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k a December Highlights and 2024 AoPS Online Class Information
jlacosta   0
Dec 2, 2024
Warmest Holiday Wishes!
As 2024 draws to a close and we look ahead to next year, we find ourselves in a perfect moment for reflection. This season gives us a chance to celebrate our successes, learn from our challenges, and express gratitude to everyone who has been part of our problem-solving journey.

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[*]Crack the code for how to participate in your first USA Computing Olympiad at our FREE math jam on December 10th.
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[*]Please note: AoPS will not be holding classes December 21 ‐ January 3 and our office will be closed December 24 - January 3.[/list]
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0 replies
jlacosta
Dec 2, 2024
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
best bounds of f(a,b,c)=4(1/a+1/b+1/c)-1/abc, when a+b+c=1, 0<a,b,c<1/2
parmenides51   7
N 9 minutes ago by iStud
Source: Balkan BMO Shortlist 2017 A4
Let $M = \{(a,b,c)\in R^3 :0 <a,b,c<\frac12$ with $a+b+c=1 \}$ and $f: M\to  R$ given as $$f(a,b,c)=4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{abc}$$Find the best (real) bounds $\alpha$ and $\beta$ such that $f(M) = \{f(a,b,c): (a,b,c)\in M\}\subseteq [\alpha,\beta]$ and determine whether any of them is achievable.
7 replies
parmenides51
Aug 1, 2019
iStud
9 minutes ago
A small NT problem
cuden   3
N 17 minutes ago by cuden
Problem ___
3 replies
cuden
Dec 2, 2024
cuden
17 minutes ago
Orthocenter on general cevian
p108771953   5
N 32 minutes ago by gnoka
Source: 2024 Mexican Mathematical Olympiad, Problem 4
Let $ABC$ an acute triangle with orthocenter $H$. Let $M$ be a point on segment $BC$. The line through $M$ and perpendicular to $BC$ intersects lines $BH$ and $CH$ in points $P$ and $Q$ respectively. Prove that the orthocenter of triangle $HPQ$ lies on the line $AM$.
5 replies
p108771953
Nov 6, 2024
gnoka
32 minutes ago
Incenter and intersection of lines making rhombus
falantrng   7
N an hour ago by Ianis
Source: Azerbaijan NMO 2023. Junior P2
Let $I$ be the incenter in the acute triangle $ABC.$ Rays $BI$ and $CI$ intersect the circumcircle of triangle $ABC$ at points $S$ and $T,$ respectively. The segment $ST$ intersects the sides $AB$ and $AC$ at points $K$ and $L,$ respectively. Prove that $AKIL$ is a rhombus.
7 replies
falantrng
Aug 24, 2023
Ianis
an hour ago
Six orthopoles lie on a circle
buratinogigle   2
N an hour ago by buratinogigle
Let $ABC$ and $A'B'C'$ be two triangles inscribed circle $(O)$. Prove that orthopoles of $B'C',C'A',A'B'$ with respect to triangle $ABC$ and orthopoles of $BC,CA,AB$ with respect to triangle $A'B'C'$ lie on a circle.
2 replies
buratinogigle
Sep 2, 2012
buratinogigle
an hour ago
Four triangles...
StewPidity   0
an hour ago
Scalene $\triangle{ABC}$ with circumcenter $O$ has orthic triangle $\triangle{A_{1}B_{1}C_{1}}$, which has tangential triangle $\triangle{A_{2}B_{2}C_{2}}$ (opposite vertices labelled correspondingly). If $H$ and $O'$ are the orthocenter and circumcenter of the triangle formed by lines $AA_{2}, BB_{2}, CC_{2}$, respectively, prove that $OO'=HO'$.
0 replies
StewPidity
an hour ago
0 replies
nice lemma
henderson   6
N an hour ago by rudransh61
Source: İMO Shortlist 2009, A4
Prove that for all $x,y\in \Bbb{R}^+
  $\[\sqrt{\frac{x^2+y^2}{2}}+\sqrt{xy}\leq \\x+y\] holds.
6 replies
henderson
Jun 2, 2015
rudransh61
an hour ago
How to do complex bash?
EaZ_Shadow   4
N an hour ago by rudransh61
How do you do complex bash and use the fact that it’s in the complex plane to make the math much easier? Also, why do people do complex bash because frankly, it doesn’t make sense to me and how do you use complex numbers in geometry. Also, how do you use it and when to use it? Thanks!
