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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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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   0
a few seconds ago
Source: Own
Let $ a ,b,c>0,a+b +c=1. $ Prove that
$$a+\frac{1}{2}\sqrt{b}+\frac{1}{3}\sqrt[3]{c}\leq \frac{491}{432}$$$$a+\frac{1}{3}\sqrt{b}+\frac{1}{4}\sqrt[4]{c} \leq  \frac{37}{36}+\frac{3}{32\sqrt[3]{2}}$$$$a+\frac{1}{2}\sqrt{b}+\frac{1}{3}\sqrt[4]{c} \leq \frac{51+2\sqrt[3]{18}}{48}$$h
0 replies
1 viewing
sqing
a few seconds ago
0 replies
one incenter and two orthocenters
jayme   1
N 25 minutes ago by WALLAH_HABIBI
Dear Mathlinkers,

an article intilted ‘‘One incenter and two orthocenters’’ been put on my website

http://jl.ayme.pagesperso-orange.fr/Docs/Un%20incentre%20et%20deux%20orthocentres.pdf

Sincerely
Jean-Louis
1 reply
1 viewing
jayme
Nov 5, 2020
WALLAH_HABIBI
25 minutes ago
Brilliant guessing game on triples
Assassino9931   4
N 33 minutes ago by Just1
Source: Al-Khwarizmi Junior International Olympiad 2025 P8
There are $100$ cards on a table, flipped face down. Madina knows that on each card a single number is written and that the numbers are different integers from $1$ to $100$. In a move, Madina is allowed to choose any $3$ cards, and she is told a number that is written on one of the chosen cards, but not which specific card it is on. After several moves, Madina must determine the written numbers on as many cards as possible. What is the maximum number of cards Madina can ensure to determine?

Shubin Yakov, Russia
4 replies
Assassino9931
May 10, 2025
Just1
33 minutes ago
Midpoint of arc black magic
MarkBcc168   15
N 39 minutes ago by Giabach298
Source: IMO Shortlist 2024 G7
Let \(ABC\) be a triangle with incenter \(I\) such that \(AB<AC<BC\). The second intersections of \(AI\), \(BI\), and \(CI\) with the circumcircle of triangle \(ABC\) are \(M_{A}\), \(M_{B}\), and \(M_{C}\), respectively. Lines \(AI\) and \(BC\) intersect at \(D\) and lines \(BM_{C}\) and \(CM_{B}\) intersect at \(X\). Suppose the circumcircle of triangles \(XM_{B}M_{C}\) and \(XBC\) intersect again at \(S\neq X\). Lines \(BX\) and \(CX\) intersect the circumcircle of triangle \(SXM_{A}\) again at \(P\neq X\) and \(Q\neq X\), respectively.

Prove that the circumcenter of triangle \(SID\) lies on \(PQ\).

Proposed by Thailand
15 replies
1 viewing
MarkBcc168
Jul 16, 2025
Giabach298
39 minutes ago
I miss Turbo
sarjinius   38
N an hour ago by Sqrt2_1.4142
Source: 2025 IMO P6
Consider a $2025\times2025$ grid of unit squares. Matilda wishes to place on the grid some rectangular tiles, possibly of different sizes, such that each side of every tile lies on a grid line and every unit square is covered by at most one tile.

Determine the minimum number of tiles Matilda needs to place so that each row and each column of the grid has exactly one unit square that is not covered by any tile.

Proposed by Zhao Yu Ma and David Lin Kewei, Singapore
38 replies
+2 w
sarjinius
Jul 16, 2025
Sqrt2_1.4142
an hour ago
Non-polynomial sequences satifying m+n|a_m+a_n?
TUAN2k8   1
N an hour ago by TUAN2k8
Source: own
Consider a sequence of integers \((a_n)_{n>0}\) such that for every pair of distinct positive integers \((m, n)\), \(m + n\) is a divisor of \(a_m + a_n\).

a) Prove that \(a_n\) is divisible by \(n\) for every positive integer \(n\).

