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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
Radiant sets
BR1F1SZ   1
N 10 minutes ago by alexheinis
Source: 2025 Francophone MO Juniors P1
A finite set $\mathcal S$ of distinct positive real numbers is called radiant if it satisfies the following property: if $a$ and $b$ are two distinct elements of $\mathcal S$, then $a^2 + b^2$ is also an element of $\mathcal S$.
[list=a]
[*]Does there exist a radiant set with a size greater than or equal to $4$?
[*]Determine all radiant sets of size $2$ or $3$.
[/list]
1 reply
BR1F1SZ
Yesterday at 11:12 PM
alexheinis
10 minutes ago
Looks like power mean, but it is not
Nuran2010   2
N 11 minutes ago by Primeniyazidayi
Source: Azerbaijan Al-Khwarizmi IJMO TST 2025
For $a,b,c$ positive real numbers satisfying $a^2+b^2+c^2 \geq 3$,show that:

$\sqrt[3]{\frac{a^3+b^3+c^3}{3}}+\frac{a+b+c}{9} \geq \frac{4}{3}$.
2 replies
+1 w
Nuran2010
an hour ago
Primeniyazidayi
11 minutes ago
ISI UGB 2025 P7
SomeonecoolLovesMaths   2
N 27 minutes ago by Primeniyazidayi
Source: ISI UGB 2025 P7
Consider a ball that moves inside an acute-angled triangle along a straight line, unit it hits the boundary, which is when it changes direction according to the mirror law, just like a ray of light (angle of incidence = angle of reflection). Prove that there exists a triangular periodic path for the ball, as pictured below.

IMAGE
2 replies
SomeonecoolLovesMaths
an hour ago
Primeniyazidayi
27 minutes ago
Knights NOT crowded on the chessboard
mshtand1   1
N 28 minutes ago by sarjinius
Source: Ukrainian Mathematical Olympiad 2025. Day 1, Problem 8.4
What is the maximum number of knights that can be placed on a chessboard of size \(8 \times 8\) such that any knight, after making 1 or 2 arbitrary moves, does not land on a square occupied by another knight?

