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k a August Highlights and 2025 AoPS Online Class Information
jwelsh   0
Friday at 2:14 PM
CONGRATULATIONS to all the competitors at this year’s International Mathematical Olympiad (IMO)! The US Team took second place with 5 gold medals and 1 silver - we are proud to say that each member of the 2025 IMO team has participated in an AoPS WOOT (Worldwide Online Olympiad Training) class!

"As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all."
— Doreen Dai, parent of IMO US Team Member Tiger Zhang

Interested to learn more about our WOOT programs? Check out the course page here or join a Free Scheduled Info Session. Early bird pricing ends August 19th!:
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There is still time to enroll in our last wave of summer camps that start in August at the Virtual Campus, our video-based platform, for math and language arts! From Math Beasts Camp 6 (Prealgebra Prep) to AMC 10/12 Prep, you can find an informative 2-week camp before school starts. Plus, our math camps don’t have homework and cover cool enrichment topics like graph theory. Our language arts courses will build the foundation for next year’s challenges, such as Language Arts Triathlon for levels 5-6 and Academic Essay Writing for high school students.

Lastly, Fall is right around the corner! 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. We’ve opened new Academy locations in San Mateo, CA, Pasadena, CA, Saratoga, CA, Johns Creek, GA, Northbrook, IL, and Upper West Side (NYC), New York.

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0 replies
jwelsh
Friday at 2:14 PM
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
hard 3 vars symetric
perfect_square   3
N a minute ago by Just1
Let $a,b,c \ge 0$ which satisfy:
$ \begin{cases}
    a+b+c=4 \\
    a^4+b^4+c^4 =18 
    \end{cases} $
Prove that: $ab+bc+ca \le 5$
3 replies
perfect_square
an hour ago
Just1
a minute ago
In a concert, 20 singers will perform
orl   34
N 6 minutes ago by ostriches88
Source: IMO Shortlist 2010, Combinatorics 1
In a concert, 20 singers will perform. For each singer, there is a (possibly empty) set of other singers such that he wishes to perform later than all the singers from that set. Can it happen that there are exactly 2010 orders of the singers such that all their wishes are satisfied?

Proposed by Gerhard Wöginger, Austria
34 replies
orl
Jul 17, 2011
ostriches88
6 minutes ago
Geometry Problem
Hopeooooo   14
N 32 minutes ago by Royal_mhyasd
Source: SRMC 2022 P1
Convex quadrilateral $ABCD$ is inscribed in circle $w.$Rays $AB$ and $DC$ intersect at $K.\ L$ is chosen on the diagonal $BD$ so that $\angle BAC= \angle DAL.\ M$ is chosen on the segment $KL$ so that $CM \mid\mid BD.$ Prove that line $BM$ touches $w.$
(Kungozhin M.)
14 replies
Hopeooooo
May 23, 2022
Royal_mhyasd
32 minutes ago
Police officers catching thief
JustPostNorthKoreaTST   0
an hour ago
Source: 2016 North Korea TST P6
There are $n$ stations in the sky, and any two stations are connected by exactly one bridge; every bridge has the same length and does not intersect with any other. $n$ police officers are standing at these stations (there may be more than one police officer at a station), and a thief is standing on a bridge. During a patrol, each police officer can choose to cross a bridge to move to the next station or remain in place, but at least one police officer must move. The thief can choose to move from one bridge to another if he is willing to. Assume that each police officer moves at the same speed and the thief moves at five times the speed of the police officers. If a thief and a police officer are in the same location at a certain moment, then the thief will be caught.
Given a positive integer $k$, show that for any initial position of the police officers, the thief has an initial position such that for any $k$ patrols, the thief has a strategy to avoid being caught during the $k$ patrols.
0 replies
JustPostNorthKoreaTST
an hour ago
0 replies
IMO ShortList 1998, combinatorics theory problem 7
orl   10
N an hour ago by legogubbe
Source: IMO ShortList 1998, combinatorics theory problem 7
A solitaire game is played on an $m\times n$ rectangular board, using $mn$ markers which are white on one side and black on the other. Initially, each square of the board contains a marker with its white side up, except for one corner square, which contains a marker with its black side up. In each move, one may take away one marker with its black side up, but must then turn over all markers which are in squares having an edge in common with the square of the removed marker. Determine all pairs $(m,n)$ of positive integers such that all markers can be removed from the board.
10 replies
orl
Oct 22, 2004
legogubbe
an hour ago
Inequality on distinct positive integers
JustPostNorthKoreaTST   0
an hour ago
Source: 2016 North Korea TST P5
Find the maximum possible value of $\lambda$, such that for any positive integer $n$ and distinct positive integers $k_1,k_2,\ldots,k_n$, we have
$$ \left(\sum_{i=1}^n \frac{1}{k_i}\right)\left(\sum_{i=1}^n \sqrt{k_i^6+k_i^3}\right)-\left(\sum_{i=1}^n k_i\right)^2 \ge \lambda n^2(n^2-1). $$
0 replies
JustPostNorthKoreaTST
an hour ago
0 replies
Perimeter bisectors
JustPostNorthKoreaTST   0
an hour ago
Source: 2016 North Korea TST P4
Given a triangle $ABC$, if the line connecting a point $X$ on a side of $\triangle ABC$ and its corresponding vertex bisects the perimeter, we write $(X \to \triangle ABC)$.

