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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.

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[*]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
1996 St. Petersburg City Mathematical Olympiad
Sadece_Threv   2
N a minute ago by reni_wee
Source: 1996 St. Petersburg City Mathematical Olympiad
Find all positive integers $n$ such that $3^{n-1}+5^{n-1}$ divides $3^{n}+5^{n}$
2 replies
Sadece_Threv
Jul 29, 2024
reni_wee
a minute ago
IMO 2010 Problem 5
mavropnevma   55
N 13 minutes ago by maromex
Each of the six boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$, $B_6$ initially contains one coin. The following operations are allowed

Type 1) Choose a non-empty box $B_j$, $1\leq j \leq 5$, remove one coin from $B_j$ and add two coins to $B_{j+1}$;

Type 2) Choose a non-empty box $B_k$, $1\leq k \leq 4$, remove one coin from $B_k$ and swap the contents (maybe empty) of the boxes $B_{k+1}$ and $B_{k+2}$.

Determine if there exists a finite sequence of operations of the allowed types, such that the five boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ become empty, while box $B_6$ contains exactly $2010^{2010^{2010}}$ coins.

Proposed by Hans Zantema, Netherlands
55 replies
mavropnevma
Jul 8, 2010
maromex
13 minutes ago
NT ineq: sum 1/a_i < (m+n)/m , {a_1,a_2,...,a_n} subset of {1,2,...,m}
parmenides51   1
N 31 minutes ago by DVDTSB
Source: 2006 MOP Homework Blue NT 6
Let $m$ and $n$ be positive integers with $m > n \ge 2$. Set $S =\{1,2,...,m\}$, and set $T = \{a_1,a_2,...,a_n\}$ is a subset of $S$ such that every element of $S$ is not divisible by any pair of distinct elements of $T$. Prove that
$$\frac{1}{a_1}+\frac{1}{a_2}+ ...+ \frac{1}{a_n} < \frac{m+n}{m}$$
1 reply
parmenides51
Apr 12, 2020
DVDTSB
31 minutes ago
Never 8
chess64   21
N an hour ago by reni_wee
Source: Canada 1970, Problem 10
Given the polynomial \[ f(x)=x^n+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots+a_{n-1}x+a_n \] with integer coefficients $a_1,a_2,\ldots,a_n$, and given also that there exist four distinct integers $a$, $b$, $c$ and $d$ such that \[ f(a)=f(b)=f(c)=f(d)=5, \] show that there is no integer $k$ such that $f(k)=8$.
21 replies
chess64
May 14, 2006
reni_wee
an hour ago
old and easy imo inequality
Valentin Vornicu   215
N 2 hours ago by cubres
Source: IMO 2000, Problem 2, IMO Shortlist 2000, A1
Let $ a, b, c$ be positive real numbers so that $ abc = 1$. Prove that
\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1.
\]
215 replies
Valentin Vornicu
Oct 24, 2005
cubres
2 hours ago
x^2-x divides by n for some n/\omega(n)+1>x>1
NO_SQUARES   1
N 2 hours ago by a_507_bc
Source: 239 MO 2025 8-9 p6
Let a positive integer number $n$ has $k$ different prime divisors. Prove that there exists a positive integer number $x \in \left(1, \frac{n}{k}+1 \right)$ such that $x^2-x$ divides by $n$.
1 reply
NO_SQUARES
4 hours ago
a_507_bc
2 hours ago
IMO Genre Predictions
ohiorizzler1434   46
N 2 hours ago by Mrcuberoot
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
46 replies
ohiorizzler1434
May 3, 2025
Mrcuberoot
2 hours ago
4 wise men and 100 hats. 3 must guess their numbers
NO_SQUARES   1
N 3 hours ago by noemiemath
Source: 239 MO 2025 10-11 p5
There are four wise men in a row, each sees only those following him in the row, i.e. the $1$st sees the other three, the $2$nd sees the $3$rd and $4$th, and the $3$rd sees only the $4$th. The devil has $100$ hats, numbered from $1$ to $100$, he puts one hat on each wise man, and hides the extra $96$ hats. After that, each wise man (in turn: first the first, then the second, etc.) loudly calls a number, trying to guess the number of his hat. The numbers mentioned should not be repeated. When all the wise men have spoken, they take off their hats and check which one of them has guessed. Can the sages to act in such a way that at least three of them knowingly guessed?
1 reply
NO_SQUARES
3 hours ago
noemiemath
3 hours ago
IMO Shortlist 2011, G4
WakeUp   126
N 3 hours ago by NuMBeRaToRiC
Source: IMO Shortlist 2011, G4
Let $ABC$ be an acute triangle with circumcircle $\Omega$. Let $B_0$ be the midpoint of $AC$ and let $C_0$ be the midpoint of $AB$. Let $D$ be the foot of the altitude from $A$ and let $G$ be the centroid of the triangle $ABC$. Let $\omega$ be a circle through $B_0$ and $C_0$ that is tangent to the circle $\Omega$ at a point $X\not= A$. Prove that the points $D,G$ and $X$ are collinear.

