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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:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]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
Already posted in HSO, too difficult
GreekIdiot   0
9 minutes ago
Source: own
Find all integer triplets that satisfy the equation $5^x-2^y=z^3$.
0 replies
GreekIdiot
9 minutes ago
0 replies
UC Berkeley Integration Bee 2025 Bracket Rounds
Silver08   25
N 13 minutes ago by Levieee
Regular Round

Quarterfinals

Semifinals

3rd Place Match

Finals
25 replies
2 viewing
Silver08
May 9, 2025
Levieee
13 minutes ago
Square on Cf
GreekIdiot   0
18 minutes ago
Let $f$ be a continuous function defined on $[0,1]$ with $f(0)=f(1)=0$ and $f(t)>0 \: \forall \: t \in (0,1)$. We define the point $X'$ to be the projection of point $X$ on the x-axis. Prove that there exist points $A, B \in C_f$ such that $ABB'A'$ is a square.
0 replies
+1 w
GreekIdiot
18 minutes ago
0 replies
Diophantine involving cube
Sadigly   2
N 18 minutes ago by Adywastaken
Source: Azerbaijan Senior NMO 2020
$a;b;c;d\in\mathbb{Z^+}$. Solve the equation: $$2^{a!}+2^{b!}+2^{c!}=d^3$$
2 replies
Sadigly
Yesterday at 10:13 PM
Adywastaken
18 minutes ago
Nice R+ FE
math_comb01   4
N 20 minutes ago by mkultra42
Source: XOOK 2025 P3
Find all functions $f : \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ so that for any $x, y \in \mathbb{R}_{>0}$, we have \[ f(xf(x^2)+yf(x))=xf(y+xf(x)). \]
Proposed by Anmol Tiwari and MathLuis
4 replies
math_comb01
Feb 9, 2025
mkultra42
20 minutes ago
2018 JBMO TST- Macedonia, problem 4
Lukaluce   4
N 29 minutes ago by Erto2011_
Source: 2018 JBMO TST- Macedonia
Determine all pairs $(p, q)$, $p, q \in \mathbb {N}$, such that

$(p + 1)^{p - 1} + (p - 1)^{p + 1} = q^q$.
4 replies
Lukaluce
May 28, 2019
Erto2011_
29 minutes ago
solve in positive integers: 3 \cdot 2^x +4 =n^2
parmenides51   4
N an hour ago by AylyGayypow009
Source: Greece JBMO TST 2019 p2
Find all pairs of positive integers $(x,n) $ that are solutions of the equation $3 \cdot 2^x +4 =n^2$.
4 replies
parmenides51
Apr 29, 2019
AylyGayypow009
an hour ago
ISI UGB 2025 P7
SomeonecoolLovesMaths   10
N an hour ago by Mathworld314
Source: ISI UGB 2025 P7
Consider a ball that moves inside an acute-angled triangle along a straight line, unit it hits the boundary, which is when it changes direction according to the mirror law, just like a ray of light (angle of incidence = angle of reflection). Prove that there exists a triangular periodic path for the ball, as pictured below.