4 replies
EaZ_Shadow
Today at 12:28 AM
rudransh61
an hour ago
Integral inequality
ohiorizzler1434   1
N an hour ago by rudransh61
Given functions f(x) and g(x) non-negative such that g(x) is decreasing, show $\int f'(x)g(x) \geq f(x)g(x)$.
1 reply
ohiorizzler1434
2 hours ago
rudransh61
an hour ago
Inequality 144
tritanngo99   1
N 2 hours ago by arqady
Cho $x,y,z$ are positive real number satisfied: $xy+yz+zx=1$. Prove that: $\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+z^2}\ge \frac{2}{3}(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}})^3$
1 reply
tritanngo99
2 hours ago
arqady
2 hours ago
Classic - sum of rows and columns nonnegative
randomusername   9
N 2 hours ago by isaacren
Source: 1961 All-Soviet Union Olympiad
Consider a table with one real number in each cell. In one step, one may switch the sign of the numbers in one row or one column simultaneously. Prove that one can obtain a table with non-negative sums in each row and each column.
9 replies
randomusername
Aug 4, 2015
isaacren
2 hours ago
Since November 4-th 2023
mihaig   6
N 3 hours ago by mihaig
Source: Own
Let $K$ be a fixed real number. Consider the set
$$S_K=\left\{(a,b,c) \in \mathbb{R}^3|~3\sum_{\text{cyc}}{a^2}+(K-1)\left(\sum_{\text{cyc}}{a}\right)^2=9\right\}.$$Find $\inf_{(a,b,c) \in S_K}abc$ and $\sup_{(a,b,c) \in S_K}abc.$
6 replies
mihaig
Dec 8, 2024
mihaig
3 hours ago
Inspired by KTOM
sqing   0
3 hours ago
Source: Own
Let $ a,b> 0 $ and $ (\sqrt{a}+1)(2\sqrt{b}+4)+b\ge 13. $ Prove that $$\frac{2a^4}{b}+\frac{b^3}{a}+6b \geq9$$
0 replies
sqing
3 hours ago
0 replies
Graph combi with weird constants
Davdav1232   1
N 3 hours ago by R8kt
Source: Israel TST 2 2025 p2
A graph with $10^{100}$ vertices satisfies the following condition: Any simple odd cycle has length > 100. Prove there is an independent set in the graph of size at least $\frac{10^{100}}{102}$
1 reply
Davdav1232
Yesterday at 8:38 PM
R8kt
3 hours ago
Every positive integer divides some number of a sequence
freemind   8
N Aug 20, 2020 by VicKmath7
Source: 1st Romanian Master in Mathematics (RMIM) 2008, Bucharest, Problem 3
Let $ a>1$ be a positive integer. Prove that every non-zero positive integer $ N$ has a multiple in the sequence $ (a_n)_{n\ge1}$, $ a_n=\left\lfloor\frac{a^n}n\right\rfloor$.
8 replies
freemind
Feb 9, 2008
VicKmath7
Aug 20, 2020
Every positive integer divides some number of a sequence
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Source: 1st Romanian Master in Mathematics (RMIM) 2008, Bucharest, Problem 3
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freemind
337 posts
#1 • 7 Y
Y by anantmudgal09, tenplusten, Adventure10, megarnie, Mango247, and 2 other users
Let $ a>1$ be a positive integer. Prove that every non-zero positive integer $ N$ has a multiple in the sequence $ (a_n)_{n\ge1}$, $ a_n=\left\lfloor\frac{a^n}n\right\rfloor$.
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Fedor Petrov
520 posts
#2 • 4 Y
Y by rafayaashary1, Adventure10, and 2 other users
Strange problem. Let $ N=N_1N_2$, where $ N_2$ and $ a$ are coprime, and $ N_1$ divides $ a^C$ for suitable positive integer $ C$. Without loss of generality, $ CN_1$ divides $ a^{C}$ (we may take $ C$ being large power of $ a$). Take large prime $ p$ such that $ p-1$ is divisible by $ \varphi(n)$. The existence of such $ p$ is well-known elementary case of the Dirichlet theorem, which may be obtained using the cyclotomic polynomials. Then take $ n=pC$. Then by Fermat's little theorem we have $ [a^{Cp}/(Cp)]=(a^{Cp}-a^C)/(Cp)$, which is divisible by $ N=N_1N_2$.
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hrkksym0
6 posts
#3 • 1 Y
Y by Adventure10
We can solve this problem without the Dirichlet theorem.

