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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
Iran TST Starter
M11100111001Y1R   2
N 2 hours ago by sami1618
Source: Iran TST 2025 Test 1 Problem 1
Let \( a_n \) be a sequence of positive real numbers such that for every \( n > 2025 \), we have:
\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]Prove that there exists a natural number \( M \) such that for all \( n > M \), the following holds:
\[
a_n < \frac{1}{1404}
\]
2 replies
1 viewing
M11100111001Y1R
Tuesday at 7:36 AM
sami1618
2 hours ago
Twin Prime Diophantine
awesomeming327.   23
N 3 hours ago by HDavisWashu
Source: CMO 2025
Determine all positive integers $a$, $b$, $c$, $p$, where $p$ and $p+2$ are odd primes and
\[2^ap^b=(p+2)^c-1.\]
23 replies
awesomeming327.
Mar 7, 2025
HDavisWashu
3 hours ago
Troublesome median in a difficult inequality
JG666   2
N 3 hours ago by navid
Source: 2022 Spring NSMO Day 2 Problem 3
Determine the minimum value of $\lambda\in\mathbb{R}$, such that for any positive integer $n$ and non-negative reals $x_1, x_2, \cdots, x_n$, the following inequality always holds:
$$\sum_{i=1}^n(m_i-a_i)^2\leqslant \lambda\cdot\sum_{i=1}^nx_i^2,$$Here $m_i$ and $a_i$ denote the median and arithmetic mean of $x_1, x_2, \cdots, x_i$, respectively.

Duanyang ZHANG, High School Affiliated to Renmin University of China
2 replies
JG666
May 22, 2022
navid
3 hours ago
D1038 : A generalization of Jensen
Dattier   4
N 3 hours ago by Dattier
Source: les dattes à Dattier
Let $f \in C^1([0,1]), g \in C^2(f([0;1]))$.

Is it true that

$$\min(|g''|)\times \min(|f'|^2) \leq 24 \times\left|\int_0^1g(f(x)) \text{d}x- g(\int_0^1 f(x) \text{d}x) \right| \leq \max(|g''|)\times \max(|f'|^2)$$?
4 replies
Dattier
Yesterday at 12:15 PM
Dattier
3 hours ago
Concurrent lines, angle bisectors
legogubbe   0
3 hours ago
Source: ???
Hi AoPS!

Let $ABC$ be an isosceles triangle with $AB=AC$, and $M$ an arbitrary point on side $BC$. The internal angle bisector of $\angle MAB$ meets the circumcircle of $\triangle ABC$ again at $P \neq A$, and the internal angle bisector of $\angle CAM$ meets it again at $Q \neq A$. Show that lines $AM$, $BQ$ and $CP$ are concurrent.
0 replies
legogubbe
3 hours ago
0 replies
Fractional Inequality
sqing   33
N 3 hours ago by Learning11
Source: Chinese Girls Mathematical Olympiad 2012, Problem 1
Let $ a_1, a_2,\ldots, a_n$ be non-negative real numbers. Prove that
$\frac{1}{1+ a_1}+\frac{ a_1}{(1+ a_1)(1+ a_2)}+\frac{ a_1 a_2}{(1+ a_1)(1+ a_2)(1+ a_3)}+$ $\cdots+\frac{ a_1 a_2\cdots a_{n-1}}{(1+ a_1)(1+ a_2)\cdots (1+ a_n)} \le 1.$
33 replies
sqing
Aug 10, 2012
Learning11
3 hours ago
Geometry angle chasing olympiads
Foxellar   1
N 3 hours ago by Ianis
Let \( \triangle ABC \) be a triangle such that \( \angle ABC = 120^\circ \). Points \( X, Y, Z \) lie on segments \( BC, CA, AB \), respectively, such that lines \( AX, BY, \) and \( CZ \) are the angle bisectors of triangle \( ABC \). Find the measure of angle \( \angle XYZ \).
1 reply
Foxellar
4 hours ago
Ianis
3 hours ago
Iran Inequality
mathmatecS   17
N 3 hours ago by Learning11
Source: Iran 1998
When $x(\ge1),$ $y(\ge1),$ $z(\ge1)$ satisfy $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2,$ prove in equality.
$$\sqrt{x+y+z}\ge\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$$
17 replies
mathmatecS
Jun 11, 2015
Learning11
3 hours ago
Problem 4
codyj   87
N 4 hours ago by ezpotd
Source: IMO 2015 #4
Triangle $ABC$ has circumcircle $\Omega$ and circumcenter $O$. A circle $\Gamma$ with center $A$ intersects the segment $BC$ at points $D$ and $E$, such that $B$, $D$, $E$, and $C$ are all different and lie on line $BC$ in this order. Let $F$ and $G$ be the points of intersection of $\Gamma$ and $\Omega$, such that $A$, $F$, $B$, $C$, and $G$ lie on $\Omega$ in this order. Let $K$ be the second point of intersection of the circumcircle of triangle $BDF$ and the segment $AB$. Let $L$ be the second point of intersection of the circumcircle of triangle $CGE$ and the segment $CA$.