Proposed by Ismail Isaev and Mikhail Isaev, Russia
126 replies
WakeUp
Jul 13, 2012
NuMBeRaToRiC
3 hours ago
<DPA+ <AQD =< QIP wanted, incircle circumcircle related
parmenides51   42
N 3 hours ago by AR17296174
Source: IMo 2019 SL G6
Let $I$ be the incentre of acute-angled triangle $ABC$. Let the incircle meet $BC, CA$, and $AB$ at $D, E$, and $F,$ respectively. Let line $EF$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\angle DPA + \angle AQD =\angle QIP$.

(Slovakia)
42 replies
1 viewing
parmenides51
Sep 22, 2020
AR17296174
3 hours ago
Help my diagram has too many points
MarkBcc168   28
N 3 hours ago by AR17296174
Source: IMO Shortlist 2023 G6
Let $ABC$ be an acute-angled triangle with circumcircle $\omega$. A circle $\Gamma$ is internally tangent to $\omega$ at $A$ and also tangent to $BC$ at $D$. Let $AB$ and $AC$ intersect $\Gamma$ at $P$ and $Q$ respectively. Let $M$ and $N$ be points on line $BC$ such that $B$ is the midpoint of $DM$ and $C$ is the midpoint of $DN$. Lines $MP$ and $NQ$ meet at $K$ and intersect $\Gamma$ again at $I$ and $J$ respectively. The ray $KA$ meets the circumcircle of triangle $IJK$ again at $X\neq K$.

Prove that $\angle BXP = \angle CXQ$.

Kian Moshiri, United Kingdom
28 replies
MarkBcc168
Jul 17, 2024
AR17296174
3 hours ago
A lot of circles
ryan17   8
N 3 hours ago by AR17296174
Source: 2019 Polish MO Finals
Denote by $\Omega$ the circumcircle of the acute triangle $ABC$. Point $D$ is the midpoint of the arc $BC$ of $\Omega$ not containing $A$. Circle $\omega$ centered at $D$ is tangent to the segment $BC$ at point $E$. Tangents to the circle $\omega$ passing through point $A$ intersect line $BC$ at points $K$ and $L$ such that points $B, K, L, C$ lie on the line $BC$ in that order. Circle $\gamma_1$ is tangent to the segments $AL$ and $BL$ and to the circle $\Omega$ at point $M$. Circle $\gamma_2$ is tangent to the segments $AK$ and $CK$ and to the circle $\Omega$ at point $N$. Lines $KN$ and $LM$ intersect at point $P$. Prove that $\sphericalangle KAP = \sphericalangle EAL$.
8 replies
ryan17
Jul 9, 2019
AR17296174
3 hours ago
NT FE from Taiwan TST
Kitayama_Yuji   13
N 3 hours ago by bin_sherlo
Source: 2024 Taiwan TST Round 2 Mock P3
Let $\mathbb{N}$ be the set of all positive integers. Find all functions $f\colon \mathbb{N}\to \mathbb{N}$ such that $mf(m)+(f(f(m))+n)^2$ divides $4m^4+n^2f(f(n))^2$ for all positive integers $m$ and $n$.
13 replies
Kitayama_Yuji
Mar 29, 2024
bin_sherlo
3 hours ago
Yet another domino problem
juckter   15
N 3 hours ago by lksb
Source: EGMO 2019 Problem 2
Let $n$ be a positive integer. Dominoes are placed on a $2n \times 2n$ board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each $n$, determine the largest number of dominoes that can be placed in this way.
(A domino is a tile of size $2 \times 1$ or $1 \times 2$. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)
15 replies
juckter
Apr 9, 2019
lksb
3 hours ago
n-gon function
ehsan2004   10
N Apr 4, 2025 by Zany9998
Source: Romanian IMO Team Selection Test TST 1996, problem 1
Let $ f: \mathbb{R}^2 \rightarrow \mathbb{R} $ be a function such that for every regular $ n $-gon $ A_1A_2 \ldots A_n $ we have $ f(A_1)+f(A_2)+\cdots +f(A_n)=0 $. Prove that $ f(x)=0 $ for all reals $ x $.
10 replies
ehsan2004
Sep 13, 2005
Zany9998
Apr 4, 2025
n-gon function
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G H BBookmark kLocked kLocked NReply
Source: Romanian IMO Team Selection Test TST 1996, problem 1
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ehsan2004
2238 posts
#1 • 3 Y
Y by Adventure10, centslordm, Mango247
Let $ f: \mathbb{R}^2 \rightarrow \mathbb{R} $ be a function such that for every regular $ n $-gon $ A_1A_2 \ldots A_n $ we have $ f(A_1)+f(A_2)+\cdots +f(A_n)=0 $. Prove that $ f(x)=0 $ for all reals $ x $.
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perfect_radio
2607 posts
#2 • 2 Y
Y by Adventure10, Mango247
ehsan2004 wrote:
Prove that $f(x)=0$ for all reals $x$.