There are positive integers $m$ and $M$ such that $GCD(m,M)=1, N\mid a^{m}M$.
We consider $M>1$(it is obvious when $M=1$)
Let $R$ is an integer such that $M^{R}>a^{m+\varphi(M)}$ and
$a^{R} mod \varphi(M), a^{R+1} mod \varphi(M), a^{R+2} mod \varphi(M), ...$
is a periodical sequence(the period is not more than $\varphi(M)$).
Then there exists an integer $0\le i\le\varphi(M)$ and
$a^{a^{R}M^{R}-m-i}\equiv a^{R}\pmod{\varphi(M)}$.
Let $k=a^{R}M^{R}-m-i$ and $n=a^{k}M^{R}$, then

$
\varphi(M^{R+1})=M^{R}\varphi(M)\mid n-k-m-i\\
\Rightarrow a^{n-k-m-i}\equiv 1\pmod{M^{R+1}}\\
\Rightarrow a^{n-k}\equiv a^{m+i}\pmod{a^{m}M^{R+1}}\\
\Rightarrow \lfloor\dfrac{a^{n-k}}{M^{R}}\rfloor=\lfloor\dfrac{a^{n}}{n}\rfloor\\
=\lfloor\dfrac{Ia^{m}M^{R+1}+a^{m+i}}{M^{R}}\rfloor (I\in\mathbb{Z})\\
=Ia^{m}M+\lfloor\dfrac{a^{m+i}}{M^{R}}\rfloor=Ia^{m}M   ( a^{m+i}\le a^{m+\varphi(M)}<M^{R})\\
\Rightarrow N\mid \lfloor\dfrac{a^{n}}{n}\rfloor\\
$
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dinoboy
2903 posts
#4 • 2 Y
Y by Adventure10, Mango247
This solution was found in collaboration with proglote. it is quite similar to Fedor Petrov's solution actually.

Write $N = bc$ where $\gcd(c,a) = \gcd(b,c) = 1$. Let $b|a^k$ and $k$ clearly exists since $b$'s prime factors are purely those of $a$ as well. Then consider $n = a^kp$ for a prime $p \equiv 1 \pmod{\text{ord}_c(a)}$ and $p > \max(c,a^k - k)$ which exists by Dirichlet. It is easy to see that: \[\frac{a^{a^kp}}{a^kp} = \frac{a^{a^kp-k}}{p}\]
Then remark that $a^{a^kp-k} \equiv a^{a^k-k} \pmod{p}$, so it follows \[a_{a^kp} = \frac{a^{a^kp - k} - a^{a^k-k}}{p}\]

It is clear that $a^{a^k(p-1)}-1$ divides the numerator. It follows that as $c|(a^{a^k(p-1)}-1)$ due to the order condition and $\gcd(c,p) = 1$ due to $p > c$ that $c$ divides $a_{a^kp}$. It is also clear that $b$ must divide $a_{a^kp}$ as $a^{a^k-k}$ divides the numerator. Thus we are done as we have shown an arbitrary $N$ divides some term of the sequence.
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JuanOrtiz
366 posts
#5 • 1 Y
Y by Adventure10
I solved it pretty ugly, no Dirichlet. Sorry I used different values. I want $k | floor(a^r/r)$

Instead of $r$ I used $ra^x$, and then I define $P$ very big and then $Q$ such that $ord_Q a=P$ and then $r$ such that $ord_r a= Q$ using two Zsigmondys.

