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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.

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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
Crossing ٍٍChords
matinyousefi   1
N 2 minutes ago by Trenod
Source: Iranian Combinatorics Olympiad 2020 P3
$1399$ points and some chords between them is given.
$a)$ In every step we can take two chords $RS,PQ$ with a common point other than $P,Q,R,S$ and erase exactly one of $RS,PQ$ and draw $PS,PR,QS,QR$ let $s$ be the minimum of chords after some steps. Find the maximum of $s$ over all initial positions.
$b)$ In every step we can take two chords $RS,PQ$ with a common point other than $P,Q,R,S$ and erase both of $RS,PQ$ and draw $PS,PR,QS,QR$ let $s$ be the minimum of chords after some steps. Find the maximum of $s$ over all initial positions.

Proposed by Afrouz Jabalameli, Abolfazl Asadi
1 reply
matinyousefi
Apr 24, 2020
Trenod
2 minutes ago
Nice NT with powers of two
oVlad   7
N 17 minutes ago by SimplisticFormulas
Source: Romania TST 2024 Day 1 P3
Let $n{}$ be a positive integer and let $a{}$ and $b{}$ be positive integers congruent to 1 modulo 4. Prove that there exists a positive integer $k{}$ such that at least one of the numbers $a^k-b$ and $b^k-a$ is divisible by $2^n.$

Cătălin Liviu Gherghe
7 replies
oVlad
Jul 31, 2024
SimplisticFormulas
17 minutes ago
Inequality in triangle
Nguyenhuyen_AG   0
41 minutes ago
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
\[\frac{1}{(a-4b)^2}+\frac{1}{(b-4c)^2}+\frac{1}{(c-4a)^2} \geqslant \frac{1}{ab+bc+ca}.\]
0 replies
Nguyenhuyen_AG
41 minutes ago
0 replies
D,E,F are collinear.
TUAN2k8   2
N an hour ago by TUAN2k8
Source: Own
Help me with this:
2 replies
TUAN2k8
May 28, 2025
TUAN2k8
an hour ago
Combinatorial identity
MehdiGolafshan   4
N an hour ago by watery
Let $n$ is a positive integer. Prove that
$$\sum_{k=0}^{n-1}\frac{1}{k+1}\binom{n-1}{k} = \frac{2^n-1}{n}.$$
4 replies
MehdiGolafshan
Jan 16, 2023
watery
an hour ago
JBMO Shortlist 2023 G7
Orestis_Lignos   7
N an hour ago by tilya_TASh
Source: JBMO Shortlist 2023, G7
Let $D$ and $E$ be arbitrary points on the sides $BC$ and $AC$ of triangle $ABC$, respectively. The circumcircle of $\triangle ADC$ meets for the second time the circumcircle of $\triangle BCE$ at point $F$. Line $FE$ meets line $AD$ at point $G$, while line $FD$ meets line $BE$ at point $H$. Prove that lines $CF, AH$ and $BG$ pass through the same point.
7 replies
Orestis_Lignos
Jun 28, 2024
tilya_TASh
an hour ago
Reflected point lies on radical axis
Mahdi_Mashayekhi   5
N 2 hours ago by Mahdi_Mashayekhi
Source: Iran 2025 second round P4
Given is an acute and scalene triangle $ABC$ with circumcenter $O$. $BO$ and $CO$ intersect the altitude from $A$ to $BC$ at points $P$ and $Q$ respectively. $X$ is the circumcenter of triangle $OPQ$ and $O'$ is the reflection of $O$ over $BC$. $Y$ is the second intersection of circumcircles of triangles $BXP$ and $CXQ$. Show that $X,Y,O'$ are collinear.
5 replies
Mahdi_Mashayekhi
Apr 19, 2025
Mahdi_Mashayekhi
2 hours ago
Find the value
sqing   18
N 2 hours ago by Yiyj
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 $f(1).$
18 replies
sqing
Jun 22, 2024
Yiyj
2 hours ago
Number Theory
fasttrust_12-mn   14
N 2 hours ago by Namisgood
Source: Pan African Mathematics Olympiad P1
Find all positive intgers $a,b$ and $c$ such that $\frac{a+b}{a+c}=\frac{b+c}{b+a}$ and $ab+bc+ca$ is a prime number
14 replies
fasttrust_12-mn
Aug 15, 2024
Namisgood
2 hours ago
find question
mathematical-forest   5
N 3 hours ago by Jupiterballs
Are there any contest questions that seem simple but are actually difficult? :-D
5 replies
mathematical-forest
Thursday at 10:19 AM
Jupiterballs
3 hours ago
Own made functional equation
Primeniyazidayi   10
N 3 hours ago by Phat_23000245
Source: own(probably)
Find all functions $f:R \rightarrow R$ such that $xf(x^2+2f(y)-yf(x))=f(x)^3-f(y)(f(x^2)-2f(x))$ for all $x,y \in \mathbb{R}$
10 replies
Primeniyazidayi
May 26, 2025
Phat_23000245
3 hours ago
Tough inequality
TUAN2k8   4
N 3 hours ago by Phat_23000245
Source: Own
Let $n \ge 2$ be an even integer and let $x_1,x_2,...,x_n$ be real numbers satisfying $x_1^2+x_2^2+...+x_n^2=n$.
Prove that
$\sum_{1 \le i < j \le n} \frac{x_ix_j}{x_i^2+x_j^2+1} \ge \frac{-n}{6}$
4 replies
TUAN2k8
May 28, 2025
Phat_23000245
3 hours ago
Guess period of function
a1267ab   9
N 3 hours ago by HamstPan38825
Source: USA TST 2025
Let $n$ be a positive integer. Ana and Banana play a game. Banana thinks of a function $f\colon\mathbb{Z}\to\mathbb{Z}$ and a prime number $p$. He tells Ana that $f$ is nonconstant, $p<100$, and $f(x+p)=f(x)$ for all integers $x$. Ana's goal is to determine the value of $p$. She writes down $n$ integers $x_1,\dots,x_n$. After seeing this list, Banana writes down $f(x_1),\dots,f(x_n)$ in order. Ana wins if she can determine the value of $p$ from this information. Find the smallest value of $n$ for which Ana has a winning strategy.

