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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 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
This question just asks if you can factorise 12 factorial or not
Sadigly   1
N 17 minutes ago by COCBSGGCTG3
Source: Azerbaijan Junior MO 2025 P1
A teacher creates a fraction using numbers from $1$ to $12$ (including $12$). He writes some of the numbers on the numerator, and writes $\times$ (multiplication) between each number. Then he writes the rest of the numbers in the denominator and also writes $\times$ between each number. There is at least one number both in numerator and denominator. The teacher ensures that the fraction is equal to the smallest possible integer possible.

What is this positive integer, which is also the value of the fraction?
1 reply
1 viewing
Sadigly
Friday at 7:34 AM
COCBSGGCTG3
17 minutes ago
Prime sums of pairs
Assassino9931   5
N 24 minutes ago by aidan0626
Source: Al-Khwarizmi Junior International Olympiad 2025 P5
Sevara writes in red $8$ distinct positive integers and then writes in blue the $28$ sums of each two red numbers. At most how many of the blue numbers can be prime?

Marin Hristov, Bulgaria
5 replies
Assassino9931
Yesterday at 9:35 AM
aidan0626
24 minutes ago
Inequality, inequality, inequality...
Assassino9931   11
N 31 minutes ago by Assassino9931
Source: Al-Khwarizmi Junior International Olympiad 2025 P6
Let $a,b,c$ be real numbers such that \[ab^2+bc^2+ca^2=6\sqrt{3}+ac^2+cb^2+ba^2.\]Find the smallest possible value of $a^2 + b^2 + c^2$.

Binh Luan and Nhan Xet, Vietnam
11 replies
+1 w
Assassino9931
Yesterday at 9:38 AM
Assassino9931
31 minutes ago
Anything real in this system must be integer
Assassino9931   2
N 38 minutes ago by Assassino9931
Source: Al-Khwarizmi International Junior Olympiad 2025 P1
Determine the largest integer $c$ for which the following statement holds: there exists at least one triple $(x,y,z)$ of integers such that
\begin{align*} x^2 + 4(y + z) = y^2 + 4(z + x) = z^2 + 4(x + y) = c \end{align*}and all triples $(x,y,z)$ of real numbers, satisfying the equations, are such that $x,y,z$ are integers.

Marek Maruin, Slovakia
2 replies
Assassino9931
Friday at 9:26 AM
Assassino9931
38 minutes ago
Iranian geometry configuration
Assassino9931   3
N an hour ago by Assassino9931
Source: Al-Khwarizmi Junior International Olympiad 2025 P7
Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$, such that $CD$ is not a diameter of its circumcircle. The lines $AD$ and $BC$ intersect at point $P$, so that $A$ lies between $D$ and $P$, and $B$ lies between $C$ and $P$. Suppose triangle $PCD$ is acute and let $H$ be its orthocenter. The points $E$ and $F$ on the lines $BC$ and $AD$, respectively, are such that $BD \parallel HE$ and $AC\parallel HF$. The line through $E$, perpendicular to $BC$, intersects $AD$ at $L$, and the line through $F$, perpendicular to $AD$, intersects $BC$ at $K$. Prove that the points $K$, $L$, $O$ are collinear.

Amir Parsa Hosseini Nayeri, Iran
3 replies
Assassino9931
Yesterday at 9:39 AM
Assassino9931
an hour ago
China South East Mathematical Olympiad 2014 Q3B
sqing   4
N an hour ago by AGCN
Source: China Zhejiang Fuyang , 27 Jul 2014
Let $p$ be a primes ,$x,y,z $ be positive integers such that $x<y<z<p$ and $\{\frac{x^3}{p}\}=\{\frac{y^3}{p}\}=\{\frac{z^3}{p}\}$.
Prove that $(x+y+z)|(x^5+y^5+z^5).$
4 replies
sqing
Aug 17, 2014
AGCN
an hour ago
P>2D
gwen01   5
N an hour ago by Binod98
Source: Baltic Way 1992 #18
Show that in a non-obtuse triangle the perimeter of the triangle is always greater than two times the diameter of the circumcircle.
5 replies
gwen01
Feb 18, 2009
Binod98
an hour ago
Inequality
Sadigly   3
N 3 hours ago by pooh123
Source: Azerbaijan Junior MO 2025 P5
For positive real numbers $x;y;z$ satisfying $0<x,y,z<2$, find the biggest value the following equation could acquire:


