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f(f(x)+y) = x+f(f(y))
NicoN9   3
N 43 minutes ago by Ntam.21
Source: own, well this is my first problem I've ever write
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that\[ f(f(x)+y) = x+f(f(y)) \]for all $x, y\in \mathbb{R}$.
3 replies
NicoN9
3 hours ago
Ntam.21
43 minutes ago
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Function from the plane to the real numbers
AndreiVila   4
N an hour ago by GreekIdiot
Source: Balkan MO Shortlist 2024 G7
Let $f:\pi\rightarrow\mathbb{R}$ be a function from the Euclidean plane to the real numbers such that $$f(A)+f(B)+f(C)=f(O)+f(G)+f(H)$$for any acute triangle $ABC$ with circumcenter $O$, centroid $G$ and orthocenter $H$. Prove that $f$ is constant.
4 replies
AndreiVila
Today at 6:50 AM
GreekIdiot
an hour ago
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Domain swept by Parabola
Kunihiko_Chikaya   1
N an hour ago by Mathzeus1024
Source: created by kunny
In the $x$-$y$ plane, given a parabola $C_t$ passing through 3 points $P(t-1,\ t),\ Q(t,\ t)$ and $R(t+1,\ t+2)$.
Let $t$ vary in the range of $-1\leq t\leq 1$, draw the domain swept out by $C_t$.
1 reply
Kunihiko_Chikaya
Jan 3, 2012
Mathzeus1024
an hour ago
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a_1 is anything but 2
EeEeRUT   4
N an hour ago by Assassino9931
Source: Thailand TSTST 2024 P4
The sequence $(a_n)_{n\in\mathbb{N}}$ is defined by $a_1=3$ and $$a_n=a_1a_2\cdots a_{n-1}-1$$Show that there exist infinitely many prime number that divide at least one number in this sequences
4 replies
EeEeRUT
Jul 18, 2024
Assassino9931
an hour ago
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Inversion exercise
Assassino9931   4
N an hour ago by ItzsleepyXD
Source: Balkan MO Shortlist 2024 G5
Let $ABC$ be an acute scalene triangle $ABC$, $D$ be the orthogonal projection of $A$ on $BC$, $M$ and $N$ are the midpoints of $AB$ and $AC$ respectively. Let $P$ and $Q$ are points on the minor arcs $\widehat{AB}$ and $\widehat{AC}$ of the circumcircle of triangle $ABC$ respectively such that $PQ \parallel BC$. Show that the circumcircles of triangles $DPQ$ and $DMN$ are tangent if and only if $M$ lies on $PQ$.
4 replies
Assassino9931
Yesterday at 10:29 PM
ItzsleepyXD
an hour ago
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A game optimization on a graph
Assassino9931   3
N an hour ago by dgrozev
Source: Bulgaria National Olympiad 2025, Day 2, Problem 6
Let \( X_0, X_1, \dots, X_{n-1} \) be \( n \geq 2 \) given points in the plane, and let \( r > 0 \) be a real number. Alice and Bob play the following game. Firstly, Alice constructs a connected graph with vertices at the points \( X_0, X_1, \dots, X_{n-1} \), i.e., she connects some of the points with edges so that from any point you can reach any other point by moving along the edges.Then, Alice assigns to each vertex \( X_i \) a non-negative real number \( r_i \), for \( i = 0, 1, \dots, n-1 \), such that $\sum_{i=0}^{n-1} r_i = 1$. Bob then selects a sequence of distinct vertices \( X_{i_0} = X_0, X_{i_1}, \dots, X_{i_k} \) such that \( X_{i_j} \) and \( X_{i_{j+1}} \) are connected by an edge for every \( j = 0, 1, \dots, k-1 \). (Note that the length $k \geq 0$ is not fixed and the first selected vertex always has to be $X_0$.) Bob wins if
\[ \frac{1}{k+1} \sum_{j=0}^{k} r_{i_j} \geq r; \]otherwise, Alice wins. Depending on \( n \), determine the largest possible value of \( r \) for which Bob has a winning strategy.
3 replies
Assassino9931
Apr 8, 2025
dgrozev
an hour ago
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Determine all the functions
Martin.s   2
N 2 hours ago by Blackbeam999


Determine all the functions $f: \mathbb{R} \to \mathbb{R}$ such that

\[ f(x^2 \cdot f(x) + f(y)) = f(f(x^3)) + y \]
for all $x, y \in \mathbb{R}$.