b) Does there exist a sequence \((a_n)_{n>0}\) that is not a polynomial in \(n\) (i.e., there does not exist a polynomial \(P(X) \in \mathbb{R}[X]\) such that \(a_n = P(n)\) for all \(n \in \mathbb{Z}_+\)) and satisfies the given condition?
1 reply
TUAN2k8
4 hours ago
TUAN2k8
an hour ago
D1053 : Set of Dirichlet
Dattier   1
N an hour ago by Dattier
Source: les dattes à Dattier
We say a set $D$ have the Dirichlet propriety, if $\forall (a,b) \in (\mathbb N^*)^2,\gcd(a,b)=1, a<b$, $\text{card}(\{ n \in\mathbb N, n \mod b=a  \} \cap D)=+\infty$.

Let $D=\{d_1,....,d_n,...\}$ with $\forall i \in \mathbb N^*,d_{i+1}>d_{i}$ subset of $\mathbb N$ and have the Dirichlet propriety.


1) Is it true that $\lim \dfrac{d_{n+1}}{d_n}=1$ ?

2) Is it true that $\liminf \dfrac{d_{n+1}}{d_n}=1$ ?
1 reply
Dattier
Yesterday at 1:23 PM
Dattier
an hour ago
orang NT
KevinYang2.71   22
N an hour ago by ray66
Source: ISL 2024 N1
Find all positive integers $n$ with the following property: for all positive divisors $d$ of $n$, we have $d+1\mid n$ or $d+1$ is prime.
22 replies
KevinYang2.71
Jul 16, 2025
ray66
an hour ago
Set of perfect powers is irreducible
Assassino9931   3
N 2 hours ago by Just1
Source: Al-Khwarizmi International Junior Olympiad 2025 P4
For two sets of integers $X$ and $Y$ we define $X\cdot Y$ as the set of all products of an element of $X$ and an element of $Y$. For example, if $X=\{1, 2, 4\}$ and $Y=\{3, 4, 6\}$ then $X\cdot Y=\{3, 4, 6, 8, 12, 16, 24\}.$ We call a set $S$ of positive integers good if there do not exist sets $A,B$ of positive integers, each with at least two elements and such that the sets $A\cdot B$ and $S$ are the same. Prove that the set of perfect powers greater than or equal to $2025$ is good.

(In any of the sets $A$, $B$, $A\cdot B$ no two elements are equal, but any two or three of these sets may have common elements. A perfect power is an integer of the form $n^k$, where $n>1$ and $k > 1$ are integers.)

Lajos Hajdu and Andras Sarkozy, Hungary
3 replies
Assassino9931
May 9, 2025
Just1
2 hours ago
shortlisted problems being used in undergraduate competition
enter16180   1
N 2 hours ago by Happycaptain
Hello, I am posting here to let know ( clarified after a post in College Math forum) that Problem 10 of Open Mathematical Olympiad for University Students ( OMOUS-2025) held at Ashgabat, Turkmenistan on 13-18 April, 2025 is found to be A6 Shortlisted Problems IMO-2024.
Following is discussion on College Math Forum
https://artofproblemsolving.com/community/c7h3551018_omous2025_team_competition_p10


Image of problem from competition for reference below.
1 reply
enter16180
3 hours ago
Happycaptain
2 hours ago
Help me prove these lemmas
dimi07   1
N 2 hours ago by blug
In the name of God, the most Merciful, the most Compassionate.
Let $a,b,c$ $\in$ $\mathbb{Z}$,then prove the following
\[
a\mid c, b\mid c \implies lcm(a,b)\mid c.
\]And also prove that
\[
c\mid a,c\mid b \implies c\mid gcd(a,b)
\]And by the help of God I finish this question.
1 reply
dimi07
5 hours ago
blug
2 hours ago
Equilateral pentagon with four right angles
Miquel-point   1
N 2 hours ago by loup blanc
Source: KoMaL B. 5396
An equilateral pentagon in the three-dimensional space has four right angles. What can be its fifth angle?

