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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)!

Be sure to mark your calendars for the following upcoming events:
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[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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
Random concyclicity in a square config
Maths_VC   3
N 33 minutes ago by Assassino9931
Source: Serbia JBMO TST 2025, Problem 1
Let $M$ be a random point on the smaller arc $AB$ of the circumcircle of square $ABCD$, and let $N$ be the intersection point of segments $AC$ and $DM$. The feet of the tangents from point $D$ to the circumcircle of the triangle $OMN$ are $P$ and $Q$ , where $O$ is the center of the square. Prove that points $A$, $C$, $P$ and $Q$ lie on a single circle.
3 replies
Maths_VC
Tuesday at 7:38 PM
Assassino9931
33 minutes ago
Basic ideas in junior diophantine equations
Maths_VC   2
N 35 minutes ago by Assassino9931
Source: Serbia JBMO TST 2025, Problem 3
Determine all positive integers $a, b$ and $c$ such that
$2$ $\cdot$ $10^a + 5^b = 2025^c$
2 replies
Maths_VC
Tuesday at 7:54 PM
Assassino9931
35 minutes ago
Prime number theory
giangtruong13   2
N 36 minutes ago by RagvaloD
Find all prime numbers $p,q$ such that: $p^2-pq-q^3=1$
2 replies
giangtruong13
an hour ago
RagvaloD
36 minutes ago
An algorithm for discovering prime numbers?
Lukaluce   2
N 38 minutes ago by Assassino9931
Source: 2025 Junior Macedonian Mathematical Olympiad P3
Is there an infinite sequence of prime numbers $p_1, p_2, ..., p_n, ...,$ such that for every $i \in \mathbb{N}, p_{i + 1} \in \{2p_i - 1, 2p_i + 1\}$ is satisfied? Explain the answer.
2 replies
Lukaluce
May 18, 2025
Assassino9931
38 minutes ago
Problem 2
delegat   147
N an hour ago by math-olympiad-clown
Source: 0
Let $n\ge 3$ be an integer, and let $a_2,a_3,\ldots ,a_n$ be positive real numbers such that $a_{2}a_{3}\cdots a_{n}=1$. Prove that
\[(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.\]

Proposed by Angelo Di Pasquale, Australia
147 replies
delegat
Jul 10, 2012
math-olympiad-clown
an hour ago
Maxi-inequality
giangtruong13   0
an hour ago
Let $a,b,c >0$ and $a+b+c=2abc$. Find max: $$P= \sum_{cyc} \frac{a+2}{\sqrt{6(a^2+2)}}$$
0 replies
giangtruong13
an hour ago
0 replies
Max and Min
Butterfly   0
3 hours ago