2 replies
Martin.s
Aug 14, 2024
Blackbeam999
2 hours ago
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Geometric inequality with Fermat point
Assassino9931   4
N 2 hours ago by ItsBesi
Source: Balkan MO Shortlist 2024 G2
Let $ABC$ be an acute triangle and let $P$ be an interior point for it such that $\angle APB = \angle BPC = \angle CPA$. Prove that
$$ \frac{PA^2 + PB^2 + PC^2}{2S} + \frac{4}{\sqrt{3}} \leq \frac{1}{\sin \alpha} + \frac{1}{\sin \beta} + \frac{1}{\sin \gamma}. $$When does equality hold?
4 replies
Assassino9931
Yesterday at 10:21 PM
ItsBesi
2 hours ago
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Iran TST P8
TheBarioBario   7
N 2 hours ago by bin_sherlo
Source: Iranian TST 2022 problem 8
In triangle $ABC$, with $AB<AC$, $I$ is the incenter, $E$ is the intersection of $A$-excircle and $BC$. Point $F$ lies on the external angle bisector of $BAC$ such that $E$ and $F$ lieas on the same side of the line $AI$ and $\angle AIF=\angle AEB$. Point $Q$ lies on $BC$ such that $\angle AIQ=90$. Circle $\omega_b$ is tangent to $FQ$ and $AB$ at $B$, circle $\omega_c$ is tangent to $FQ$ and $AC$ at $C$ and both circles pass through the inside of triangle $ABC$. if $M$ is the Midpoint od the arc $BC$, which does not contain $A$, prove that $M$ lies on the radical axis of $\omega_b$ and $\omega_c$.

