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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
ISI UGB 2025 P4
SomeonecoolLovesMaths   2
N 2 minutes ago by ZeroAlephZeta
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 : $2025 = 3^4 \times 5^2$)
2 replies
SomeonecoolLovesMaths
3 hours ago
ZeroAlephZeta
2 minutes ago
ISI UGB 2025 P7
SomeonecoolLovesMaths   6
N 3 minutes ago by ZeroAlephZeta
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
6 replies
SomeonecoolLovesMaths
3 hours ago
ZeroAlephZeta
3 minutes ago
R to R, with x+f(xy)=f(1+f(y))x
NicoN9   4
N 8 minutes ago by CM1910
Source: Own.
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that\[
x+f(xy)=f(1+f(y))x
\]for all $x, y\in \mathbb{R}$.
4 replies
NicoN9
6 hours ago
CM1910
8 minutes ago
JBMO Shortlist 2021 G5
Lukaluce   7
N 26 minutes ago by africanboy
Source: JBMO Shortlist 2021
Let $ABC$ be an acute scalene triangle with circumcircle $\omega$. Let $P$ and $Q$ be interior points of the sides $AB$ and $AC$, respectively, such that $PQ$ is parallel to $BC$. Let $L$ be a point on $\omega$ such that $AL$ is parallel to $BC$. The segments $BQ$ and $CP$ intersect at $S$. The line $LS$ intersects $\omega$ at $K$. Prove that $\angle BKP = \angle CKQ$.

Proposed by Ervin Macić, Bosnia and Herzegovina
7 replies
Lukaluce
Jul 2, 2022
africanboy
26 minutes ago
D1028 : A strange result about linear algebra
Dattier   2
N 4 hours ago by ysharifi
Source: les dattes à Dattier
Let $p>3$ a prime number, with $H \subset M_p(\mathbb R), \dim(H)\geq 2$ and $H-\{0\} \subset GL_p(\mathbb R)$, $H$ vector space.

Is it true that $H-\{0\}$ is a group?
2 replies
Dattier
Yesterday at 1:49 PM
ysharifi
4 hours ago
Mathematical expectation 1
Tricky123   0
5 hours ago
X is continuous random variable having spectrum
$(-\infty,\infty) $ and the distribution function is $F(x)$ then
$E(X)=\int_{0}^{\infty}(1-F(x)-F(-x))dx$ and find the expression of $V(x)$

Ans:- $V(x)=\int_{0}^{\infty}(2x(1-F(x)+F(-x))dx-m^{2}$

How to solve help me
0 replies
Tricky123
5 hours ago
0 replies
Double integrals
fermion13pi   1
N Today at 8:11 AM by Svyatoslav
Source: Apostol, vol 2
Evaluate the double integral by converting to polar coordinates:

\[
\int_0^1 \int_{x^2}^x (x^2 + y^2)^{-1/2} \, dy \, dx
\]
Change the order of integration and then convert to polar coordinates.

1 reply
fermion13pi
Yesterday at 1:58 PM
Svyatoslav
Today at 8:11 AM
Roots of a polynomial not in the disc of unity
Fatoushima   1
N Today at 7:59 AM by alexheinis
Show that the polynomial $p_n(z)=\sum_{k=1}^nkz^{n-k}$ has no roots in the disc of unity.
1 reply
Fatoushima
Today at 1:48 AM
alexheinis
Today at 7:59 AM
Integration Bee Kaizo
Calcul8er   61
N Today at 6:36 AM by Svyatoslav
Hey integration fans. I decided to collate some of my favourite and most evil integrals I've written into one big integration bee problem set. I've been entering integration bees since 2017 and I've been really getting hands on with the writing side of things over the last couple of years. I hope you'll enjoy!
61 replies
Calcul8er
Mar 2, 2025
Svyatoslav
Today at 6:36 AM
Japanese Olympiad
parkjungmin   2
N Today at 5:26 AM by parkjungmin
It's about the Japanese Olympiad

I can't solve it no matter how much I think about it.

If there are people who are good at math

Please help me.
2 replies
parkjungmin
Yesterday at 6:51 PM
parkjungmin
Today at 5:26 AM
Japanese high school Olympiad.
parkjungmin   0
Today at 5:25 AM
It's about the Japanese high school Olympiad.

If there are any students who are good at math, try solving it.
0 replies
parkjungmin
Today at 5:25 AM
0 replies
Marginal Profit
NC4723   2
N Today at 4:35 AM by Juno_34
Please help me solve this
2 replies
NC4723
Dec 11, 2015
Juno_34
Today at 4:35 AM
D1029 : A story of equivalent and increasing sequence
Dattier   1
N Yesterday at 8:13 PM by Phorphyrion
Source: les dattes à Dattier
Let $(a_n) \in (\mathbb R^*_+) ^\mathbb N$ an increasing sequence, with $\forall (b_n) \in (\mathbb R^*_+) ^\mathbb N$, if $\lim \dfrac {a_n}{b_n}=1$ then $(b_n)$ increasing, from a certain rank.

