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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Yesterday at 11:16 PM
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
Yesterday at 11:16 PM
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
Familiar cyclic quad config
Rijul saini   11
N 3 minutes ago by ihategeo_1969
Source: India IMOTC Practice Test 1 Problem 2
Let $ABCD$ be a convex cyclic quadrilateral with circumcircle $\omega$. Let $BA$ produced beyond $A$ meet $CD$ produced beyond $D$, at $L$. Let $\ell$ be a line through $L$ meeting $AD$ and $BC$ at $M$ and $N$ respectively, so that $M,D$ (respectively $N,C$) are on opposite sides of $A$ (resp. $B$). Suppose $K$ and $J$ are points on the arc $AB$ of $\omega$ not containing $C,D$ so that $MK, NJ$ are tangent to $\omega$. Prove that $K,J,L$ are collinear.

Proposed by Rijul Saini
11 replies
Rijul saini
May 31, 2024
ihategeo_1969
3 minutes ago
Inspired by Nice inequality
sqing   1
N 5 minutes ago by sqing
Source: Own
Let $  a,b,c >0  $. Show that
$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+1\right)^2 \geq \frac{16}{k+3}\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}+k\right)$$Where $ k\geq 1.$
$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+1\right)^2 \geq  \frac{b}{a}+\frac{c}{b}+\frac{a}{c}+13$$$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+1\right)^2 \geq \frac{16}{5}\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}+2\right)$$
1 reply
1 viewing
sqing
5 hours ago
sqing
5 minutes ago
Hard inequality
ys33   5
N 13 minutes ago by sqing
Let $a, b, c, d>0$. Prove that
$\sqrt[3]{ab}+ \sqrt[3]{cd} < \sqrt[3]{(a+b+c)(b+c+d)}$.
5 replies
ys33
4 hours ago
sqing
13 minutes ago
China Northern MO 2009 p4 CNMO
parkjungmin   1
N 20 minutes ago by exoticc
Source: China Northern MO 2009 p4 CNMO
China Northern MO 2009 p4 CNMO

The problem is too difficult.
Is there anyone who can help me?
1 reply
parkjungmin
Apr 30, 2025
exoticc
20 minutes ago
Hojoo Lee problem 73
Leon   25
N 23 minutes ago by sqing
Source: Belarus 1998
Let $a$, $b$, $c$ be real positive numbers. Show that \[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a+b}{b+c}+\frac{b+c}{a+b}+1\]
25 replies
Leon
Aug 21, 2006
sqing
23 minutes ago
Almost Squarefree Integers
oVlad   3
N 30 minutes ago by Primeniyazidayi
Source: Romania Junior TST 2025 Day 1 P1
A positive integer $n\geqslant 3$ is almost squarefree if there exists a prime number $p\equiv 1\bmod 3$ such that $p^2\mid n$ and $n/p$ is squarefree. Prove that for any almost squarefree positive integer $n$ the ratio $2\sigma(n)/d(n)$ is an integer.
3 replies
+1 w
oVlad
Apr 12, 2025
Primeniyazidayi
30 minutes ago
Math camp combi
ErTeeEs06   3
N 31 minutes ago by genius_007
Source: BxMO 2025 P2
Let $N\geq 2$ be a natural number. At a mathematical olympiad training camp the same $N$ courses are organised every day. Each student takes exactly one of the $N$ courses each day. At the end of the camp, every student has takes each course exactly once, and any two students took the same course on at least one day, but took different courses on at least one other day. What is, in terms of $N$, the largest possible number of students at the camp?
3 replies
ErTeeEs06
Apr 26, 2025
genius_007
31 minutes ago
Benelux fe
ErTeeEs06   10
N 40 minutes ago by genius_007
Source: BxMO 2025 P1
Does there exist a function $f:\mathbb{R}\to \mathbb{R}$ such that $$f(x^2+f(y))=f(x)^2-y$$for all $x, y\in \mathbb{R}$?
10 replies
ErTeeEs06
Apr 26, 2025
genius_007
40 minutes ago
IMO Shortlist Problems
ABCD1728   0
an hour ago
Source: IMO official website
Where can I get the official solution for ISL before 2005? The official website only has solutions after 2006. Thanks :)
0 replies
ABCD1728
an hour ago
0 replies
Geometric inequality in quadrilateral
BBNoDollar   0
an hour ago
Source: Romanian Mathematical Gazette 2025
Let ABCD be a convex quadrilateral with angles BAD and BCD obtuse, and let the points E, F ∈ BD, such that AE ⊥ BD and CF ⊥ BD.
Prove that 1/(AE*CF) ≥ 1/(AB*BC) + 1/(AD*CD) .
0 replies
BBNoDollar
an hour ago
0 replies
A coincidence about triangles with common incenter
flower417477   2
N an hour ago by flower417477
$\triangle ABC,\triangle ADE$ have the same incenter $I$.Prove that $BCDE$ is concyclic iff $BC,DE,AI$ is concurrent
2 replies
flower417477
Wednesday at 2:08 PM
flower417477
an hour ago
Function equation
LeDuonggg   5
N 2 hours ago by luutrongphuc
Find all functions $f: \mathbb{R^+} \rightarrow \mathbb{R^+}$ , such that for all $x,y>0$:
\[ f(x+f(y))=\dfrac{f(x)}{1+f(xy)}\]
5 replies
LeDuonggg
Yesterday at 2:59 PM
luutrongphuc
2 hours ago
Consecutive sum of integers sum up to 2020
NicoN9   2
N 2 hours ago by NicoN9
Source: Japan Junior MO Preliminary 2020 P2
Let $a$ and $b$ be positive integers. Suppose that the sum of integers between $a$ and $b$, including $a$ and $b$, are equal to $2020$.
All among those pairs $(a, b)$, find the pair such that $a$ achieves the minimum.
2 replies
NicoN9
Today at 6:09 AM
NicoN9
2 hours ago
Range of a^3+b^3-3c
Kunihiko_Chikaya   1
N 2 hours ago by Mathzeus1024
Let $a,\ b,\ c$ be real numbers such that $b<\frac{1}{c}<a$ and

