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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.

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0 replies
jlacosta
May 1, 2025
0 replies
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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
J
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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
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Orthocenter lies on circumcircle
whatshisbucket   89
N 15 minutes ago by Mathandski
Source: 2017 ELMO #2
Let $ABC$ be a triangle with orthocenter $H,$ and let $M$ be the midpoint of $\overline{BC}.$ Suppose that $P$ and $Q$ are distinct points on the circle with diameter $\overline{AH},$ different from $A,$ such that $M$ lies on line $PQ.$ Prove that the orthocenter of $\triangle APQ$ lies on the circumcircle of $\triangle ABC.$

Proposed by Michael Ren
89 replies
whatshisbucket
Jun 26, 2017
Mathandski
15 minutes ago
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Hard math inequality
noneofyou34   5
N 23 minutes ago by JARP091
If a,b,c are positive real numbers, such that a+b+c=1. Prove that:
(b+c)(a+c)/(a+b)+ (b+a)(a+c)/(c+b)+(b+c)(a+b)/(a+c)>= Sqrt.(6(a(a+c)+b(a+b)+c(b+c)) +3
5 replies
noneofyou34
Sunday at 2:00 PM
JARP091
23 minutes ago
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Interesting inequalities
sqing   0
26 minutes ago
Source: Own
Let $ a,b>0 $. Prove that
$$\frac{ab-1} {ab(a+b+2)} \leq \frac{1} {8}$$$$\frac{2ab-1} {ab(a+b+1)} \leq 6\sqrt 3-10$$
0 replies
sqing
26 minutes ago
0 replies
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Inspired by SXJX (12)2022 Q1167
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b,c>0 $. Prove that$$\frac{kabc-1} {abc(a+b+c+8(2k-1))}\leq \frac{1}{16 }$$Where $ k>\frac{1}{2}.$
1 reply
sqing
Yesterday at 4:01 AM
sqing
an hour ago
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Algebra manipulation excercise
Marinchoo   3
N an hour ago by compoly2010
Source: 2007 Bulgarian Autumn Math Competition, Problem 9.2
Let $a$, $b$, $c$ be real numbers, such that $a+b+c=0$ and $a^4+b^4+c^4=50$. Determine the value of $ab+bc+ca$.
3 replies
Marinchoo
Mar 17, 2022
compoly2010
an hour ago
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Numbers on a circle
navi_09220114   2
N an hour ago by ja.
Source: TASIMO 2025 Day 1 Problem 1
For a given positive integer $n$, determine the smallest integer $k$, such that it is possible to place numbers $1,2,3,\dots, 2n$ around a circle so that the sum of every $n$ consecutive numbers takes one of at most $k$ values.
2 replies
navi_09220114
Yesterday at 11:35 AM
ja.
an hour ago
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Inspired by 2007 Bulgarian
sqing   0
2 hours ago
Source: Own
Let $a$, $b$, $c$ be real numbers such that $a+b+c=0$ and $a^2+b^2+c^2+a^4+b^4+c^4=2$. Prove that $$ab+bc+ca=\frac{1-\sqrt 5}{2}$$Let $a$, $b$, $c$ be real numbers such that $a+b+c=0$ and $ab+bc+ca+a^2+b^2+c^2+a^4+b^4+c^4=2$. Prove that $$ab+bc+ca=\frac{1-\sqrt{17}}{4}$$
0 replies
sqing
2 hours ago
0 replies
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Gives typical russian combinatorics vibes
Sadigly   4
N 3 hours ago by lbd4203
Source: Azerbaijan Senior MO 2025 P3
You are given a positive integer $n$. $n^2$ amount of people stand on coordinates $(x;y)$ where $x,y\in\{0;1;2;...;n-1\}$. Every person got a water cup and two people are considered to be neighbour if the distance between them is $1$. At the first minute, the person standing on coordinates $(0;0)$ got $1$ litres of water, and the other $n^2-1$ people's water cup is empty. Every minute, two neighbouring people are chosen that does not have the same amount of water in their water cups, and they equalize the amount of water in their water cups.