Proposed by Amirmahdi Mohseni
7 replies
TheBarioBario
Apr 2, 2022
bin_sherlo
2 hours ago
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Parallel lines with incircle
buratinogigle   1
N 2 hours ago by luutrongphuc
Source: Own, test for the preliminary team of HSGS 2025
Let $ABC$ be a triangle with incircle $(I)$, which touches sides $CA$ and $AB$ at points $E$ and $F$, respectively. Choose points $M$ and $N$ on the line $EF$ such that $BM = BF$ and $CN = CE$. Let $P$ be the intersection of lines $CM$ and $BN$. Define $Q$ and $R$ as the intersections of $PN$ and $PM$ with lines $IC$ and $IB$, respectively. Assume that $J$ is the intersection of $QR$ and $BC$. Prove that $PJ \parallel MN$.
1 reply
buratinogigle
Yesterday at 11:23 AM
luutrongphuc
2 hours ago
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Triple Sum
P162008   1
N Yesterday at 10:09 PM by ysharifi
Evaluate $\Omega = \sum_{k=1}^{\infty} \sum_{n=k}^{\infty} \sum_{m=1}^{n} \frac{1}{n(n+1)(n+2)km^2}$
1 reply
P162008
Apr 26, 2025
ysharifi
Yesterday at 10:09 PM
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integral
Svyatoslav   5
N Nov 16, 2024 by tobiSALT
How do you find
$$I(a,b)=\int_0^\infty\frac{\ln|b-\cos ax|}{1+x^2}\,dx,\,\,b\in[0;1]\,\,?$$
5 replies
Svyatoslav
Nov 14, 2024
tobiSALT
Nov 16, 2024
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integral
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integration
calculus
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Svyatoslav
540 posts
Svyatoslav
#1 h  Nov 14, 2024, 6:05 AM • 1 Y
Y by Martin.s
How do you find
$$I(a,b)=\int_0^\infty\frac{\ln|b-\cos ax|}{1+x^2}\,dx,\,\,b\in[0;1]\,\,?$$
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tobiSALT
81 posts
tobiSALT
#2 h  Nov 15, 2024, 6:43 AM • 1 Y
Y by StarLex1
Let the integral be
$ I(a, b) = \int_0^{\infty} \frac{\ln |b - \cos(ax)|}{1 + x^2} \, dx, \quad b \in [0, 1] \text{ and } a > 0. $
Let $b = \cos \theta$ for some $\theta \in [0, \pi/2]$. Then we can rewrite the integral as
$ I(a, b) = \int_0^{\infty} \frac{\ln |\cos \theta - \cos(ax)|}{1 + x^2} \, dx. $
We have the identity $\cos \theta - \cos(ax) = 2 \sin \frac{ax + \theta}{2} \sin \frac{ax - \theta}{2}$. Thus,
$ \ln |\cos \theta - \cos(ax)| = \ln |2 \sin \frac{ax+\theta}{2} \sin \frac{ax-\theta}{2}| = \ln 2 + \ln |\sin \frac{ax+\theta}{2}| + \ln |\sin \frac{ax-\theta}{2}|. $
Also, we know that
$ \int_0^{\infty} \frac{\ln 2}{1 + x^2} \, dx = \ln 2 \int_0^{\infty} \frac{dx}{1 + x^2} = \ln 2 \arctan(x) \Big|_0^{\infty} = \ln 2 \cdot \frac{\pi}{2} = \frac{\pi}{2} \ln 2. $
We use the known result
$ \int_0^{\infty} \frac{\ln|\sin(rx)|}{1 + x^2} \, dx = \frac{\pi}{2} \ln \left( \frac{1 - e^{-2r}}{2} \right). $
Then
$ \int_0^{\infty} \frac{\ln |\sin \frac{ax+\theta}{2}|}{1 + x^2} \, dx = \int_0^{\infty} \frac{\ln |\sin \frac{a}{2} (x+\theta/a)|}{1 + x^2} \, dx $
and
$ \int_0^{\infty} \frac{\ln |\sin \frac{ax-\theta}{2}|}{1 + x^2} \, dx = \int_0^{\infty} \frac{\ln |\sin \frac{a}{2} (x-\theta/a)|}{1 + x^2} \, dx. $
So we have
$ I(a, b) = \int_0^{\infty} \frac{\ln |b - \cos(ax)|}{1 + x^2} \, dx = \int_0^{\infty} \frac{\ln |\cos \theta - \cos(ax)|}{1 + x^2} \, dx = \frac{\pi}{2} \ln 2 + \int_0^{\infty} \frac{\ln |\sin(\frac{ax+\theta}{2})| + \ln |\sin(\frac{ax-\theta}{2})|}{1 + x^2} \, dx. $
If $\theta = 0$, i.e., $b = 1$, then
$ \int_0^{\infty} \frac{\ln(1 - \cos(ax))}{1 + x^2} \, dx = \int_0^{\infty} \frac{2 \ln |\sin(ax/2)| + \ln 2}{1 + x^2} \, dx = \frac{\pi}{2} \ln 2 + 2 \int_0^{\infty} \frac{\ln |\sin(ax/2)|}{1 + x^2} \, dx. $
Using the formula above, we get
$ \frac{\pi}{2} \ln 2 + 2 \frac{\pi}{2} \ln \left( \frac{1 - e^{-a}}{2} \right) = \frac{\pi}{2} (\ln 2 + 2 \ln (1 - e^{-a}) - 2 \ln 2) = \pi \ln(1 - e^{-a}) - \frac{\pi}{2} \ln 2. $
Thus
$ I(a, 1) = \pi \ln(1 - e^{-a}) - \frac{\pi}{2} \ln 2. $
Therefore, our final answer is
$ I(a, b) = \pi \ln \left( \frac{1 + \sqrt{1-b^2} - e^{-a}}{2} \right). $
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Svyatoslav
540 posts
Svyatoslav
#3 h  Nov 15, 2024, 12:54 PM • 1 Y
Y by tobiSALT
Thank you for your solution, but I think the answer is not correct. At $b=0\,$ and $\,a=0$
$$\int_0^\infty\frac{\ln (1)}{1+x^2}dx=0\neq\,\pi\ln\frac12$$There is also a known result (I.S. Gradshteyn and I.M. Ryzhik, GW (338)(28a), p 569)
$$\int_0^\infty\frac{\ln\cos^2ax}{1+x^2}dx=\pi\ln\frac{1+e^{-2a}}2$$
This post has been edited 1 time. Last edited by Svyatoslav, Nov 15, 2024, 6:42 PM
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tobiSALT
81 posts
tobiSALT
#4 h  Nov 15, 2024, 1:08 PM
Y by
True. I´ll try again later today, this integral is really hard.
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Svyatoslav
540 posts
Svyatoslav
#5 h  Nov 15, 2024, 1:20 PM • 1 Y
Y by Martin.s
Thank you. I got
$$I(a,b)=\frac\pi2\ln\frac{1-2be^{-a}+e^{-2a}}2, \,b\in [-1;1];\,a\geqslant0$$what gives the correct answer for one of the options: $b=0$ or $b=1$ or $a=0$. My problem is that WA (free option) is not able to evaluate the integral numerically, and I cannot get a reliable numeric check.

PS The answer is indeed correct - please see the solution.
This post has been edited 2 times. Last edited by Svyatoslav, Nov 16, 2024, 1:18 AM
Reason: slight correction
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tobiSALT
81 posts
tobiSALT
#7 h  Nov 16, 2024, 5:07 PM • 1 Y
Y by Svyatoslav
Great! Good solution
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