Then I can control $x$ in mod $d=ord_{k_1} a$ (where $k_1$ is biggest divisor of $k$ coprime to $a$), mod $ORD_d a$ (where $ORD$ is like the $ord$ function but without caring about $a$ and the modulus coprime), mod $P$ and mod $Q$ and use Chinese Theorem to finish, choosing $a^{ra^x-x} \equiv a^{c+i}$ (mod $kr$), where $c$ is such that $g | a^c$ ($g=k/k_1$) and $i<d$).

$i$ is useful since $ORD_da$ and $d$ aren't necessarily coprime.
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anantmudgal09
1978 posts
#6 • 1 Y
Y by Adventure10
Probably i am mistaken or this is on the easier end of RMM. Indeed let $p$ be a prime such that $p>a$ and $p$ is congruent to 1 modulo $\phi(N)$.(Clearly by the properties of cyclotomic polynomials infinitely many such $p$ exist or just use Weak Dirichlet's theorem) Now clearly $N$ divides $a_p$
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Shaddoll
688 posts
#7 • 4 Y
Y by guangzhou-2015, canhhoang30011999, tenplusten, Adventure10
Solution
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MNJ2357
643 posts
#8 • 2 Y
Y by Adventure10, Mango247
I'll just post this cuz it's different from the above solutions.
(Actually, #6 and #7 are really nice, but proving that there are infinitely many primes in the $nk+1$ takes a lot of time.)

My main idea is Power Towers.

We'll prove that $n=b_m-1$ works for sufficiently large $m>2$. One can easily check that
$$a_n=\left\lfloor\frac{a^n}n\right\rfloor=\left\lfloor\frac{a^{a^{b_{m-1}}-1}}{a^{a^{b_{m-2}}}-1}\right\rfloor=\frac{a^{a^{b_{m-1}}-1}-a^{a^{b_{m-2}}-1}}{a^{a^{b_{m-2}}}-1}=a^{b_{m-1}-1}\cdot\frac{a^{b_m-b_{m-1}}-1}{a^{b_{m-1}}-1}$$Now let $N=N_1N_2$, where $N_2$ is the largest divisor of $N_2$ such that $(a,N_2)=1$. We'll choose $m$ large enough so that $N_1|a^{b_{m-1}-1}$. Observe that for each $p|N_2$, the sequence $v_p(a^{b_m}-1)$ is bounded, so let $x_p$ be the upper bound. Using the well-known property of power-towers, find some $M$ such that
$$m>M\Longrightarrow \phi(N_2\cdot\prod_{p|N_2}p^{x_p})|{b_m-b_{m-1}}$$, and we're done. :)
This post has been edited 1 time. Last edited by MNJ2357, Feb 8, 2019, 7:49 AM
Reason: (N,N_2)=1 -> (a,N_2)=1
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VicKmath7
1375 posts
#9
Y by
Same as above solutions, posting for storage.
Solution. Take a prime $p$, such that $p>a$ (1) and $p=k.\phi{N}$ (*) and $(p,N)=1$ (**)(existence is guaranteed by Dirichlet`s theorem). By FLT, $a^p-a$ is divisible by $p$ (2).From (1) and (2), we have that $a_p=\lfloor\frac{a^p}{p}\rfloor=\frac {a^p-a}{p}$. Using (*), (**) and Euler`s theorem, we immediately see that $a_p$ is divisible by $N$, which is what we need. $\blacksquare $
This post has been edited 1 time. Last edited by VicKmath7, Aug 20, 2020, 12:06 PM
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