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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.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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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
weird conditions in geo
Davdav1232   2
N 9 minutes ago by teoira
Source: Israel TST 7 2025 p1
Let \( \triangle ABC \) be an isosceles triangle with \( AB = AC \). Let \( D \) be a point on \( AC \). Let \( L \) be a point inside the triangle such that \( \angle CLD = 90^\circ \) and
\[
CL \cdot BD = BL \cdot CD.
\]Prove that the circumcenter of triangle \( \triangle BDL \) lies on line \( AB \).
2 replies
Davdav1232
May 8, 2025
teoira
9 minutes ago
Long FE with f(0)=0
Fysty   4
N an hour ago by MathLuis
Source: Own
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ satisfying $f(0)=0$ and
$$f(f(x)+xf(y)+y)+xf(x+y)+f(y^2)=x+f(f(y))+(f(x)+y)(f(y)+x)$$for all $x,y\in\mathbb{R}$.
4 replies
Fysty
May 23, 2021
MathLuis
an hour ago
Inspired by old results
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b> 0. $ Prove that
$$  \frac{a^3}{b^3+ab^2}+ \frac{4b^3}{a^3+b^3+2ab^2}\geq \frac{3}{2}$$$$\frac{a^3}{b^3+(a+b)^3}+ \frac{b^3}{a^3+(a+b)^3}+ \frac{(a+b)^2}{a^2+b^2+ab} \geq \frac{14}{9}$$
1 reply
sqing
2 hours ago
sqing
an hour ago
Quadruple isogonal conjugate inside cyclic quad
Noob_at_math_69_level   8
N 2 hours ago by awesomeming327.
Source: DGO 2023 Team & Individual P3
Let $ABCD$ be a cyclic quadrilateral with $M_1,M_2,M_3,M_4$ being the midpoints of segments $AB,BC,CD,DA$ respectively. Suppose $E$ is the intersection of diagonals $AC,BD$ of quadrilateral $ABCD.$ Define $E_1$ to be the isogonal conjugate point of point $E$ in $\triangle{M_1CD}.$ Define $E_2,E_3,E_4$ similarly. Suppose $E_1E_3$ intersects $E_2E_4$ at a point $W.$ Prove that: The Newton-Gauss line of quadrilateral $ABCD$ bisects segment $EW.$

Proposed by 土偶 & Paramizo Dicrominique
8 replies
Noob_at_math_69_level
Dec 18, 2023
awesomeming327.
2 hours ago
Interesting inequality
sqing   3
N 2 hours ago by sqing
Source: Own
Let $ a,b,c\geq  0 , a^2+b^2+c^2 =3.$ Prove that
$$ a^4+ b^4+c^4+6abc\leq9$$$$ a^3+ b^3+  c^3+3( \sqrt{3}-1)abc\leq 3\sqrt 3$$
3 replies
sqing
Yesterday at 2:54 AM
sqing
2 hours ago
2-var inequality
sqing   12
N 2 hours ago by sqing
Source: Own
Let $ a,b>0 , a^2+b^2-ab\leq 1 . $ Prove that
$$a^3+b^3 -\frac{a^4}{b+1}  -\frac{b^4}{a+1} \leq 1 $$
12 replies
1 viewing
sqing
May 27, 2025
sqing
2 hours ago
Sum of whose elements is divisible by p
nntrkien   46
N 3 hours ago by Jackson0423
Source: IMO 1995, Problem 6, Day 2, IMO Shortlist 1995, N6
Let $ p$ be an odd prime number. How many $ p$-element subsets $ A$ of $ \{1,2,\dots,2p\}$ are there, the sum of whose elements is divisible by $ p$?
46 replies
nntrkien
Aug 8, 2004
Jackson0423
3 hours ago
Graph Theory
achen29   4
N 4 hours ago by ABCD1728
Are there any good handouts or even books in Graph Theory for a beginner in it? Preferable handouts which are extensive!
4 replies
achen29
Apr 24, 2018
ABCD1728
4 hours ago
Guess period of function
a1267ab   10
N 4 hours ago by cosmicgenius
Source: USA TST 2025
Let $n$ be a positive integer. Ana and Banana play a game. Banana thinks of a function $f\colon\mathbb{Z}\to\mathbb{Z}$ and a prime number $p$. He tells Ana that $f$ is nonconstant, $p<100$, and $f(x+p)=f(x)$ for all integers $x$. Ana's goal is to determine the value of $p$. She writes down $n$ integers $x_1,\dots,x_n$. After seeing this list, Banana writes down $f(x_1),\dots,f(x_n)$ in order. Ana wins if she can determine the value of $p$ from this information. Find the smallest value of $n$ for which Ana has a winning strategy.