IMAGE
10 replies
1 viewing
SomeonecoolLovesMaths
Yesterday at 11:28 AM
Mathworld314
an hour ago
Trigo relation in a right angled. ISIBS2011P10
Sayan   12
N an hour ago by SomeonecoolLovesMaths
Show that the triangle whose angles satisfy the equality
\[\frac{\sin^2A+\sin^2B+\sin^2C}{\cos^2A+\cos^2B+\cos^2C} = 2\]
is right angled.
12 replies
Sayan
Mar 31, 2013
SomeonecoolLovesMaths
an hour ago
Cute matrix equation
RobertRogo   2
N an hour ago by BlueSyrup
Source: "Traian Lalescu" student contest 2025, Section A, Problem 2
Find all matrices $A \in \mathcal{M}_n(\mathbb{Z})$ such that $$2025A^{2025}=A^{2024}+A^{2023}+\ldots+A$$Edit: Proposed by Marian Vasile (congrats!).
2 replies
RobertRogo
May 9, 2025
BlueSyrup
an hour ago
Six variables
Nguyenhuyen_AG   4
N an hour ago by Sunjee
Let $a,\,b,\,c,\,x,\,y,\,z$ be six positive real numbers. Prove that
$$\frac{a}{b+c} \cdot \frac{y+z}{x} + \frac{b}{c+a} \cdot \frac{z+x}{y} + \frac{c}{a+b} \cdot \frac{x+y}{z} \geqslant 2+\sqrt{\frac{8abc}{(a+b)(b+c)(c+a)}}.$$
4 replies
Nguyenhuyen_AG
Yesterday at 5:09 AM
Sunjee
an hour ago
Expressing polynomial as product of two polynomials
Sadigly   3
N 2 hours ago by Sadigly
Source: Azerbaijan Senior NMO 2021
Define $P(x)=((x-a_1)(x-a_2)...(x-a_n))^2 +1$, where $a_1,a_2...,a_n\in\mathbb{Z}$ and $n\in\mathbb{N^+}$. Prove that $P(x)$ couldn't be expressed as product of two non-constant polynomials with integer coefficients.
3 replies
Sadigly
Yesterday at 9:10 PM
Sadigly
2 hours ago
Product of consecutive terms divisible by a prime number
BR1F1SZ   2
N 2 hours ago by bin_sherlo
Source: 2025 Francophone MO Seniors P4
Determine all sequences of strictly positive integers $a_1, a_2, a_3, \ldots$ satisfying the following two conditions:
[list]
[*]There exists an integer $M > 0$ such that, for all indices $n \geqslant 1$, $0 < a_n \leqslant M$.
[*]For any prime number $p$ and for any index $n \geqslant 1$, the number
\[
a_n a_{n+1} \cdots a_{n+p-1} - a_{n+p}
\]is a multiple of $p$.
[/list]


2 replies
BR1F1SZ
Yesterday at 12:09 AM
bin_sherlo
2 hours ago
Pentagon with given diameter, ratio desired
bin_sherlo   2
N 2 hours ago by Umudlu
Source: Türkiye 2025 JBMO TST P7
$ABCDE$ is a pentagon whose vertices lie on circle $\omega$ where $\angle DAB=90^{\circ}$. Let $EB$ and $AC$ intersect at $F$, $EC$ meet $BD$ at $G$. $M$ is the midpoint of arc $AB$ on $\omega$, not containing $C$. If $FG\parallel DE\parallel CM$ holds, then what is the value of $\frac{|GE|}{|GD|}$?
2 replies
bin_sherlo
Yesterday at 7:21 PM
Umudlu
2 hours ago
ecuation real numbers
MihaiT   3
N Apr 2, 2025 by MihaiT
Let $f,g:R->R $be continuous, and $f$ is increasing and $g $is decreasing. If $ f,g $ have the same horizontal asymptote at $+\infty$,which is the number of solutions of the equation $f(x)=g(x)$ ?(Or can this number not be specified?)
3 replies
MihaiT
Mar 31, 2025
MihaiT
Apr 2, 2025
ecuation real numbers
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MihaiT
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Let $f,g:R->R $be continuous, and $f$ is increasing and $g $is decreasing. If $ f,g $ have the same horizontal asymptote at $+\infty$,which is the number of solutions of the equation $f(x)=g(x)$ ?(Or can this number not be specified?)
This post has been edited 1 time. Last edited by MihaiT, Mar 31, 2025, 6:56 PM
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Etkan
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#2 • 1 Y
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MihaiT wrote:
Let $f,g:R->R $be continuous, and $f$ is increasing and $g $is decreasing. If $ f,g $ have the same horizontal asymptote at $+\infty$,which is the number of solutions of the equation $f(x)=g(x)$ ?(Or can this number not be specified?)

It depends on whether you consider "increasing" as $x<y\implies f(x)<f(y)$ or as $x<y\implies f(x)\leq f(y)$.

In the first case, the answer is $0$. Indeed, having the same horizontal asymptote at $\infty$ means that $\lim \limits _{x\to \infty}f(x)=a=\lim \limits _{x\to \infty}g(x)$, for some $a\in \mathbb{R}$. Since $f$ is increasing, we have $f(x)<a$ for all $x\in \mathbb{R}$, and since $g$ is decreasing, we have $g(x)>a$ for all $x\in \mathbb{R}$. Hence $f(x)<a<g(x)$ for all $x\in \mathbb{R}$, and so the number of solutions to $f(x)=g(x)$ is $0$.

