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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Jul 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
Problem 16
SlovEcience   4
N 30 minutes ago by SunnyEvan
Find the smallest positive integer \( k \) such that the following inequality holds:
\[
x^k y^k z^k (x^3 + y^3 + z^3) \leq 3
\]for all positive real numbers \( x, y, z \) satisfying the condition \( x + y + z = 3 \).
4 replies
SlovEcience
Jul 19, 2025
SunnyEvan
30 minutes ago
Inspired by old results
sqing   4
N 34 minutes ago by RagvaloD
Source: Own
Let $ a,b,c\geq 1 . $ Prove that
$$  (k- \frac {1} {a}) (k- \frac {1} {b}) (k- \frac {1} {c})  \leq (k-1)^3a^2b^2c^2  $$Where $ k\geq \frac{3}{2}$
$$  (2- \frac {1} {a}) (2- \frac {1} {b}) (2- \frac {1} {c})  \leq a^2b^2c^2  $$$$  (3- \frac {2} {a}) (3- \frac {2} {b}) (3- \frac {2} {c})  \leq a^2b^2c^2  $$
4 replies
sqing
2 hours ago
RagvaloD
34 minutes ago
D1053 : Set of Dirichlet
Dattier   1
N 42 minutes ago by Dattier
Source: les dattes à Dattier
We say a set $D$ have the Dirichlet propriety, if $\forall (a,b) \in (\mathbb N^*)^2,\gcd(a,b)=1, a<b$, $\text{card}(\{ n \in\mathbb N, n \mod b=a  \} \cap D)=+\infty$.

Let $D=\{d_1,....,d_n,...\}$ with $\forall i \in \mathbb N^*,d_{i+1}>d_{i}$ subset of $\mathbb N$ and have the Dirichlet propriety.


1) Is it true that $\lim \dfrac{d_{n+1}}{d_n}=1$ ?

2) Is it true that $\liminf \dfrac{d_{n+1}}{d_n}=1$ ?
1 reply
Dattier
Tuesday at 1:23 PM
Dattier
42 minutes ago
Minimum Length for Upper or Lower Subsequences
steven_zhang123   0
an hour ago
Source: 2025 Hope League Test 3 P14
For a sequence of distinct real numbers \( a_1, a_2, \cdots, a_n \), a subsequence \( a_{i_1}, a_{i_2}, \cdots, a_{i_k} \) (\( 1 \leq i_1 < i_2 < \cdots < i_k \leq n \)) is called:
- An upper subsequence if \( a_{i_1} < \cdots < a_{i_k} \) and there exists no index \( m \) with \( i_j < m < i_{j+1} \) such that
\[a_m > \frac{i_{j+1} - m}{i_{j+1} - i_j} \cdot a_{i_j} + \frac{m - i_j}{i_{j+1} - i_j} \cdot a_{i_{j+1}};\]- A lower subsequence if \( a_{i_1} > \cdots > a_{i_k} \) and there exists no index \( m \) with \( i_j < m < i_{j+1} \) such that
\[a_m > \frac{i_{j+1} - m}{i_{j+1} - i_j} \cdot a_{i_j} + \frac{m - i_j}{i_{j+1} - i_j} \cdot a_{i_{j+1}}.\]Find the smallest positive integer \( N \) such that in any sequence of \( N \) distinct real numbers, there is either an upper subsequence of length $20$ or a lower subsequence of length $25$.
Proposed by Dong Zichao and Wu Zhuo
0 replies
steven_zhang123
an hour ago
0 replies
Structure of the group $(\mathbb{Z}/p\mathbb{Z})^{\times}$ and its application t
nayr   0
an hour ago
Let $\mathbb{F}_p^{\times} = (\mathbb{Z} / p\mathbb{Z})^{\times}$ be the unit group of $\mathbb{F}_p$. It is well known that this group is cyclic. Let $g$ be a generator of this group and consider the map $\varphi : \mathbb{F}_p^{\times} \rightarrow \mathbb{F}_p^{\times}, x\mapsto x^k$ for a fixed positive integer $k$. I know that the kernel $\ker \varphi$ has oder $d:= (p-1, k)$. By the first isomorphism theorem, $\mathbb{F}_p^{\times} / \ker \varphi \cong \operatorname{im} \varphi$. Since $\mathbb{F}_p^{\times}$ is cyclic, so are its subgroups and hence $\operatorname{im} \varphi$ is cyclic of oder $\frac{p-1}{d}$. Let $H = \operatorname{im} \varphi$. Then $\mathbb{F}_p^{\times}/H$ is cyclic too and hence we have the partition:

$$\mathbb{F}_p = \{0\} \sqcup H \sqcup s^2H \sqcup \cdots \sqcup s^{d-1}H$$
for any $s\notin H$ (for example $g$).

