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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
n-variable inequality
bakkune   0
14 minutes ago
Source: Own
Prove that the following inequality holds for all positive integer $n$ and all real numbers $x_1, x_2, \dots, x_n\neq 0$:
$$
\sum_{1\leq i < j \leq n} \dfrac{x_ix_j}{x_i^2 + x_j^2} \ge -\dfrac{n}{4}.
$$
0 replies
bakkune
14 minutes ago
0 replies
Arbitrary point on BC and its relation with orthocenter
falantrng   35
N 18 minutes ago by Giant_PT
Source: Balkan MO 2025 P2
In an acute-angled triangle \(ABC\), \(H\) be the orthocenter of it and \(D\) be any point on the side \(BC\). The points \(E, F\) are on the segments \(AB, AC\), respectively, such that the points \(A, B, D, F\) and \(A, C, D, E\) are cyclic. The segments \(BF\) and \(CE\) intersect at \(P.\) \(L\) is a point on \(HA\) such that \(LC\) is tangent to the circumcircle of triangle \(PBC\) at \(C.\) \(BH\) and \(CP\) intersect at \(X\). Prove that the points \(D, X, \) and \(L\) lie on the same line.

Proposed by Theoklitos Parayiou, Cyprus
35 replies
falantrng
Apr 27, 2025
Giant_PT
18 minutes ago
Kosovo MO 2021 Grade 12, Problem 2
bsf714   7
N 24 minutes ago by Bardia7003
Find all functions $f:\mathbb R\to\mathbb R$ so that the following relation holds for all $x, y\in\mathbb R$.

$$f(f(x)f(y)-1) = xy - 1$$
7 replies
bsf714
Feb 27, 2021
Bardia7003
24 minutes ago
Calculus
youochange   4
N an hour ago by youochange
Find the area enclosed by the curves $e^x,e^{-x},x^2+y^2=1$
4 replies
youochange
Yesterday at 2:38 PM
youochange
an hour ago
5-th powers is a no-go - JBMO Shortlist
WakeUp   9
N an hour ago by Namisgood
Prove that there are are no positive integers $x$ and $y$ such that $x^5+y^5+1=(x+2)^5+(y-3)^5$.

Note
9 replies
WakeUp
Oct 30, 2010
Namisgood
an hour ago
IMO 2008, Question 1
orl   156
N an hour ago by Siddharthmaybe
Source: IMO Shortlist 2008, G1
Let $ H$ be the orthocenter of an acute-angled triangle $ ABC$. The circle $ \Gamma_{A}$ centered at the midpoint of $ BC$ and passing through $ H$ intersects the sideline $ BC$ at points $ A_{1}$ and $ A_{2}$. Similarly, define the points $ B_{1}$, $ B_{2}$, $ C_{1}$ and $ C_{2}$.

Prove that the six points $ A_{1}$, $ A_{2}$, $ B_{1}$, $ B_{2}$, $ C_{1}$ and $ C_{2}$ are concyclic.

Author: Andrey Gavrilyuk, Russia
156 replies
orl
Jul 16, 2008
Siddharthmaybe
an hour ago
APMO 2015 P1
aditya21   63
N an hour ago by Siddharthmaybe
Source: APMO 2015
Let $ABC$ be a triangle, and let $D$ be a point on side $BC$. A line through $D$ intersects side $AB$ at $X$ and ray $AC$ at $Y$ . The circumcircle of triangle $BXD$ intersects the circumcircle $\omega$ of triangle $ABC$ again at point $Z$ distinct from point $B$. The lines $ZD$ and $ZY$ intersect $\omega$ again at $V$ and $W$ respectively.
Prove that $AB = V W$

