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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
1 viewing
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 qrxz17
sqing   4
N 3 minutes ago by sqing
Source: Own
Let $a, b,c>0 ,(a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 27 $. Prove that $$a+b+c\geq 3\sqrt {3}$$
4 replies
sqing
3 hours ago
sqing
3 minutes ago
Inspired by qrxz17
sqing   3
N 5 minutes ago by sqing
Source: Own
Let $ a,b,c $ be reals such that $ (a^2+b^2)^2 + (b^2+c^2)^2 +(c^2+a^2)^2 = 28 $ and $  (a^2+b^2+c^2)^2 =16. $ Find the value of $ a^2(a^2-1) + b^2(b^2-1)+c^2(c^2-1).$
3 replies
+1 w
sqing
4 hours ago
sqing
5 minutes ago
Hardest in ARO 2008
discredit   30
N 7 minutes ago by Phat_23000245
Source: ARO 2008, Problem 11.8
In a chess tournament $ 2n+3$ players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least $ n$ next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
30 replies
discredit
Jun 11, 2008
Phat_23000245
7 minutes ago
Find the value
sqing   12
N 10 minutes ago by Phat_23000245
Source: 2024 China Fujian High School Mathematics Competition
Let $f(x)=a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0,$ $a_i\in\{-1,1\} ,i=0,1,2,\cdots,6 $ and $f(2)=-53 .$ Find the value of $f(1).$
12 replies
sqing
Jun 22, 2024
Phat_23000245
10 minutes ago
Midpoints of arcs form a similar triangle
v_Enhance   19
N 10 minutes ago by Adywastaken
Source: USA TSTST 2013, Problem 1
Let $ABC$ be a triangle and $D$, $E$, $F$ be the midpoints of arcs $BC$, $CA$, $AB$ on the circumcircle. Line $\ell_a$ passes through the feet of the perpendiculars from $A$ to $DB$ and $DC$. Line $m_a$ passes through the feet of the perpendiculars from $D$ to $AB$ and $AC$. Let $A_1$ denote the intersection of lines $\ell_a$ and $m_a$. Define points $B_1$ and $C_1$ similarly. Prove that triangle $DEF$ and $A_1B_1C_1$ are similar to each other.
19 replies
v_Enhance
Aug 13, 2013
Adywastaken
10 minutes ago
find question
mathematical-forest   4
N 11 minutes ago by GreekIdiot
Are there any contest questions that seem simple but are actually difficult? :-D
4 replies
mathematical-forest
2 hours ago
GreekIdiot
11 minutes ago
4 var inequality
SunnyEvan   1
N 13 minutes ago by SunnyEvan
Source: Own
Let $ x,y,z,t \in R^+ ,$ such that : $ (x+y+z+t)^2 = x+y+z+t + (x+z)(y+t) $ and $ x \geq y \geq z \geq t .$
Try to prove or disprove : $$ \frac{2 \sqrt{x+y+z+t +(x+t)(y+z)}}{x^2+y^2+z^2+t^2 +3xz+3yt+xt+yz} \geq \frac{11(x+y)(z+t)-(x+y+z+t)}{x+y+z+t +(x+z)(y+t)} $$
1 reply
SunnyEvan
5 hours ago
SunnyEvan
13 minutes ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   14
N 44 minutes ago by math90
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
14 replies
OgnjenTesic
May 22, 2025
math90
44 minutes ago
Three Distinct Divisors Sum to 2022
ike.chen   31
N an hour ago by Jupiterballs
Source: ISL 2022/N1
A number is called Norwegian if it has three distinct positive divisors whose sum is equal to $2022$. Determine the smallest Norwegian number.
(Note: The total number of positive divisors of a Norwegian number is allowed to be larger than $3$.)
31 replies
ike.chen
Jul 9, 2023
Jupiterballs
an hour ago
MOP 2012 Inequality
holdmyquadrilateral   3
N an hour ago by Adywastaken
Source: MOP 2012
For $a,b,c>0$, prove that
\[\frac{a^3}{(b-c)^2+bc}+\frac{b^3}{(c-a)^2+ca}+\frac{c^3}{(a-b)^2+ab}\ge a+b+c.\]
3 replies
holdmyquadrilateral
Mar 11, 2023
Adywastaken
an hour ago
strange geometry problem
Zavyk09   2
N an hour ago by Zavyk09
Source: own
Let $ABC$ be a triangle with circumcenter $O$ and internal bisector $AD$. Let $AD$ cuts $(O)$ again at $M$ and $MO$ cuts $(O)$ again at $N$. Point $L$ lie on $AD$ such that $(AD, LM) = -1$. The line pass through $L$ and perpendicular to $AD$ intersects $NC, NB$ at $P, Q$ respectively. Let circumcircle of $\triangle NPQ$ cuts $(O)$ at $G \ne N$. Prove that $\angle AGD = 90^{\circ}$.
2 replies
Zavyk09
Yesterday at 4:32 PM
Zavyk09
an hour ago
exponential diophantine with factorials
skellyrah   5
N an hour ago by skellyrah
find all non negative integers (x,y) such that $$ x! + y! = 2025^x + xy$$
5 replies
skellyrah
Feb 24, 2025
skellyrah
an hour ago
A second final attempt to make a combinatorics problem
JARP091   2
N 2 hours ago by JARP091
Source: At the time of writing this problem I do not know the source if any
Arthur Morgan is playing a game.

