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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.

Our full course list for upcoming classes is below:
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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
Inspired by elim and Sam Tong
sqing   3
N a minute ago by sqing
Source: Own
Let $ a ,b,c>0,a+b +c=1. $ Prove that
$$6a+3\sqrt{b}+2\sqrt[3]{c}\leq \frac{491}{72}$$$$12a+4\sqrt{b}+3\sqrt[4]{c} \leq  \frac{37}{3}+\frac{9}{8\sqrt[3]{2}}$$$$6a+3\sqrt{b}+2\sqrt[4]{c} \leq \frac{51+2\sqrt[3]{18}}{8}$$h
3 replies
1 viewing
sqing
Yesterday at 12:53 PM
sqing
a minute ago
Gregoriano Numbers
Supertinito   1
N a minute ago by Synchrone
Source: Colombia Junior Olympiad 2025 P1
A positive integer $n$ is called gregoriano if it is divisible by any subsequence of consecutive digits of $n$. For example, 48 is gregoriano because it is divisible by 4, 8 and 48, but 312 isn't because 312 isn't divisible by 31. Determine all gregoriano numbers.

Proposed by Mateo Matijasevick
1 reply
Supertinito
Jun 27, 2025
Synchrone
a minute ago
old product!
teomihai   8
N 2 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 $n$.?
8 replies
+1 w
teomihai
Yesterday at 3:51 PM
teomihai
2 minutes ago
functional equation reals
danepale   35
N 9 minutes ago by TigerOnion
Source: Croatia TST 2016
Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that for all real $x,y$:
$$ f(x^2) + xf(y) = f(x) f(x + f(y)) \, . $$
35 replies
danepale
Apr 25, 2016
TigerOnion
9 minutes ago
Inspired by old results
sqing   6
N 11 minutes ago by pooh123
Source: Own
Let $ a,b,c\geq 1 . $ Prove that
$$  (k- \frac {1} {a}) (k- \frac {1} {b}) (k- \frac {1} {c})  \leq (k-1)^3a^2b^2c^2  $$Where $ k\geq \frac{3}{2}$
$$  (2- \frac {1} {a}) (2- \frac {1} {b}) (2- \frac {1} {c})  \leq a^2b^2c^2  $$$$  (3- \frac {2} {a}) (3- \frac {2} {b}) (3- \frac {2} {c})  \leq a^2b^2c^2  $$
6 replies
1 viewing
sqing
5 hours ago
pooh123
11 minutes ago
Algebraic Number Theory book (from an olympiad perspective)
azerluigi   29
N 17 minutes ago by Not__Infinity
Hi!

This summer, I wrote a book about algebraic number theory from an olympiad perspective. There's basically no prerequisite, but it's better to have done some amount of olympiad number theory and polynomials beforehand.

I originally planned to write one more chapter and add hints, but uni is busier than I thought (I have so many classes, and most of them are not maths because France...) so I'm putting this off for a while. The most realistic scenario is that this last chapter will be written during the end-of-october holidays, and the hints during the Christmas holidays.

Well, that's about it I guess. For more info, you can see the table of content of the book and the read the foreword. https://www.dropbox.com/sh/lribbpkfooq96gs/AABJsHF8MuGZCkP8J0FxslD7a?dl=0