Proposed by Warut Suksompong, Thailand
63 replies
aditya21
Mar 30, 2015
Siddharthmaybe
an hour ago
AD=BE implies ABC right
v_Enhance   118
N an hour ago by Siddharthmaybe
Source: European Girl's MO 2013, Problem 1
The side $BC$ of the triangle $ABC$ is extended beyond $C$ to $D$ so that $CD = BC$. The side $CA$ is extended beyond $A$ to $E$ so that $AE = 2CA$. Prove that, if $AD=BE$, then the triangle $ABC$ is right-angled.
118 replies
v_Enhance
Apr 10, 2013
Siddharthmaybe
an hour ago
Cyclic Quads and Parallel Lines
gracemoon124   17
N an hour ago by Siddharthmaybe
Source: 2015 British Mathematical Olympiad?
Let $ABCD$ be a cyclic quadrilateral. Let $F$ be the midpoint of the arc $AB$ of its circumcircle which does not contain $C$ or $D$. Let the lines $DF$ and $AC$ meet at $P$ and the lines $CF$ and $BD$ meet at $Q$. Prove that the lines $PQ$ and $AB$ are parallel.
17 replies
gracemoon124
Aug 16, 2023
Siddharthmaybe
an hour ago
Problem 1 (First Day)
Valentin Vornicu   137
N an hour ago by Siddharthmaybe
1. Let $ABC$ be an acute-angled triangle with $AB\neq AC$. The circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$ respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $R$. Prove that the circumcircles of the triangles $BMR$ and $CNR$ have a common point lying on the side $BC$.
137 replies
Valentin Vornicu
Jul 12, 2004
Siddharthmaybe
an hour ago
Concentric Circles
MithsApprentice   62
N an hour ago by Siddharthmaybe
Source: USAMO 1998
Let ${\cal C}_1$ and ${\cal C}_2$ be concentric circles, with ${\cal C}_2$ in the interior of ${\cal C}_1$. From a point $A$ on ${\cal C}_1$ one draws the tangent $AB$ to ${\cal C}_2$ ($B\in {\cal C}_2$). Let $C$ be the second point of intersection of $AB$ and ${\cal C}_1$, and let $D$ be the midpoint of $AB$. A line passing through $A$ intersects ${\cal C}_2$ at $E$ and $F$ in such a way that the perpendicular bisectors of $DE$ and $CF$ intersect at a point $M$ on $AB$. Find, with proof, the ratio $AM/MC$.
62 replies
MithsApprentice
Oct 9, 2005
Siddharthmaybe
an hour ago
four points lie on a circle
pohoatza   77
N an hour ago by Siddharthmaybe
Source: IMO Shortlist 2006, Geometry 2, AIMO 2007, TST 1, P2
Let $ ABCD$ be a trapezoid with parallel sides $ AB > CD$. Points $ K$ and $ L$ lie on the line segments $ AB$ and $ CD$, respectively, so that $AK/KB=DL/LC$. Suppose that there are points $ P$ and $ Q$ on the line segment $ KL$ satisfying \[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]Prove that the points $ P$, $ Q$, $ B$ and $ C$ are concyclic.

Proposed by Vyacheslev Yasinskiy, Ukraine
77 replies
pohoatza
Jun 28, 2007
Siddharthmaybe
an hour ago
Hard combi
EeEApO   6
N an hour ago by aidan0626
In a quiz competition, there are a total of $100 $questions, each with $4$ answer choices. A participant who answers all questions correctly will receive a gift. To ensure that at least one member of my family answers all questions correctly, how many family members need to take the quiz?

Now, suppose my spouse and I move into a new home. Every year, we have twins. Starting at the age of $16$, each of our twin children also begins to have twins every year. If this pattern continues, how many years will it take for my family to grow large enough to have the required number of members to guarantee winning the quiz gift?
6 replies
EeEApO
May 8, 2025
aidan0626
an hour ago
The familiar right angle from the orthocenter
buratinogigle   0
an hour ago
Source: Own, HSGSO P6
Let $ABC$ be a triangle inscribed in a circle $\omega$ with orthocenter $H$ and altitude $BE$. Let $M$ be the midpoint of $AH$. Line $BM$ meets $\omega$ again at $P$. Line $PE$ meets $\omega$ again at $Q$. Let $K$ be the orthogonal projection of $E$ on the line $BC$. Line $QK$ meets $\omega$ again at $G$. Prove that $GA\perp GH$.
0 replies
buratinogigle
an hour ago
0 replies
Apple sharing in Iran
mojyla222   3
N Apr 23, 2025 by math-helli
Source: Iran 2025 second round p6
Ali is hosting a large party. Together with his $n-1$ friends, $n$ people are seated around a circular table in a fixed order. Ali places $n$ apples for serving directly in front of himself and wants to distribute them among everyone. Since Ali and his friends dislike eating alone and won't start unless everyone receives an apple at the same time, in each step, each person who has at least one apple passes one apple to the first person to their right who doesn't have an apple (in the clockwise direction).