You wanted to say "$f(A)=0$ for all points $A$", right?
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ehsan2004
2238 posts
#3 • 2 Y
Y by Adventure10, Mango247
perfect_radio wrote:
ehsan2004 wrote:
Prove that $f(x)=0$ for all reals $x$.

You wanted to say "$f(A)=0$ for all points $A$", right?

excuse me, my meant was $f(x)\equiv 0$
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perfect_radio
2607 posts
#4 • 2 Y
Y by Adventure10, Mango247
Take $A \neq B$. Let $\ell$ be the perpendicular bisector of $AB$. Construct a rhombus $ACBD$, with $C,D \in \ell$ and $\measuredangle DAC = \measuredangle DBC = \dfrac{\pi}{3}$. This yields $f(A)+f(C)+f(D)=0=f(B)+f(C)+f(D)$, so $f(A)=f(B)=t$, $\forall A \neq B$.

Therefore $nt=0$, so $t=0$.

Have I done something wrong? It looks too good to be true :blush: (because I used the condition given only for $n=3$)
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enescu
741 posts
#5 • 2 Y
Y by Adventure10, Mango247
Actually, $n$ is fixed in the original statement.
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perfect_radio
2607 posts
#6 • 2 Y
Y by Adventure10, Mango247
enescu wrote:
Actually, $n$ is fixed in the original statement.
Oops... sorry :( . do you know the solution for $n \geq 4$?
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enescu
741 posts
#7 • 7 Y
Y by Batominovski, wateringanddrowned, Adventure10, Upwgs_2008, Mango247, and 2 other users
Yes. Let $A$ be an arbitrary point. Consider a regular $n-$gon $AA_{1}A_{2}\ldots A_{n-1}.$ Let $k$ be an integer, $0\leq k\leq n-1.$ A rotation with center $A$ of angle $\dfrac{2k\pi}{n}$ sends the polygon $AA_{1}A_{2}\ldots A_{n-1}$ to $A_{k0}A_{k1}\ldots A_{k,n-1},$ where $A_{k0}=A$ and $A_{ki}$ is the image of $A_{i}$, for all $i=1,2,\ldots,n-1.$

From the condition of the statement, we have
\[ \sum_{k=0}^{n-1} \sum_{i=0}^{n-1}{f(A_{ki})}=0.  \]
Observe that in the sum the number $f(A)$ appears $n$ times, therefore
\[ nf(A)+\sum_{k=0}^{n-1} \sum_{i=1}^{n-1}{f(A_{ki})}=0.  \]
On the other hand, we have
\[ \sum_{k=0}^{n-1} \sum_{i=1}^{n-1}{f(A_{ki})}=\sum_{i=1}^{n-1} \sum_{k=0}^{n-1}{f(A_{ki})}=0,  \]
since the polygons $A_{0i}A_{1i}\ldots A_{n-1,i}$ are all regular $n-$gons. From the two equalities above we deduce $f(A)=0,$ hence $f$ is the zero function.
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TheUltimate123
1740 posts
#8 • 1 Y
Y by MS_asdfgzxcvb
We may assume \(n\) is even, since for \(n\) odd, the sum of the vertices in any \(2n\)-gon is zero.

Now let \(A_1\cdots A_n\) be a regular \(n\)-gon. For each \(i\) and \(j\), let \(M_{ij}\) be the midpoint of \(\overline{A_iA_j}\) (so in particular, \(M_iM_i=A_i\)), and let \(O\) be the center of the \(n\)-gon.

We know since \(M_{i1}M_{i2}\cdots M_{in}\) and \(M_{1,1+i}M_{2,2+i}\cdots M_{n,n+i}\) are regular \(n\)-gons that \begin{align*}     0=\sum_i\sum_jf(M_{ij})     =n\cdot f(O)+\sum_j\sum_{\substack{i<n\\ i\ne n/2}}f(M_{j,j+i})     &=n\cdot f(O) \end{align*}
This post has been edited 1 time. Last edited by TheUltimate123, Jun 30, 2021, 8:40 AM
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jasperE3
11293 posts
#9
Y by
ehsan2004 wrote:
Let $ f: \mathbb{R}^2 \rightarrow \mathbb{R} $ be a function such that for every regular $ n $-gon $ A_1A_2 \ldots A_n $ we have $ f(A_1)+f(A_2)+\cdots +f(A_n)=0 $. Prove that $ f(x)=0 $ for all reals $ x $.

The claim for just $n=4$:
https://aops.com/community/p1703551
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AshAuktober
1004 posts
#10
Y by
For $n$ even, draw a lot of $n$-gons with diametres the segments through the respective vertices and the centre of some $n$-gon, and the calculation works out to give $f(\text{ centre }) = 0$, so we're done.
For $n$ odd, notice that the statement then holds for $2n$ as well, so we're done.
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Zany9998
11 posts
#11
Y by
I wonder if there exists a coloring proof. i.e. label all negative points red, all 0 points yellow and all positive points green. I’ve proved that both red and green are dense in R^2 if whole board is not yellow. Is this sufficient to prove that there exists a regular n-gon whose vertices have at least one red and no green or vice versa? The condition seems to be strong enough.
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