$$\begin{cases}a+b+c=1 \ \\ a^2+b^2+c^2=23	

\end{cases}$$
Find the range of $a^3+b^3-3c.$


Proposed by Kunihiko Chikaya/September 23, 2020
1 reply
Kunihiko_Chikaya
Sep 23, 2020
Mathzeus1024
2 hours ago
Outcome related combinatorics problem
egxa   1
N Apr 29, 2025 by iliya8788
Source: All Russian 2025 10.7
A competition consists of $25$ sports, each awarding one gold medal to a winner. $25$ athletes participate, each in all $25$ sports. There are also $25$ experts, each of whom must predict the number of gold medals each athlete will win. In each prediction, the medal counts must be non-negative integers summing to $25$. An expert is called competent if they correctly guess the number of gold medals for at least one athlete. What is the maximum number \( k \) such that the experts can make their predictions so that at least \( k \) of them are guaranteed to be competent regardless of the outcome?
1 reply
egxa
Apr 18, 2025
iliya8788
Apr 29, 2025
Outcome related combinatorics problem
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G H BBookmark kLocked kLocked NReply
Source: All Russian 2025 10.7
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egxa
209 posts
#1
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A competition consists of $25$ sports, each awarding one gold medal to a winner. $25$ athletes participate, each in all $25$ sports. There are also $25$ experts, each of whom must predict the number of gold medals each athlete will win. In each prediction, the medal counts must be non-negative integers summing to $25$. An expert is called competent if they correctly guess the number of gold medals for at least one athlete. What is the maximum number \( k \) such that the experts can make their predictions so that at least \( k \) of them are guaranteed to be competent regardless of the outcome?
This post has been edited 1 time. Last edited by egxa, Apr 18, 2025, 5:21 PM
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iliya8788
8 posts
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We claim that the maximum $k$ is equal to $24$.
We correspond each athlete to a number from $1$ to $25$.
Define Podium as the $n$-tuple $(a_{1},a_{2},...,a_{25})$ where $a_{i}$ is the amount of medals won by athlete $i$.
First notice that none of the experts can predict the $n$-tuple $(1,1,...,1)$ because then if the podium is equal to $(25,0,0,...,0)$ then the most possible amount of judges predicting correctly is equal to $24$. From the previous statement it follows that one of the judges predicts a $n$-tuple which has a $0$. So we proceed with the following algorithm. Pick one of the prediction $n$-tuples and convert it with the following algorithm to a new $n$-tuple: turn anything that isn't equal to $0$ to $0$ and then spread the 25 medals among the zero's.
Now if the podium $n$-tuple is the same as this $n$-tuple then at least one of the experts isn't competent and so maximum $k$ is equal to $24$.
Now we will prove that no matter what the podium looks like we can achieve $k=24$.
So we propose the following predictions: $(24,1,0,0,...,0),(24,0,1,0,0,...,0),(24,0,0,1,0,0,...0),...,(24,0,0,...,1),(1,1,...,1,1,1)$.
Define the count of zeros of the podium as $C$.
Now 3 cases might happen:
$C>2$: then it's easy to notice that every expert except the last one is competent.
$C=2$: an expert isn't competent if and only if both zeros are in the same place as the $1$ and $24$. notice that no $2$ of the predictions have these 2 numbers in the same spot so worst case possible $24$ of the experts are competent.
$C=1$: notice that the podium $n$-tuple must have $23$, $1$'s and a single $2$ and a single $0$. an expert isn't competent if and only if one of the following 2 cases occur:
First case: the $1$ in the prediction is in the same place as the $0$ of the podium.
Second case: the $24$ in the prediction is in the same place as the $0$ of the podium and the $1$ in the prediction is in the same place as the $2$ of the podium.
notice that the position of the $1$ differs among the predictions so for a fixed podium only one expert can be not competent because of the first case. The same applies for the second case as well. Since the union of the positions of $1$ and $24$ differs among the predictions there can't be 2 experts that are incompetent because of the second case. Also both cases can't make experts incompetent simultaneously since in case of both cases happening the $1$ and $24$ must basically switch places but in each of the first $24$ predictions $24$ is not in the first spot and all of the $1$'s are in the first spot. So only one of these $24$ experts can be incompetent and the last expert is competent so at least $24$ of the experts are competent. $\blacksquare$
This post has been edited 5 times. Last edited by iliya8788, Apr 29, 2025, 12:11 PM
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