Is it true $\exists M >1, \exists N \in \mathbb N, \forall n>N, \dfrac {a_{n+1}}{a_n} \geq M$ ?
1 reply
Dattier
Yesterday at 5:47 PM
Phorphyrion
Yesterday at 8:13 PM
Hello, I'm a math Olympiad question for a Japanese high school. I'm asking here
parkjungmin   0
Yesterday at 6:41 PM
This is very difficult, can anyone solve it?

The percentage of correct answers is low
0 replies
parkjungmin
Yesterday at 6:41 PM
0 replies
easy functional
B1t   13
N Apr 30, 2025 by AshAuktober
Source: Mongolian TST 2025 P1.
Denote the set of real numbers by $\mathbb{R}$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$,
\[
f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]
13 replies
B1t
Apr 26, 2025
AshAuktober
Apr 30, 2025
easy functional
G H J
G H BBookmark kLocked kLocked NReply
Source: Mongolian TST 2025 P1.
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B1t
24 posts
#1 • 1 Y
Y by farhad.fritl
Denote the set of real numbers by $\mathbb{R}$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$,
\[
f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]
This post has been edited 2 times. Last edited by B1t, Apr 26, 2025, 7:01 AM
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NicoN9
154 posts
#2
Y by
$f(x)f(x)$ means $f(x)^2$? or is it a typo?
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Ilikeminecraft
627 posts
#3
Y by
$x = y = z = 0$ implies $f(0) = 0$
$x = -y$ implies $f(z) = f(z) - xf(x) + f(x)^2,$ so $f\in\{0, x\}.$
assume $f(a ) = 0, f(b ) = b.$
take $y = 0,$ and we get $f(xf(x) + z) = f(z) + f(x)^2.$
if we take $x = b, y = a - b, z = a,$ we get $0 = ab.$ this implies $f \equiv 0, x.$
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B1t
24 posts
#5
Y by
NicoN9 wrote:
$f(x)f(x)$ means $f(x)^2$? or is it a typo?

I wrote it wrong. im sorry
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Ilikeminecraft
627 posts
#6
Y by
suppose $f$ is not identically $0$
suppose $x_0$ satisfies $f(x_0)\neq 0,$ and if we pick $x = x_0$ and range the values of $y,$ we obtain that $f$ is surjective. ignore the definition of $x_0$
take $y = 0$ and we get $f(xf(x) + z) = f(xf(x)) + f(z)$
take $x = - y$ yields $f(xf(x))=xf(x).$
thus, $f(f(xf(x)) + z) = f(z) + f(f(xf(x)))$ which implies that the function is cauchy
plugging back into original equation, one gets $f(xf(y)) = f(x)y$
pick $x = 1$ to get $f(f(y)) = y,$ which is well-known to imply that $|f(y)| = |y|.$

assume $A$ is the set of $x$ such that $f(x) = x$ and $B$ is the set of $x$ such that $f(x) = -x.$
if $x, y \in A,$ then $f(xy) = xy,$ so $xy \in A$
if $x\in A, y\in B,$ then $f(xy) = -xy,$ so $xy\in B$
if $x \in B, y \in A,$ then $f(xy) = -xy,$ so $xy \in B$
if $x \in B, y\in B,$ then $f(xy) = xy,$ so $xy\in A$