Suppose that the lines $FK$ and $GL$ are different and intersect at the point $X$. Prove that $X$ lies on the line $AO$.

Proposed by Greece
87 replies
codyj
Jul 11, 2015
ezpotd
4 hours ago
Prove the statement
Butterfly   13
N 4 hours ago by oty
Given an infinite sequence $\{x_n\} \subseteq  [0,1]$, there exists some constant $C$, for any $r>0$, among the sequence $x_n$ and $x_m$ could be chosen to satisfy $|n-m|\ge r $ and $|x_n-x_m|<\frac{C}{|n-m|}$.
13 replies
Butterfly
May 7, 2025
oty
4 hours ago
3rd AKhIMO for University Students, P2
UzbekMathematician   1
N 4 hours ago by grupyorum
Source: AKhIMO 2025, P2
Find all possible values of $gcd(a^{2m}+1, a^n+1)$, where $a, m, n$ are positive integers and $n$ is odd.
1 reply
UzbekMathematician
Yesterday at 1:48 PM
grupyorum
4 hours ago
IMO96/2 [the lines AP, BD, CE meet at a point]
Arne   47
N 5 hours ago by Bridgeon
Source: IMO 1996 problem 2, IMO Shortlist 1996, G2
Let $ P$ be a point inside a triangle $ ABC$ such that
\[ \angle APB - \angle ACB = \angle APC - \angle ABC.
\]
Let $ D$, $ E$ be the incenters of triangles $ APB$, $ APC$, respectively. Show that the lines $ AP$, $ BD$, $ CE$ meet at a point.
47 replies
1 viewing
Arne
Sep 30, 2003
Bridgeon
5 hours ago
A sharp one with 3 var (3)
mihaig   4
N 5 hours ago by aaravdodhia
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$a^2+b^2+c^2+5abc\geq8.$$
4 replies
mihaig
Tuesday at 5:17 PM
aaravdodhia
5 hours ago
Problem 2, Grade 12th RMO Shortlist - Year 2002
sticknycu   5
N Yesterday at 4:07 PM by P_Fazioli
Let $A \in M_2(C), A \neq O_2, A \neq I_2, n \in \mathbb{N}^*$ and $S_n = \{ X \in M_2(C) | X^n = A \}$.
Show:
a) $S_n$ with multiplication of matrixes operation is making an isomorphic-group structure with $U_n$.
b) $A^2 = A$.

Marian Andronache
5 replies
sticknycu
Jan 3, 2020
P_Fazioli
Yesterday at 4:07 PM
Are these functions invertible?
Levieee   5
N Apr 9, 2025 by Levieee
Which one of these functions are invertible? or both are invertible?
Let $f : (1, \infty) \to \mathbb{R}$ and $g : \mathbb{R} \to \mathbb{R}$ be defined as

$f(x) = \frac{x}{x - 1}$,
$\quad g(x) = 7 - x^3.$

I think both of them are invertible, for $g$ its trivial for $f$ is where it gets confusing.
The condition for invertibility of a function according to Wikipedia is

>The function $f$ is invertible if and only if it is bijective. This is because the condition
$g(f(x))=x\quad\forall x\in X$ implies that $f$ is injective, and the condition
$f(g(y))=y\quad\forall y\in Y$ implies that $f$ is surjective.

