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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
5 hours ago
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)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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
5 hours ago
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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Bogus Proof Marathon
pifinity   7615
N 12 minutes ago by happyfish0922
Hi!
I'd like to introduce the Bogus Proof Marathon.

In this marathon, simply post a bogus proof that is middle-school level and the next person will find the error. You don't have to post the real solution :P

Use classic Marathon format: 
[hide=P#]a1b2c3[/hide]
[hide=S#]a1b2c3[/hide]


Example posts:

P(x)
-----
S(x)
P(x+1)
-----
Let's go!! Just don't make it too hard!
7615 replies
pifinity
Mar 12, 2018
happyfish0922
12 minutes ago
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Queue geo
vincentwant   4
N an hour ago by vincentwant
Let $ABC$ be an acute scalene triangle with circumcenter $O$. Let $Y, Z$ be the feet of the altitudes from $B, C$ to $AC, AB$ respectively. Let $D$ be the midpoint of $BC$. Let $\omega_1$ be the circle with diameter $AD$. Let $Q\neq A$ be the intersection of $(ABC)$ and $\omega$. Let $H$ be the orthocenter of $ABC$. Let $K$ be the intersection of $AQ$ and $BC$. Let $l_1,l_2$ be the lines through $Q$ tangent to $\omega,(AYZ)$ respectively. Let $I$ be the intersection of $l_1$ and $KH$. Let $P$ be the intersection of $l_2$ and $YZ$. Let $l$ be the line through $I$ parallel to $HD$ and let $O'$ be the reflection of $O$ across $l$. Prove that $O'P$ is tangent to $(KPQ)$.
4 replies
1 viewing
vincentwant
Wednesday at 3:54 PM
vincentwant
an hour ago
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Do not try to bash on beautiful geometry
ItzsleepyXD   9
N an hour ago by Captainscrubz
Source: Own , Mock Thailand Mathematic Olympiad P9
Let $ABC$be triangle with point $D,E$ and $F$ on $BC,AB,CA$
such that $BE=CF$ and $E,F$ are on the same side of $BC$
Let $M$ be midpoint of segment $BC$ and $N$ be midpoint of segment $EF$
Let $G$ be intersection of $BF$ with $CE$ and $\dfrac{BD}{DC}=\dfrac{AC}{AB}$
Prove that $MN\parallel DG$
9 replies
ItzsleepyXD
Wednesday at 9:30 AM
Captainscrubz
an hour ago
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A sequence containing every natural number exactly once
Pomegranat   1
N an hour ago by YaoAOPS
Source: Own
Does there exist a sequence \( \{a_n\}_{n=1}^{\infty} \), which is a permutation of the natural numbers (that is, each natural number appears exactly once), such that for every \( n \in \mathbb{N} \), the sum of the first \( n \) terms is divisible by \( n \)?
1 reply
Pomegranat
an hour ago
YaoAOPS
an hour ago
J
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Math with Connect4 Boards
Math-lover1   5
N an hour ago by Soupboy0
Hi! So I was playing Connect4 with my friends the other day and I wondered: how many "legal" arrangements of Connect4 can be reached at the ending position?

We assume that we do not stop the game when there is a four in a row, and we have 21 red pieces and 21 yellow pieces. We also drop the pieces one by one into a standard 7 by 6 board. We can start the game with any color piece.

https://en.wikipedia.org/wiki/Connect_Four

Initial Thoughts
Attempt to use one-to-one correspondences
5 replies
1 viewing
Math-lover1
Yesterday at 1:58 AM
Soupboy0
an hour ago
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centroid lies outside of triangle (not clickbait)
Scilyse   1
N 2 hours ago by LoloChen
Source: 数之谜 January (CHN TST Mock) Problem 5
Let $P$ be a convex polygon with centroid $G$, and let $\mathcal P$ be the set of vertices of $P$. Let $\mathcal X$ be the set of triangles with vertices all in $\mathcal P$. We sort the elements $\triangle ABC$ of $\mathcal X$ into the following three types:
[list]
[*] (Type 1) $G$ lies in the strict interior of $\triangle ABC$; let $\mathcal A$ be the set of triangles of this type.
[*] (Type 2) $G$ lies in the strict exterior of $\triangle ABC$; let $\mathcal B$ be the set of triangles of this type.
[*] (Type 3) $G$ lies on the boundary of $\triangle ABC$.
[/list]
For any triangle $T$, denote by $S_T$ the area of $T$. Prove that \[\sum_{T \in \mathcal A} S_T \geq \sum_{T \in \mathcal B} S_T.\]
1 reply
Scilyse
Jan 26, 2025
LoloChen
2 hours ago
J
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Berkeley mini Math Tournament Online is June 7
BerkeleyMathTournament   6
N 2 hours ago by jb2015007
Berkeley mini Math Tournament is a math competition hosted for middle school students once a year. Students compete in multiple rounds: individual round, team round, puzzle round, and relay round.