Let $a_1,a_2,\cdots,a_n$ be an arrangement of $\{1,2,3,\cdots,n\}$. Find the maximum and minimum values of $$\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots+\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}.$$
0 replies
Butterfly
3 hours ago
0 replies
Inequality conjecture
RainbowNeos   0
4 hours ago
Show (or deny) that there exists an absolute constant $C>0$ that, for all $n$ and $n$ positive real numbers $x_i ,1\leq i \leq n$, there is
\[\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\geq C \ln n\left(\prod_{i=1}^n x_i\right)^{\frac{1}{n}}\]
0 replies
RainbowNeos
4 hours ago
0 replies
inequality 2905
pennypc123456789   0
4 hours ago
Consider positive real numbers \( x, y, z \) that satisfy the condition
\[
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3.
\]Find the maximum value of the expression
\[
P = \dfrac{yz}{\sqrt[3]{3y^2z^2+ 3x^2y^2z^2+ x^2z^2 + x^2y^2}}
+ \frac{xz}{\sqrt[3]{3x^2z^2 + 3x^2y^2z^2 + x^2y^2 + y^2z^2}}
+ \frac{xy}{\sqrt[3]{3x^2y^2 + 3x^2y^2z^2 +y^2z^2 + x^2z^2}}.
\]
0 replies
pennypc123456789
4 hours ago
0 replies
Inspired by m4thbl3nd3r
sqing   3
N 4 hours ago by sqing
Source: Own
Let $  a, b,c>0,b+c>a$. Prove that$$\sqrt{\frac{a}{b+c-a}}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)}\geq 1$$$$\frac{a}{b+c-a}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)} \geq  \frac{4\sqrt 2}{3}-1$$
3 replies
sqing
Today at 3:43 AM
sqing
4 hours ago
Inspired by qrxz17
sqing   7
N 4 hours ago by sqing
Source: Own
Let $a, b,c>0 ,(a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 27 $. Prove that $$a+b+c\geq 3\sqrt {3}$$
7 replies
sqing
Today at 8:50 AM
sqing
4 hours ago
4 var inequality
SunnyEvan   1
N 5 hours ago by SunnyEvan
Source: Own
Let $ x,y,z,t \in R^+ ,$ such that : $ (x+y+z+t)^2 = x+y+z+t + (x+z)(y+t) $ and $ x \geq y \geq z \geq t .$
Try to prove or disprove : $$ \frac{2 \sqrt{x+y+z+t +(x+t)(y+z)}}{x^2+y^2+z^2+t^2 +3xz+3yt+xt+yz} \geq \frac{11(x+y)(z+t)-(x+y+z+t)}{x+y+z+t +(x+z)(y+t)} $$
1 reply
SunnyEvan
Today at 7:07 AM
SunnyEvan
5 hours ago
MOP 2012 Inequality
holdmyquadrilateral   3
N 6 hours ago by Adywastaken
Source: MOP 2012
For $a,b,c>0$, prove that
\[\frac{a^3}{(b-c)^2+bc}+\frac{b^3}{(c-a)^2+ca}+\frac{c^3}{(a-b)^2+ab}\ge a+b+c.\]
3 replies
holdmyquadrilateral
Mar 11, 2023
Adywastaken
6 hours ago
Inspired by a9opsow_
sqing   3
N Today at 8:25 AM by sqing
Source: Own
Let $ a,b > 0  .$ Prove that
$$ \frac{(ka^2 - kab-b)^2 + (kb^2 - kab-a)^2 + (ab-ka-kb )^2}{ (ka+b)^2 + (kb+a)^2+(a - b)^2 }\geq  \frac {1}{(k+1)^2}$$Where $ k\geq 0.37088 .$
$$\frac{(a^2 - ab-b)^2 + (b^2 - ab-a)^2 + ( ab-a-b)^2}{a^2 +b^2+(a - b)^2 } \geq 1$$$$ \frac{(2a^2 - 2ab-b)^2 + (2b^2 - 2ab-a)^2 + (ab-2a-2b )^2}{ (2a+b)^2 + (2b+a)^2+(a - b)^2 }\geq  \frac 19$$
3 replies
sqing
Today at 1:49 AM
sqing
Today at 8:25 AM
prove |a-b| is a square, given a-b=a^2c-b^2d
Alpha314159   4
N Apr 17, 2025 by Leman_Nabiyeva
Source: Macau Inter High School Competition
Let $a, b$ be integers such that there are consecutive integers $c,d$ satisfy $$a-b=a^2 c-b^2 d$$.
Prove : $|a-b|$ is a perfect square.
4 replies
Alpha314159
Mar 7, 2020
Leman_Nabiyeva
Apr 17, 2025
prove |a-b| is a square, given a-b=a^2c-b^2d
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G H BBookmark kLocked kLocked NReply
Source: Macau Inter High School Competition
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Alpha314159
196 posts
#1
Y by
Let $a, b$ be integers such that there are consecutive integers $c,d$ satisfy $$a-b=a^2 c-b^2 d$$.
Prove : $|a-b|$ is a perfect square.
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MessingWithMath
340 posts
#2 • 1 Y
Y by Mango247
WLOG $a>b$ or we can just switch $c$ and $d$ around to make it work. Then, let $a-b=k>0$ as $0$ is already a perfect square. We see quickly that $d=c+1$ or else the $LHS\geq a^2-b^2 > a-b$ except for a few trivial cases.Then, solving for $c$ in terms of $k$ and $b$, we have $c=\frac{k+b^2}{k(2b+k)}$. Assume that a prime $p$ divides $k$ an odd number of times (call this power $m$). Then, since $b^2\equiv 0 \pmod{k}$, we can assume $p^n|b$. Thus, we have two cases, when $2n>m$ or $2n<m$. If $2n>m$, then $v_p(k+b^2)=m$ but $v_p(k(2b+k))>m$ so contradiction. Else, if $2n<m$, then we have $v_p(k+b^2)=2n<m$ but $v_p(k(2b+k))>m$ so contradiction. Thus, we prove that this can't happen, so all prime factors of $k$ have even power. This means that $k$ is a perfect square.
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Math-wiz
6107 posts
#3
Y by
Alpha314159 wrote:
Let $a, b$ be integers such that there are consecutive integers $c,d$ satisfy $$a-b=a^2 c-b^2 d$$.
Prove : $|a-b|$ is a perfect square.

This problem is also there in Mathematical Olympiad Challenges by Titu Andreescu :)
Z K Y
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Alpha314159
196 posts
#4
Y by
Thanks for the the help :)
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Leman_Nabiyeva
3 posts
#5 • 2 Y
Y by Omerking, aqusha_mlp12
Alpha314159 wrote:
Let $a, b$ be integers such that there are consecutive integers $c,d$ satisfy $$a-b=a^2 c-b^2 d$$.
Prove : $|a-b|$ is a perfect square.

Let $gcd(a;b)=x$ then $a=xy$, $b=xk$ where $gcd(y;k)=1$ $c=d+1$ So $y-k=xy^2d+xy^2-xk^2d$ .
We get $x|y-k$.
$y-k=xd(y-k)(y+k)+xy^2$ $xy^2=(y-k)(1-xdy-xdk)$
$gcd(y;y-k)=1$
$gcd(1-dxy-dxk;x)=1$
We get $y-k= \pm x$
$|a-b|=x^2$ finished $d=c+1$ case is analogical
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