numbers. Prove that
\[a^2+b^2+c^2+\frac{4(ax+by+cz)\sqrt{ab+bc+ca}}{x+y+z} \geqslant 2(ab+bc+ca).\]
2 replies
Nguyenhuyen_AG
Jun 5, 2025
arqady
2 hours ago
Prove that OA and RA are perpendicular
MellowMelon   93
N 2 hours ago by Kempu33334
Source: USA TSTST 2011/2012 P4
Acute triangle $ABC$ is inscribed in circle $\omega$. Let $H$ and $O$ denote its orthocenter and circumcenter, respectively. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively. Rays $MH$ and $NH$ meet $\omega$ at $P$ and $Q$, respectively. Lines $MN$ and $PQ$ meet at $R$. Prove that $OA\perp RA$.
93 replies
MellowMelon
Jul 26, 2011
Kempu33334
2 hours ago
The Most Difficult Functional Equation in the World
EthanWYX2009   1
N 2 hours ago by RunboLi
Source: 2023 September 谜之竞赛-3, lemma aops.com/community/c6h3525285p34245733
Determine all functions $f:\mathbb N_+\to\mathbb N_+$, such that for any positive integers $x$, $y$,
\[f(x)^2+y^2\mid\sum_{i=0}^{2023}(xf(x))^{2023-i}\left(f^{(i)}(y)\right)^{2i}.\]Created by Yuxing Ye
1 reply
EthanWYX2009
Jul 12, 2025
RunboLi
2 hours ago
Concurrency in Parallelogram
amuthup   96
N 3 hours ago by LeYohan
Source: 2021 ISL G1
Let $ABCD$ be a parallelogram with $AC=BC.$ A point $P$ is chosen on the extension of ray $AB$ past $B.$ The circumcircle of $ACD$ meets the segment $PD$ again at $Q.$ The circumcircle of triangle $APQ$ meets the segment $PC$ at $R.$ Prove that lines $CD,AQ,BR$ are concurrent.
96 replies
amuthup
Jul 12, 2022
LeYohan
3 hours ago
in n^2-9 has 6 positive divisors than GCD (n-3, n+3)=1
parmenides51   11
N 3 hours ago by Just1
Source: Greece JBMO TST 2016 p3
Positive integer $n$ is such that number $n^2-9$ has exactly $6$ positive divisors. Prove that GCD $(n-3, n+3)=1$
11 replies
parmenides51
Apr 29, 2019
Just1
3 hours ago
AM+HM compare with GM
Teng   0
3 hours ago
Let $a,b,c\in\mathbb{R}^+$. Prove that $\dfrac{a+b+c}{3}+\dfrac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\geq5\sqrt[3]{\dfrac{abc}{16}}$
I found out that the equality holds when $(a,b,c)=(k,4k,4k)$ where $k\in\mathbb{R}^+$ and its cyclic permutation. One of my solution is directly guessing $a=16$ and $b=c=x^3$ by homogeneity. But I want a more elegant proof (use AM-GM, Cauchy, Schur, etc).
0 replies
Teng
3 hours ago
0 replies
Combinatorics
AlexCenteno2007   2
N May 11, 2025 by Royal_mhyasd
Adrian and Bertrand take turns as follows: Adrian starts with a pile of ($n\geq 3$) stones. On their turn, each player must divide a pile. The player who can make all piles have at most 2 stones wins. Depending on n, determine which player has a winning strategy.
2 replies
AlexCenteno2007
May 9, 2025
Royal_mhyasd
May 11, 2025
Combinatorics
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AlexCenteno2007
166 posts
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Adrian and Bertrand take turns as follows: Adrian starts with a pile of ($n\geq 3$) stones. On their turn, each player must divide a pile. The player who can make all piles have at most 2 stones wins. Depending on n, determine which player has a winning strategy.
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AlexCenteno2007
166 posts
#2
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Could you help me?
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Royal_mhyasd
89 posts
#3 • 1 Y
Y by AlexCenteno2007
Nice one. There's quite a few solutions as well but I'll just leave the one I thought of
We're going to prove that Adrian wins for n=3 and n=even and Bertrand wins for n=odd, n>3.
n=3 is obvious - A wins by dividing it into a pile of 1 and a pile of 2.
For n>=4, we're gonna use induction. (our induction is that A wins when n=3 and n=even and B wins when n=odd, n>3)
If n=even, A can simply divide his pile into a pile of 1 and a pile of n-1. n-1 is odd, so the player who is about to make a move (player B) is gonna lose, so A wins.
If n=odd, A is gonna divide the pile into two piles of a and b stones, a<b, a+b = n so either a is odd and b is even or a is even and b is odd.
If b-a isn't 3 :
Bertrand is gonna divide the current piles into 3 piles of : b-a, a, a stones. We define "the first zone" as any move that can be traced back to the pile of b-a stones and "the second zone" as any love that can be traced back to the piles of a stones. If A makes a move in the second zone, B will simply mirror him in the other sub-zone (the one that can be traced back to the other pile of a stones). Therefore, it's pretty obvious that B will make the last move in the second zone, seeing as he can always mirror A. b-a is odd because a and b arent equal mod 2, so if A makes a move in the first zone then B can simply apply his winning strategy for b-a (it's odd and it isn't 3 so the second player wins), ensuring that he doesn't lose. Also, as long as the second zone still hasn't been reduced to piles of at most 2 stones, moves can still be made in the first zone even if B already technically won since there will still be piles of 2 stones. But it's impossible for A to make the last move in the first zone since that would mean that an odd number of total moves were made, so there's an even number of piles of 1 and no other piles, so b-a is even which isn't true. Therefore, B makes the last move in the first zone. So B wins.
If b-a = 3:
If the odd number out of a and b isn't 3 :
Then B can simply split the 2 piles into 2 piles of x/2, x/2 and y where x is the even number and y is the odd number. He can apply the same strategy as before since y isn't 3.
If the odd number out of a and b is 3:
Then we have n=9, b=6, a=3.
In this case, B will divide the pile of 6 into two piles of 3. This forces A to divide one of the piles of 3 into 2 piles of 1 and 2 so we will have
2 3 3 (the pile of 1 doesn't matter)
B divides the pile of 2 into 2 piles of 1,so we have
3 3
A divides one of the piles of 3 into 2 piles of 1 and 2,and then B will do the same for the other pile of 3 so B wins, which concludes our proof.
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