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k a June Highlights and 2025 AoPS Online Class Information
jlacosta   0
Jun 2, 2025
Congratulations to all the mathletes who competed at National MATHCOUNTS! If you missed the exciting Countdown Round, you can watch the video at this link. Are you interested in training for MATHCOUNTS or AMC 10 contests? How would you like to train for these math competitions in half the time? We have accelerated sections which meet twice per week instead of once starting on July 8th (7:30pm ET). These sections fill quickly so enroll today!

[list][*]MATHCOUNTS/AMC 8 Basics
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[*]AMC 10 Problem Series[/list]
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0 replies
jlacosta
Jun 2, 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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Inspired by current year (2025)
Rijul saini   4
N a few seconds ago by Rg230403
Source: India IMOTC 2025 Day 4 Problem 1
Let $k>2$ be an integer. We call a pair of integers $(a,b)$ $k-$good if \[0\leqslant a<k,\hspace{0.2cm} 0<b \hspace{1cm} \text{and} \hspace{1cm} (a+b)^2=ka+b\]Prove that the number of $k-$good pairs is a power of $2$.

Proposed by Prithwijit De and Rohan Goyal
4 replies
1 viewing
Rijul saini
Yesterday at 6:46 PM
Rg230403
a few seconds ago
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Write down sum or product of two numbers
Rijul saini   2
N 12 minutes ago by Rg230403
Source: India IMOTC Practice Test 2 Problem 3
Suppose Alice's grimoire has the number $1$ written on the first page and $n$ empty pages. Suppose in each of the next $n$ seconds, Alice can flip to the next page, and write down the sum or product of two numbers (possibly the same) which are already written in her grimoire.

Let $F(n)$ be the largest possible number such that for any $k < F(n)$, Alice can write down the number $k$ on the last page of her grimoire. Prove that there exists a positive integer $N$ such that for all $n>N$, we have that \[n^{0.99n}\leqslant F(n)\leqslant n^{1.01n}.\]
Proposed by Rohan Goyal and Pranjal Srivastava
2 replies
Rijul saini
Yesterday at 6:56 PM
Rg230403
12 minutes ago
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"all of the stupid geo gets sent to tst 2/5" -allen wang
pikapika007   27
N 41 minutes ago by HamstPan38825
Source: USA TST 2024/2
Let $ABC$ be a triangle with incenter $I$. Let segment $AI$ intersect the incircle of triangle $ABC$ at point $D$. Suppose that line $BD$ is perpendicular to line $AC$. Let $P$ be a point such that $\angle BPA = \angle PAI = 90^\circ$. Point $Q$ lies on segment $BD$ such that the circumcircle of triangle $ABQ$ is tangent to line $BI$. Point $X$ lies on line $PQ$ such that $\angle IAX = \angle XAC$. Prove that $\angle AXP = 45^\circ$.

Luke Robitaille
27 replies
pikapika007
Dec 11, 2023
HamstPan38825
41 minutes ago
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Interior point of ABC
Jackson0423   2
N an hour ago by Diamond-jumper76
Let D be an interior point of the acute triangle ABC with AB > AC so that ∠DAB = ∠CAD. The point E on the segment AC satisfies ∠ADE = ∠BCD, the point F on the segment AB satisfies ∠F DA = ∠DBC, and the point X on the line AC satisfies CX = BX. Let O1 and O2 be the circumcenters of the triangles ADC and EXD, respectively. Prove that the lines BC, EF, and O1O2 are concurrent
2 replies
Jackson0423
Today at 2:17 PM
Diamond-jumper76
an hour ago
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IMO ShortList 2008, Number Theory problem 2
April   41
N an hour ago by shendrew7
Source: IMO ShortList 2008, Number Theory problem 2, German TST 2, P2, 2009
Let $ a_1$, $ a_2$, $ \ldots$, $ a_n$ be distinct positive integers, $ n\ge 3$. Prove that there exist distinct indices $ i$ and $ j$ such that $ a_i + a_j$ does not divide any of the numbers $ 3a_1$, $ 3a_2$, $ \ldots$, $ 3a_n$.

Proposed by Mohsen Jamaali, Iran
41 replies
April
Jul 9, 2009
shendrew7
an hour ago
J
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Cute Geometry
EthanWYX2009   1
N an hour ago by Funcshun840
In triangle \( X_AX_BX_C \), let \( X \) and \( Y \) be a pair of isogonal conjugate points. The line \( XX_A \) intersects \( X_BX_C \) at \( P \), and the line \( XY \) intersects \( X_BX_C \) at \( Q \). Let the circumcircle of \( XX_BX_C \) and the circumcircle of \( XPQ \) intersect again at \( R \) (other than \( X \)). Prove that the line \( RX \) bisects \( \angle PRX_A \).
IMAGE
1 reply
EthanWYX2009
Today at 2:19 PM
Funcshun840
an hour ago
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Cheese??? - I'm definitely doing smth wrong
Sid-darth-vater   0
an hour ago
Source: European Girls Math Olympiad 2013/1
The problem is attached. So is my diagram which has a couple of markings on it for clarity :)

So basically, I found a solution which I am 99% confident that I am doing smth wrong, I just can't find the error. Any help would be appreciated!