Anthony Wang
10 replies
a1267ab
Dec 14, 2024
cosmicgenius
4 hours ago
interesting geo config (2/3)
Royal_mhyasd   1
N 5 hours ago by Royal_mhyasd
Source: own
Let $\triangle ABC$ be an acute triangle and $H$ its orthocenter. Let $P$ be a point on the parallel through $A$ to $BC$ such that $\angle APH = |\angle ABC-\angle ACB|$. Define $Q$ and $R$ as points on the parallels through $B$ to $AC$ and through $C$ to $AB$ similarly. If $P,Q,R$ are positioned around the sides of $\triangle ABC$ as in the given configuration, prove that $P,Q,R$ are collinear.
1 reply
Royal_mhyasd
5 hours ago
Royal_mhyasd
5 hours ago
interesting geo config (1\3)
Royal_mhyasd   0
6 hours ago
Source: own
Let $\triangle ABC$ be an acute triangle with $AC > AB$, $H$ its orthocenter and $O$ it's circumcenter. Let $P$ be a point on the parallel through $A$ to $BC$ such that $\angle APH = \angle ABC - \angle ACB$ and $P$ and $C$ are on different sides of $AB$. Denote by $S$ the intersection of the circumcircle of $\triangle ABC$ and $PA'$, where $A'$ is the reflection of $H$ over $BC$, $M$ the midpoint of $PH$, $Q$ the intersection of $OA$ and the parallel through $M$ to $AS$, $R$ the intersection of $MS$ and the perpendicular through $O$ to $PS$ and $N$ a point on $AS$ such that $NT \parallel PS$, where $T$ is the midpoint of $HS$. Prove that $Q, N, R$ lie on a line.

fiy it's 2am and i'm bored so i decided to look further into this interesting config that i had already made some observations on, maybe this problem is trivial from some theorem so if that's the case then i'm sorry lol :P i'll probably post 2 more problems related to it soon, i'd say they're easier than this though
0 replies
Royal_mhyasd
6 hours ago
0 replies
Parallel lines..
ts0_9   9
N 6 hours ago by OutKast
Source: Kazakhstan National Olympiad 2014 P3 D1 10 grade
The triangle $ABC$ is inscribed in a circle $w_1$. Inscribed in a triangle circle touchs the sides $BC$ in a point $N$. $w_2$ — the circle inscribed in a segment $BAC$ circle of $w_1$, and passing through a point $N$. Let points $O$ and $J$ — the centers of circles $w_2$ and an extra inscribed circle (touching side $BC$) respectively. Prove, that lines $AO$ and $JN$ are parallel.
9 replies
ts0_9
Mar 26, 2014
OutKast
6 hours ago
KMN and PQR are tangent at a fixed point
hal9v4ik   4
N 6 hours ago by OutKast
Let $ABCD$ be cyclic quadrilateral. Let $AC$ and $BD$ intersect at $R$, and let $AB$ and $CD$ intersect at $K$. Let $M$ and $N$ are points on $AB$ and $CD$ such that $\frac{AM}{MB}=\frac{CN}{ND}$. Let $P$ and $Q$ be the intersections of $MN$ with the diagonals of $ABCD$. Prove that circumcircles of triangles $KMN$ and $PQR$ are tangent at a fixed point.
4 replies
hal9v4ik
Mar 19, 2013
OutKast
6 hours ago
one cyclic formed by two cyclic
CrazyInMath   40
N 6 hours ago by HamstPan38825
Source: EGMO 2025/3
Let $ABC$ be an acute triangle. Points $B, D, E$, and $C$ lie on a line in this order and satisfy $BD = DE = EC$. Let $M$ and $N$ be the midpoints of $AD$ and $AE$, respectively. Suppose triangle $ADE$ is acute, and let $H$ be its orthocentre. Points $P$ and $Q$ lie on lines $BM$ and $CN$, respectively, such that $D, H, M,$ and $P$ are concyclic and pairwise different, and $E, H, N,$ and $Q$ are concyclic and pairwise different. Prove that $P, Q, N,$ and $M$ are concyclic.
40 replies
CrazyInMath
Apr 13, 2025
HamstPan38825
6 hours ago
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
920 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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