$$(2x-yz)(2y-zx)(2z-xy)$$
3 replies
Sadigly
Friday at 7:59 AM
pooh123
3 hours ago
Calculus
youochange   2
N 3 hours ago by youochange
Find the area enclosed by the curves $e^x,e^{-x},x^2+y^2=1$
2 replies
youochange
Yesterday at 2:38 PM
youochange
3 hours ago
A strong inequality problem
hn111009   0
3 hours ago
Source: Somewhere
Let $a,b,c$ be the positive number satisfied $a^2+b^2+c^2=3.$ Find the minimum of $$P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{3abc}{2(ab+bc+ca)}.$$
0 replies
hn111009
3 hours ago
0 replies
help me please,thanks
tnhan.129   0
3 hours ago
find f: R+ -> R such that:
f(x)/x + f(y)/y = (1/x + 1/y).f(sqrt(xy))
0 replies
tnhan.129
3 hours ago
0 replies
Easy divisibility
a_507_bc   2
N 3 hours ago by TUAN2k8
Source: ARO Regional stage 2023 9.4~10.4
Let $a, b, c$ be positive integers such that no number divides some other number. If $ab-b+1 \mid abc+1$, prove that $c \geq b$.
2 replies
a_507_bc
Feb 16, 2023
TUAN2k8
3 hours ago
Inspired by old results
sqing   0
3 hours ago
Source: Own
Let $a,b,c,d$ be real numbers such that $a^2+b^2+c^2 =3$. Prove that$$\frac{9}{5}>(a-b)(b-c)(2a-1)(2c-1)\geq -16$$
0 replies
sqing
3 hours ago
0 replies
integer functional equation
ABCDE   149
N 3 hours ago by ezpotd
Source: 2015 IMO Shortlist A2
Determine all functions $f:\mathbb{Z}\rightarrow\mathbb{Z}$ with the property that \[f(x-f(y))=f(f(x))-f(y)-1\]holds for all $x,y\in\mathbb{Z}$.
149 replies
ABCDE
Jul 7, 2016
ezpotd
3 hours ago
Inequality with 3 variables and a special condition
Nuran2010   8
N May 2, 2025 by sqing
Source: Azerbaijan Al-Khwarizmi IJMO TST 2024
For positive real numbers $a,b,c$ we have $3abc \geq ab+bc+ca$.
Prove that:

$\frac{1}{a^3+b^3+c}+\frac{1}{b^3+c^3+a}+\frac{1}{c^3+a^3+b} \leq \frac{3}{a+b+c}$.

Determine the equality case.
8 replies
Nuran2010
Apr 29, 2025
sqing
May 2, 2025
Inequality with 3 variables and a special condition
G H J
Source: Azerbaijan Al-Khwarizmi IJMO TST 2024
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Nuran2010
85 posts
#1
Y by
For positive real numbers $a,b,c$ we have $3abc \geq ab+bc+ca$.
Prove that:

$\frac{1}{a^3+b^3+c}+\frac{1}{b^3+c^3+a}+\frac{1}{c^3+a^3+b} \leq \frac{3}{a+b+c}$.

Determine the equality case.
Z K Y
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arqady
30244 posts
#2 • 3 Y
Y by MuradSafarli, Nuran2010, FarrukhBurzu
Nuran2010 wrote:
For positive real numbers $a,b,c$ we have $3abc \geq ab+bc+ca$.
Prove that:

$\frac{1}{a^3+b^3+c}+\frac{1}{b^3+c^3+a}+\frac{1}{c^3+a^3+b} \leq \frac{3}{a+b+c}$.