BmMT 2025 Online will be held on June 7th, and registration is OPEN! Registration is $8 per student. Our website https://berkeley.mt/events/bmmt-2025-online/ has more details about the event, past tests to practice with, and frequently asked questions. We look forward to building community and inspiring students as they explore the world of math!

3 out of 4 of the rounds are completed with a team, so it’s a great opportunity for students to work together. Beyond getting more comfortable with math and becoming better problem solvers, our team is preparing some fun post-competition activities!

Registration is open to students in grades 8 or below. You do not have to be local to the Bay Area or California to register for BmMT Online. Students may register as a team of 1, but it is beneficial to compete on a team of at least 3 due to our scoring guideline and for the experience.

We hope you consider attending, or if you are a parent or teacher, that you encourage your students to think about attending BmMT. Thank you, and once again find more details/register at our website,https://berkeley.mt.
6 replies
BerkeleyMathTournament
Yesterday at 7:37 AM
jb2015007
2 hours ago
J
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4 lines concurrent
Zavyk09   6
N 2 hours ago by hectorleo123
Source: Homework
Let $ABC$ be triangle with circumcenter $(O)$ and orthocenter $H$. $BH, CH$ intersect $(O)$ again at $K, L$ respectively. Lines through $H$ parallel to $AB, AC$ intersects $AC, AB$ at $E, F$ respectively. Point $D$ such that $HKDL$ is a parallelogram. Prove that lines $KE, LF$ and $AD$ are concurrent at a point on $OH$.
6 replies
Zavyk09
Apr 9, 2025
hectorleo123
2 hours ago
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What Are The Chances?
IbrahimNadeem   66
N 2 hours ago by RabtejKalra
Hello, I'm curious to have honest advice on how far I can make it (by 11th-12th grade-ish);

If I have:

- Started AMC 8 study in 6th grade
- Started AMC 10 study in 7th grade
- Started practicing harder & went from 60 to around 100 on AMC 10 (on practice tests with official conditions)
- Started AMC 12 study in 8th grade
- Currently (fall of 8th grade) getting ~120 on AMC 10/12 & 7-10 while practicing AIME

At this rate, what are the chances of me making the USA(J)MO, for example, by ~11th grade?

Please be completely honest and don't hold back; This can be useful to see if I have the need to practice harder.
66 replies
IbrahimNadeem
Oct 31, 2021
RabtejKalra
2 hours ago
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No More than √㏑x㏑㏑x Digits
EthanWYX2009   4
N 2 hours ago by tom-nowy
Source: 2024 April 谜之竞赛-3
Let $f(x)\in\mathbb Z[x]$ have positive integer leading coefficient. Show that there exists infinte positive integer $x,$ such that the number of digit that doesn'r equal to $9$ is no more than $\mathcal O(\sqrt{\ln x\ln\ln x}).$

Created by Chunji Wang, Zhenyu Dong
4 replies
1 viewing
EthanWYX2009
Mar 24, 2025
tom-nowy
2 hours ago
J
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Old hard problem
ItzsleepyXD   1
N 3 hours ago by ItzsleepyXD
Source: IDK
Let $ABC$ be a triangle and let $O$ be its circumcenter and $I$ its incenter.
Let $P$ be the radical center of its three mixtilinears and let $Q$ be the isogonal conjugate of $P$.
Let $G$ be the Gergonne point of the triangle $ABC$.
Prove that line $QG$ is parallel with line $OI$ .
1 reply
ItzsleepyXD
Apr 25, 2025
ItzsleepyXD
3 hours ago
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Existence of a solution of a diophantine equation
syk0526   5
N 3 hours ago by cursed_tangent1434
Source: North Korea Team Selection Test 2013 #6
Show that $ x^3 + x+ a^2 = y^2 $ has at least one pair of positive integer solution $ (x,y) $ for each positive integer $ a $.
5 replies
syk0526
May 17, 2014
cursed_tangent1434
3 hours ago
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Inequality with 3 variables
sqing   0
3 hours ago
Source: Own
Let $ a,b,c\geq 0 ,a^3b^3+b^3c^3+c^3a^3+2abc\geq 1 . $ Prove that$$a+b+c\geq 2 $$Let $ a,b,c\geq 0 ,a^3b^3+b^3c^3+c^3a^3+6abc\geq 9 . $ Prove that$$a+b+c\geq 2\sqrt 3 $$Let $ a,b,c\geq 0 ,a^3b+b^3c+c^3a+6abc\geq 9 . $ Prove that$$a+b+c\geq 3 $$Let $ a,b,c\geq 0 ,a^3b+b^3c+c^3a+3abc\geq 3 . $ Prove that$$a+b+c\geq \frac{4}{\sqrt 3} $$
0 replies
sqing
3 hours ago
0 replies
J
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Inequality with 3 variables and a special condition
Nuran2010   5
N 3 hours ago by sqing
Source: Azerbaijan Al-Khwarizmi IJMO TST 2024
For positive real numbers $a,b,c$ we have $3abc \geq ab+bc+ca$.
Prove that:

