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Question on Balkan SL
Fmimch   3
N a few seconds ago by GreekIdiot
Does anyone know where to find the Balkan MO Shortlist 2024? If you have the file, could you send in this thread? Thank you!
3 replies
Fmimch
Today at 12:13 AM
GreekIdiot
a few seconds ago
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Functional equation
Amin12   17
N 19 minutes ago by bin_sherlo
Source: Iran 3rd round 2017 first Algebra exam
Find all functions $f:\mathbb{R^+}\rightarrow\mathbb{R^+}$ such that
$$\frac{x+f(y)}{xf(y)}=f(\frac{1}{y}+f(\frac{1}{x}))$$for all positive real numbers $x$ and $y$.
17 replies
Amin12
Aug 7, 2017
bin_sherlo
19 minutes ago
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Generating Functions
greenplanet2050   6
N 5 hours ago by ohiorizzler1434
So im learning generating functions and i dont really understand why $1+2x+3x^2+4x^3+5x^4+…=\dfrac{1}{(1-x)^2}$

can someone help

thank you :)
6 replies
greenplanet2050
Yesterday at 10:42 PM
ohiorizzler1434
5 hours ago
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9 Physical or online
wimpykid   0
5 hours ago
Do you think the AoPS print books or the online books are better?

0 replies
wimpykid
5 hours ago
0 replies
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Three variables inequality
Headhunter   6
N 6 hours ago by lbh_qys
$\forall a\in R$ ,$~\forall b\in R$ ,$~\forall c \in R$
Prove that at least one of $(a-b)^{2}$, $(b-c)^{2}$, $(c-a)^{2}$ is not greater than $\frac{a^{2}+b^{2}+c^{2}}{2}$.

I assume that all are greater than it, but can't go more.
6 replies
Headhunter
Apr 20, 2025
lbh_qys
6 hours ago
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Sequence
lgx57   8
N Today at 5:08 AM by Vivaandax
$a_1=1,a_{n+1}=a_n+\frac{1}{a_n}$. Find the general term of $\{a_n\}$.
8 replies
lgx57
Apr 27, 2025
Vivaandax
Today at 5:08 AM
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Geometric inequality
ReticulatedPython   3
N Today at 4:27 AM by ItalianZebra
Let $A$ and $B$ be points on a plane such that $AB=n$, where $n$ is a positive integer. Let $S$ be the set of all points $P$ such that $\frac{AP^2+BP^2}{(AP)(BP)}=c$, where $c$ is a real number. The path that $S$ traces is continuous, and the value of $c$ is minimized. Prove that $c$ is rational for all positive integers $n.$
3 replies
ReticulatedPython
Apr 22, 2025
ItalianZebra
Today at 4:27 AM
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Transformation of a cross product when multiplied by matrix A
Math-lover1   1
N Today at 1:02 AM by greenturtle3141
I was working through AoPS Volume 2 and this statement from Chapter 11: Cross Products/Determinants confused me.
[quote=AoPS Volume 2]A quick comparison of $|\underline{A}|$ to the cross product $(\underline{A}\vec{i}) \times (\underline{A}\vec{j})$ reveals that a negative determinant [of $\underline{A}$] corresponds to a matrix which reverses the direction of the cross product of two vectors.[/quote]
I understand that this is true for the unit vectors $\vec{i} = (1 \ 0)$ and $\vec{j} = (0 \ 1)$, but am confused on how to prove this statement for general vectors $\vec{v}$ and $\vec{w}$ although its supposed to be a quick comparison.

How do I prove this statement easily with any two 2D vectors?
1 reply
Math-lover1
Yesterday at 10:29 PM
greenturtle3141
Today at 1:02 AM
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Geometry books
T.Mousavidin   4
N Today at 12:10 AM by compoly2010
Hello, I wanted to ask if anybody knows some good books for geometry that has these topics in:
Desargues's Theorem, Projective geometry, 3D geometry,
4 replies
T.Mousavidin
Yesterday at 4:25 PM
compoly2010
Today at 12:10 AM
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trigonometric functions
VivaanKam   3
N Yesterday at 10:08 PM by aok
Hi could someone explain the basic trigonometric functions to me like sin, cos, tan etc.
Thank you!
3 replies
VivaanKam
Yesterday at 8:29 PM
aok
Yesterday at 10:08 PM
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Inequalities
sqing   16
N Yesterday at 5:25 PM by martianrunner
Let $ a,b \in [0 ,1] . $ Prove that
$$\frac{a}{ 1-ab+b }+\frac{b }{ 1-ab+a } \leq 2$$$$ \frac{a}{ 1+ab+b^2 }+\frac{b }{ 1+ab+a^2 }+\frac{ab }{2+ab } \leq 1$$$$\frac{a}{ 1-ab+b^2 }+\frac{b }{ 1-ab+a^2 }+\frac{1 }{1+ab }\leq \frac{5}{2}$$$$\frac{a}{ 1-ab+b^2 }+\frac{b }{ 1-ab+a^2 }+\frac{1 }{1+2ab }\leq \frac{7}{3}$$$$\frac{a}{ 1+ab+b^2 }+\frac{b }{ 1+ab+a^2 } +\frac{ab }{1+ab }\leq \frac{7}{6 }$$
16 replies
sqing
Apr 25, 2025
martianrunner
Yesterday at 5:25 PM
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Geometry Angle Chasing
Sid-darth-vater   6
N Yesterday at 2:18 PM by sunken rock
Is there a way to do this without drawing obscure auxiliary lines? (the auxiliary lines might not be obscure I might just be calling them obscure)

