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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   12
N 41 minutes ago by atdaotlohbh
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
12 replies
OgnjenTesic
May 22, 2025
atdaotlohbh
41 minutes ago
Geometry with fix circle
falantrng   33
N an hour ago by zuat.e
Source: RMM 2018 Problem 6
Fix a circle $\Gamma$, a line $\ell$ to tangent $\Gamma$, and another circle $\Omega$ disjoint from $\ell$ such that $\Gamma$ and $\Omega$ lie on opposite sides of $\ell$. The tangents to $\Gamma$ from a variable point $X$ on $\Omega$ meet $\ell$ at $Y$ and $Z$. Prove that, as $X$ varies over $\Omega$, the circumcircle of $XYZ$ is tangent to two fixed circles.
33 replies
falantrng
Feb 25, 2018
zuat.e
an hour ago
Brilliant Problem
M11100111001Y1R   2
N an hour ago by Davdav1232
Source: Iran TST 2025 Test 3 Problem 3
Find all sequences \( (a_n) \) of natural numbers such that for every pair of natural numbers \( r \) and \( s \), the following inequality holds:
\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]
2 replies
M11100111001Y1R
Yesterday at 7:28 AM
Davdav1232
an hour ago
USAMO 2001 Problem 2
MithsApprentice   54
N an hour ago by lpieleanu
Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2=BD_1$ and $CE_2=AE_1$, and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$. Circle $\omega$ intersects segment $AD_2$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ=D_2P$.
54 replies
MithsApprentice
Sep 30, 2005
lpieleanu
an hour ago
No more topics!
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
400 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
400 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
Y by
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
46 posts
#92
Y by
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
Y by
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
Y by
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
Y by
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
665 posts
#96
Y by
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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