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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
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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
J
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usamOOK geometry
KevinYang2.71   106
N an hour ago by jasperE3
Source: USAMO 2025/4, USAJMO 2025/5
Let $H$ be the orthocenter of acute triangle $ABC$, let $F$ be the foot of the altitude from $C$ to $AB$, and let $P$ be the reflection of $H$ across $BC$. Suppose that the circumcircle of triangle $AFP$ intersects line $BC$ at two distinct points $X$ and $Y$. Prove that $C$ is the midpoint of $XY$.
106 replies
KevinYang2.71
Mar 21, 2025
jasperE3
an hour ago
J
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Croatian mathematical olympiad, day 1, problem 2
Matematika   6
N 2 hours ago by Cqy00000000
There were finitely many persons at a party among whom some were friends. Among any $4$ of them there were either $3$ who were all friends among each other or $3$ who weren't friend with each other. Prove that you can separate all the people at the party in two groups in such a way that in the first group everyone is friends with each other and that all the people in the second group are not friends to anyone else in second group. (Friendship is a mutual relation).
6 replies
Matematika
Apr 10, 2011
Cqy00000000
2 hours ago
J
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Geo #3 EQuals FReak out
Th3Numb3rThr33   106
N 2 hours ago by BS2012
Source: 2018 USAJMO #3
Let $ABCD$ be a quadrilateral inscribed in circle $\omega$ with $\overline{AC} \perp \overline{BD}$. Let $E$ and $F$ be the reflections of $D$ over lines $BA$ and $BC$, respectively, and let $P$ be the intersection of lines $BD$ and $EF$. Suppose that the circumcircle of $\triangle EPD$ meets $\omega$ at $D$ and $Q$, and the circumcircle of $\triangle FPD$ meets $\omega$ at $D$ and $R$. Show that $EQ = FR$.
106 replies
1 viewing
Th3Numb3rThr33
Apr 18, 2018
BS2012
2 hours ago
J
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Game
Pascual2005   27
N 2 hours ago by HamstPan38825
Source: Colombia TST, IMO ShortList 2004, combinatorics problem 5
$A$ and $B$ play a game, given an integer $N$, $A$ writes down $1$ first, then every player sees the last number written and if it is $n$ then in his turn he writes $n+1$ or $2n$, but his number cannot be bigger than $N$. The player who writes $N$ wins. For which values of $N$ does $B$ win?

Proposed by A. Slinko & S. Marshall, New Zealand
27 replies
Pascual2005
Jun 7, 2005
HamstPan38825
2 hours ago
J
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Aime ll 2022 problem 5
Rook567   4
N 4 hours ago by Rook567
I don’t understand the solution. I got 220 as answer. Why does it insist, for example two primes must add to the third, when you can take 2,19,19 or 2,7,11 which for drawing purposes is equivalent to 1,1,2 and 2,7,9?
4 replies
Rook567
Thursday at 9:08 PM
Rook567
4 hours ago
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Lines concur on bisector of BAC
Invertibility   2
N 4 hours ago by NO_SQUARES
Source: Slovenia 2025 TST 3 P2
Let $\Omega$ be the circumcircle of a scalene triangle $ABC$. Let $\omega$ be a circle internally tangent to $\Omega$ in $A$. Tangents from $B$ touch $\omega$ in $P$ and $Q$, such that $P$ lies in the interior of $\triangle{}ABC$. Similarly, tangents from $C$ touch $\omega$ in $R$ and $S$, such that $R$ lies in the interior of $\triangle{}ABC$.

Prove that $PS$ and $QR$ concur on the bisector of $\angle{}BAC$.
2 replies
Invertibility
4 hours ago
NO_SQUARES
4 hours ago
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Why is the old one deleted?
EeEeRUT   16
N 5 hours ago by ravengsd
Source: EGMO 2025 P1
For a positive integer $N$, let $c_1 < c_2 < \cdots < c_m$ be all positive integers smaller than $N$ that are coprime to $N$. Find all $N \geqslant 3$ such that $$\gcd( N, c_i + c_{i+1}) \neq 1$$for all $1 \leqslant i \leqslant m-1$

Here $\gcd(a, b)$ is the largest positive integer that divides both $a$ and $b$. Integers $a$ and $b$ are coprime if $\gcd(a, b) = 1$.