In the second case, the answer is that the number of solutions is either $0$ or infinite (more precisely, either $0$ or $|\mathbb{R}|$). The same argument as above gives $f(x)\leq a$ for all $x\in \mathbb{R}$ and $g(x)\geq a$ for all $x\in \mathbb{R}$. If $f(x)<a$ for all $x\in \mathbb{R}$ or $g(x)>a$ for all $x\in \mathbb{R}$, there are no solutions to $f(x)=g(x)$. Else there exist $x_f,x_g\in \mathbb{R}$ such that $f(x_f)=a$ and $g(x_g)=a$, so $f(x)=a$ for all $x\geq x_f$ (because $f$ is increasing and $f(x)\leq a$ for all $x\in \mathbb{R}$) and $g(x)=a$ for all $x\geq x_g$ (because $g$ is decreasing and $g(x)\geq a$ for all $x\in \mathbb{R}$). Hence $f(x)=a=g(x)$ for all $x\geq \max \{x_f,x_g\}$, and so the number of solutions to $f(x)=g(x)$ is $|\mathbb{R}|$.
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MihaiT
750 posts
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Etkan wrote:
MihaiT wrote:
Let $f,g:R->R $be continuous, and $f$ is increasing and $g $is decreasing. If $ f,g $ have the same horizontal asymptote at $+\infty$,which is the number of solutions of the equation $f(x)=g(x)$ ?(Or can this number not be specified?)

It depends on whether you consider "increasing" as $x<y\implies f(x)<f(y)$ or as $x<y\implies f(x)\leq f(y)$.

In the first case, the answer is $0$. Indeed, having the same horizontal asymptote at $\infty$ means that $\lim \limits _{x\to \infty}f(x)=a=\lim \limits _{x\to \infty}g(x)$, for some $a\in \mathbb{R}$. Since $f$ is increasing, we have $f(x)<a$ for all $x\in \mathbb{R}$, and since $g$ is decreasing, we have $g(x)>a$ for all $x\in \mathbb{R}$. Hence $f(x)<a<g(x)$ for all $x\in \mathbb{R}$, and so the number of solutions to $f(x)=g(x)$ is $0$.

In the second case, the answer is that the number of solutions is either $0$ or infinite (more precisely, either $0$ or $|\mathbb{R}|$). The same argument as above gives $f(x)\leq a$ for all $x\in \mathbb{R}$ and $g(x)\geq a$ for all $x\in \mathbb{R}$. If $f(x)<a$ for all $x\in \mathbb{R}$ or $g(x)>a$ for all $x\in \mathbb{R}$, there are no solutions to $f(x)=g(x)$. Else there exist $x_f,x_g\in \mathbb{R}$ such that $f(x_f)=a$ and $g(x_g)=a$, so $f(x)=a$ for all $x\geq x_f$ (because $f$ is increasing and $f(x)\leq a$ for all $x\in \mathbb{R}$) and $g(x)=a$ for all $x\geq x_g$ (because $g$ is decreasing and $g(x)\geq a$ for all $x\in \mathbb{R}$). Hence $f(x)=a=g(x)$ for all $x\geq \max \{x_f,x_g\}$, and so the number of solutions to $f(x)=g(x)$ is $|\mathbb{R}|$.

Thanks very much! :)
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MihaiT
750 posts
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MihaiT
746 posts
#1VPMMar 31, 2025, 7:43 AM
Given $m_1$ weights, each weighing $k_1$ and another $m_2$ weights with $k_2$ each. Write a algorithm that determines the ways in which a scale can be balanced with a weight $X$ on the left pan, and display the number of possible solutions. (The weights can be placed on both pans and the program starts with the numbers $m_1,k_1,m_2,k_2,X$. What will be displayed after three successive runs: 5,2,5,1,4 | 5,2,5,1,11 | 5,2,5,1,20?

One answer is possible:
a)10;5;0;
b)20;7;0;
c)20;7;1;
d)10;10;0;
e)10;7;0;
f)20;5;0,

if is possible one hint
Thanks very much!
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