He has $n$ eggs, each with a hardness value $k_1, k_2, \dots, k_n$, where $\{k_1, k_2, \dots, k_n\}$ is a permutation of the set $\{1, 2, \dots, n\}$. He is throwing the eggs from an $m$-floor building.

When the $i$-th egg is dropped from the $j$-th floor, its new hardness becomes
\[
\left\lfloor \frac{k_i}{j+1} \right\rfloor.
\]If $\left\lfloor \frac{k_i}{j+1} \right\rfloor = 0$, then the egg breaks and cannot be used again.

Arthur can drop each egg from a particular floor at most once.
For which values of $n$ and $m$ can Arthur always determine the correct ordering of the eggs according to their initial hardness values?
Note: The problem might be wrong or too easy
2 replies
JARP091
May 25, 2025
JARP091
2 hours ago
Iran 3rd Round Geo
M11100111001Y1R   12
N 2 hours ago by Mathgloggers
Source: Iran 2024 3rd Round Test 1 P3
Consider an acute scalene triangle $\triangle{ABC}$. The interior bisector of $A$ intersects $BC$ at $E$ and the minor arc of $\overarc {BC}$ in circumcircle of $\triangle{ABC}$ at $M$. Suppose that $D$ is a point on the minor arc of $\overarc {BC}$ such that $ED=EM$. $P$ is a point on the line segment of $AD$ such that $\angle ABP=\angle ACP \not= 0$. $O$ is the circumcenter of $\triangle{ABC}$. Prove that $OP \perp AM$.
12 replies
M11100111001Y1R
Aug 23, 2024
Mathgloggers
2 hours ago
Connecting chaos in a grid
Assassino9931   3
N Apr 25, 2025 by dgrozev
Source: Bulgaria National Olympiad 2025, Day 1, Problem 2
Exactly \( n \) cells of an \( n \times n \) square grid are colored black, and the remaining cells are white. The cost of such a coloring is the minimum number of white cells that need to be recolored black so that from any black cell \( c_0 \), one can reach any other black cell \( c_k \) through a sequence \( c_0, c_1, \ldots, c_k \) of black cells where each consecutive pair \( c_i, c_{i+1} \) are adjacent (sharing a common side) for every \( i = 0, 1, \ldots, k-1 \). Let \( f(n) \) denote the maximum possible cost over all initial colorings with exactly \( n \) black cells. Determine a constant $\alpha$ such that
\[
\frac{1}{3}n^{\alpha} \leq f(n) \leq 3n^{\alpha}
\]for any $n\geq 100$.
3 replies
Assassino9931
Apr 8, 2025
dgrozev
Apr 25, 2025
Connecting chaos in a grid
G H J
Source: Bulgaria National Olympiad 2025, Day 1, Problem 2
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Assassino9931
1374 posts
#1 • 1 Y
Y by cubres
Exactly \( n \) cells of an \( n \times n \) square grid are colored black, and the remaining cells are white. The cost of such a coloring is the minimum number of white cells that need to be recolored black so that from any black cell \( c_0 \), one can reach any other black cell \( c_k \) through a sequence \( c_0, c_1, \ldots, c_k \) of black cells where each consecutive pair \( c_i, c_{i+1} \) are adjacent (sharing a common side) for every \( i = 0, 1, \ldots, k-1 \). Let \( f(n) \) denote the maximum possible cost over all initial colorings with exactly \( n \) black cells. Determine a constant $\alpha$ such that
\[
\frac{1}{3}n^{\alpha} \leq f(n) \leq 3n^{\alpha}
\]for any $n\geq 100$.
This post has been edited 1 time. Last edited by Assassino9931, Apr 9, 2025, 8:54 AM
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internationalnick123456
135 posts
#2 • 1 Y
Y by cubres
Answer
Solution
This post has been edited 3 times. Last edited by internationalnick123456, Apr 9, 2025, 10:15 AM
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Assassino9931
1374 posts
#3 • 1 Y
Y by internationalnick123456
Answer

