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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)!

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
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
Stability of Additive Cauchy Equation
doanquangdang   1
N an hour ago by jasperE3
Show that if $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies
$$
|f(x+y)-f(x)-f(y)-x y| \leq \varepsilon\left(|x|^p+|y|^p\right)
$$for some $\varepsilon>0,$ $p \in[0,1)$ and for all $x, y \in \mathbb{R}$, then there exists a unique solution $a: \mathbb{R} \rightarrow \mathbb{R}$ of the functional equation $a(x+y)=$ $a(x)+a(y)$ for all $x, y \in \mathbb{R}$ such that
$$
\left|f(x)-a(x)-\frac{1}{2} x^2\right| \leq \frac{2}{2-2^p} \varepsilon|x|^p
$$for all $x \in \mathbb{R}$.
1 reply
doanquangdang
Aug 16, 2024
jasperE3
an hour ago
Polynomials with common roots and coefficients
VicKmath7   10
N an hour ago by math-olympiad-clown
Source: Balkan MO SL 2020 A3
Let $P(x), Q(x)$ be distinct polynomials of degree $2020$ with non-zero coefficients. Suppose that they have $r$ common real roots counting multiplicity and $s$ common coefficients. Determine the maximum possible value of $r + s$.

Demetres Christofides, Cyprus
10 replies
VicKmath7
Sep 9, 2021
math-olympiad-clown
an hour ago
Serbian selection contest for the IMO 2025 - P3
OgnjenTesic   1
N an hour ago by korncrazy
Source: Serbian selection contest for the IMO 2025
Find all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that:
- $f$ is strictly increasing,
- there exists $M \in \mathbb{N}$ such that $f(x+1) - f(x) < M$ for all $x \in \mathbb{N}$,
- for every $x \in \mathbb{Z}$, there exists $y \in \mathbb{Z}$ such that
\[
            f(y) = \frac{f(x) + f(x + 2024)}{2}.
        \]Proposed by Pavle Martinović
1 reply
OgnjenTesic
Yesterday at 4:06 PM
korncrazy
an hour ago
Interesting inequalities
sqing   0
an hour ago
Source: Own
Let $ a,b,c \geq  0 , a+b+c+abc = 4.$ Prove that
$$ a+ab^2+\frac{15}{4}ab^2c^3  \leq \frac{2(100+13\sqrt{13})}{27}$$$$ 2a+ab^2+ 4ab^2c^3\leq \frac{4(68+5\sqrt{10})}{27}$$$$ 3a+ab^2+ \frac{9}{2}ab^2c^3\leq \frac{2(172+7\sqrt{7})}{27}$$$$a+ab^2+ 3.75982ab^2c^3 \leq \frac{2(100+13\sqrt{13})}{27}$$$$ 2a+ab^2+ 4.21981ab^2c^3\leq \frac{4(68+5\sqrt{10})}{27}$$$$ 3a+ab^2+4.73626ab^2c^3\leq \frac{2(172+7\sqrt{7})}{27}$$
0 replies
sqing
an hour ago
0 replies
Inspired by old results
sqing   0
an hour ago
Source: Own
Let $ a,b,c \geq  0 , a+b+c+abc = 4.$ Prove that
$$ a+ab+2ab^2c^3  \leq \frac{25}{4}$$$$ 2a+ab+\frac{29}{10}ab^2c^3 \leq 9$$$$3a+ab+4ab^2c^3 \leq \frac{49}{4}$$$$ 2a+ab+2.9746371ab^2c^3 \leq 9$$$$3a+ab+4.062494ab^2c^3 \leq \frac{49}{4}$$
0 replies
sqing
an hour ago
0 replies
Interesting inequalities
sqing   3
N 2 hours ago by sqing
Source: Own
Let $ a,b,c,d\geq  0 , a+b+c+d \leq 4.$ Prove that
$$a(kbc+bd+cd)  \leq \frac{64k}{27}$$$$a (b+c) (kb c+  b d+  c d) \leq \frac{27k}{4}$$Where $ k\geq 2. $
3 replies
sqing
Yesterday at 12:44 PM
sqing
2 hours ago
Functional equation with powers
tapir1729   14
N 2 hours ago by Mathandski
Source: TSTST 2024, problem 6
Determine whether there exists a function $f: \mathbb{Z}_{> 0} \rightarrow \mathbb{Z}_{> 0}$ such that for all positive integers $m$ and $n$,
\[f(m+nf(m))=f(n)^m+2024! \cdot m.\]Jaedon Whyte
14 replies
tapir1729
Jun 24, 2024
Mathandski
2 hours ago
Inspired by lbh_qys
sqing   3
N 2 hours ago by sqing
Source: Own
Let $ a, b $ be real numbers such that $ (a-3)(b-3)(a-b)\neq 0 $ and $ a + b =6 . $ Prove that
$$ \left( \frac{a+k-1}{b - 3} + \frac{b+k-1}{3 - a} + \frac{k+2}{a - b} \right)^2 + 2(a^2 + b^2 )\geq6(k+8)$$Where $ k\in N^+.$
3 replies
sqing
May 20, 2025
sqing
2 hours ago
Number theory for people who love theory
Assassino9931   2
N 2 hours ago by MathLuis
Source: Bulgaria RMM TST 2019
Prove that there is no positive integer $n$ such that $2^n + 1$ divides $5^n-1$.
2 replies
Assassino9931
Jul 31, 2024
MathLuis
2 hours ago
RMM 2021 Problem 3
VicKmath7   14
N 3 hours ago by hectorleo123
Source: RMM 2021/3
A number of $17$ workers stand in a row. Every contiguous group of at least $2$ workers is a $\textit{brigade}$. The chief wants to assign each brigade a leader (which is a member of the brigade) so that each worker’s number of assignments is divisible by $4$. Prove that the number of such ways to assign the leaders is divisible by $17$.

