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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)!

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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
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
Inspired by KhuongTrang
sqing   7
N 17 minutes ago by TNKT
Source: Own
Let $a,b,c\ge 0 $ and $ a+b+c=3.$ Prove that
$$\frac{1}{ab+2}+\frac{1}{bc+2}+\frac{1}{ca+2}\ge \frac{6}{abc+5}$$$$\frac{1}{ab+2}+\frac{1}{bc+2}+\frac{1}{ca+2}\ge \frac{7}{abc+6}$$$$\frac{1}{ab+2}+\frac{1}{bc+2}+\frac{1}{ca+2}\ge \frac{21}{17(abc+1)}$$$$\frac{1}{ab+2}+\frac{1}{bc+2}+\frac{1}{ca+2}\ge \frac{42}{17(abc+2)}$$
7 replies
1 viewing
sqing
Jan 21, 2024
TNKT
17 minutes ago
Cycle in a graph with a minimal number of chords
GeorgeRP   5
N 21 minutes ago by Photaesthesia
Source: Bulgaria IMO TST 2025 P3
In King Arthur's court every knight is friends with at least $d>2$ other knights where friendship is mutual. Prove that King Arthur can place some of his knights around a round table in such a way that every knight is friends with the $2$ people adjacent to him and between them there are at least $\frac{d^2}{10}$ friendships of knights that are not adjacent to each other.
5 replies
GeorgeRP
Wednesday at 7:51 AM
Photaesthesia
21 minutes ago
Sum of bad integers to the power of 2019
mofumofu   8
N 32 minutes ago by Orthocenter.THOMAS
Source: China TST 2019 Test 2 Day 2 Q6
Given coprime positive integers $p,q>1$, call all positive integers that cannot be written as $px+qy$(where $x,y$ are non-negative integers) bad, and define $S(p,q)$ to be the sum of all bad numbers raised to the power of $2019$. Prove that there exists a positive integer $n$, such that for any $p,q$ as described, $(p-1)(q-1)$ divides $nS(p,q)$.
8 replies
mofumofu
Mar 11, 2019
Orthocenter.THOMAS
32 minutes ago
Collinearity with orthocenter
liberator   181
N 35 minutes ago by Giant_PT
Source: IMO 2013 Problem 4
Let $ABC$ be an acute triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by $\omega_1$ is the circumcircle of $BWN$, and let $X$ be the point on $\omega_1$ such that $WX$ is a diameter of $\omega_1$. Analogously, denote by $\omega_2$ the circumcircle of triangle $CWM$, and let $Y$ be the point such that $WY$ is a diameter of $\omega_2$. Prove that $X,Y$ and $H$ are collinear.

Proposed by Warut Suksompong and Potcharapol Suteparuk, Thailand
181 replies
liberator
Jan 4, 2016
Giant_PT
35 minutes ago
Simple but hard
Lukariman   0
39 minutes ago
Given triangle ABC. Outside the triangle, construct rectangles ACDE and BCFG with equal areas. Let M be the midpoint of DF. Prove that CM passes through the center of the circle circumscribing triangle ABC.
0 replies
Lukariman
39 minutes ago
0 replies
Interesting inequalities
sqing   11
N an hour ago by sqing
Source: Own
Let $ a,b,c\geq 0 , (a+k )(b+c)=k+1.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2k-3+2\sqrt{k+1}}{3k-1}$$Where $ k\geq \frac{2}{3}.$
Let $ a,b,c\geq 0 , (a+1)(b+c)=2.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 2\sqrt{2}-1$$Let $ a,b,c\geq 0 , (a+3)(b+c)=4.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{7}{4}$$Let $ a,b,c\geq 0 , (3a+2)(b+c)= 5.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2(2\sqrt{15}-5)}{3}$$
11 replies
sqing
May 10, 2025
sqing
an hour ago
Interesting inequalities
sqing   1
N an hour ago by sqing
Source: Own
Let $a,b,c \geq 0 $ and $ abc+2(ab+bc+ca) =32.$ Show that
$$ka+b+c\geq 8\sqrt k-2k$$Where $0<k\leq 4. $
$$ka+b+c\geq 8 $$Where $ k\geq 4. $
$$a+b+c\geq 6$$$$2a+b+c\geq 8\sqrt 2-4$$
1 reply
+1 w
sqing
Yesterday at 2:51 PM
sqing
an hour ago
4-vars inequality
xytunghoanh   3
N an hour ago by lbh_qys
For $a,b,c,d \ge 0$ and $a\ge c$, $b \ge d$. Prove that
$$a+b+c+d+ac+bd+8 \ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{cd}+\sqrt{da}+\sqrt{ac}+\sqrt{bd})$$.
3 replies
xytunghoanh
Yesterday at 2:10 PM
lbh_qys
an hour ago
Interesting inequality of sequence
GeorgeRP   2
N 2 hours ago by oty
Source: Bulgaria IMO TST 2025 P2
Let $d\geq 2$ be an integer and $a_0,a_1,\ldots$ is a sequence of real numbers for which $a_0=a_1=\cdots=a_d=1$ and:
$$a_{k+1}\geq a_k-\frac{a_{k-d}}{4d}, \forall_{k\geq d}$$Prove that all elements of the sequence are positive.
2 replies
GeorgeRP
Wednesday at 7:47 AM
oty
2 hours ago
Coloring cubes in a large cube with restrictions
emi3.141592   1
N 2 hours ago by venhancefan777
Source: Problem 3 from Regional Olympiad of Mexico Southeast 2024
A large cube of size \(4 \times 4 \times 4\) is made up of 64 small unit cubes. Exactly 16 of these small cubes must be colored red, subject to the following condition:

In each block of \(1 \times 1 \times 4\), \(1 \times 4 \times 1\), and \(4 \times 1 \times 1\) cubes, there must be exactly one red cube.