this implies $f$ is multiplicative
hence, since $f$ is both multiplicative and additive, $f$ is identity
This post has been edited 1 time. Last edited by Ilikeminecraft, Apr 26, 2025, 3:22 PM
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Haris1
77 posts
#7
Y by
I wont write the solution , i will just write the steps.
$1.$ Prove that the function is additive
$2.$ Prove that its either constant or bijective
$3.$ Prove that its multiplicative and finish
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cazanova19921
552 posts
#8 • 1 Y
Y by farhad.fritl
B1t wrote:
Denote the set of real numbers by $\mathbb{R}$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$,
\[
P(x,y,z):\, f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]
$P(0,0,0)$ $\implies$ $f(0)=0$.
$P(x,-x,0)$ $\implies$ $f(xf(x))=xf(x)$
So $P(x,y-x,z): \, f(xf(y)+z)=f(z)+yf(x)$ (new $P$)
- If $f(t)=0$ for some $t \neq 0$, then $P(x,t,0)$ $\implies$ $\boxed{f=0}$ which is a solution.
- Suppose $f(t)=0$ $\iff$ $t=0$.
$P(1,x,0)$ $\implies$ $f(f(x))=xf(1)$. hence $f$ is bijective, replace $x=1$ in this equation, we get $f(f(1))=f(1)$ so $f(1)=1$.
Therefore $f(f(x))=x$ for all $x$.
$P(x, f(y), 0)$ $\implies$ $f(xy)=f(x)f(y)$ for all $x, y$
$P(x, 1, y)$ $\implies$ $f(x+y)=f(x)+f(y)$ for all $x, y$
So $f$ is additive and multiplicative, hence $\boxed{f=\mathrm{Id}}$ which is also a valid solution.
This post has been edited 1 time. Last edited by cazanova19921, Apr 26, 2025, 1:55 PM
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MathLuis
1525 posts
#9
Y by
Denote $P(x,y,z)$ the assertion of the given F.E.
Notice from $P(0,y,z)$ we can inmediately get $f(0)=0$ and also $P(x,-x,z)$ gives that $f(xf(x))=xf(x)$ but also $P(x,y,0)$ now gives that $f(xf(x+y))=(x+y)f(x)$ so in fact shifting gives $Q(x,y)$ which is $f(xf(y))=yf(x)$ now notice if ther existed some $c \ne 0$ for which $f(c)=0$ then $Q(x,c)$ gives that $f(x)=0$ for all reals $x$ so its either that or $f$ is injective at zero, but now basically note that our F.E. may now by re-written as $R(x,y,z)$ to be $f(xf(y)+z)=f(z)+yf(x)$ for all reals $x,y,z$ but also take $y \ne 0$ and shift $x$ to get that $f$ is additive, but also from $Q(x, f(x))$ we get that $f(f(1)x^2)=f(x)^2$ and therefore shifting $x \to f(x)$ gives $f(f(1)f(x)^2)=f(1)^2 \cdot x^2$ so for example $f$ is surjective on all non-negative reals also from $Q(x,y)$ we have $f$ injective trivially when setting $x \ne 0$ and this take $f(d)=1$ and $Q(x,d)$ to get that $d=1$ and thus $f(1)=1$ so $Q(1,x)$ gives $f$ is an involution so by $Q(x,f(y))$ we get $f$ multiplicative so addivitive+multiplicative means $f$ is the identity function or constant (later case gives $f$ is zero everywhere so doesn't count), so $f(x)=0,x$ for all reals $x$ are the only two solutions that work thus we are done :cool:.
This post has been edited 1 time. Last edited by MathLuis, Apr 27, 2025, 3:32 PM
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jasperE3
11320 posts
#10
Y by
B1t wrote:
Denote the set of real numbers by $\mathbb{R}$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$,
\[
f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]

Let $P(x,y)$ be the assertion $f(xf(x+y)+z)=f(z)+yf(x)+f(xf(x))$.
Note that $\boxed{f(x)=0}$ is a solution, else there is some $j$ with $f(j)\ne0$.
$P(0,0,0)\Rightarrow f(0)=0\Rightarrow j\ne0$
$P(j,-j,0)\Rightarrow f(jf(j))=jf(j)$
$P(j,0,x)\Rightarrow f(x+jf(j))=f(x)+jf(j)$
$P(x,y+jf(j),z)\Rightarrow f(xf(x+y)+jf(j)x+z)=f(z)+yf(x)+f(xf(x))+jf(j)f(x)$
Comparing this last line with $P(x,y)$ we get:
$$f(xf(x+y)+jf(j)x+z)=f(xf(x+y)+z)+jf(j)f(x)$$and setting $z=-xf(x+y)$ and $x=1$ this is $jf(j)(f(1)-1)=0$, so $f(1)=1$.
$P(1,x-1,y)\Rightarrow f(f(x)+y)=x+f(y)$
Call this assertion $Q(x,y)$.
$Q(x,0)\Rightarrow f(f(x))=x$
$Q(f(x),y)\Rightarrow f(x+y)=f(x)+f(y)$
$P(x,0,y)\Rightarrow f(xf(x)+y)=f(xf(x))+f(y)$
$P(x,-x,0)\Rightarrow f(xf(x))=xf(x)$
Now $P(x,y)$ becomes:
\begin{align*}
xf(x)+f(z)+f(xf(y))&=f(xf(x))+f(z)+f(xf(y))\\
&=f(xf(x)+xf(y)+z)\\
&=f(xf(x+y)+z)\\
&=f(z)+yf(x)+f(xf(x))\\
&=f(z)+yf(x)+xf(x)
\end{align*}and so $f(xf(y))=yf(x)$. Taking $y\mapsto f(y)$ we have $f(xy)=f(x)f(y)$, well-known that the only additive and multiplicative functions are $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$ which both work.
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GreekIdiot
222 posts
#11
Y by
why is mongolian tst so easy?
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B1t
24 posts
#12
Y by
GreekIdiot wrote:
why is mongolian tst so easy?

true
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MuradSafarli
110 posts
#13 • 2 Y
Y by B1t, Nuran2010
Interesting problem!