Anthony Wang
9 replies
a1267ab
Dec 14, 2024
HamstPan38825
3 hours ago
Inequality with abc=1
tenplusten   11
N 4 hours ago by sqing
Source: JBMO 2011 Shortlist A7
$\boxed{\text{A7}}$ Let $a,b,c$ be positive reals such that $abc=1$.Prove the inequality $\sum\frac{2a^2+\frac{1}{a}}{b+\frac{1}{a}+1}\geq 3$
11 replies
tenplusten
May 15, 2016
sqing
4 hours ago
USAMO 2002 Problem 4
MithsApprentice   90
N Apr 25, 2025 by Ilikeminecraft
Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that \[ f(x^2 - y^2) = x f(x) - y f(y)  \] for all pairs of real numbers $x$ and $y$.
90 replies
MithsApprentice
Sep 30, 2005
Ilikeminecraft
Apr 25, 2025
USAMO 2002 Problem 4
G H J
G H BBookmark kLocked kLocked NReply
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KevinYang2.71
428 posts
#80 • 1 Y
Y by Orthogonal.
Solved with Orthogonal..

We claim the only functions are $\boxed{f(x)\equiv cx}$ for $c\in\mathbb{R}$. It is easy to check that these work.

Let $P(x,y)$ denote the given assertion. $P(0,0)$ gives $f(0)=0$ and $P(x,0)$ gives $f(x^2)=xf(x)$. From $P(x,x)$ we get $f(x)=-f(-x)$. Thus $P(x,y)$ becomes $f(x^2-y^2)=f(x^2)+f(-y^2)$. It follows that
\[
f(x+y)=f(x)+f(y)\ \ \ \ \ \ \ \forall(x,y)\in\mathbb{R}^+\times\mathbb{R}^-.\tag{*}
\]
Claim. We have $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$.