I am trying to use this fact to solve the following question: Show that $3x^3+4y^3+5z^3 \equiv 0 \pmod{p}$ have non-trivial solution for all primes $p$. Here is my attempt:

For simplicity, we rewrite the original equation for $p>3$, as $x^3+Ay^3+Bz^3\equiv 0 \pmod{p}$ (the case $p=2,3$ is easy).

If $p\equiv 2\pmod{3}$, then everything is a cube (since the cubing map $x\,mapsto x^3$ is an anutomorphism by above) and the equation is solvable.

If $p\equiv 1\pmod{3}$, let $H:=\{x^3|x\in \mathbb{F}_p^{\times}\}$ and $sH, s^2H$ be the cosets where $s \notin H$, then we have the following cases:

Case 1: $A \in H$ or $B\in H$, Without loss of generality, assume $A=4/3$ is a cube, then $4/3=a^3$ or $4=3a^3$ and we may take $(x,y,z)=(a,-1,0)$ as our solution.

Case 2: $A \in sH$ and $B\in sH$, then $A=sa^3$ and $B=sb^3$ and we may take $(x,y,z)=(0,b,-a)$ as our solution.

Case 3: $A \in s^2H$ and $B\in s^2H$, then $A=s^2a^3$ and $B=s^2b^3$ and we may take $(x,y,z)=(0,b,-a)$ as our solution.

Case 4: $A \in sH$ and $B\in s^2H$, then $A=sa^3$ and $B=s^2b^3$. This is the case I am stuck with. If we have $s^3=1$, then we may take $(x,y,z)=(ab,b,a)$ as our solution since $1+s+s^2=0$ for $s^3=1$ and $s$ is not $1$). But it is not always possible to have both $s^3=1$ and $s\notin H$. For example, I can take $s=g^{\frac{p-1}{3}}$, then $s^3=1$, but $g^{\frac{p-1}{3}}\notin H$ iff $9\nmid p-1$.