Determine the equality case.
Use C-S and AM-GM.
Z K Y
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Assassino9931
1340 posts
#3 • 1 Y
Y by Nuran2010
Reduce to a common denominator via Cauchy-Schwarz exercise (similar to problems from JBMO 2019 and 2023 shortlists). We have $(a^3 + b^3 + c)\left(\frac{1}{a} + \frac{1}{b} + c\right) \geq (a+b+c)^2$ and the analogous ones, so the original left-hand side does not exceed
$$ \frac{2\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) + a+b+c}{(a+b+c)^2}  $$and we wish the latter to not exceed $\frac{3}{a+b+c}$. Equivalently, we want $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq a+b+c$. But actually, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq 3$ from the problem condition, while $3abc \geq ab + bc + ca \geq 3\sqrt[3]{a^2b^2c^2}$ by AM-GM implies $abc \geq 1$, thus $a+b+c \geq 3\sqrt[3]{abc} \geq 3$ by AM-GM and we are done. The AM-GM-s and $abc \geq 1$ require $a=b=c=1$ for equality and substituting in the initial inequality indeed shows that it is an equality in this case.
This post has been edited 1 time. Last edited by Assassino9931, Apr 29, 2025, 10:54 PM
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sqing
42107 posts
#4
Y by
Prove that for all non-negative real numbers $x,y,z$, not all equal to $0$, the following inequality holds
$$\displaystyle \dfrac{2x^2-x+y+z}{x+y^2+z^2}+\dfrac{2y^2+x-y+z}{x^2+y+z^2}+\dfrac{2z^2+x+y-z}{x^2+y^2+z}\geq 3$$Let $a, b, c $ be positive real numbers such that $abc = \frac {2} {3}. $ Prove that:
$$\frac {ab}{a + b} + \frac {bc} {b + c} + \frac {ca} {c + a} \geqslant  \frac {a+b+c} {a^3+b ^ 3 + c ^ 3}.$$(JBMO 2019 and 2023 shortlists)
This post has been edited 1 time. Last edited by sqing, May 1, 2025, 8:59 AM
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Assassino9931
1340 posts
#5 • 1 Y
Y by ehuseyinyigit
@Sqing No, I meant A5 in 2019 and A4 in 2023
This post has been edited 1 time. Last edited by Assassino9931, May 2, 2025, 12:57 AM
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sqing
42107 posts
#6 • 1 Y
Y by ehuseyinyigit
Assassino9931 wrote:
@Sqing No, I meant A5 in 2019 and A4 in 2023
Let $a, b, c, d$ be positive real numbers such that $abcd = 1$. Prove the inequality
$\frac{1}{a^3 + b + c + d} +\frac{1}{a + b^3 + c + d}+\frac{1}{a + b + c^3 + d} +\frac{1}{a + b + c + d^3} \leq \frac{a+b+c+d}{4}$
Let $a,b,c,d$ be positive real numbers with $abcd=1$. Prove that
$$\sqrt{\frac{a}{b+c+d^2+a^3}}+\sqrt{\frac{b}{c+d+a^2+b^3}}+\sqrt{\frac{c}{d+a+b^2+c^3}}+\sqrt{\frac{d}{a+b+c^2+d^3}} \leq 2$$
This post has been edited 1 time. Last edited by sqing, May 2, 2025, 1:26 AM
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mihaig
7361 posts
#7
Y by
arqady wrote:
Nuran2010 wrote:
For positive real numbers $a,b,c$ we have $3abc \geq ab+bc+ca$.
Prove that:

$\frac{1}{a^3+b^3+c}+\frac{1}{b^3+c^3+a}+\frac{1}{c^3+a^3+b} \leq \frac{3}{a+b+c}$.

Determine the equality case.
Use C-S and AM-GM.

Beautiful
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sqing
42107 posts
#8
Y by
Let $ a,b>0 $ and $a^3+1= a^2(b+1). $ Prove that
$$\frac{1}{a^3+b^3+1}+\frac{1}{a^3+b+1}+\frac{1}{b^3+a+1} \leq \frac{3}{ a+b+1} $$
This post has been edited 1 time. Last edited by sqing, May 2, 2025, 10:06 AM
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sqing
42107 posts
#9
Y by
Let $ a,b>0 $ and $a^2+b^2+ab=\frac{3}{4}. $ Prove that
$$\frac{1}{a^3+b^3+\frac{1}{2}}+\frac{1}{a^3+b+\frac{1}{8}}+\frac{1}{b^3+a+\frac{1}{8}} \leq \frac{6}{ a+b+\frac{1}{2}} $$
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