Proposed by Paulius Aleknavičius, Lithuania
16 replies
EeEeRUT
Apr 16, 2025
ravengsd
5 hours ago
J
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angle chasing with 2 midpoints, equal angles given and wanted
parmenides51   5
N 5 hours ago by breloje17fr
Source: Ukrainian Geometry Olympiad 2017, IX p1, X p1, XI p1
In the triangle $ABC$, ${{A}_{1}}$ and ${{C}_{1}} $ are the midpoints of sides $BC $ and $AB$ respectively. Point $P$ lies inside the triangle. Let $\angle BP {{C}_{1}} = \angle PCA$. Prove that $\angle BP {{A}_{1}} = \angle PAC $.
5 replies
parmenides51
Dec 11, 2018
breloje17fr
5 hours ago
J
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Problem 4 of Finals
GeorgeRP   2
N 6 hours ago by Assassino9931
Source: XIII International Festival of Young Mathematicians Sozopol 2024, Theme for 10-12 grade
The diagonals \( AD \), \( BE \), and \( CF \) of a hexagon \( ABCDEF \) inscribed in a circle \( k \) intersect at a point \( P \), and the acute angle between any two of them is \( 60^\circ \). Let \( r_{AB} \) be the radius of the circle tangent to segments \( PA \) and \( PB \) and internally tangent to \( k \); the radii \( r_{BC} \), \( r_{CD} \), \( r_{DE} \), \( r_{EF} \), and \( r_{FA} \) are defined similarly. Prove that
\[ r_{AB}r_{CD} + r_{CD}r_{EF} + r_{EF}r_{AB} = r_{BC}r_{DE} + r_{DE}r_{FA} + r_{FA}r_{BC}. \]
2 replies
GeorgeRP
Sep 10, 2024
Assassino9931
6 hours ago
J
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Interesting functional equation with geometry
User21837561   3
N Yesterday at 6:34 PM by Double07
Source: BMOSL 2024 G7
For an acute triangle $ABC$, let $O$ be the circumcentre, $H$ be the orthocentre, and $G$ be the centroid.
Let $f:\pi\rightarrow\mathbb R$ satisfy the following condition:
$f(A)+f(B)+f(C)=f(O)+f(G)+f(H)$
Prove that $f$ is constant.
3 replies
User21837561
Yesterday at 8:14 AM
Double07
Yesterday at 6:34 PM
J
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greatest volume
hzbrl   1
N Yesterday at 6:32 PM by hzbrl
Source: purple comet
A large sphere with radius 7 contains three smaller balls each with radius 3 . The three balls are each externally tangent to the other two balls and internally tangent to the large sphere. There are four right circular cones that can be inscribed in the large sphere in such a way that the bases of the cones are tangent to all three balls. Of these four cones, the one with the greatest volume has volume $n \pi$. Find $n$.
1 reply
hzbrl
Thursday at 9:56 AM
hzbrl
Yesterday at 6:32 PM
J
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(n+1)2^n, (n+3)2^{n+2} not perfect squares for the same n
parmenides51   3
N Yesterday at 6:31 PM by AylyGayypow009
Source: Greece JBMO TST 2015 p3
Prove that there is not a positive integer $n$ such that numbers $(n+1)2^n, (n+3)2^{n+2}$ are both perfect squares.
3 replies
parmenides51
Apr 29, 2019
AylyGayypow009
Yesterday at 6:31 PM
J
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IMO 2010 Problem 3
canada   59
N Yesterday at 6:23 PM by pi271828
Find all functions $g:\mathbb{N}\rightarrow\mathbb{N}$ such that \[\left(g(m)+n\right)\left(g(n)+m\right)\] is a perfect square for all $m,n\in\mathbb{N}.$

