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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
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0 replies
jlacosta
Mar 2, 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
IZHO 2017 Functional equations
user01   51
N 18 minutes ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
user01
Jan 14, 2017
lksb
18 minutes ago
chat gpt
fuv870   2
N 20 minutes ago by fuv870
The chat gpt alreadly knows how to solve the problem of IMO USAMO and AMC?
2 replies
1 viewing
fuv870
32 minutes ago
fuv870
20 minutes ago
Inequality with wx + xy + yz + zw = 1
Fermat -Euler   23
N 22 minutes ago by hgomamogh
Source: IMO ShortList 1990, Problem 24 (THA 2)
Let $ w, x, y, z$ are non-negative reals such that $ wx + xy + yz + zw = 1$.
Show that $ \frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$.
23 replies
1 viewing
Fermat -Euler
Nov 2, 2005
hgomamogh
22 minutes ago
Waiting for a dm saying me again "old geometry"
drmzjoseph   0
an hour ago
Source: Idk easy
Given $ABCD$ a tangencial quadrilateral that is not a rhombus, let $a,b,c,d$ be lengths of tangents from $A,B,C,D$ to the incircle of the quadrilateral which center is $I$. Let $M,N$ be the midpoints of $AC,BD$ resp. Prove that
\[ \frac{MI}{IN}=\frac{a+c}{b+d} \]
0 replies
drmzjoseph
an hour ago
0 replies
Finally hard NT on UKR MO from NT master
mshtand1   2
N an hour ago by IAmTheHazard
Source: Ukrainian Mathematical Olympiad 2025. Day 1, Problem 11.4
A pair of positive integer numbers \((a, b)\) is given. It turns out that for every positive integer number \(n\), for which the numbers \((n - a)(n + b)\) and \(n^2 - ab\) are positive, they have the same number of divisors. Is it necessarily true that \(a = b\)?

Proposed by Oleksii Masalitin
2 replies
mshtand1
Mar 13, 2025
IAmTheHazard
an hour ago
IMOC 2017 G5 (<A=120 => E, F, Y,Z are concyclic, incenter related)
parmenides51   4
N an hour ago by ehuseyinyigit
Source: https://artofproblemsolving.com/community/c6h1740077p11309077
We have $\vartriangle ABC$ with $I$ as its incenter. Let $D$ be the intersection of $AI$ and $BC$ and define $E, F$ in a similar way. Furthermore, let $Y = CI \cap DE, Z = BI \cap DF$. Prove that if $\angle BAC = 120^o$, then $E, F, Y,Z$ are concyclic.
IMAGE
4 replies
parmenides51
Mar 20, 2020
ehuseyinyigit
an hour ago
Bosnia and Herzegovina JBMO TST 2013 Problem 1
gobathegreat   3
N 2 hours ago by DensSv
Source: Bosnia and Herzegovina Junior Balkan Mathematical Olympiad TST 2013
It is given $n$ positive integers. Product of any one of them with sum of remaining numbers increased by $1$ is divisible with sum of all $n$ numbers. Prove that sum of squares of all $n$ numbers is divisible with sum of all $n$ numbers
3 replies
gobathegreat
Sep 16, 2018
DensSv
2 hours ago
D1015 : A strange EF for polynomials
Dattier   0
2 hours ago
Source: les dattes à Dattier
Find all $P \in \mathbb R[x,y]$ with $P \not\in \mathbb R[x] \cup \mathbb R[y]$ and $\forall g,f$ homeomorphismes of $\mathbb R$, $P(f,g)$ is an homoemorphisme too.
0 replies
1 viewing
Dattier
2 hours ago
0 replies
P, Q,R collinear and U, R, O, V concyclic wanted, cyclic ABCD, circumcenters
parmenides51   2
N 2 hours ago by DensSv
Source: 2012 Romania JBMO TST2 P4
The quadrilateral $ABCD$ is inscribed in a circle centered at $O$, and $\{P\} = AC \cap BD, \{Q\} = AB \cap CD$. Let $R$ be the second intersection point of the circumcircles of the triangles $ABP$ and $CDP$.
a) Prove that the points $P, Q$, and $R$ are collinear.
b) If $U$ and $V$ are the circumcenters of the triangles $ABP$, and $CDP$, respectively, prove that the points $U, R, O, V$ are concyclic.
2 replies
parmenides51
May 29, 2020
DensSv
2 hours ago
Unsolved Diophantine(I think)
Nuran2010   1
N 2 hours ago by Nuran2010
Find all solutions for the equation $2^n=p+3^p$ where $n$ is a positive integer and $p$ is a prime.(Don't get mad at me,I've used the search function and did not see a correct and complete solution anywhere.)
1 reply
Nuran2010
Mar 14, 2025
Nuran2010
2 hours ago
2^a + 3^b + 1 = 6^c
togrulhamidli2011   1
N 2 hours ago by CM1910
Find all positive integers (a, b, c) such that:

