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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
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0 replies
jlacosta
Mar 2, 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
Is it fake? how can someone score 12 in AIME but can't qualify RMO
Bruce_wayne123   3
N a few seconds ago by Bruce_wayne123
Source: https://www.reddit.com/r/JEENEETards/comments/1jgduci/op_qualified_for_usamo/#lightbox
He also claims to have scored 93 percentile in JEEM maths another thing which makes it more doubtful and also he didn't got any letter from MAA
3 replies
2 viewing
Bruce_wayne123
23 minutes ago
Bruce_wayne123
a few seconds ago
2 var inquality
sqing   4
N a minute ago by sqing
Source: Own
Let $ a,b $ be nonnegative real numbers such that $ a^2+ab+b^2+a+b=1. $ Prove that
$$  (ab+1)(a+b)\leq \frac{ 20}{27}  $$$$ (ab+1)(a+b-1)\leq  - \frac{ 10}{27}  $$Let $ a,b $ be nonnegative real numbers such that $ a^2+b^2+a+b=1. $ Prove that
$$  (ab+1)(a+b)\leq \frac{ 5\sqrt 3-7}{2}  $$$$ (ab+1)(a+b-1)\leq 3\sqrt 3- \frac{ 11}{2}  $$
4 replies
+1 w
sqing
Yesterday at 3:00 PM
sqing
a minute ago
AZE JBMO TST
IstekOlympiadTeam   9
N 3 minutes ago by DensSv
Source: AZE JBMO TST
Prove that there are not intgers $a$ and $b$ with conditions,
i) $16a-9b$ is a prime number.
ii) $ab$ is a perfect square.
iii) $a+b$ is also perfect square.
9 replies
IstekOlympiadTeam
May 2, 2015
DensSv
3 minutes ago
Iterates of function give distinct residues
Tintarn   6
N 9 minutes ago by AshAuktober
Source: Abelkonkurransen Finale 2024, Problem 1b
Find all functions $f:\mathbb{Z} \to \mathbb{Z}$ such that the numbers
\[n, f(n),f(f(n)),\dots,f^{m-1}(n)\]are distinct modulo $m$ for all integers $n,m$ with $m>1$.
(Here $f^k$ is defined by $f^0(n)=n$ and $f^{k+1}(n)=f(f^{k}(n))$ for $k \ge 0$.)
6 replies
Tintarn
Mar 8, 2024
AshAuktober
9 minutes ago
Interesting NT function
amogususususus   5
N 11 minutes ago by AshAuktober
Source: Indonesia TSTST - Number Theory
Find all functions $f:\mathbb{N} \rightarrow \mathbb{N}$ such that for every prime number $p$ and natural number $x$,
$$\{ x,f(x),\cdots f^{p-1}(x) \} $$is a complete residue system modulo $p$. With $f^{k+1}(x)=f(f^k(x))$ for every natural number $k$ and $f^1(x)=f(x)$.