My friends whom I discussed this problem with say that $f$ isn't invertible because $f$ isn't surjective since $f$ never reaches values $<1$
since when I make $\mathbb{R}$ in $f^{-1}$ the domain it won't produce values for for $\mathbb{R}^{-}$, but I think that argument is wrong but I can't point it out. The only argument I could provide was

\begin{align*}
& e^x : \mathbb{R} \to \mathbb{R} \\
& e^x \text{ has an inverse} \\
& \ln x \\
& \text{which only takes values in } (0, \infty) \\
& \text{Yes, the range of } e^x \text{ is } (0, \infty) \text{ but can’t we still write the codomain as } \mathbb{R}? \\
& \text{Similarly here, the range of } f(x) \text{ is } (1, \infty) \text{ but the codomain is still } \mathbb{R}
\end{align*}

Is it necessary for surjectivity here? and if so why is it contradicting the defintion? where am i going wrong?


The question asks, "Is it invertible?", which I interpret as, "Will an inverse exist?"
YES, if I restrict the codomain to the range.
NO, if I keep the codomain as it is.
It never said anything about whether I can restrict the codomain or not, so I should be allowed to restrict it.

Restricting the codomain doesn’t change the actual input-output behavior of the function — it just changes how we describe the function.

The mapping rule remains the same. For example, if $f(x) = x^2$, then $f(2) = 4$ and $f(-2) = 4$, regardless of whether the codomain is $\mathbb{R}$ or $[0, \infty)$.


It is valid to restrict the codomain to the range when discussing invertibility.

This doesn't alter the nature of the function — it just makes the description precise and allows an inverse to exist by making the function surjective. Many textbooks do this without issue.

If restricting the domain or codomain changes the function, then technically those functions don't have inverses.
But we came up with functions like $\log x$ or $\sqrt{x}$ precisely to define inverses — so maybe it's fair to come up with an inverse here too.

Yes, strictly speaking, changing the codomain defines a different function in the formal sense.
But since the input-output rule doesn't change, and the goal is to determine if an inverse exists, it's mathematically acceptable — and often necessary by many textbooks — to restrict the codomain to the range.

Since the question was to analyze invertibility, it makes sense to say the function can be made invertible in that context.


5 replies
Levieee
Apr 9, 2025
Levieee
Apr 9, 2025
Are these functions invertible?
G H J
G H BBookmark kLocked kLocked NReply
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Levieee
244 posts
#1
Y by
Which one of these functions are invertible? or both are invertible?
Let $f : (1, \infty) \to \mathbb{R}$ and $g : \mathbb{R} \to \mathbb{R}$ be defined as

$f(x) = \frac{x}{x - 1}$,
$\quad g(x) = 7 - x^3.$

I think both of them are invertible, for $g$ its trivial for $f$ is where it gets confusing.
The condition for invertibility of a function according to Wikipedia is

>The function $f$ is invertible if and only if it is bijective. This is because the condition
$g(f(x))=x\quad\forall x\in X$ implies that $f$ is injective, and the condition
$f(g(y))=y\quad\forall y\in Y$ implies that $f$ is surjective.

My friends whom I discussed this problem with say that $f$ isn't invertible because $f$ isn't surjective since $f$ never reaches values $<1$
since when I make $\mathbb{R}$ in $f^{-1}$ the domain it won't produce values for for $\mathbb{R}^{-}$, but I think that argument is wrong but I can't point it out. The only argument I could provide was

\begin{align*}
& e^x : \mathbb{R} \to \mathbb{R} \\
& e^x \text{ has an inverse} \\
& \ln x \\
& \text{which only takes values in } (0, \infty) \\
& \text{Yes, the range of } e^x \text{ is } (0, \infty) \text{ but can’t we still write the codomain as } \mathbb{R}? \\
& \text{Similarly here, the range of } f(x) \text{ is } (1, \infty) \text{ but the codomain is still } \mathbb{R}
\end{align*}

Is it necessary for surjectivity here? and if so why is it contradicting the defintion? where am i going wrong?


The question asks, "Is it invertible?", which I interpret as, "Will an inverse exist?"
YES, if I restrict the codomain to the range.
NO, if I keep the codomain as it is.
It never said anything about whether I can restrict the codomain or not, so I should be allowed to restrict it.

Restricting the codomain doesn’t change the actual input-output behavior of the function — it just changes how we describe the function.

The mapping rule remains the same. For example, if $f(x) = x^2$, then $f(2) = 4$ and $f(-2) = 4$, regardless of whether the codomain is $\mathbb{R}$ or $[0, \infty)$.


It is valid to restrict the codomain to the range when discussing invertibility.