Prove that, no matter what, the person standing on the coordinates $(x;y)$ will not have more than $\frac1{x+y+1}$ litres of water.
4 replies
Sadigly
May 8, 2025
lbd4203
3 hours ago
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Product of Sum
shobber   4
N 3 hours ago by alexanderchew
Source: CGMO 2006
Given that $x_{i}>0$, $i = 1, 2, \cdots, n$, $k \geq 1$. Show that: \[\sum_{i=1}^{n}\frac{1}{1+x_{i}}\cdot \sum_{i=1}^{n}x_{i}\leq \sum_{i=1}^{n}\frac{x_{i}^{k+1}}{1+x_{i}}\cdot \sum_{i=1}^{n}\frac{1}{x_{i}^{k}}\]
4 replies
shobber
Aug 9, 2006
alexanderchew
3 hours ago
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Prove that two different boards can be obtained
hectorleo123   1
N 4 hours ago by Joalro178
Source: 2014 Peru Ibero TST P2
Let $n\ge 4$ be an integer. You have two $n\times n$ boards. Each board contains the numbers $1$ to $n^2$ inclusive, one number per square, arbitrarily arranged on each board. A move consists of exchanging two rows or two columns on the first board (no moves can be made on the second board). Show that it is possible to make a sequence of moves such that for all $1 \le i \le n$ and $1 \le j \le n$, the number that is in the $i-th$ row and $j-th$ column of the first board is different from the number that is in the $i-th$ row and $j-th$ column of the second board.
1 reply
hectorleo123
Sep 15, 2023
Joalro178
4 hours ago
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Italian WinterCamps test07 Problem4
mattilgale   90
N 4 hours ago by mathwiz_1207
Source: ISL 2006, G3, VAIMO 2007/5
Let $ ABCDE$ be a convex pentagon such that
\[ \angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE. \]The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$.

Proposed by Zuming Feng, USA
90 replies
1 viewing
mattilgale
Jan 29, 2007
mathwiz_1207
4 hours ago
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Iran TST P8
TheBarioBario   8
N 5 hours ago by Mysteriouxxx
Source: Iranian TST 2022 problem 8
In triangle $ABC$, with $AB<AC$, $I$ is the incenter, $E$ is the intersection of $A$-excircle and $BC$. Point $F$ lies on the external angle bisector of $BAC$ such that $E$ and $F$ lieas on the same side of the line $AI$ and $\angle AIF=\angle AEB$. Point $Q$ lies on $BC$ such that $\angle AIQ=90$. Circle $\omega_b$ is tangent to $FQ$ and $AB$ at $B$, circle $\omega_c$ is tangent to $FQ$ and $AC$ at $C$ and both circles pass through the inside of triangle $ABC$. if $M$ is the Midpoint od the arc $BC$, which does not contain $A$, prove that $M$ lies on the radical axis of $\omega_b$ and $\omega_c$.

Proposed by Amirmahdi Mohseni
8 replies
TheBarioBario
Apr 2, 2022
Mysteriouxxx
5 hours ago
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IMO 2010 Problem 6
mavropnevma   42
N 5 hours ago by awesomeming327.
Let $a_1, a_2, a_3, \ldots$ be a sequence of positive real numbers, and $s$ be a positive integer, such that
\[a_n = \max \{ a_k + a_{n-k} \mid 1 \leq k \leq n-1 \} \ \textrm{ for all } \ n > s.\]
Prove there exist positive integers $\ell \leq s$ and $N$, such that
\[a_n = a_{\ell} + a_{n - \ell} \ \textrm{ for all } \ n \geq N.\]