Proposed by IndoMathXdZ
5 replies
amogususususus
Feb 29, 2024
AshAuktober
11 minutes ago
Find min
hunghd8   1
N 13 minutes ago by Mathzeus1024
Let $a,b,c$ be nonnegative real numbers such that $ a+b+c\geq 2+abc $. Find min
$$P=a^2+b^2+c^2.$$
1 reply
hunghd8
2 hours ago
Mathzeus1024
13 minutes ago
4 free variables ineq
RainbowNeos   0
16 minutes ago
Given $0\leq a,b,c,d\leq 1$, show that
\[abc\sqrt{1-d}+bcd\sqrt{1-a}+cda\sqrt{1-b}+dab\sqrt{1-c}\leq 1.\]
0 replies
RainbowNeos
16 minutes ago
0 replies
An interesting inequality
JK1603JK   0
29 minutes ago
Source: unknown
Let a,b,c>=0 and a^2+b^2+c^2+abc=4 then prove \frac{1}{a+b+2}+\frac{1}{b+c+2}+\frac{1}{c+a+2} \le \frac{6-(a+b+c)}{4}
When does equality occur?
0 replies
JK1603JK
29 minutes ago
0 replies
Romania Junior TST 2021 Day 3 P2
oVlad   4
N an hour ago by DensSv
Let $O$ be the circumcenter of triangle $ABC$ and let $AD$ be the height from $A$ ($D\in BC$). Let $M,N,P$ and $Q$ be the midpoints of $AB,AC,BD$ and $CD$ respectively. Let $\mathcal{C}_1$ and $\mathcal{C}_2$ be the circumcircles of triangles $AMN$ and $POQ$. Prove that $\mathcal{C}_1\cap \mathcal{C}_2\cap AD\neq \emptyset$.
4 replies
oVlad
Jun 7, 2021
DensSv
an hour ago
A lot of z
Anulick   4
N an hour ago by quasar_lord
Source: CMI 2024
(a) FInd the number of complex roots of $Z^6 = Z + \bar{Z}$
(b) Find the number of complex solutions of $Z^n = Z + \bar{Z}$ for $n \in \mathbb{Z}^+$
4 replies
Anulick
May 19, 2024
quasar_lord
an hour ago
Inspired by Titu Andreescu
sqing   2
N an hour ago by sqing
Source: Own
Let $ a,b,c>0 $ and $ a+b+c\geq 3abc . $ Prove that
$$a^2+b^2+c^2+1\geq \frac{4}{3}(ab+bc+ca) $$
2 replies
sqing
5 hours ago
sqing
an hour ago
INMO Problem 1
JetFire008   0
an hour ago
Source: INMO 2007
Problem 1: In a triangle $ABC$ right-angled at $C$, the median through $B$ bisects the angle between $BA$ and the bisector of $\angle B$. Prove that $\frac{5}{2}<\frac{AB}{BC}<3$.
0 replies
JetFire008
an hour ago
0 replies
Show these 2 circles are tangent to each other.
MTA_2024   0
an hour ago
A, B, C, and O are four points in the plane such that
\(\angle ABC > 90^\circ\)
and
\( OA = OB = OC \).

Let \( D \) be a point on \( (AB) \), and let \( (d) \) be a line passing through \( D \) such that
\( (AC) \perp (DC) \)
and
\( (d) \perp (AO) \).

The line \( (d) \) intersects \( (AC) \) at \( E \) and the circumcircle of triangle \( ABC \) at \( F \) (\( F \neq A \)).

Show that the circumcircles of triangles \( BEF \) and \( CFD \) are tangent at \( F \).
0 replies
MTA_2024
an hour ago
0 replies
CMI just asks Schur directly
Anulick   6
N an hour ago by quasar_lord
(a) For non negetive $a,b,c, r$ prove that
\[a^r(a-b)(a-c) + b^r(b-a)(b-c) + c^r (c-a)(c-b) \geq 0 \](b) Find an inequality for non negative $a,b,c$ with $a^4+b^4+c^4 + abc(a+b+c)$ on the greater side.
(c) Prove that if $abc = 1$ for non negative $a,b,c$, $a^4+b^4+c^4+a^3+b^3+c^3+a+b+c \geq \frac{a^2+b^2}{c}+\frac{b^2+c^2}{a}+\frac{c^2+a^2}{b}+3$
6 replies
Anulick
May 19, 2024
quasar_lord
an hour ago
degree of f=2^k
Sayan   15
N Mar 18, 2025 by Gejabsk
Source: ISI 2012 #8
Let $S = \{1,2,3,\ldots,n\}$. Consider a function $f\colon S\to S$. A subset $D$ of $S$ is said to be invariant if for all $x\in D$ we have $f(x)\in D$. The empty set and $S$ are also considered as invariant subsets. By $\deg (f)$ we define the number of invariant subsets $D$ of $S$ for the function $f$.

i) Show that there exists a function $f\colon S\to S$ such that $\deg (f)=2$.

