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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
Olympiad book reading help
Enes040612   1
N 44 minutes ago by haohao6688
Hello, does anyone else struggle with reading math olympiad books or am I just the only one? Whenever i try to study any different books I often get confused or overwhelmed very easily. This makes the process of studying very hard for me. Do you guys have any tips, or techniques you used? Any good videos you know?
1 reply
Enes040612
Jan 4, 2025
haohao6688
44 minutes ago
Sums Of Polynomials
oVlad   16
N an hour ago by N3bula
Source: IZhO 2022 Day 2 Problem 5
A polynomial $f(x)$ with real coefficients of degree greater than $1$ is given. Prove that there are infinitely many positive integers which cannot be represented in the form \[f(n+1)+f(n+2)+\cdots+f(n+k)\]where $n$ and $k$ are positive integers.
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+1 w
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Source: 2024 China Round 1 (Gao Lian)
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1 Real number $m>1$ satisfies $\log_9 \log_8 m =2024$. Find the value of $\log_3 \log_2 m$.
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N an hour ago by sami1618
Source: Turkey TST 2025 P6
Let $ABC$ be a scalene triangle with incenter $I$ and incircle $\omega$. Let the tangency points of $\omega$ to $BC,AC\text{ and } AB$ be $D,E,F$ respectively. Let the line $EF$ intersect the circumcircle of $ABC$ at the points $G, H$. Assume that $E$ lies between the points $F$ and $G$. Let $\Gamma$ be a circle that passes through $G$ and $H$ and that is tangent to $\omega$ at the point $M$ which lies on different semi-planes with $D$ with respect to the line $EF$. Let $\Gamma$ intersect $BC$ at points $K$ and $L$ and let the second intersection point of the circumcircle of $ABC$ and the circumcircle of $AKL$ be $N$. Prove that the intersection point of $NM$ and $AI$ lies on the circumcircle of $ABC$ if and only if the intersection point of $HB$ and $GC$ lies on $\Gamma$.
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AlperenINAN
Yesterday at 6:44 AM
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No more topics!
International Zhautykov Olympiad 2011 - Problem 2
wangsacl   14
N Jan 2, 2022 by Assassino9931
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ which satisfy the equality,
\[f(x+f(y))=f(x-f(y))+4xf(y)\]for any $x,y\in\mathbb{R}$.
14 replies
wangsacl
Jan 16, 2011
Assassino9931
Jan 2, 2022
International Zhautykov Olympiad 2011 - Problem 2
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wangsacl
182 posts
#1 • 2 Y
Y by Adventure10, Mango247
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ which satisfy the equality,
\[f(x+f(y))=f(x-f(y))+4xf(y)\]for any $x,y\in\mathbb{R}$.
This post has been edited 2 times. Last edited by Amir Hossein, May 4, 2018, 2:50 AM
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pco
23444 posts
#2 • 15 Y
Y by aahmeetface, Ankoganit, mhq, G.K-1.618, adilbek, Atpar, Jalil_Huseynov, qwertyboyfromalotoftime, Assassino9931, WinterSecret, Adventure10, Mango247, and 3 other users
wangsacl wrote:
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all $x,y\in\mathbb{R}$, then
\[f(x+f(y))=f(x-f(y))+4xf(y)\]
where $\mathbb{R}$ denote the set of real numbers.
Let $P(x,y)$ be the assertion $f(x+f(y))=f(x-f(y))+4xf(y)$

The function $f(x)=0$ $\forall x$ is a solution. Let us from now look for non all zero solutions.
Let $u$ such that $f(u)\ne 0$

$P(\frac x{8f(u)},u)$ $\implies$ $f(\frac x{8f(u)}+f(u))=f(\frac x{8f(u)}-f(u))+\frac x2$

And so $x=2f(\frac x{8f(u)}+f(u))-2f(\frac x{8f(u)}-f(u))$ and so any real $x$ may be written as $2f(y)-2f(z)$ for some $y,z$

$P(-f(z),z)$ $\implies$ $f(0)=f(-2f(z))-4f(z)^2$
$P(f(y)-2f(z),y)$ $\implies$ $f(2f(y)-2f(z))=f(-2f(z))+4f(y)^2-8f(y)f(z)$
Subtracting these two lines, we get : $f(2f(y)-2f(z))=(2f(y)-2f(z))^2+f(0)$

And so $f(x)=x^2+f(0)$ $\forall x$ which indeed is a solution.