In a convex quadrilateral $ABCD$, let $X,Y,Z,M$ be points on sides $AB,AD,DC,CB$, respectively, such that $BM=CM$, $AX=AY$, $YD=DZ$, $ZC=BX$, and $(X \to \triangle ABM)$, $(Y \to \triangle AMD)$, $(Z \to \triangle CDM)$.

Prove that $\triangle ABM \cong \triangle MDA \cong \triangle DMC$.
0 replies
JustPostNorthKoreaTST
an hour ago
0 replies
Classic inequality in retrospect
JustPostNorthKoreaTST   0
an hour ago
Source: 2016 North Korea TST P3
Let $a_1,a_2,\ldots,a_n$ be positive real numbers, and denote $a_{n+1}=a_1$. Prove that
$$ \sum_{i=1}^n \frac{a_{i+1}}{a_i} \ge \sum_{i=1}^n \sqrt{\frac{a_{i+1}^2+1}{a_i^2+1}}. $$
0 replies
JustPostNorthKoreaTST
an hour ago
0 replies
Infinitely many prime divisors of a recurrence sequence
JustPostNorthKoreaTST   0
an hour ago
Source: 2016 North Korea TST P2
Given a sequence $\{a_n\}_{n \ge 1}$ of positive integers, such that $a_1 \in \mathbb{N}_+$, and for $n \ge 1$,
$$ a_{n+1}=\sum_{i=2}^{n+1} \lfloor \sqrt[i]{a_n} \rfloor. $$Show that for any prime $p$, there are infinitely many terms in $\{a_n\}_{n \ge 1}$ that are divisible by $p$.
0 replies
JustPostNorthKoreaTST
an hour ago
0 replies
Find sum over permutations
JustPostNorthKoreaTST   0
an hour ago
Source: 2016 North Korea TST P1
Given an odd positive integer $n$. Find the value of
$$ \sum_\pi \prod_{i=1}^n (\pi(i)-i), $$where $\{\pi(i)\}_{i=1}^n$ is a permutation of $\{1,2,\ldots,n\}$, and the summation runs over all such permutations.
0 replies
JustPostNorthKoreaTST
an hour ago
0 replies
Set family with special conditions
JustPostNorthKoreaTST   0
an hour ago
Source: 2015 North Korea Mathematical Olympiad P6
Let $S$ be a set with $n$ ($n \ge 3$) elements. Find all possible integers $n$ such that there exists a family $\mathcal{F}$ of three-element subsets of $S$, satisfying the following conditions:
(1) For any $a,b \in S$, $a \neq b$, exactly one element of $\mathcal{F}$ contains $\{a,b\}$.
(2) For any $a,b,c,x,y,z \in S$, if $\{a,b,z\}, \{b,c,x\}, \{c,a,y\} \in \mathcal{F}$, then $\{x,y,z\} \in \mathcal{F}$.
0 replies
JustPostNorthKoreaTST
an hour ago
0 replies
inequality
aktyw19   1
N 2 hours ago by Mathzeus1024
Let $x,y>0$ and $xy<1$. Prove $\left(\frac{2x}{1+x^{2}}\right)^{2}+\left(\frac{2y}{1+y^{2}}\right)^{2}\le\frac{1}{1-xy}$.
1 reply
aktyw19
Dec 19, 2012
Mathzeus1024
2 hours ago
Easy geometry with orthocenter
JustPostNorthKoreaTST   0
2 hours ago
Source: 2015 North Korea Mathematical Olympiad P4
Let $ABC$ be a scalene triangle with circumcircle $\odot O$ and orthocenter $H$, and let $AH,BH,CH$ intersects $\odot O$ at $A_1,B_1,C_1$, respectively. The line passing through $A_1$ and parallel to $BC$ intersects $\odot O$ at $A_2$. Define $B_2,C_2$ similarly. Let $AC_2$ intersects $BC_1$ at $M$, $BA_2$ intersects $CA_1$ at $N$, and $CB_2$ intersects $AB_1$ at $P$. Prove that $\angle MNB=\angle AMP$.
0 replies
JustPostNorthKoreaTST
2 hours ago
0 replies
Tricolor complete graphs
JustPostNorthKoreaTST   0
2 hours ago
Source: 2015 North Korea Mathematical Olympiad P3
Consider a complete graph $K_n$ on $n$ vertices, where $n \ge 3$. Each edge is colored with one of three colors, and each color is used on at least one edge. Find the minimum positive integer $k$ such that for any such edge coloring and any color $C$ chosen from the three colors, it is possible to recolor at most $k$ edges to color $C$ so that the subgraph consisting of all edges of color $C$ is connected.
0 replies
JustPostNorthKoreaTST
2 hours ago
0 replies
partitioning 1 to p-1 into several a+b=c (mod p)
capoouo   5
N May 28, 2025 by NerdyNashville
Source: own
Given a prime number $p$, a set is said to be $p$-good if the set contains exactly three elements $a, b, c$ and $a + b \equiv c \pmod{p}$.
Find all prime number $p$ such that $\{ 1, 2, \cdots, p-1 \}$ can be partitioned into several $p$-good sets.