ii) Show that for every $1\leq k\leq n$ there exists a function $f\colon S\to S$ such that $\deg (f)=2^{k}$.
15 replies
Sayan
May 13, 2012
Gejabsk
Mar 18, 2025
degree of f=2^k
G H J
Source: ISI 2012 #8
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Sayan
2130 posts
#1 • 2 Y
Y by mathbuzz, Adventure10
Let $S = \{1,2,3,\ldots,n\}$. Consider a function $f\colon S\to S$. A subset $D$ of $S$ is said to be invariant if for all $x\in D$ we have $f(x)\in D$. The empty set and $S$ are also considered as invariant subsets. By $\deg (f)$ we define the number of invariant subsets $D$ of $S$ for the function $f$.

i) Show that there exists a function $f\colon S\to S$ such that $\deg (f)=2$.

ii) Show that for every $1\leq k\leq n$ there exists a function $f\colon S\to S$ such that $\deg (f)=2^{k}$.
This post has been edited 3 times. Last edited by Sayan, May 16, 2012, 7:18 AM
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iarnab_kundu
866 posts
#2 • 3 Y
Y by mathbuzz, Adventure10, and 1 other user
pls post it in combinatorics; I wrote it fully but someone erased everything :furious:

writing again :

lets take f to be a permutation. consider a permutation where

(n-t+2,...,n) are fixed points. and consider the biggest of the remaining elements; assign the biggest i.e (n-t+1) to f(1) ; and the smallest to f(2), second smallest to f(3)(if it is possible ) and so on

this works (easy to prove)


for example the required permutation of (1,2,..,5) with degree 2^3 is (3,1,2,4,5)
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mavropnevma
15142 posts
#3 • 6 Y
Y by Abhinandan18, opptoinfinity, math_and_me, Adventure10, Mango247, PRMOisTheHardestExam
As a first observation, one must allow $\emptyset$ to be (vacuously) considered as an invariant subset, for any $f$ (otherwise point ii) would fail for $k=n$).

Now, the idea from above can be written more simply, with the permutation $f_k = (1)(2)\cdots(k-1)(k,k+1,\ldots,n)$, written as a product of cycles, having $\deg(f) = 2^{k}$ for all $1\leq k \leq n$.
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iarnab_kundu
866 posts
#4 • 2 Y
Y by Adventure10, Mango247
In the actual paper it was mentioned that the empty set and the set {1,...,n} are ''invariant'' subset. So I request Sayan to edit it.
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mathbuzz
803 posts
#5 • 4 Y
Y by ISI4, agirlhasnoname, Adventure10, Mango247
well, for part (1) , we may take a cycle permutation.i.e. take f(1)=2 , f(2)=3 ,...,f(n-1)=n,f(n)=1.this clearly satisfies deg.(f)=2
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Irfan
389 posts
#6 • 2 Y
Y by Adventure10, Mango247
Can we simply not use f(1)=1 f(2)=4.... argument for (1) ?
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JackXD
151 posts
#7 • 2 Y
Y by Adventure10, Mango247
ignore this post
This post has been edited 1 time. Last edited by JackXD, May 9, 2017, 1:00 PM
Reason: faulty post in another forum
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stranger_02
337 posts
#9 • 2 Y
Y by Madhavi, PRMOisTheHardestExam
Sayan wrote:
Let $S = \{1,2,3,\ldots,n\}$. Consider a function $f\colon S\to S$. A subset $D$ of $S$ is said to be invariant if for all $x\in D$ we have $f(x)\in D$. The empty set and $S$ are also considered as invariant subsets. By $\deg (f)$ we define the number of invariant subsets $D$ of $S$ for the function $f$.

i) Show that there exists a function $f\colon S\to S$ such that $\deg (f)=2$.

ii) Show that for every $1\leq k\leq n$ there exists a function $f\colon S\to S$ such that $\deg (f)=2^{k}$.

For the second part, we can simply consider a function defined piecewise as
$f(x)=x$ when $x\leq{k}$
$f(x)=1$ $($or any number in this range$)$, otherwise

Therefore, we see that $f(x)$ has exactly $2^k$ subsets..