Find all values of $n$ such that after some number of steps, the situation reaches a point where each person has exactly one apple.
3 replies
mojyla222
Apr 20, 2025
math-helli
Apr 23, 2025
Apple sharing in Iran
G H J
G H BBookmark kLocked kLocked NReply
Source: Iran 2025 second round p6
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mojyla222
103 posts
#1 • 1 Y
Y by sami1618
Ali is hosting a large party. Together with his $n-1$ friends, $n$ people are seated around a circular table in a fixed order. Ali places $n$ apples for serving directly in front of himself and wants to distribute them among everyone. Since Ali and his friends dislike eating alone and won't start unless everyone receives an apple at the same time, in each step, each person who has at least one apple passes one apple to the first person to their right who doesn't have an apple (in the clockwise direction).

Find all values of $n$ such that after some number of steps, the situation reaches a point where each person has exactly one apple.
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YaoAOPS
1541 posts
#2 • 2 Y
Y by sami1618, jannatiar
Very nice problem. Sketch I will clean up later:

$n$ which are powers of $2$ work inductively as it goes from $2^k$ to two copies $2^{k-1}$ which are apart, this decays into all ones.

$n$ which are equal to $2^k + r$ turn into a $2^k$ and $r$ component with $2^k - 1$ and $r - 1$ zeros before them. The $2^k$ acts like an inch worm which jumps every $2^k$ so it can't ever hit the $r$ from one direction. The $r = 2^a + s$ decays the same way so we can finish inductively to get that it never is all ones. Thus this ends up becoming $2^i$ inch worms in different states which never have the same all $1$ time which gives the result.
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sami1618
908 posts
#3 • 1 Y
Y by jannatiar
Answer: $n=2^k$ for all non-negative integers $k$.

Solution: We will show that if $n$ is a power of $2$ then eventually each person will have exactly one apple, and if $n$ is not a power of $2$ this will not happen.

Assume that $n=2^k$. We claim that after $2^k-1$ steps, everyone will have exactly one apple. We proceed by induction on $k$. The base case $k=0$ is trivial. For the induction step, assume that the result holds for $k$ and we will show it holds for $k+1$. Notice that for each of the first $n-1$ steps the range of people that have an apple will expand by one in the clockwise direction. Thus no apple will make its way around the circle in the first $n-1$ moves, so we can imagine "cutting" the circle to the left of Ali and only considering the passing in straight line. By the inductive hypothesis, after $2^k-1$ moves, Ali will be left with $2^k+1$ apples, the $2^k-1$ friends to the right of Ali will have exactly $1$ apple, and no one else has apples yet. After $1$ more step, Ali will be left with $2^k$ apples, the friend $2^k$ spots to the right of Ali will also have $2^k$ apples, and no one else will have apples. Thus by the inductive step, after another $2^k-1$ steps all $2^{k+1}$ people will have exactly $1$ apple. This completes this part of the solution.

Now we prove two claims.

Claim 1. All such $n\neq 1$ are even.
Proof. Assume $n\neq 1$ works. Consider the situation one step before everyone gets an apple. Everybody having at least one apple must have exactly $2$ apples in order to end up with just $1$ apple after the step. Then $2|n$, as claimed.

Claim 2. If $n=2k$ works, then $n=k$ also works.
Proof. Consider the party with $n=2k$ people. Let $A$ denote the set of $k$ people which are an even number of seats away from Ali and let $B$ denote the set of the other $k$ people. We claim that after every two steps, only the people in $A$ will have apples, and each of them will have an even number of them. Additionally, the people in $A$ function as a party of $k$ people where every two steps it is as if they pass with $2$ apples instead of $1$. Notice that this is true from the beginning. Now consider a person in $A$ that has no apples and is adjacent (to the left) to a block of friends in $A$ with apples. After the first step all the people in $B$ in front of a person from the block will receive $1$ apple. The person to the left of the block still does not have an apple so after the second move all the apples received by people in $B$ plus one additional apple from each person from the block of friends in $A$ will go to the person in consideration. Thus effectively, after two steps, the people in $B$ just helped "passing" the apples and returned to having no apples, while the people in $A$ functioned as a sub-party with $k$ people and twice as many apples. This only stops when everyone in $A$ has exactly $2$ apples, in which case we can not consider a person in $A$ that has no apples and thus after one more step, everyone will have an apple. But by examining our sub-party, this means that $n=k$ must also work, as claimed.

Now if $n$ is not a power of $2$, then express $n$ as $2^k\cdot m$ for a non-negative integer $k$ and an odd integer $m\geq 3$. By Claim 2, if $n$ works then $m$ must also work. But by Claim 1, this is a contradiction, as desired.
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math-helli
13 posts
#4
Y by
Here you can find some solutions
https://t.me/matholampiad123
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