Proposed by capoouo
5 replies
capoouo
Apr 21, 2024
NerdyNashville
May 28, 2025
partitioning 1 to p-1 into several a+b=c (mod p)
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G H BBookmark kLocked kLocked NReply
Source: own
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capoouo
1 post
#1
Y by
Given a prime number $p$, a set is said to be $p$-good if the set contains exactly three elements $a, b, c$ and $a + b \equiv c \pmod{p}$.
Find all prime number $p$ such that $\{ 1, 2, \cdots, p-1 \}$ can be partitioned into several $p$-good sets.

Proposed by capoouo
This post has been edited 1 time. Last edited by capoouo, Apr 21, 2024, 5:42 PM
Reason: remove awkward line-breaks; add "proposed by ..."
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Acclab
33 posts
#3 • 1 Y
Y by sami1618
For storage

The answer is all $3 | p-1$. Let $1, ..., p-1$ rewrite into $e, g, g^2, ..., g^{3k-1}$. We can verify that $e + g^k + g^{2k} = 0$ as $g^{3k} = e$, giving $g^{c} + g^{k+c} = g^{-2k-c} = g^{k-c}$. As $c$ runs through $0, 1, ..., k-1$ we achieve $k$ $p-good$ sets. The contrary is obvious.
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DTforever
6 posts
#4
Y by
11 also satifsfies with sets $(1,3,9)$, $(2,4,6)$,$(5,7,8,10)$ can someone post a correct solution?
This post has been edited 1 time. Last edited by DTforever, Apr 8, 2025, 10:30 AM
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Quidditch
818 posts
#5
Y by
DTforever wrote:
11 also satifsfies with sets $(1,3,9)$, $(2,4,6)$,$(5,7,8,10)$ can someone post a correct solution?

the set has to contain exactly three elements
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NerdyNashville
19 posts
#6
Y by
Acclab wrote:
For storage

The answer is all $3 | p-1$. Let $1, ..., p-1$ rewrite into $e, g, g^2, ..., g^{3k-1}$. We can verify that $e + g^k + g^{2k} = 0$ as $g^{3k} = e$, giving $g^{c} + g^{k+c} = g^{-2k-c} = g^{k-c}$. As $c$ runs through $0, 1, ..., k-1$ we achieve $k$ $p-good$ sets. The contrary is obvious.

Hmmm I am little confused about ur solution,
How u got $g^c + g^{k+c} \equiv g^{-2k-c} \pmod{p}$?
I believe it should be $g^c + g^{k+c} \equiv -g^{2k+c} \equiv g^{k/2+c} \pmod{p}$
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NerdyNashville
19 posts
#7
Y by
Solution
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N Quick Reply
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