Q.E.D $\square$
This post has been edited 1 time. Last edited by stranger_02, Jun 18, 2020, 3:25 PM
Reason: Latex
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Titttu
7 posts
#10
Y by
I can give a hint sumation of binomial coffecient wii helpful.. breka as required in 2,3,4,...n parts the total set and for every break u get the required 2^k
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Titttu
7 posts
#11
Y by
I can give a hint sumation of binomial coffecient wii helpful.. breka as required in 2,3,4,...n parts the total set and for every break u get the required 2^k
I gave one example which help you .
Let, k=3 then break S in 3 parts.
The break is going to happen like this=
f(x)=x+1 and f(b) =a ( where a is the first element of the partion and b is the last element of the partion . and for rest of the elemnets of the element it will tak care by the f(x) =x+1 )
Now if we choose one partition we can do it by 3c1 ways , we can choose 2 partition by 3c2 ways and 3 partition(which is total set for this case) 3c3 ways . And we can choose no partition in 3co ways ( which is empty set) .
The sum of all the cases is 2^3 .And you can do it 4 partition(u get 2^4) and so on ..upto n partition (after some partition u have to consider 1 element as partition & always shuffle the partition like that inside the partition there is no kind of subset which can satisfy our requirements .


Sorry guys I can't use the latex format .
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ftheftics
651 posts
#12 • 1 Y
Y by Gerninza
It will enough to show a construction of a function to prove the existence.

For a fixed positive integer $k \in [1,n]$ consider the function

\[f(x) = \begin{cases}  
 x + 1 &  \text{for} ,  1\le x\le n-k-1
\\
1 & \text{for}    ,  x\ge n-k \end{cases} \]
Now consider $A =\{ 1 ,2,\cdots , n-k \}$ and $B =\{n-k+1,\cdots ,n\}$

Now Any subset $D$ of $S$ with $D= A \cup R$ (Where $R$ is subset of $B$) is invariant Under $f$ .

Since there is $2^k$ subsets of $B$ thus , There can have $2^k$ subset $D$ of $S$ for which $D$ is invariant under $f$ .

By definition $ \deg (f) =2^k$


Now for part (i) just take $k=1$
This post has been edited 1 time. Last edited by ftheftics, May 20, 2021, 1:07 PM
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allofme
2 posts
#13
Y by
@ftheftics the subsets you have considered as AUR where R is subset of B, all of them essentially consist A and some additional elements. So we get 2^k subsets all containing at least A. But apart from this there is phi which does not include anything making the deg(f)=2^k+1. Kindly enlighten me on this aspect.
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idkwhatthisis
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#14
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I believe induction should work, for $\deg(f) = 2$ just take a cycle, and to get to a permutation with $\deg(f) = 2^m$ from $n$ to $n+1$ just take $f(n+1) = n+1$ and a permutation $g$ of $\{1,\ldots,n\}$ such that $\deg(g) = 2^{m-1}$ and we should be done.

EDIT: I just realised that by fixing $k-1$ points and applying the same construction in (i) to the remaining numbers then everything is cool
This post has been edited 2 times. Last edited by idkwhatthisis, May 28, 2021, 11:51 AM
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green_leaf
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#15
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For each function $f$, consider a directed graph where vertices are numbers in $S$ and there is an edge $A \to B$ iff $f(A) = B$.
i) This is equivalent to showing there is a function such that there are no directed cycles of length $k < n$. This can simply be achieved by taking $f(i) = i+1 ~~~ \forall 1 \leq i \leq n-1$ and $f(n) = 1$.

ii) Choose a function that splits $S$ into $k$ disjoint directed cycles. The number of invariant subsets are equivalent to the subsets of the set whose elements are the $k$ cycles, since the sets of cycles can be combined to form new sets which are still valid invariant sets. But the number of subsets of an set with $k$ elements is simply $2^k$ (including the empty set).
This post has been edited 1 time. Last edited by green_leaf, May 27, 2021, 3:07 PM
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quasar_lord
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#17
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soln
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Gejabsk
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#18
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If we're mapping one subset of k elements to itself and dearranging the rest, but S in invariant too , so total would be 2^k + 1
for deg(f) = 2^k , the invariants excluding S must be 2^k -1 , which doesn't seem possible ...
This post has been edited 1 time. Last edited by Gejabsk, Mar 18, 2025, 6:00 PM
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