How should I resolve case 4?
0 replies
nayr
an hour ago
0 replies
Angle Equality in Cyclic Quadrilateral
steven_zhang123   0
an hour ago
Source: 2025 Hope League Test 3 P13
Quadrilateral \( ABCD \) is inscribed in circle \( \Omega \). The tangent to \( \Omega \) at \( A \) intersects line \( BC \) at point \( E \). The perpendicular from \( A \) to line \( AD \) intersects line \( CD \) at point \( F \). Let \( O \) be the circumcenter of \( \triangle BCF \). The perpendicular from \( E \) to line \( AO \) meets it at point \( K \). Prove that \( \angle EKB \) and \( \angle BAD \) are either equal or supplementary.
Proposed by Li Tianqin
0 replies
steven_zhang123
an hour ago
0 replies
Minimum Constant for Triple Sum Products
steven_zhang123   0
an hour ago
Source: 2025 Hope League Test 3 P12
Find the smallest positive real number \( M \) with the following property: For any six non-negative real numbers \( a_1, \dots, a_6 \) summing to $1$, there exist six non-negative real numbers \( b_1, \dots, b_6 \) summing to $1$ such that for any \( 1 \leq i < j < k \leq 6 \),
\[(a_i + a_j + a_k)(b_i + b_j + b_k) \leq M.\]Proposed by Wu Zhuo and Xu Wenchang
0 replies
steven_zhang123
an hour ago
0 replies
Minimizing the Sum of a Function with Inequality Constraint
steven_zhang123   0
an hour ago
Source: 2025 Hope League Test 2 P6
A function \( f : \{1, 2, \cdots, 301\} \rightarrow \{1, 2, \cdots, 301\} \) satisfies that for any positive integer \( x \leq 301 \),
\[f(f(x)) + f(x) \geq 301 + x.\]Find the minimum possible value of \( S = f(1) + f(2) + \cdots + f(301) \).
Proposed by Chen Yiyi and Wu Zhuo
0 replies
steven_zhang123
an hour ago
0 replies
Minimum Area for Monochromatic Triangles
steven_zhang123   0
an hour ago
Source: 2025 Hope League Test 2 P5
Given a positive integer \( n \), find the smallest real number \( S \) such that no matter how each integer point in the coordinate plane \( xOy \) is colored with one of \( n \) different colors, there always exist three non-collinear points \( A, B, C \) of the same color such that the area of \(\triangle ABC\) is at most \( S \).
Proposed by Li Tianqin
0 replies
steven_zhang123
an hour ago
0 replies
Digit Sum Equality in Multiple Bases
steven_zhang123   0
an hour ago
Source: 2025 Hope League Test 2 P4
Let \( S_q(a) \) denote the sum of the digits of the positive integer \( a \) in base \( q \), where \( q \) is an integer greater than or equal to 2. Find the largest positive integer \( k \) such that for any infinite subset \( S \) of positive integers, there exist integers \( 1 < q_1 < q_2 < \cdots < q_{100} \) and \( k \) distinct positive integers \( a_1, a_2, \cdots, a_k \) in \( S \) satisfying \( S_{q_i}(a_1) = S_{q_i}(a_2) = \cdots = S_{q_i}(a_k) \) for each \( i = 1, 2, \cdots, 100 \).
Proposed by Wu Zhuo)
0 replies
steven_zhang123
an hour ago
0 replies
Wordy Geometry in Taiwan TST
ckliao914   10
N an hour ago by ErTeeEs06
Source: 2023 Taiwan TST Round 3 Mock Exam 6
Given triangle $ABC$ with $A$-excenter $I_A$, the foot of the perpendicular from $I_A$ to $BC$ is $D$. Let the midpoint of segment $I_AD$ be $M$, $T$ lies on arc $BC$(not containing $A$) satisfying $\angle BAT=\angle DAC$, $I_AT$ intersects the circumcircle of $ABC$ at $S\neq T$. If $SM$ and $BC$ intersect at $X$, the perpendicular bisector of $AD$ intersects $AC,AB$ at $Y,Z$ respectively, prove that $AX,BY,CZ$ are concurrent.
10 replies
ckliao914
Apr 29, 2023
ErTeeEs06
an hour ago
Group Theory resources
JerryZYang   3
N 5 hours ago by JerryZYang
Can someone give me some resources for group theory. ;)
3 replies
JerryZYang
Yesterday at 8:38 PM
JerryZYang
5 hours ago
Find max(a+√b+∛c) where 0< a, b, c < 1= a+b+c.
elim   7
N Today at 2:25 AM by sqing
Find $\max_{a,\,b,\,c>0\atop a+b+c=1}(a+\sqrt{b}+\sqrt[3]{c})$
7 replies
elim
Feb 7, 2020
sqing
Today at 2:25 AM
Are all solutions normal ?
loup blanc   11
N Yesterday at 9:02 PM by GreenKeeper
This post is linked to this one
https://artofproblemsolving.com/community/c7t290f7h3608120_matrix_equation
Let $Z=\{A\in M_n(\mathbb{C}) ; (AA^*)^2=A^4\}$.
If $A\in Z$ is a normal matrix, then $A$ is unitarily similar to $diag(H_p,S_{n-p})$,
where $H$ is hermitian and $S$ is skew-hermitian.
But are there other solutions? In other words, is $A$ necessarily normal?
I don't know the answer.
11 replies
loup blanc
Jul 17, 2025
GreenKeeper
Yesterday at 9:02 PM
real analysis
ay19bme   4
N Apr 8, 2025 by ay19bme
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4 replies
ay19bme
Apr 7, 2025
ay19bme
Apr 8, 2025
real analysis
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ay19bme
287 posts
#1
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alexheinis
10757 posts
#2
Y by
With $1/k=a$ we can rewrite as $\sum_2^\infty {1\over {n(\ln n)^a}}$.
This converges iff $\int_p^\infty {{dx}\over {x(\ln x)^a}}=\int_q^\infty {{dt}\over {t^a}}$ converges.
This is true iff $a>1$ hence iff $k<1$.

@below: we have $n^{a\ln \ln n/\ln n}=(e^{\ln n})^{a\ln \ln n/\ln n}=e^{a\ln \ln n}=(e^{\ln \ln n})^a=(\ln n)^a$.
This post has been edited 1 time. Last edited by alexheinis, Apr 7, 2025, 8:16 PM
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ay19bme
287 posts
#3
Y by
Can you please explain a little how the given sum turns out to be equal with your sum.
Thanks
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Svenskerhaor
11 posts
#4
Y by
ay19bme wrote:
Can you please explain a little how the given sum turns out to be equal with your sum.
Thanks

\begin{align*}
n^{1+(ln(ln(n))/ln(n))}=nn^{ln(ln(n))/ln(n)}=n(e^{ln n})^{ln(ln(n))/ln(n)}=ne^{ln(ln(n))}=n(ln(n))
\end{align*}
This post has been edited 1 time. Last edited by Svenskerhaor, Apr 7, 2025, 9:50 PM
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ay19bme
287 posts
#5
Y by
what about the lower bound of $k$
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