We claim that triangle $BAC$ is right angled (for clarity, $<BAC = 90$). Define $S$ as a point on line $AC$ such that $SD$ is parallel to $AB$. Additionally, since $BC = DC$, $\triangle BAC \cong \triangle DSC$ meaning $<BAC = <CSD$, $AC = CS$, and $AB = SD$. Also, since $BE = AD$, by SSS, we have $\triangle BEA \cong DAS$ meaning $\angle EAB= \angle CSD$. Since $\angle EAS + \angle BAC = 180$, we have $2\angle ASD = 180$ or $\angle ASD = \angle BAC = 90$ and we are done.
0 replies
Sid-darth-vater
an hour ago
0 replies
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IMO ShortList 2002, algebra problem 1
orl   132
N an hour ago by SomeonecoolLovesMaths
Source: IMO ShortList 2002, algebra problem 1
Find all functions $f$ from the reals to the reals such that

\[f\left(f(x)+y\right)=2x+f\left(f(y)-x\right)\]

for all real $x,y$.
132 replies
orl
Sep 28, 2004
SomeonecoolLovesMaths
an hour ago
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All good sequences are degenerate
v4913   13
N an hour ago by Jack_w
Source: EGMO 2022/3
An infinite sequence of positive integers $a_1, a_2, \dots$ is called $good$ if
(1) $a_1$ is a perfect square, and
(2) for any integer $n \ge 2$, $a_n$ is the smallest positive integer such that $$na_1 + (n-1)a_2 + \dots + 2a_{n-1} + a_n$$is a perfect square.
Prove that for any good sequence $a_1, a_2, \dots$, there exists a positive integer $k$ such that $a_n=a_k$ for all integers $n \ge k$.
(reposting because the other thread didn't get moved)
13 replies
v4913
Apr 10, 2022
Jack_w
an hour ago
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One of the lines is tangent
Rijul saini   6
N 2 hours ago by bin_sherlo
Source: LMAO 2025 Day 2 Problem 2
Let $ABC$ be a scalene triangle with incircle $\omega$. Denote by $N$ the midpoint of arc $BAC$ in the circumcircle of $ABC$, and by $D$ the point where the $A$-excircle touches $BC$. Suppose the circumcircle of $AND$ meets $BC$ again at $P \neq D$ and intersects $\omega$ at two points $X$, $Y$.

Prove that either $PX$ or $PY$ is tangent to $\omega$.

Proposed by Sanjana Philo Chacko
6 replies
Rijul saini
Yesterday at 7:02 PM
bin_sherlo
2 hours ago
J
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Tricky coloured subgraphs
bomberdoodles   1
N 2 hours ago by bomberdoodles
Consider a graph with nine vertices, with the vertices labelled 1 through 9. An
edge is drawn between each pair of vertices.

Sally picks any edge of her choice, and colours that edge either red or blue. She keeps repeating
this process, choosing any uncoloured edge, and colouring that edge either red or blue.
The only rule is that she is never allowed to colour an edge either red or blue so that one
of these scenarios occurs:

(i) There exist three numbers $a, b, c$, with $1 \le a < b < c \le 9$, for which the edges $ab, bc, ac$ are
all coloured red.

(ii) There exist four numbers $p, q, r, s,$ with $1 \le p < q < r < s \le 9$, for which the edges $pq, pr, ps, qr, qs, rs$ are all coloured blue.

For example, suppose Sally starts by choosing edges 14 and 34, and colouring both of these
edges red. Then if she picks edge 13, she must colour this edge blue, because she cannot colour
it red.

What is the maximum number of edges that Sally can colour?
1 reply
bomberdoodles
2 hours ago
bomberdoodles
2 hours ago
J
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Permutation guessing game
Rijul saini   1
N 2 hours ago by asbodke
Source: India IMOTC Day 3 Problem 3
Let $n$ be a positive integer. Alice and Bob play the following game. Alice considers a permutation $\pi$ of the set $[n]=\{1,2, \dots, n\}$ and keeps it hidden from Bob. In a move, Bob tells Alice a permutation $\tau$ of $[n]$, and Alice tells Bob whether there exists an $i \in [n]$ such that $\tau(i)=\pi(i)$. Bob wins if he ever tells Alice the permutation $\pi$. Prove that Bob can win the game in at most $n \log_2(n) + 2025n$ moves.