This doesn't alter the nature of the function — it just makes the description precise and allows an inverse to exist by making the function surjective. Many textbooks do this without issue.

If restricting the domain or codomain changes the function, then technically those functions don't have inverses.
But we came up with functions like $\log x$ or $\sqrt{x}$ precisely to define inverses — so maybe it's fair to come up with an inverse here too.

Yes, strictly speaking, changing the codomain defines a different function in the formal sense.
But since the input-output rule doesn't change, and the goal is to determine if an inverse exists, it's mathematically acceptable — and often necessary by many textbooks — to restrict the codomain to the range.

Since the question was to analyze invertibility, it makes sense to say the function can be made invertible in that context.
Z K Y
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oz.the.wizard
3 posts
#2
Y by
Explanation
This post has been edited 2 times. Last edited by oz.the.wizard, Apr 9, 2025, 5:55 PM
Reason: typo
Z K Y
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Levieee
244 posts
#3
Y by
oz.the.wizard wrote:
Explanation

no I'm saying that the question asks "is it invertible" that can very well mean "by any way is it invertible?" yes it can be if we restrict the codomain
Z K Y
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Etkan
1576 posts
#4
Y by
Levieee wrote:
oz.the.wizard wrote:
Explanation

no I'm saying that the question asks "is it invertible" that can very well mean "by any way is it invertible?" yes it can be if we restrict the codomain

No, it does not mean that, because if we change the codomain, we change the function.
To be more precise, the functions\begin{align*}f:(1,\infty ) & \to \mathbb{R} \\
x & \mapsto f(x)=\frac{x}{x-1}
\end{align*}and\begin{align*}\widetilde f:(1,\infty ) & \to (1,\infty ) \\
x & \mapsto \widetilde f(x)=\frac{x}{x-1}
\end{align*}are two different functions. That happens because the domain and the codomain are part of the definition of the function.
Hence the answer to the question "Is the function\begin{align*}f:(1,\infty ) & \to \mathbb{R} \\
x & \mapsto f(x)=\frac{x}{x-1}
\end{align*}invertible?" is "No, it isn't.", and there's no ambiguity in the question. You can make a function surjective by restricting its codomain and you can make it injective by restricting its domain, but that doesn't mean the original function is either. Else we could conclude that\begin{align*}h:\mathbb{R} & \to \mathbb{R} \\
x & \mapsto h(x)=x^2
\end{align*}is bijective, because we make it surjective by restricting its codomain to $[0,\infty )$ and we make it injective by restricting its domain to $[0,\infty )$.
Z K Y
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Levieee
244 posts
#5
Y by
Etkan wrote:
Levieee wrote:
oz.the.wizard wrote:
Explanation

no I'm saying that the question asks "is it invertible" that can very well mean "by any way is it invertible?" yes it can be if we restrict the codomain

No, it does not mean that, because if we change the codomain, we change the function.
To be more precise, the functions\begin{align*}f:(1,\infty ) & \to \mathbb{R} \\
x & \mapsto f(x)=\frac{x}{x-1}
\end{align*}and\begin{align*}\widetilde f:(1,\infty ) & \to (1,\infty ) \\
x & \mapsto \widetilde f(x)=\frac{x}{x-1}
\end{align*}are two different functions. That happens because the domain and the codomain are part of the definition of the function.
Hence the answer to the question "Is the function\begin{align*}f:(1,\infty ) & \to \mathbb{R} \\
x & \mapsto f(x)=\frac{x}{x-1}
\end{align*}invertible?" is "No, it isn't.", and there's no ambiguity in the question. You can make a function surjective by restricting its codomain and you can make it injective by restricting its domain, but that doesn't mean the original function is either. Else we could conclude that\begin{align*}h:\mathbb{R} & \to \mathbb{R} \\
x & \mapsto h(x)=x^2
\end{align*}is bijective, because we make it surjective by restricting its codomain to $[0,\infty )$ and we make it injective by restricting its domain to $[0,\infty )$.

yea but often a times I've seen real analysis text books restrict a co domain in a proof to get some result as long as the behaviour of the function doesn't change,so if that changes the definition of a function then those proofs are invalid?
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Levieee
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https://math.stackexchange.com/questions/5054233/is-fx-fracxx-1-invertible-when-f-1-infty-to-mathbbr
This post has been edited 2 times. Last edited by Levieee, Apr 9, 2025, 8:25 PM
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