Mikhail Antipov, Russia
14 replies
VicKmath7
Oct 13, 2021
hectorleo123
3 hours ago
2021 KMO 1st round Gauss Combinatoric Problem
kwan2010   3
N 3 hours ago by MathIQ.
Find the number of functions that satisfy both of the following conditions.
(i) f(1)≤f(2)≤...≤f(9)
(ii) The number of elements in the range of the composition function f∘f is 7.

The answer is Click to reveal hidden text
3 replies
kwan2010
Feb 16, 2025
MathIQ.
3 hours ago
IMO Shortlist 2014 G6
hajimbrak   30
N 4 hours ago by awesomeming327.
Let $ABC$ be a fixed acute-angled triangle. Consider some points $E$ and $F$ lying on the sides $AC$ and $AB$, respectively, and let $M$ be the midpoint of $EF$ . Let the perpendicular bisector of $EF$ intersect the line $BC$ at $K$, and let the perpendicular bisector of $MK$ intersect the lines $AC$ and $AB$ at $S$ and $T$ , respectively. We call the pair $(E, F )$ $\textit{interesting}$, if the quadrilateral $KSAT$ is cyclic.
Suppose that the pairs $(E_1 , F_1 )$ and $(E_2 , F_2 )$ are interesting. Prove that $\displaystyle\frac{E_1 E_2}{AB}=\frac{F_1 F_2}{AC}$
Proposed by Ali Zamani, Iran
30 replies
hajimbrak
Jul 11, 2015
awesomeming327.
4 hours ago
sum of 4 primes with 5 <p <q <r <s <p + 10 is divisible by 60
parmenides51   4
N 4 hours ago by MathIQ.
Source: 2019 Austrian Mathematical Olympiad Junior Regional Competition , Problem 4
Let $p, q, r$ and $s$ be four prime numbers such that $$5 <p <q <r <s <p + 10.$$Prove that the sum of the four prime numbers is divisible by $60$.

(Walther Janous)
4 replies
parmenides51
Dec 18, 2020
MathIQ.
4 hours ago
2player game, adding numbers, whoever reaches no >= 2019 wins
parmenides51   2
N 4 hours ago by MathIQ.
Source: 2019 Austrian Mathematical Olympiad Junior Regional Competition , Problem 3
Alice and Bob are playing a year number game.
There will be two game numbers $19$ and $20$ and one starting number from the set $\{9, 10\}$ used. Alice chooses independently her game number and Bob chooses the starting number. The other number is given to Bob. Then Alice adds her game number to the starting number, Bob adds his game number to the result, Alice adds her number of games to the result, etc. The game continues until the number $2019$ is reached or exceeded.
Whoever reaches the number $2019$ wins. If $2019$ is exceeded, the game ends in a draw.
$\bullet$ Show that Bob cannot win.
$\bullet$ What starting number does Bob have to choose to prevent Alice from winning?

(Richard Henner)
2 replies
parmenides51
Dec 18, 2020
MathIQ.
4 hours ago
Geometry with iff
SP0SkopjeMK   1
N Dec 31, 2015 by FabrizioFelen
Source: 2015 JBMO TST - Macedonia, Problem 4
Let $\triangle ABC$ be an acute angled triangle and let $k$ be its circumscribed circle. A point $O$ is given in the interior of the triangle, such that $CE=CF$, where $E$ and $F$ are on $k$ and $E$ lies on $AO$ while $F$ lies on $BO$. Prove that $O$ is on the angle bisector of $\angle ACB$ if and only if $AC=BC$.
1 reply
SP0SkopjeMK
Dec 31, 2015
FabrizioFelen
Dec 31, 2015
Geometry with iff
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G H BBookmark kLocked kLocked NReply
Source: 2015 JBMO TST - Macedonia, Problem 4
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SP0SkopjeMK
52 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $\triangle ABC$ be an acute angled triangle and let $k$ be its circumscribed circle. A point $O$ is given in the interior of the triangle, such that $CE=CF$, where $E$ and $F$ are on $k$ and $E$ lies on $AO$ while $F$ lies on $BO$. Prove that $O$ is on the angle bisector of $\angle ACB$ if and only if $AC=BC$.
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FabrizioFelen
241 posts
#2 • 2 Y
Y by Adventure10, Mango247
My solution:
Part a): If $AC=BC$
Since $AC=BC$ and $CE=CF$ $\Longrightarrow$ $\triangle CEA\cong \triangle CFB$ since $O=BE\cap AF$ $\Longrightarrow$ $\angle FAB=\angle FCB=\angle ECA=\angle EBA$ $\Longrightarrow$ $OA=OB$ $\Longrightarrow$ $CO$ is bisector of $\angle ACB$
Part b): If $CO$ is bisector of $\angle ACB$
Since $CO$ is bisector of $\angle ACB$ $\Longrightarrow$ $\angle OCA=\angle OCB$ and is easy to see that: $\angle CAE=\angle FAC=\angle CBF=\angle CBE$ since $CE=CF$ $\Longrightarrow$ $\triangle OCA\cong \triangle OCB$ since $\angle CAO=\angle CBO$ and $\angle OCA=\angle OCB$... :P
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