Proposed by Gabriel Carroll, USA
59 replies
canada
Jul 7, 2010
pi271828
Yesterday at 6:23 PM
J
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USAJMO problem 2: Side lengths of an acute triangle
BOGTRO   59
N Yesterday at 4:41 PM by ostriches88
Source: Also USAMO problem 1
Find all integers $n \geq 3$ such that among any $n$ positive real numbers $a_1, a_2, \hdots, a_n$ with $\text{max}(a_1,a_2,\hdots,a_n) \leq n \cdot \text{min}(a_1,a_2,\hdots,a_n)$, there exist three that are the side lengths of an acute triangle.
59 replies
BOGTRO
Apr 24, 2012
ostriches88
Yesterday at 4:41 PM
J
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J
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USA(J)MO Statistics Out
BS2012   53
N Apr 24, 2025 by maxamc
Source: MAA edvistas page
https://maa.edvistas.com/eduview/report.aspx?view=1561&mode=6
who were the 2 usamo perfects
53 replies
BS2012
Apr 23, 2025
maxamc
Apr 24, 2025
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USA(J)MO Statistics Out
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Reveal topic tags
USA(J)MO
USAJMO
USAMO
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G H BBookmark kLocked kLocked NReply
Source: MAA edvistas page
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blueprimes
353 posts
blueprimes
#48 h  Apr 24, 2025, 9:00 PM • 2 Y
Y by Arcticturn, elasticwealth
joebub56 wrote:
I would argue mts is more fair than amc. Why is a multiple choice test being factored into oly qualification?? And honestly cheating isnt even a big issue because of the way mts is structured. The whole point is for the competition to be more open and not under time pressure

This has to be ragebait...

AMC is multiple choice and Scantron graded because there are tens of thousands of people taking AMC ever year. Nobody is grading thousands of proofs. And as said before, USAMTS is discontinuous over a month and allows all resources whereas olympiad is 4.5 hours so it is terrible at discerning olympiad skill
This post has been edited 3 times. Last edited by blueprimes, Apr 24, 2025, 9:03 PM
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joebub56
14 posts
joebub56
#49 h  Apr 24, 2025, 9:04 PM
Y by
blueprimes wrote:
joebub56 wrote:
I would argue mts is more fair than amc. Why is a multiple choice test being factored into oly qualification?? And honestly cheating isnt even a big issue because of the way mts is structured. The whole point is for the competition to be more open and not under time pressure

This has to be ragebait...

AMC is multiple choice and Scantron graded because there are tens of thousands of people taking AMC ever year. Nobody is grading thousands of proofs. And as said before, USAMTS is discontinuous over a month and allows all resources whereas olympiad is 4.5 hours so it is terrible at discerning olympiad skill

I

This doesn’t disprove my point? At the end of the day a multiple choice test is a horrible indicator of olympiad skill. Also its not like mts directly determines oly qualification either theres still aime
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ethan2011
318 posts
ethan2011
#50 h  Apr 24, 2025, 9:06 PM • 1 Y
Y by elasticwealth
joebub56 wrote:
I would argue mts is more fair than amc. Why is a multiple choice test being factored into oly qualification?? And honestly cheating isnt even a big issue because of the way mts is structured. The whole point is for the competition to be more open and not under time pressure

Why is a proof test that allows AI and coding factored into oly qualification?
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joebub56
14 posts
joebub56
#51 h  Apr 24, 2025, 9:17 PM
Y by
ethan2011 wrote:
joebub56 wrote:
I would argue mts is more fair than amc. Why is a multiple choice test being factored into oly qualification?? And honestly cheating isnt even a big issue because of the way mts is structured. The whole point is for the competition to be more open and not under time pressure

Why is a proof test that allows AI and coding factored into oly qualification?

its not like ai and coding can do the whole thing for you. and even if you solve a problem with the help of ai, thats still better than doing multiple choice
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jellybeanzzz
497 posts
jellybeanzzz
#52 h  Apr 24, 2025, 9:18 PM
Y by
Honestly I felt that anyone capable of solving all the problems would do it in a few hours. (For example I spent ~15 hours across all 3 contests which was 15 questions + writeups.) So at least for me, the “you get 1 month” thing doesn’t make it different from oly time constraints.

Also several things
1. AI is not allowed
2. The hard step of solving problems with code isn’t the coding itself, it’s proving that your code is rigorous. Most problems this year weren’t solvable with code either.
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joebub56
14 posts
joebub56
#53 h  Apr 24, 2025, 9:21 PM
Y by
jellybeanzzz wrote:
Honestly I felt that anyone capable of solving all the problems would do it in a few hours. (For example I spent ~15 hours across all 3 contests which was 15 questions + writeups.) So at least for me, the “you get 1 month” thing doesn’t make it different from oly time constraints.