Hence the two solutions :
$f(x)=0$ $\forall x$
$f(x)=x^2+a$ $\forall x$
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mavropnevma
15142 posts
#3 • 2 Y
Y by Adventure10, Mango247
Notice the uncanny resemblance (if not utter equivalence) with Problem 3 at the 2007 Balkan Mathematical Olympiad.
See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=827178&sid=0d0afe7c258254c48c003fe03fe58e44#p827178.
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Utkirstudios
56 posts
#4 • 1 Y
Y by Adventure10
Can anybody post other problems of first day?
Thanks!!!
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lasha
204 posts
#5 • 2 Y
Y by Adventure10 and 1 other user
$f(x+f(y)) = f(x-f(y))+4xf(y) $. (1)
First note that $f(x)=0$ for any $x$ is solution to the given equation. Suppose there exist some $u$ such that $f(u)$ doesn't equal to $0$. Then, $4xf(u)$ can get equal to any real number, but by (1) it's the difference of two f-s. It means that any real can be expressed as a $f(a)-f(b)$. (2)
Denote $f(0)=c$. Than, setting $x=f(y)$ in (1), we get:
$f(2f(y))=(2f(y))^{2}+c $ (3).
Furthermore, setting $f(x)$ instead of $x$ in (1), $f(f(x)-f(y)=f(f(x)+f(y))-4f(x)f(y)$. So, as the right hand side is symmetric, $f(f(x)-f(y))=f(f(y)-f(x))$. Hence, $f$ is odd, because $f(x)-f(y)$ gets any real value by (2).
Denote by $A$ the set of all $a$, satisfying $f(a)=a^{2}+c$.
From (3), $2f(x)$ belong to $A$ and as $f$ is odd, also $-2f(x)$ belong to that set.
From (1), it's clear that if $a-b=2f(x)$ for some $x,a,b$, than $f(a)-f(b)=a^{2}-b^{2}$. So, if $a$ is in $A$, than so does $a+2f(x)$. From the last fact, as $-2f(a)$ is in$A$ for any $a$, than for any $b$, $-2f(a)+2f(b)=2(f(b)-f(a))$ is in $A$. But $f(b)-f(a)$ gets any real value, as well as $2(f(b)-f(a))$. So, $A=R$. It means, from the definition of $A$, that $f(x)=x^{2}+c$ holds for any $x$.
Finally, the only solutions are $f(x)=0$ and $f(x)=x^{2}+c$.
P.S. The problem is not easy, but the approach is quite well known :)
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MatjazL
36 posts
#6 • 2 Y
Y by Adventure10, Mango247
lasha wrote:
P.S. The problem is not easy, but the approach is quite well known :)

What about this approach is well known?

(I don't see what the main idea, that is said to be well known is?)
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jjax
108 posts
#7 • 2 Y
Y by Adventure10, Mango247
The main idea is that the expression $f(a)-f(b)$ may take on any real value.
Such a step is certainly not easy to motivate. Indeed, it is the key step of IMO 1999 Q6 (so it's quite well known). However, after seeing a solution using this method, the method is easy to mimic for similar questions.
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waver123
142 posts
#8 • 1 Y
Y by Adventure10
If we put $f(x)=x^2 + g(x)$, then we get LHS=RHS for all $x,y$.

Does that mean anything?
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mavropnevma
15142 posts
#9 • 1 Y
Y by Adventure10
If you do that, you get $g(x+y^2 + g(y)) = g(x-y^2 - g(y))$. It remains for you to prove that this functional equation has as only solutions $g(x)=-x^2$ (leading to $f \equiv 0$), and $g(x) = c$ (constant, leading to $f(x) = x^2+c$).
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lasha
204 posts
#10 • 2 Y
Y by Adventure10, Mango247
MatjazL wrote:
lasha wrote:
P.S. The problem is not easy, but the approach is quite well known :)

What about this approach is well known?

(I don't see what the main idea, that is said to be well known is?)



Sorry for a late response, I've just came across it :) Anyway, jjax anwered to you correctly. That approach has been used in IMO 1999/6 (But not only, I've solved several problems with that, just don't remember the sources).
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liimr
34 posts
#11 • 1 Y
Y by Adventure10
waver123 wrote:
If we put $f(x)=x^2 + g(x)$, then we get LHS=RHS for all $x,y$.

Does that mean anything?

solution with $g(x)$ is pretty good, you may see in
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=827178&sid=f14af484eddd9e2dc421aec7daa5ed21#p827178

for solution by Anto
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DanDumitrescu
196 posts
#12 • 3 Y
Y by Sharingan06, Adventure10, Mango247
I don't know if it's good but i think to create the function $g(x)=f(x)-x^2$ we can write $4xf(y)=(x+f(y))^2-(x-f(y))^2$ and we have that $g(x+f(y))=g(x-f(y)) $ and we can put y=0 and note that f(0)=a and we have that $f(k)=k^2+g(f(0))=k^2+f(a)-a^2$
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imn8128
13 posts
#13 • 2 Y
Y by Adventure10, Mango247
$P(x+f(y),z)$ yields $f(x+f(y)+f(z))=f(x+f(y)-f(z))+4(x+f(y))f(z)$
$P(x-f(y),z)$ yields $f(x-f(y)+f(z))=f(x-f(y)-f(z))+4(x-f(y))f(z)$
adding the two equations yields
$f(x+f(y)+f(z))-f(x+f(y)-f(z))+f(x-f(y)+f(z))-f(x-f(y)-f(z))=8xf(z)$
changing $y$ and $z$ yields
$f(x+f(y)+f(z))+f(x+f(y)-f(z))-f(x-f(y)+f(z))-f(x-f(y)-f(z))=8xf(y)$
subtracting the two equations yields
$f(x+f(y)-f(z))-f(x-(f(y)-f(z)))=4x(f(y)-f(z))$
since $f(a)-f(b)$ runs through the entire real numbers(when $f$ is nontrivial) we can write $$f(x+y)-f(x-y)=4xy$$thus $f(x)=x^2+c$, $f=0$
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hungvuong
42 posts
#14
Y by
Let P(x,y) be the assertion f(x-f(y)) = f(x) -f(y)
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Assassino9931
1197 posts
#15
Y by
liimr wrote:
waver123 wrote:
If we put $f(x)=x^2 + g(x)$, then we get LHS=RHS for all $x,y$.

Does that mean anything?

solution with $g(x)$ is pretty good, you may see in
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=827178&sid=f14af484eddd9e2dc421aec7daa5ed21#p827178

for solution by Anto

@liimr Are you sure that this solution can be used for this problem? We can get $g$ to be periodic but then in the $x=y+T$ equation $g(y)$ does not cancel and so we cannot work with $y = (-z-2T^2)/4T$ (or a similar expression) here.
This post has been edited 1 time. Last edited by Assassino9931, Jan 2, 2022, 5:41 PM
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