$\frac{1}{a^3+b^3+c}+\frac{1}{b^3+c^3+a}+\frac{1}{c^3+a^3+b} \leq \frac{3}{a+b+c}$.

Determine the equality case.
5 replies
Nuran2010
Apr 29, 2025
sqing
3 hours ago
J
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J
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easy olympiad problem
kjhgyuio   7
N Apr 23, 2025 by Charizard_637
Find all positive integer values of \( x \) such that
\[ \sqrt{x - 2011} + \sqrt{2011 - x} + 10 \]is an integer.
7 replies
kjhgyuio
Apr 17, 2025
Charizard_637
Apr 23, 2025
J
easy olympiad problem
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kjhgyuio
53 posts
kjhgyuio
#1 h  Apr 17, 2025, 2:00 PM
Y by
Find all positive integer values of \( x \) such that
\[ \sqrt{x - 2011} + \sqrt{2011 - x} + 10 \]is an integer.
This post has been edited 2 times. Last edited by kjhgyuio, Apr 17, 2025, 2:01 PM
Reason: nil
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Mathdreams
1469 posts
Mathdreams
#2 h  Apr 17, 2025, 2:14 PM
Y by
Solution
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Charizard_637
110 posts
Charizard_637
#3 h  Apr 17, 2025, 7:44 PM
Y by
$x-2011$ and $2011-x$ are each other's negatives, and you can't take the square root of a negative number while having an integer solution. Therefore the only solution that would work is if they were both non-negative, and based on this they must be zero because one positive number will lead to one negative number. Since 0 is neither negative or positive it's what's under both square roots. Therefore, $\sqrt{x-2011} = 0$. Squaring both sides gives $x-2011 = 0$, hence $x = 2011$. This is the only solution.

Edit: I know it's verbose but it's an "olympiad problem"
This post has been edited 2 times. Last edited by Charizard_637, Apr 21, 2025, 5:44 PM
Reason: e
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vincentwant
1367 posts
vincentwant
#4 h  Apr 18, 2025, 1:25 AM
Y by
Charizard_637 wrote:
$x-2011$ and $2011-x$ are each other's negatives, and you can't take the square root of a negative number while having an integer solution. Therefore the only solution that would work is if they were both non-negative, and based on this they must be zero because one positive number will lead to one negative number. Since 0 is neither negative or positive it's what's under both square roots. Therefore, $\sqrt{x-2011} = 0$. Squaring both sides gives $x-2011 = 0$, hence $x = 2011$. This is the only solution.

Edit: I know it's verbose but it's an olympiad problem

you dont have to do this, this is enough

Notice that for the expression to be real, $x-2011\geq0$ and $2011-x\geq 0$, otherwise the imaginary part of the expression would be positive. Thus no solutions other than $x=2011$ exist, and inspection gives that $x=2011$ works. Thus the answer is $x=2011$.
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deduck
217 posts
deduck
#5 h  Apr 18, 2025, 1:59 AM
Y by
this is a past amc8 problem
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Roger.Moore
5 posts
Roger.Moore
#6 h  Apr 18, 2025, 6:01 PM
Y by
The both the roots there are reals only if x=2011
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Maxklark
6 posts
Maxklark
#7 h  Apr 22, 2025, 6:18 PM
Y by
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Charizard_637
110 posts
Charizard_637
#8 h  Apr 23, 2025, 2:32 PM
Y by
deduck wrote:
this is a past amc8 problem

oh
my
:rotfl:
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