\[
2^a + 3^b + 1 = 6^c
\]
1 reply
togrulhamidli2011
Today at 12:34 PM
CM1910
2 hours ago
Prove that OA and RA are perpendicular
MellowMelon   90
N 3 hours ago by ehuseyinyigit
Source: USA TSTST 2011/2012 P4
Acute triangle $ABC$ is inscribed in circle $\omega$. Let $H$ and $O$ denote its orthocenter and circumcenter, respectively. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively. Rays $MH$ and $NH$ meet $\omega$ at $P$ and $Q$, respectively. Lines $MN$ and $PQ$ meet at $R$. Prove that $OA\perp RA$.
90 replies
MellowMelon
Jul 26, 2011
ehuseyinyigit
3 hours ago
find all f
mecrazywong   8
N 3 hours ago by HamstPan38825
Source: Chinese team training 2004
Find all $f:\mathbb N\rightarrow\mathbb Z$ satisfying both of the following properties:
(1)If a,b are positive integers, then $f(ab)+f(a^2+b^2)=f(a)+f(b)$.
(2)If a,b are positive integers and a|b, then $f(a)\ge f(b)$.
8 replies
mecrazywong
Feb 16, 2005
HamstPan38825
3 hours ago
Sequence 1994
Jan   5
N 3 hours ago by AshAuktober
Source: IMO Shortlist 1994, A1
Let $ a_{0} = 1994$ and $ a_{n + 1} = \frac {a_{n}^{2}}{a_{n} + 1}$ for each nonnegative integer $ n$. Prove that $ 1994 - n$ is the greatest integer less than or equal to $ a_{n}$, $ 0 \leq n \leq 998$
5 replies
Jan
Dec 26, 2006
AshAuktober
3 hours ago
n is prime if for every divisor d of n, d + 1 is a divisor of n + 1
parmenides51   5
N Jun 17, 2022 by megarnie
Source: Dutch NMO 2021 p5
We consider an integer $n > 1$ with the following property: for every positive divisor $d$ of $n$ we have that $d + 1$ is a divisor of$ n + 1$. Prove that $n$ is a prime number.
5 replies
parmenides51
Dec 28, 2021
megarnie
Jun 17, 2022
n is prime if for every divisor d of n, d + 1 is a divisor of n + 1
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G H BBookmark kLocked kLocked NReply
Source: Dutch NMO 2021 p5
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parmenides51
30628 posts
#1
Y by
We consider an integer $n > 1$ with the following property: for every positive divisor $d$ of $n$ we have that $d + 1$ is a divisor of$ n + 1$. Prove that $n$ is a prime number.
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cadaeibf
700 posts
#2
Y by
Let $p$ be the smallest prime dividing $n$ so that $n=pk$. We must have $k|pk\implies k+1|pk+1\implies k+1|pk+1-pk-p=1-p\implies k+1|p-1\implies k\leq p-2$. However if $k>1$ there must be at least a prime divisor of $k$ which is $\geq p$, and so $k\geq p>p-2\geq k$ which is a contradiction. If instead $k=1$, we are done.
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Tintarn
9022 posts
#3 • 2 Y
Y by parmenides51, GuoGuoBellevue
parmenides51 wrote:
We consider an integer $n > 1$ with the following property: for every positive divisor $d$ of $n$ we have that $d + 1$ is a divisor of$ n + 1$. Prove that $n$ is a prime number.
Had you used the search function, you would have found that this problem was already used in Bay Area 2004 and the problem was already posted on this forum two years ago here, indeed by a certain guy called parmenides51.
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parmenides51
30628 posts
#4
Y by
indeed, at least now we got a solution

(it was posted in Bay Area forum so a search within forum HSO, might not have found that)
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Tintarn
9022 posts
#5 • 1 Y
Y by parmenides51
Well, of course the problem has also been posted on HSO several times before (it is really a classic), and of course also with a solution. See e.g. here or here (Romanian Star of Mathematics 2019).
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megarnie
5531 posts
#6
Y by
Suppose not and let $p$ be the smallest prime divisor of $n=pk$, where $k>1$.

If $n$ satisfy the problem conditions then $k+1\mid pk+1\implies k+1\mid pk+1-p(k+1)=1-p$, so $k+1\mid p-1$.

Since $k+1>0$ and $p-1>0$, we have $k+1\le p-1\implies k<p$.

However, since $n$ isn't prime, $k>1$, so $k$ must have a prime divisor. However, this prime factor is less than $p$ which contradicts the fact that $p$ is the smallest prime divisor of $n$.
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