Also several things
1. AI is not allowed
2. The hard step of solving problems with code isn’t the coding itself, it’s proving that your code is rigorous. Most problems this year weren’t solvable with code either.

ai is allowed but only for certain things. like you cant just give it the whole problem but you can ask things like "useful formulas for combinatorics". and thats all ai is good for anyways so yeah its not even a big deal
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ethan2011
318 posts
ethan2011
#54 h  Apr 24, 2025, 9:32 PM
Y by
joebub56 wrote:
jellybeanzzz wrote:
Honestly I felt that anyone capable of solving all the problems would do it in a few hours. (For example I spent ~15 hours across all 3 contests which was 15 questions + writeups.) So at least for me, the “you get 1 month” thing doesn’t make it different from oly time constraints.

Also several things
1. AI is not allowed
2. The hard step of solving problems with code isn’t the coding itself, it’s proving that your code is rigorous. Most problems this year weren’t solvable with code either.

ai is allowed but only for certain things. like you cant just give it the whole problem but you can ask things like "useful formulas for combinatorics". and thats all ai is good for anyways so yeah its not even a big deal

you can use AI and they won't notice
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jellybeanzzz
497 posts
jellybeanzzz
#55 h  Apr 24, 2025, 9:35 PM
Y by
ethan2011 wrote:
you can use AI and they won't notice

They noticed actually! They banned dozens of people in just the first round. That’s dozens more bans than AMC cheaters got!
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LostRiptide
29 posts
LostRiptide
#56 h  Apr 24, 2025, 9:40 PM
Y by
jellybeanzzz wrote:
ethan2011 wrote:
you can use AI and they won't notice

They noticed actually! They banned dozens of people in just the first round. That’s dozens more bans than AMC cheaters got!

Makes a very cheatable contest
Bans some of the many cheaters
Many cheaters still remain
SEE! WE'RE BETTER THAN YOU!!! ah argument
Also AMC cheaters also get mass banned bro
This post has been edited 2 times. Last edited by LostRiptide, Apr 24, 2025, 9:46 PM
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BS2012
1036 posts
BS2012
#57 h  Apr 24, 2025, 9:40 PM
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also who said that the only way to cheat on usamts is to use ai
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xTimmyG
265 posts
xTimmyG
#58 h  Apr 24, 2025, 10:27 PM • 7 Y
Y by hsuya1, blueprimes, MathPerson12321, elasticwealth, megarnie, Andyluo, lpieleanu
Anyone who thinks USAMTS cannot be cheated is extremely wrong. People can easily ask AI for a solution, then put it in their own words, so that it is undetectable. Contestants can also just groupsolve or share answers. Something needs to be done about this, as this allows for people who would've gotten a terrible score on the amc (like 50) to easily qualify for AIME.
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bwu_2022
53 posts
bwu_2022
#59 h  Apr 24, 2025, 10:52 PM
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We will see about it next year. I bet the number of people who attend USAMTA will increase significantly.

Look at the leaderboard: https://www.usamts.org/leaderboard/. 6 people got full points last year, and this year, it was 54.
This post has been edited 1 time. Last edited by bwu_2022, Apr 25, 2025, 1:30 AM
Reason: Accuracy
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TiguhBabeHwo
469 posts
TiguhBabeHwo
#60 h  Apr 24, 2025, 11:12 PM
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Well yeah obviously because it's the most widespread information in all of jmo qualification threads at this point
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bachkieu
137 posts
bachkieu
#61 h  Apr 24, 2025, 11:13 PM
Y by
bwu_2022 wrote:
We will see about it next year. I bet the number of people who attend USAMTA will increase significantly.

Look at the leaderboard: https://www.usamts.org/leaderboard/. 6 people got full points, and this year, it was 54.

i would like to point out i got full last year and 71 this year :skull:
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maxamc
573 posts
maxamc
#62 h  Apr 24, 2025, 11:32 PM
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bwu_2022 wrote:
We will see about it next year. I bet the number of people who attend USAMTA will increase significantly.

Look at the leaderboard: https://www.usamts.org/leaderboard/. 6 people got full points, and this year, it was 54.

What is USAMTA?
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