Construction for upper bound

Argument for lower bound

By accident, the constants actually matter!
This post has been edited 1 time. Last edited by Assassino9931, Apr 9, 2025, 8:46 PM
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dgrozev
2466 posts
#4 • 1 Y
Y by Miquel-point
Here is an argument using a spannong tree for the lower bound.

We may think of the configuration as a graph $G$ with $n^2$ vertices (cells) and edges that connect each two adjacent cells (sharing a side). We want to choose $m$ such that a grid with distance $m$ between two consecutive columns/rows contains at least $n$ cells. So $m$ must satisfy:
$$\left(\left\lfloor \frac{n-1}{m}\right\rfloor+1\right)^2\ge n.$$An easy calculation shows that it's enough to take $m:=\left\lfloor\sqrt{n}-\frac{1}{\sqrt{n}}\right\rfloor$. Let $V'$ be the set of $n$ cells, any two of them of distance at least $m$ between them. Let $T(V)$ be a connected subgraph of $G$ with a vertex-set $V$ such that $V'\subset V$. In what follows below, given an edge $e=uv$, by |uv| we denote its length, that is, the number of hops needed to reach $v$ from $u$, or, in other words, the so called manhattan distance between $u$ and $v$. |T| denotes the total length of $T$, that is, the sum of lengths of its edges.

Clearly $T$ is a tree, because we can remove the redundant edges until $T$ is still connected. Note that the claim is trivial in case $V'=V$, because in this case $|V|=n$ and each edge of $T$ has a length at least $m$, hence $|T|\ge (n-1)m$. (There was a glitch in the originally proposed solution, since it took for granted that $V'=V$, which cannot be guaranteed).

So, a bit more care is needed in case $V'\subsetneq V$. Let us refer to the vertices in $V\setminus V'$ as to the white vertices and to the vertices in $V'$ as to the black vertices. Let $v_0$ be a white vertex of $T$. We refer further to it as the root of $T$. We apply the following

Procedure. We delete all white leaves of $T$ (except of $v_0$). Take a leaf $v$ of $T$ with the largest number of vertices between $v$ and $v_0$. Let $v'$ be the neighbor of $v$. If $v'$ is black, we delete the edge $v'v$ whose length is at least $m$ and start the procedure from the beginning.

Assume now, $v'$ is white. If $\deg(v')=2$, we can assume that there is no need of $v'$ and just prolong the two edges that $v'$ is incident with. Let $v_1=v, v_2,\ldots,v_k$ be all the leaves that $v'$ is connected to, and $k\ge 2$. We have:
$$|v'v_i|+|v'v_{i+1}|\ge |v_iv_{i+1}|, i=1,2,\ldots,k,$$where we assume $v_{k+1}=v_1$. Note that $|v_iv_{i+1}|\ge m$ since all $v_i$ are black. Summing up these inequalities, we get:
$$\sum_{i=1}^k|v'v_i|\ge \frac{1}{2}\sum_{i=1}^k \left |v_i v_{i+1}\right|\ge \frac{km}{2}.$$Now, we delete the vertex $v'$. With this, we delete edges with length at least $mk/2$.
Repeat the procedure till we are left with the root $v_0$, which is white.

Note that each time we delete a set of $k$ black vertices, the corresponding deleted edges have total length at least $mk/2$, that is, the number of cells not in $V'$ that cover these edges is at least $mk/2-k=k(m-2)/2$. Therefore, the total number of cells not in $V'$ that cover $T$ is at least:
$$|V'|\cdot (m-2)/2=\frac{1}{2}n \cdot (m-2) \ge \frac{n(\sqrt{n}-3-1/\sqrt{n})}{2}\ge \frac{n^{3/2}}{3},$$as $n\ge 100$.
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