Proposed by Siddharth Choppara and Shantanu Nene
1 reply
Rijul saini
Yesterday at 6:43 PM
asbodke
2 hours ago
J
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Tricky FE
Rijul saini   10
N 2 hours ago by MathematicalArceus
Source: LMAO 2025 Day 1 Problem 1
Let $\mathbb{R}$ denote the set of all real numbers. Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that
$$f(xy) + f(f(y)) = f((x + 1)f(y))$$for all real numbers $x$, $y$.

Proposed by MV Adhitya and Kanav Talwar
10 replies
Rijul saini
Yesterday at 6:58 PM
MathematicalArceus
2 hours ago
J
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IMO ShortList 2002, number theory problem 6
orl   32
N 2 hours ago by Rayvhs
Source: IMO ShortList 2002, number theory problem 6
Find all pairs of positive integers $m,n\geq3$ for which there exist infinitely many positive integers $a$ such that \[ \frac{a^m+a-1}{a^n+a^2-1} \] is itself an integer.

Laurentiu Panaitopol, Romania
32 replies
orl
Sep 28, 2004
Rayvhs
2 hours ago
J
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J
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BMO 2024 SL A3
MuradSafarli   6
N May 1, 2025 by quacksaysduck

A3.
Find all triples \((a, b, c)\) of positive real numbers that satisfy the system:
\[ \begin{aligned} 11bc - 36b - 15c &= abc \\ 12ca - 10c - 28a &= abc \\ 13ab - 21a - 6b &= abc. \end{aligned} \]
6 replies
MuradSafarli
Apr 27, 2025
quacksaysduck
May 1, 2025
J
BMO 2024 SL A3
G H J
Reveal topic tags
algebra
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G H BBookmark kLocked kLocked NReply
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MuradSafarli
115 posts
MuradSafarli
#1 h  Apr 27, 2025, 12:42 PM
Y by
A3.
Find all triples \((a, b, c)\) of positive real numbers that satisfy the system:
\[ \begin{aligned} 11bc - 36b - 15c &= abc \\ 12ca - 10c - 28a &= abc \\ 13ab - 21a - 6b &= abc. \end{aligned} \]
Z K Y
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MR.1
138 posts
MR.1
#2 h  Apr 27, 2025, 12:46 PM
Y by
also georgia bmo tst p1
Z K Y
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Tamam
23 posts
Tamam
#3 h  Apr 27, 2025, 1:37 PM • 3 Y
Y by tom-nowy, ehuseyinyigit, Lahmacuncu
Divide first equation by $bc$ second by $ac$ third by $ab$. You get $$a=11-\frac{36}{c}-\frac{15}{b}$$$$b=12-\frac{10}{a}-\frac{28}{c}$$$$c=13-\frac{21}{b}-\frac{6}{a}$$Sum them up then we get $a+\frac{16}{a}+b+\frac{36}{b}+c+\frac{64}{c}=36$. From AM-GM we get $(a,b,c)=(4,6,8)$
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Nuran2010
99 posts
Nuran2010
#4 h  Apr 27, 2025, 2:30 PM • 1 Y
Y by TDVOLIMPTEAM
MuradSafarli wrote:
A3.
Find all triples \((a, b, c)\) of positive real numbers that satisfy the system:
\[ \begin{aligned} 11bc - 36b - 15c &= abc \\ 12ca - 10c - 28a &= abc \\ 13ab - 21a - 6b &= abc. \end{aligned} \]

How is this even BMO SL A3?
Z K Y
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GreekIdiot
277 posts
GreekIdiot
#5 h  Apr 27, 2025, 2:33 PM
Y by
Because there is no other way to 'get there' other than AMGM
Z K Y
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Nuran2010
99 posts
Nuran2010
#6 h  Apr 30, 2025, 9:57 PM • 1 Y
Y by TDVOLIMPTEAM
GreekIdiot wrote:
Because there is no other way to 'get there' other than AMGM

Nice problem, but, anyways ,I think it is too absurd to be A3
This post has been edited 1 time. Last edited by Nuran2010, Apr 30, 2025, 9:59 PM
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quacksaysduck
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quacksaysduck
#7 h  May 1, 2025, 2:59 AM
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One idea solve :<

Divide both sides of the first equation by $bc$, we get \[11-\frac{36}c-\frac{15}b=a.\]Similarly we have \[12-\frac{10}a-\frac{28}c=b,\]\[13-\frac{21}b-\frac6a=c.\]Summing all 3 equations together we get $36=a+\frac{16}a+b+\frac{36}b+c+\frac{64}c$, which by $AM-GM$ on $(a,\frac{16}a)$, $(b,\frac{36}b)$, $(c,\frac{64}c)$ we get $(a,b,c)$ = $(4,6,8)$, which works when plugging back.
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