Let \( P(x, -x, y) \) denote the assertion of the functional equation:
\[
xf(x) = f(xf(x)) \tag{1}
\]
Now, consider the following:

- \( P(1, -1, x) \) gives:
\[
f(1) = f(f(1)).
\]- \( P(1, x-1, 0) \) gives:
\[
f(f(x)) = x \cdot f(1). \tag{2}
\]
Now, we consider two cases:

---

**Case 1:** \( f(1) = 0 \)

From equation (2), we have:
\[
f(f(x)) = 0 \quad \text{for all } x.
\]
Applying \( f \) to both sides of equation (1):
\[
xf(x) = f(xf(x)) = f(f(xf(x))) = 0,
\]thus implying:
\[
f(x) = 0 \quad \text{for all } x.
\]
---

**Case 2:** \( f(1) \neq 0 \)

From equation (2), we can conclude that \( f \) is bijective.

Suppose there exists some \( k \) such that \( f(k) = 1 \).
Applying \( P(k, -k, x) \) gives:
\[
k = 1,
\]thus \( k = 1 \).

Moreover, from (2), we obtain:
\[
f(f(x)) = x.
\]
Now, consider \( P(1, f(x) - 1, y) \). We get:
\[
f(x) + f(y) = f(x+y),
\]meaning \( f \) is **additive**.

Since \( f \) is additive and satisfies \( f(f(x)) = x \), we deduce that \( f(x) = cx \) for some constant \( c \). Substituting back into the original functional equation shows that \( c = 1 \), and thus:
\[
f(x) = x.
\]
---

**Final answer:**
1. \( f(x) = 0 \) for all real numbers \( x \), or
2. \( f(x) = x \) for all real numbers \( x \).
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complex2math
7 posts
#14
Y by
Denote by $P(x, y, z)$ the assertion of the given functional equation. Then $P(0, 0, 0)$ and $P(x, -x, 0)$ gives the following

Claim 1. $f(0) = 0$ and $f(xf(x)) = xf(x)$.

Now we can rewrite the given functional equation as
\[
f(xf(x + y) + z) = f(z) + f(x)(x + y) \qquad (\heartsuit)
\]and let $P(x, y, z)$ re-denote the assertion of equation $(\heartsuit)$ and set $f(1) = a$.

Claim 2. If $a = 0$, then $f(x) \equiv 0$.

Proof. $P(x, 1 - x, z)$ implies $f(z) = f(ax + z) = f(z) + f(x)$, so $f(x) = 0$ for all $x \in \mathbb{R}$.

In what follows, we always assume $a \ne 0$.

Claim 3. $f(x)$ is bijective when $a \ne 0$.

Proof. From $P(1, y, 0)$ we get $f(f(1 + y)) = a(y + 1) \implies f(f(y)) = ay$. This is a bijection whenever $a \ne 0$.

Claim 4. $a = f(1) = 1$ and $f(x)$ is additive, i.e. $f(x + y) = f(x) + f(y)$ holds.

Proof. Since $xf(x) = f(xf(x))$, substituting $x = 1$ we obtain $f(1) = f(f(1)) \implies f(1) = 1$ as $f$ is injective. Then $P(x, 1 - x, z)$ gives $f(x + z) = f(z) + f(x)$.

Claim 5. $f(x)$ is multiplicative, i.e. $f(xy) = f(x)f(y)$ holds.

Proof. We have $f(f(y)) = ay = y$ so $P(x, y, 0)$ gives $f(xf(x + y)) = f(x)(x + y) = f(x)\cdot f(f(x + y))$. Then note that
\[
S_x := \{f(x + y): y \in \mathbb{R}\} = \mathbb{R}
\]for fixed $x \in \mathbb{R}$ since $f$ is surjective.

It's well-known that if $f(x)$ is both additive and multiplicative, then either $f(x) \equiv 0$ or $f(x) = x$.
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AshAuktober
1006 posts
#15
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The claims I made in order to solve:
1) $f(0) = 0$.
2) $f(xf(x)) = xf(x)$.
3) $f(xf(y)) = yf(x)$.
4) Either $f \equiv 0$ or $f(x) = 0 \implies x = 0$.
(Now onwards taking second case...)
5) $f$ is an involution
6) $f$ is additive
7) $f$ is multiplicative
And done!
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