Proof. If $x$ and $y$ are both positive, $f(x+y)-f(x)=f(x+y)+f(-x)=f(y)$ so we are done. If exactly one of $x$ and $y$ are positive, we are done by $(*)$. If $x$ and $y$ are both negative,
\[
f(x+y)=-f(-x-y)=-f(-x)-f(-y)=f(x)+f(y)
\]by case $1$ so we are done. $\square$

Now
\begin{align*}
xf(x)+xf(1)+f(x)+f(1)&=(x+1)f(x+1)\\
&=f((x+1)^2)\\
&=f(x^2+2x+1)\\
&=xf(x)+2f(x)+f(1)
\end{align*}so $f(x)=f(1)x$, as desired. $\square$
This post has been edited 1 time. Last edited by KevinYang2.71, May 1, 2024, 6:01 PM
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Jndd
1417 posts
#81
Y by
We claim that the solutions are $f(x)=cx$ for any constant $c$, and it's easy to see that this satisfies the equation.

Plugging in $y=0$, we get $f(x^2)=xf(x)$, and plugging in $x=0$, we get $f(-y^2)=-yf(y)$. From this, we get $f(-x^2)=-xf(x)=-f(x^2)$, giving $-f(x)=f(-x)$.

Now, using the first two equations we got by plugging in $x=0$ and then $y=0$, we get $f(x^2-y^2)=f(x^2)-f(y^2)$, which further becomes $f(x^2-y^2)=f(x^2)+f(-y^2)$ since $-f(x)=f(-x)$. This implies that if $x\geq 0, y\leq 0$ or $x\leq 0, y\geq 0$, we have $f(x)+f(y)=f(x+y)$.

However, we can also get $f(x^2-y^2)+f(y^2)=f(x^2)$, and setting $x>y$ gives us that $f(x)+f(y)=f(x+y)$ also holds true when both $x,y\geq 0$. Finally, by negating everything and using $-f(x)=f(-x)$, this also holds true when both $x,y\leq 0$. Hence, $f$ satisfies Cauchy.

Then, we plug in $x+1$ into $f(x^2)=xf(x)$ and use the fact that $f$ satisfies Cauchy to get \[f((x+1)^2)=f(x^2)+f(x)+f(x+1)=(x+1)f(x+1)=xf(x)+xf(1)+f(x)+f(1),\]and by using $f(x^2)=xf(x)$, we can cancel things to get $f(x+1)=f(1)(x+1)$, giving $f(x)=cx$ where $c=f(1)$.
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Markas
150 posts
#82
Y by
We will show that the answer is $f(x) = cx$. These work since $(x^2 - y^2).c = x.cx - y.cy$. Now we need to show these are the only solutions. Let x = 0, we get $f(-y^2) = -yf(y)$. Let y = 0, we get $f(x^2) = xf(x)$. By these two we get that f is odd and also that $f(0) = 0$. So now using what we got we can write the starting equation as $f(x^2 - y^2) + f(y^2) = f(x^2)$. Using this and the fact that f is odd we get that f is additive. Now plugging in x = x + 1 in $f(x^2) = xf(x)$ and using that f is additive we get $f((x+1)^2 ) = (x+1)f(x+1)$ $\Rightarrow$ $f(x^2 + 2x + 1) = f(x^2) + 2f(x) + f(1) = (x+1)f(x) + (x+1)f(1)$ which after clearing things up gives us that $f(x) = f(1)x = cx$ $\Rightarrow$ there are no other solutions and we are ready.
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fearsum_fyz
56 posts
#83 • 2 Y
Y by alexanderhamilton124, poirasss
We claim that the only solution is $\boxed{f(x) = cx}$ for some constant $c$. It is easy to verify that this satisfies the given condition. Now we will prove that it is the only solution.

$\underline{P}(x, x) \implies \boxed{f(0) = 0}$
$\underline{P}(x, y) - \underline{P}(y, x) \implies f(y^2 - x^2) = - f(x^2 - y^2) \implies \boxed{f \text{ is odd}}$.