Proposed by Bogdan Rublov
1 reply
mshtand1
Mar 13, 2025
sarjinius
28 minutes ago
ISI UGB 2025 P5
SomeonecoolLovesMaths   2
N 41 minutes ago by SomeonecoolLovesMaths
Source: ISI UGB 2025 P5
Let $a,b,c$ be nonzero real numbers such that $a+b+c \neq 0$. Assume that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a+b+c}$$Show that for any odd integer $k$, $$\frac{1}{a^k} + \frac{1}{b^k} + \frac{1}{c^k} = \frac{1}{a^k+b^k+c^k}.$$
2 replies
SomeonecoolLovesMaths
2 hours ago
SomeonecoolLovesMaths
41 minutes ago
A bit tricky invariant with 98 numbers on the board.
Nuran2010   0
an hour ago
Source: Azerbaijan Al-Khwarizmi IJMO TST 2025
The numbers $\frac{50}{1},\frac{50}{2},...\frac{50}{97},\frac{50}{98}$ are written on the board.In each step,two random numbers $a$ and $b$ are chosen and deleted.Then,the number $2ab-a-b-1$ is written instead.What will be the number remained on the board after the last step.
0 replies
Nuran2010
an hour ago
0 replies
Taking antipode on isosceles triangle's circumcenter
Nuran2010   0
an hour ago
Source: Azerbaijan Al-Khwarizmi IJMO TST 2025
In isosceles triangle, the condition $AB=AC>BC$ is satisfied. Point $D$ is taken on the circumcircle of $ABC$ such that $\angle CAD=90^{\circ}$.A line parallel to $AC$ which passes from $D$ intersects $AB$ and $BC$ respectively at $E$ and $F$.Show that circumcircle of $ADE$ passes from circumcenter of $DFC$.
0 replies
Nuran2010
an hour ago
0 replies
R to R, with x+f(xy)=f(1+f(y))x
NicoN9   3
N an hour ago by EeEeRUT
Source: Own.
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that\[
x+f(xy)=f(1+f(y))x
\]for all $x, y\in \mathbb{R}$.
3 replies
NicoN9
4 hours ago
EeEeRUT
an hour ago
find angle
TBazar   7
N an hour ago by TBazar
Given $ABC$ triangle with $AC>BC$. We take $M$, $N$ point on AC, AB respectively such that $AM=BC$, $CM=BN$. $BM$, $AN$ lines intersect at point $K$. If $2\angle AKM=\angle ACB$, find $\angle ACB$
7 replies
TBazar
May 8, 2025
TBazar
an hour ago
ISI UGB 2025 P4
SomeonecoolLovesMaths   0
an hour ago
Source: ISI UGB 2025 P4
Let $S^1 = \{ z \in \mathbb{C} \mid |z| =1 \}$ be the unit circle in the complex plane. Let $f \colon S^1 \longrightarrow S^2$ be the map given by $f(z) = z^2$. We define $f^{(1)} \colon = f$ and $f^{(k+1)} \colon = f \circ f^{(k)}$ for $k \geq 1$. The smallest positive integer $n$ such that $f^{(n)}(z) = z$ is called the period of $z$. Determine the total number of points in $S^1$ of period $2025$.
(Hint : $2025 = 3^4 \times 5^2$)
0 replies
SomeonecoolLovesMaths
an hour ago
0 replies
ISI UGB 2025 P8
SomeonecoolLovesMaths   0
an hour ago
Source: ISI UGB 2025 P8
Let $n \geq 2$ and let $a_1 \leq a_2 \leq \cdots \leq a_n$ be positive integers such that $\sum_{i=1}^{n} a_i = \prod_{i=1}^{n} a_i$. Prove that $\sum_{i=1}^{n} a_i \leq 2n$ and determine when equality holds.
0 replies
SomeonecoolLovesMaths
an hour ago
0 replies
ISI UGB 2025 P6
SomeonecoolLovesMaths   0
an hour ago
Source: ISI UGB 2025 P6
Let $\mathbb{N}$ denote the set of natural numbers, and let $\left( a_i, b_i \right)$, $1 \leq i \leq 9$, be nine distinct tuples in $\mathbb{N} \times \mathbb{N}$. Show that there are three distinct elements in the set $\{ 2^{a_i} 3^{b_i} \colon 1 \leq i \leq 9 \}$ whose product is a perfect cube.
0 replies
SomeonecoolLovesMaths
an hour ago
0 replies
ISI UGB 2025 P2
SomeonecoolLovesMaths   0
2 hours ago
Source: ISI UGB 2025 P2
If the interior angles of a triangle $ABC$ satisfy the equality, $$\sin ^2 A + \sin ^2 B + \sin^2  C = 2 \left( \cos ^2 A + \cos ^2 B + \cos ^2 C \right),$$prove that the triangle must have a right angle.
0 replies
SomeonecoolLovesMaths
2 hours ago
0 replies
Six variables
Nguyenhuyen_AG   1
N 2 hours ago by TNKT
Let $a,\,b,\,c,\,x,\,y,\,z$ be six positive real numbers. Prove that
$$\frac{a}{b+c} \cdot \frac{y+z}{x} + \frac{b}{c+a} \cdot \frac{z+x}{y} + \frac{c}{a+b} \cdot \frac{x+y}{z} \geqslant 2+\sqrt{\frac{8abc}{(a+b)(b+c)(c+a)}}.$$
1 reply
Nguyenhuyen_AG
Today at 5:09 AM
TNKT
2 hours ago
FE solution too simple?
Yiyj1   9
N Apr 23, 2025 by jasperE3
Source: 101 Algebra Problems from the AMSP
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that the equality $$f(f(x)+y) = f(x^2-y)+4f(x)y$$holds for all pairs of real numbers $(x,y)$.

My solution

I feel like my solution is too simple. Is there something I did wrong or something I missed?
9 replies
Yiyj1
Apr 9, 2025
jasperE3
Apr 23, 2025
FE solution too simple?
G H J
G H BBookmark kLocked kLocked NReply
Source: 101 Algebra Problems from the AMSP
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Yiyj1
1266 posts
#1
Y by
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that the equality $$f(f(x)+y) = f(x^2-y)+4f(x)y$$holds for all pairs of real numbers $(x,y)$.