Proposed by Péter Dombi, Pécs
1 reply
Miquel-point
Jun 11, 2024
loup blanc
2 hours ago
2024 PMO Part II #2
orangefronted   3
N 2 hours ago by BinariouslyRandom
Determine all positive integers $k$ less than 2024 for which $4n+1$ and $kn+1$ are relatively prime for all integers $n$.
3 replies
orangefronted
Jan 16, 2024
BinariouslyRandom
2 hours ago
Right tetrahedron of fixed volume and min perimeter
Miquel-point   1
N 3 hours ago by Mathzeus1024
Source: Romanian IMO TST 1981, Day 4 P3
Determine the lengths of the edges of a right tetrahedron of volume $a^3$ so that the sum of its edges' lengths is minumum.

1 reply
Miquel-point
Apr 6, 2025
Mathzeus1024
3 hours ago
A nice and easy gem off of StackExchange
NamelyOrange   2
N May 3, 2025 by Royal_mhyasd
Source: https://math.stackexchange.com/questions/3818796/
Define $S$ as the set of all numbers of the form $2^i5^j$ for some nonnegative $i$ and $j$. Find (with proof) all pairs $(m,n)$ such that $m,n\in S$ and $m-n=1$.


Rephrased: Find (with proof) all solutions to $2^a5^b-2^c5^d=1$ over $(\mathbb{N}_0)^4$.
2 replies
NamelyOrange
May 2, 2025
Royal_mhyasd
May 3, 2025
A nice and easy gem off of StackExchange
G H J
G H BBookmark kLocked kLocked NReply
Source: https://math.stackexchange.com/questions/3818796/
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
NamelyOrange
566 posts
#1
Y by
Define $S$ as the set of all numbers of the form $2^i5^j$ for some nonnegative $i$ and $j$. Find (with proof) all pairs $(m,n)$ such that $m,n\in S$ and $m-n=1$.


Rephrased: Find (with proof) all solutions to $2^a5^b-2^c5^d=1$ over $(\mathbb{N}_0)^4$.
This post has been edited 3 times. Last edited by NamelyOrange, May 3, 2025, 8:45 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
NamelyOrange
566 posts
#2
Y by
Bump. I rephrased the question. It's easier than it looks :D
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Royal_mhyasd
134 posts
#3
Y by
Let's start by noticing that either a or c has to be 0. If this isn't true, then they're both >=1, so 2^a and 2^c are even, so 1 is even which is a contradiction. Let's analyze c=0 :
2^a * 5^b - 5^d = 1
We also notice that either b or d has to be 0, since otherwise we'd have 5^b is divisible by 5 and 5^d is divisible by 5, so 1 is divisible by 5 which is false
Let's analyze d=0 :
2^a * 5^b = 2. The only solution is evidently a=1 and b=0, so we have a=1, b=0, c=0, d=0.
If b=0 :
2^a = 1 + 5^d. a=0 yields no solution, for a=1 we have d=0 which we already found in the previous case, and for a>=2 we have 2^a == 0 (mod 4), but 5^d + 1 == 2 (mod 4), contradiction.
If a=0 :
5^b - 2^c * 5^d = 1. This time we obviously can't have b=0, so d=0.
5^b - 2^c = 1, 5^b = 2^c + 1.
If c=odd we have 2^c == -1 (mod 3), so 2^c + 1 == 0 (mod 3), so 5^b is divisible by 3, contradiction. Therefore we have c=even. Let's denote c = 2x, where x is a natural number.
5^b = 4^x + 1. b=1 obviously yields x=1, so c=2, so a=0, b=1, c=2, d=0.
For b>=2 :
If x was even, we'd have 4^x == (-1)^x == 1 (mod 5), so 4^x+1 == 2 (mod 5), contradiction. Therefore x is odd.
5^b = 4^x + 1^x, 5 | 4 + 1 and x is odd, (4;5)=1, (1;5)=1 means we can use L.T.E, so v5(5^b) = v5(4+1) + v5(x), so b = 1 + v5(x), so v5(x) = b - 1
x>= 5^(b-1)
Since b>=2 we have 5^(b-1) > 2b (easy to prove through induction) so x > 2b.
Therefore, 4^x > 4^2b = 16^b, so 4^x + 1 > 16^b + 1, 5^b > 16^b + 1, contradiction.

So S = { (1, 0, 0, 0), (0, 1, 2, 0)}
Z K Y
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