$\underline{P}(x, 0) \implies \boxed{f(x^2) = x f(x)}$
$\implies f(x^2 - y^2) = f(x^2) - f(y^2)$
$\implies \boxed{f(x - y) = f(x) - f(y) \text{ for positive } x, y} \ldots \textcircled{1}$.

Now choose some positive $x$.
$\underline{P}(x - 1, x) \implies f(1 - 2x) = (x - 1)f(x - 1) - xf(x)$
$\overset{\textcircled{1}}{\implies} f(1) - f(2x) = (x - 1)(f(x) - f(1)) - xf(x)$
$\implies f(2x) - f(x) = xf(1)$
$\overset{\textcircled{1}}{\implies} f(x) = xf(1)$

So $f(x) = cx$ for positive $x$. Since $f$ is odd, this is also true for negative $x$.
We are done.
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alexanderhamilton124
401 posts
#85
Y by
From $f(x^2) = xf(x)$, and $f$ is odd, we can directly use Cauchy to finish :wink:
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Siddharthmaybe
119 posts
#86
Y by
alexanderhamilton124 wrote:
From $f(x^2) = xf(x)$, and $f$ is odd, we can directly use Cauchy to finish :wink:

wait what?
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alexanderhamilton124
401 posts
#87 • 1 Y
Y by S_14159
I don't know if it's well known (somebody told me it was in BJV (?)), but look at the last property here
This post has been edited 1 time. Last edited by alexanderhamilton124, Dec 23, 2024, 7:22 PM
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ItsBesi
147 posts
#89 • 2 Y
Y by alexanderhamilton124, S_14159
alexanderhamilton124 wrote:
I don't know if it's well known (somebody told me it was in BJV (?)), but look at the last property here

Here is another problem you can use this trick. The problem is from 2022 Kosovo TST P1

https://artofproblemsolving.com/community/c6h2797116p24625420
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mathwiz_1207
105 posts
#90
Y by
We claim the solutions are either $\boxed{f \equiv 0}$ and $\boxed{f = cx}$. It is easy to verify these both work. Now, we will show that these are the only functions satisfying the given conditions. Plugging in $x = 0$,
\[f(x^2) = xf(x)\]Plugging in $y = 0$,
\[f(-y^2) = -yf(y) = -f(y^2)\]So, $f(a) = -f(-a)$ for all $a$, thus it is odd. Now, we prove the following claim:

$f$ is additive.
We can rewrite the condition as
\[f(x^2 - y^2) = f(x^2) - f(y^2) \implies f(a - b) = f(a) - f(b)\]for all nonnegative reals $a, b$. Now, set $a = x + y, b = y$, for nonnegative $x, y$. This gives
\[f(y) = f(x+y) - f(x) \implies f(x) + f(y) = f(x + y)\]for all $x, y \geq 0$. Since $f$ is odd, we have
\[-f(-x-y) = f(x + y) = f(x) + f(y) = -f(-x) - f(-y) \implies f(-x-y) = f(-x) + f(-y)\]so $f$ is also additive over all the nonpositive reals. Now, rewrite
\[f(x) - f(-y) = f(x + y) \implies f(x) = f(x + y) + f(-y)\]We can set $x = -y + a$ for any $a \geq y \geq 0$, therefore the above equation rewrites as
\[f(a -y) = f(a) + f(-y)\]So, $f(x) + f(y) = f(x + y)$ is also true for all pairs $(a, b)$ with $a \geq 0 \geq b$.