My solution

I feel like my solution is too simple. Is there something I did wrong or something I missed?
Z K Y
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InterLoop
280 posts
#2 • 1 Y
Y by Yiyj1
You cannot immediately "cancel" the $f$ without further conclusions.

For example $f(3) = f(2) = 1$ is possible for a function - this does not mean that $3 = 2$.
The property that leads you to $f(a) = f(b) \implies a = b$ or the "cancellation" of $f$ is called injectivity. You have to prove the function is injective first before cancellation.

Another example is simply the fact that you have not "excluded" the solution $f(x) \equiv 0$ from the equation $f(f(x)) =f(x^2)$ in any way - so $f(x) = x^2$ is wrong for that function as well. (thus $f(x) \equiv 0$ is not injective)
This post has been edited 2 times. Last edited by InterLoop, Apr 9, 2025, 3:39 AM
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Yiyj1
1266 posts
#3
Y by
InterLoop wrote:
You cannot immediately "cancel" the $f$ without further conclusions.

For example $f(3) = f(2) = 1$ is possible for a function - this does not mean that $3 = 2$.
The property that leads you to $f(a) = f(b) \implies a = b$ or the "cancellation" of $f$ is called injectivity. You have to prove the function is injective first before cancellation.

ahh ic. I'll try to prove the injectivity. ty!
Z K Y
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AshAuktober
1006 posts
#4
Y by
This is in fact from Iran TST.
Z K Y
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davichu
8 posts
#5
Y by
Clearly, $f(x)\equiv0$ is a trivial solution, from now on, we assume it is not the case
Let $P(x,y)$ denote the assertion $f(f(x)+y) = f(x^2-y)+4f(x)y$
$$P(x,-f(x))\rightarrow f(0)=f(x^2+f(x))-4f(x)^2$$$$P(x,x^2)\rightarrow f(x^2+f(x))=f(0)+4f(x)x^2$$Adding these two together we get:
$4f(x)^2=4f(x)x^2\rightarrow f(x)^2=f(x)x^2$
Since $f(x)\neq0$,we can divide by $f(x)$ on both sides to get $f(x)=x^2$
so the only solutions are $f(x)\equiv0$ and $f(x)=x^2\forall x \in \mathbb{R}$
Z K Y
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Primeniyazidayi
101 posts
#6
Y by
davichu wrote:
Since $f(x)\neq0$,we can divide by $f(x)$ on both sides to get $f(x)=x^2$

You must at first prove that $f(x) =0 \text{ iff } x=0$(or simply avoid pointwise trap).
This post has been edited 1 time. Last edited by Primeniyazidayi, Apr 22, 2025, 11:12 AM
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Primeniyazidayi
101 posts
#7
Y by
The finish for @2above(hopefully correct):We will avoid pointwise trap.We of course have $f(0) =0$.Let $f(t) =0$ for $t \neq 0$.$P(t,y)$ gives $f(y) =f(t^2-y)$.Take some $u$ such that $f(u) =u^2 \neq 0$.Then we have $u^2=t^2(t^2-2u) +u^2$ or $u=\frac{t^2}{2}$.But $P(0, x) $ gives that $f$ is even which means $\frac{t^2}{2}=-\frac{t^2}{2}$ or $t=0$, contradiction. Thus we are done.
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ariopro1387
16 posts
#8
Y by
Let $P(x,y)$ be the assertion of the problem.
$P(x,\frac{x^2-f(x)}{2});$ $\frac{x^2-f(x)}{2}.f(x) = 0$
$\forall x \in \mathbb{R}$
1. $f(x)\equiv0$
2. $f(x)=x^2$
we have to just check that both won't happen:
if $f(x_{1}) = 0:$
$P(x_{1},y);$ $f(y) = f(x_{1}^2-y)$
then by changing $y$ value we get that $x_{1} = 0$ or $f(x)\equiv C$ (Just $C=0$ works).
This post has been edited 1 time. Last edited by ariopro1387, Apr 22, 2025, 4:06 PM
Reason: edit
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lksb
172 posts
#9 • 1 Y
Y by Yiyj1
one-liner
This post has been edited 1 time. Last edited by lksb, Apr 22, 2025, 7:15 PM
Reason: typo
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jasperE3
11320 posts
#10
Y by
lksb wrote:
one-liner

pointwise trap
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N Quick Reply
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