Determine how many different ways it is possible to choose the 16 small cubes to be colored red.

Note: Two colorings are considered different even if one can be obtained from the other by rotations or symmetries of the cube.
1 reply
emi3.141592
Sep 29, 2024
venhancefan777
2 hours ago
IMO 2009, Problem 5
orl   91
N 2 hours ago by maromex
Source: IMO 2009, Problem 5
Determine all functions $ f$ from the set of positive integers to the set of positive integers such that, for all positive integers $ a$ and $ b$, there exists a non-degenerate triangle with sides of lengths
\[ a, f(b) \text{ and } f(b + f(a) - 1).\]
(A triangle is non-degenerate if its vertices are not collinear.)

Proposed by Bruno Le Floch, France
91 replies
orl
Jul 16, 2009
maromex
2 hours ago
What is thiss
EeEeRUT   4
N 2 hours ago by lksb
Source: Thailand MO 2025 P6
Find all function $f: \mathbb{R}^+ \rightarrow \mathbb{R}$,such that the inequality $$f(x) + f\left(\frac{y}{x}\right) \leqslant \frac{x^3}{y^2} + \frac{y}{x^3}$$holds for all positive reals $x,y$ and for every positive real $x$, there exist positive reals $y$, such that the equality holds.
4 replies
EeEeRUT
Wednesday at 6:45 AM
lksb
2 hours ago
problem about equation
jred   2
N 3 hours ago by Truly_for_maths
Source: China south east mathematical Olympiad 2006 problem4
Given any positive integer $n$, let $a_n$ be the real root of equation $x^3+\dfrac{x}{n}=1$. Prove that
(1) $a_{n+1}>a_n$;
(2) $\sum_{i=1}^{n}\frac{1}{(i+1)^2a_i} <a_n$.
2 replies
jred
Jul 4, 2013
Truly_for_maths
3 hours ago
number theory and combinatoric sets of integers relations
trying_to_solve_br   40
N 3 hours ago by MathLuis
Source: IMO 2021 P6
Let $m\ge 2$ be an integer, $A$ a finite set of integers (not necessarily positive) and $B_1,B_2,...,B_m$ subsets of $A$. Suppose that, for every $k=1,2,...,m$, the sum of the elements of $B_k$ is $m^k$. Prove that $A$ contains at least $\dfrac{m}{2}$ elements.
40 replies
trying_to_solve_br
Jul 20, 2021
MathLuis
3 hours ago
Petya and vasya are playing with ones
egxa   1
N Apr 20, 2025 by oolite
Source: All Russian 2025 11.6
$100$ ones are written in a circle. Petya and Vasya take turns making \( 10^{10} \) moves each. In each move, Petya chooses 9 consecutive numbers and decreases each by $2$. Vasya chooses $10$ consecutive numbers and increases each by $1$. They alternate turns, starting with Petya. Prove that Vasya can act in such a way that after each of his moves, there are always at least five positive numbers, regardless of how Petya plays.
1 reply
egxa
Apr 18, 2025
oolite
Apr 20, 2025
Petya and vasya are playing with ones
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G H BBookmark kLocked kLocked NReply
Source: All Russian 2025 11.6
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egxa
211 posts
#1
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$100$ ones are written in a circle. Petya and Vasya take turns making \( 10^{10} \) moves each. In each move, Petya chooses 9 consecutive numbers and decreases each by $2$. Vasya chooses $10$ consecutive numbers and increases each by $1$. They alternate turns, starting with Petya. Prove that Vasya can act in such a way that after each of his moves, there are always at least five positive numbers, regardless of how Petya plays.
This post has been edited 2 times. Last edited by egxa, Apr 18, 2025, 5:20 PM
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oolite
344 posts
#2
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Vasya pays attention only to the ten special positions $9,18,\ldots,90$.

Let $(a_1,b_1),\ldots,(a_5,b_5)$ be the numbers at those special positions.

Whenever Petya plays, either:
  • he changes exactly one of $(a_k,b_k)$ by $-2$ for some $k$,
  • to which Vasya should respond by changing both $a_k$ and $b_k$ by $+1$;
or
  • he misses all ten special numbers,
  • in which case Vasya could play anywhere, but let's say that he also misses all ten special numbers.

Thus in each round of two moves, a pair of special numbers $(a_k,b_k)$ changes by $(0,0), (1,-1)$ or $(-1,1)$, and so is the form $(1+x,1-x)$ after $10^{10}$ rounds. At least one of those two numbers is positive.

Since there are five special pairs, that's at least five positive numbers in total$\quad\square$
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