Proposed by Morteza Saghafiyan, Iran
42 replies
mavropnevma
Jul 8, 2010
awesomeming327.
5 hours ago
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PJ // AC iff BC^2 = AC· QC
parmenides51   1
N 6 hours ago by FrancoGiosefAG
Source: Mexican Mathematical Olympiad 1998 OMM P5
The tangents at points $B$ and $C$ on a given circle meet at point $A$. Let $Q$ be a point on segment $AC$ and let $BQ$ meet the circle again at $P$. The line through $Q $ parallel to $AB$ intersects $BC$ at $J$. Prove that $PJ$ is parallel to $AC$ if and only if $BC^2 = AC\cdot QC$.
1 reply
parmenides51
Jul 28, 2018
FrancoGiosefAG
6 hours ago
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Number Theory
fasttrust_12-mn   5
N Apr 23, 2025 by GreekIdiot
Source: Pan African Mathematics Olympiad p6
Find all integers $n$ for which $n^7-41$ is the square of an integer
5 replies
fasttrust_12-mn
Aug 16, 2024
GreekIdiot
Apr 23, 2025
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Number Theory
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Reveal topic tags
number theory
PAMO 2024
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G H BBookmark kLocked kLocked NReply
Source: Pan African Mathematics Olympiad p6
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fasttrust_12-mn
118 posts
fasttrust_12-mn
#1 h  Aug 16, 2024, 10:21 AM • 1 Y
Y by cheikhennahoui
Find all integers $n$ for which $n^7-41$ is the square of an integer
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a_507_bc
678 posts
a_507_bc
#2 h  Aug 16, 2024, 10:43 AM • 2 Y
Y by cheikhennahoui, Monica_Twigg_239
We can rewrite it as $n^7+2^7=m^2+13^2$ and use the approach from USA TST 2008 P4.
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DylanN
194 posts
DylanN
#4 h  Aug 16, 2024, 2:42 PM • 1 Y
Y by Monica_Twigg_239
I personally probably would have waited until the paper was finished being written before posting the problems online.

To fill in the details for the other response, we note that $m^2 + 13^2$ is a sum of two squares, and so the only possible primes factors are $2$, prime factors of $\gcd(m, 13)$, and primes that are congruent to $1$ modulo $4$. If $2$ were one of the prime factors, then $n$ would be even. But then we would have that $n^7 - 41 \equiv 3 \pmod 4$, and so it would not be a square. Thus every prime factor of $n^7 + 2^7$ is congruent to $1$ modulo $4$. (Since $13$ is itself congruent to $1$ modulo $4$.) Since $n + 2$ is a factor of $n^7 + 2^7$, this implies that every prime factor of $n + 2$ is congruent to $1$ modulo $4$, and so $n + 2 \equiv 1 \pmod 4 \implies n \equiv -1 \pmod 4$. But then $n^7 + 2^7 \equiv (-1)^7 \equiv 3 \pmod 4$, and so $n^7 + 2^7$ would have a prime factor congruent to $3$ modulo $4$, a contradiction.
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Mrcuberoot
28 posts
Mrcuberoot
#5 h  Aug 16, 2024, 5:32 PM • 1 Y
Y by Monica_Twigg_239
yeah same old trick as USATST 2008 p4
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Rayanelba
20 posts
Rayanelba
#6 h  Apr 23, 2025, 10:31 PM • 1 Y
Y by ATM_
Motivation (USATST p4 trick:$n^7-7=m^2$ :)
Notice that : $41=13^2-2^7$
So : $n^7+2^7=13^2$
And : $n+2|n^7+2^7\implies n+2|m^2+13^2$
By Fermat's Chrismats theoreme : $n+2\equiv 1[4]\implies n\equiv -1[4]\implies n^7+2^7\equiv -1[4]\implies m^2+13^2\equiv -1[4]$
Hence : $m^2\equiv 2[4]$
Which is impossible because quadratique residus mod 4 are 0,1
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GreekIdiot
247 posts
GreekIdiot
#7 h  Apr 23, 2025, 10:33 PM
Y by
known trick, like BMO 1998
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