In conclusion, $f$ is additive over all the real numbers. Thus, we have
\[f((x + 1)^2) = f(x^2) + f(2x) + f(1) = xf(x) + 2f(x) + f(1)\]\[f((x + 1)^2) = (x + 1)f(x + 1) = (x + 1)f(x) + xf(1) + f(1)\]Setting the two equations equal, we get
\[(x + 2)f(x) = (x + 1)f(x) + xf(1) \implies f(x) = xf(1) \implies f(x) = cx\]for some $c \in \mathbb{R}$, so we are done.
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eg4334
636 posts
#91
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First, $f(0)=0$ from $P(0, 0)$. Also $y=0$ gives $f(x^2)=xf(x)$ so we have $f(x^2-y^2)=f(x^2)-f(y^2)$. This also tells us that $f$ is odd. In other words $f(a+b)=f(a)+f(b)$ for one of $a, b$ being positive. We can naturally extend this to the other sign cases. Now we consider $f((x+1)^2) = (x+1)(f(x)+f(1)) = f(x^2+2x+1)=xf(x)+2f(x)+f(1)$. This tells us $f(x)=xf(1)$, or $f(x) = kx$.
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ray66
48 posts
#92
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Let $P(x,y)$ be the assertion that the functional equation holds for $(x,y)$. Plugging $(0,0)$ gives $f(0)=0$ and plugging $P(x,-y)$ gives $f$ is odd. Plugging $P(x,0)$ gives $f(x^2)=xf(x)$ for all real $x$. So for nonnegative real $x$ $P(\sqrt{x},0)$ gives $f(x)=\sqrt{x} f(\sqrt{x}) = x^{\frac{3}{4}}f(x^{\frac{1}{4}}) = \ldots = xf(1)$, and we can extend this to all negatives because $f$ is odd. So $f(x)=cx$ for all real $x$. Plugging this into the original equation gives $c(x^2-y^2)=cx^2-cy^2$, so $f(x)=cx$ is the only solution.
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Marcus_Zhang
980 posts
#93
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Cool ig
This post has been edited 1 time. Last edited by Marcus_Zhang, Mar 14, 2025, 1:46 AM
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Maximilian113
575 posts
#94
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Let $P(x, y)$ denote the assertion. Then $P(x, 0) \implies f(x^2)=xf(x) \implies f(x)=-f(-x).$ Therefore $$P(x) \iff P(x-y)=P(x)-P(y)$$for all nonnegative $x, y.$ Therefore, for positive $x, y$ we have $$f(y)=f(x+y-x)=f(x+y)-f(x) \implies f(x+y)=f(x)+f(y).$$As $f(x)$ is odd, it follows that $f(x)$ is additive. Hence $$(x+1)(f(x)+f(1))=(x+1)f(x+1)=f((x+1)^2) = f(x^2)+2f(x)+f(1) \implies f(x)=xf(1).$$This solution clearly works.
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blueprimes
362 posts
#95
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We claim the answer is $f(x) \equiv cx$ for any constant $c$ which clearly works. Now we prove they are the only ones.

Note that the assertions $(x, 0)$ and $(-x, 0)$ for nonzero $x$ implies $f$ is odd. Now $y = 0$ gives $f(x^2) = x f(x)$ so in fact $f(x^2 - y^2) + f(y^2) = f(x^2)$. Imposing $x^2 > y^2$, replacing $x^2 - y^2 \mapsto x, y^2 \mapsto y$ yields $f(x) + f(y) = f(x + y)$ for all $x, y \ge 0$.

We use this fact to our advantage to decompose terms: Consider $(x, y) \mapsto (x + 1, x)$ for $x \ge 0$, we have
\begin{align*}
2 f(x) + f(1) &= f(2x + 1) \\
&= (x + 1) f(x + 1) - x f(x) \\
&= (x + 1) [f(x) + f(1)] - x f(x) \\
&= f(x) + x f(1) + f(1) \\
\end{align*}so $f(x) \equiv x f(1)$ for $x \ge 0$. But $f$ is odd, so $f(x) \equiv cx$ for some constant $x$ as needed.
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Ilikeminecraft
674 posts
#96
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I claim that $f(x) = ax$.

By taking $x = 0, y = 0$ as seperate cases, we see that $f(x^2) = xf(x), f(-y^2)= - yf(y).$ Thus, we also see that $f(x) = -f(-x).$

Thus, we can rewrite our equation as $f(x^2 - y^2) = f(x^2) - f(y^2),$ or $f(x^2 - y^2) + f(y^2) = f(x^2).$ For $x \geq y,$ we see that the domain is all positive, and thus, $(x + 1.434)^2 + (f(x))^2 > 1.$ Now we substitute $x = u + 1$ to get that $f(x) = xf(1). $ Thus, we are done.
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