Tell me if the dropbox link somehow dies (it shouldn't). Also, feel free to send me any suggestion or comment you have. I would be particularly grateful if you could send me all the typos and mistakes you find, because I'm sure there are a lot.
29 replies
azerluigi
Oct 5, 2021
Not__Infinity
17 minutes ago
2025 IMO Results
ilikemath247365   8
N 35 minutes ago by Cats_on_a_computer
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!
8 replies
ilikemath247365
Yesterday at 4:48 PM
Cats_on_a_computer
35 minutes ago
A lies on the radical axis of BQX and CPX
a_507_bc   38
N 36 minutes ago by Royal_mhyasd
Source: APMO 2024 P1
Let $ABC$ be an acute triangle. Let $D$ be a point on side $AB$ and $E$ be a point on side $AC$ such that lines $BC$ and $DE$ are parallel. Let $X$ be an interior point of $BCED$. Suppose rays $DX$ and $EX$ meet side $BC$ at points $P$ and $Q$, respectively, such that both $P$ and $Q$ lie between $B$ and $C$. Suppose that the circumcircles of triangles $BQX$ and $CPX$ intersect at a point $Y \neq X$. Prove that the points $A, X$, and $Y$ are collinear.
38 replies
a_507_bc
Jul 29, 2024
Royal_mhyasd
36 minutes ago
EF bisects KL
Giahuytls2326   1
N an hour ago by Royal_mhyasd
Source: my teacher
Let \( ABC \) be an acute triangle inscribed in a circle \( (O) \). The altitudes \( BE \) and \( CF \) intersect at \( H \). Let \( M \) be a point on the minor arc \( BC \) . The lines \( MC \) and \( MB \) intersect the lines \( BE \) and \( CF \) at \( L \) and \( K \), respectively. Prove that the line \( EF \) passes through the midpoint of segment \( KL \).
1 reply
Giahuytls2326
4 hours ago
Royal_mhyasd
an hour ago
APMO 2025
motannoir   0
an hour ago
When will the subjects of apmo 2025 be published ? It usually appears right after imo.
0 replies
motannoir
an hour ago
0 replies
The refinement of GMA 567
mihaig   0
an hour ago
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
$$\left(\sum_{i=1}^{n}{a_i}\right)^2+\frac{2n-1}{(n-1)^3}\cdot\sum_{1\leq i<j\leq n}{\left(a_i-a_j\right)^2}\geq n^2.$$
0 replies
mihaig
an hour ago
0 replies
Useless identity
mashumaro   1
N an hour ago by mashumaro
Source: Own
Let $a_1$, $a_2$, $\dots$, $a_6$ be reals, and let $f(m, n) = \sum_{i=m}^{n} a_i$. Show that
\[ f(5,5)f(5,6) = f(2,4)f(3,4) = f(1,4)f(4,4) \implies f(1,1)f(1,2)f(4,5)f(4,6) = f(2,3)f(3,3)f(1,5)f(1,6) \]
1 reply
mashumaro
Yesterday at 6:01 AM
mashumaro
an hour ago
Inequality
SunnyEvan   1
N an hour 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
1 reply
SunnyEvan
an hour ago
SunnyEvan
an hour ago
A feasible refinement of GMA 567
Rhapsodies_pro   3
N an hour ago by mihaig
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: \[\frac1{n-1}\sum_{1\leqslant i<j\leqslant n}{\left(a_i-a_j\right)}^2\geqslant{\left({\left(n^2-2n-1\right)}\sum_{k=1}^na_k-n{\left(n-1\right)}{\left(n-3\right)}\right)}{\left(n-\sum_{k=1}^na_k\right)}\textnormal.\]
3 replies
Rhapsodies_pro
Jul 21, 2025
mihaig
an hour ago
system in R+, four equations/variables
jasperE3   2
N May 19, 2025 by Yiyj
Source: Bulgaria 1972 P2
Solve the system of equations:
$$\begin{cases}\sqrt{\frac{y(t-y)}{t-x}-\frac4x}+\sqrt{\frac{z(t-z)}{t-x}-\frac4x}=\sqrt x\\\sqrt{\frac{z(t-z)}{t-y}-\frac4y}+\sqrt{\frac{x(t-x)}{t-y}-\frac4y}=\sqrt y\\\sqrt{\frac{x(t-x)}{t-z}-\frac4z}+\sqrt{\frac{y(t-y)}{t-z}-\frac4z}=\sqrt z\\x+y+z=2t\end{cases}$$if the following conditions are satisfied: $0<x<t$, $0<y<t$, $0<z<t$.

H. Lesov
2 replies
jasperE3
Jun 21, 2021
Yiyj
May 19, 2025
system in R+, four equations/variables
G H J
G H BBookmark kLocked kLocked NReply
Source: Bulgaria 1972 P2
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jasperE3
11446 posts
#1
Y by
Solve the system of equations:
$$\begin{cases}\sqrt{\frac{y(t-y)}{t-x}-\frac4x}+\sqrt{\frac{z(t-z)}{t-x}-\frac4x}=\sqrt x\\\sqrt{\frac{z(t-z)}{t-y}-\frac4y}+\sqrt{\frac{x(t-x)}{t-y}-\frac4y}=\sqrt y\\\sqrt{\frac{x(t-x)}{t-z}-\frac4z}+\sqrt{\frac{y(t-y)}{t-z}-\frac4z}=\sqrt z\\x+y+z=2t\end{cases}$$if the following conditions are satisfied: $0<x<t$, $0<y<t$, $0<z<t$.

H. Lesov
Z K Y
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AlexCenteno2007
180 posts
#2
Y by
\WLOG x = y = z = a.
We have to give us the equation
\[
x + y + z = 3a = 2t \Rightarrow a = \dfrac{2t}{3}
\]
Substituting in(1)
\[
\sqrt{\dfrac{a(t-a)}{t-a} - \dfrac{4}{a}} + \sqrt{\dfrac{a(t-a)}{t-a} - \dfrac{4}{a}} = \sqrt{a}
\]\[
\Rightarrow 2\sqrt{a - \dfrac{4}{a}} = \sqrt{a}
\Rightarrow 4\left(a - \dfrac{4}{a}\right) = a \Rightarrow 4a - \dfrac{16}{a} = a
\Rightarrow 3a = \dfrac{16}{a} \Rightarrow 3a^2 = 16 \Rightarrow a^2 = \dfrac{16}{3}
\Rightarrow a = \dfrac{4\sqrt{3}}{3}
\]
Then:
\[
x = y = z = \dfrac{4\sqrt{3}}{3}, \quad t = \dfrac{3a}{2} = 2\sqrt{3}
\]
Checking the conditions
\[
0 < x = \dfrac{4\sqrt{3}}{3} < t = 2\sqrt{3} \quad \text{(If it clearly complies )}
\]

\[
\boxed{
x = y = z = \dfrac{4\sqrt{3}}{3}, \quad t = 2\sqrt{3}
}
\]
This post has been edited 3 times. Last edited by AlexCenteno2007, May 19, 2025, 6:36 PM
Reason: Error
Z K Y
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Yiyj
434 posts
#3
Y by
4 years later…
Z K Y
N Quick Reply
G
H
=
a