For example I tried rotating triangle MBC 80 degrees around point C (so the BC line segment would now lie on segment AC) but I couldn't get any results. Any help would be appreciated!
6 replies
Sid-darth-vater
Apr 21, 2025
sunken rock
Yesterday at 2:18 PM
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Two Functional Inequalities
Mathdreams   7
N Apr 7, 2025 by John_Mgr
Source: 2025 Nepal Mock TST Day 2 Problem 2
Determine all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(x) \le x^3$$and $$f(x + y) \le f(x) + f(y) + 3xy(x + y)$$for any real numbers $x$ and $y$.

(Miroslav Marinov, Bulgaria)
7 replies
Mathdreams
Apr 6, 2025
John_Mgr
Apr 7, 2025
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Two Functional Inequalities
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functional inequalities
algebra
functional equation
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Source: 2025 Nepal Mock TST Day 2 Problem 2
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Mathdreams
1469 posts
Mathdreams
#1 h  Apr 6, 2025, 1:34 PM • 4 Y
Y by PikaPika999, AlexCenteno2007, cubres, khan.academy
Determine all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(x) \le x^3$$and $$f(x + y) \le f(x) + f(y) + 3xy(x + y)$$for any real numbers $x$ and $y$.

(Miroslav Marinov, Bulgaria)
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grupyorum
1416 posts
grupyorum
#2 h  Apr 6, 2025, 1:45 PM • 2 Y
Y by PikaPika999, cubres
I will show $f(x)=x^3$ for all $x$, which clearly works.

Taking $x=y=0$ in both, we find $f(0)\le 0$ and $f(0)\ge 0$, so $f(0)=0$. Taking $y=-x$ in the second, we find $0=f(0)\le f(x)+f(-x)$. Likewise, $f(x)\le x^3$ and $f(-x)\le (-x)^3=-x^3$. So, $f(x)+f(-x)\le 0$. Combining, $f(x)+f(-x)=0$ for all $x$. Lastly, using $f(-x)=-f(x)$, we get $-x^3 \ge f(-x)=-f(x)$, so $f(x)\ge x^3$ too. Thus, $f(x)=x^3$ for all $x$.
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kokcio
69 posts
kokcio
#3 h  Apr 6, 2025, 1:46 PM • 2 Y
Y by PikaPika999, cubres
Putting $x=y=0$, we have $f(0)\leq 2f(0)$, but $f(0)\leq 0$, so we have to have that $f(0)=0$.
Putting now $y=-x$, we have $0\leq f(x)+f(-x) \leq x^3 + (-x)^3 = 0$, but this means that we have to have $f(x)=x^3$, because of inequalities.
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John_Mgr
67 posts
John_Mgr
#4 h  Apr 6, 2025, 4:33 PM • 2 Y
Y by PikaPika999, cubres
Claim: $f(x) \equiv x^3$ is the only function $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfies the given conditions.
Let P(x,y) be the assertion for the convenience.
i.e $f(x+y) \le f(x)+f(y)+3xy(x+y)$
P(0,0): $f(0)\ge0$ and $f(0)\le0$, $\Rightarrow$ $f(0)=0$,
Assume $f(x)=g(x)+x^3$, $\implies$ $g(x+y)\le g(x)+g(y)$
This tells us that g is subadditive.
$g(x)\le 0$ as $f(x)\le x^3$
Let Q(x,y) be the assertion for $g(x+y)\le g(x)+g(y)$
Q(x,-x): $g(0)\le g(x)+g(-x)$$\rightarrow g(x)+g(-x)\ge 0$ as $f(0)=0$ but $g(x)+g(-x)\le 0$
So, $g(x)+g(-x)=0$$\Rightarrow g(-x)=-g(x) \ge 0$
The only way both $g(x)\le x$ and $g(-x)\ge 0$ hold is if $g(x)=0$, $\forall$ $x\in \mathbb{R}$
Therefore $\boxed{f(x)=x^3}$, $\forall$ $x\in \mathbb{R}$
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jasperE3
11279 posts
jasperE3
#5 h  Apr 6, 2025, 7:01 PM • 5 Y
Y by PikaPika999, Assassino9931, megarnie, navier3072, cubres
Let $g(x)=f(x)-x^3+x$, it reduces to this old problem:
https://artofproblemsolving.com/community/c6h257315p1403063
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Maximilian113
568 posts
Maximilian113
#6 h  Apr 6, 2025, 7:16 PM • 2 Y
Y by PikaPika999, cubres
Note that $x=y=0 \implies 0 \leq f(0) \leq 0 \implies f(0)=0.$ Now $$y=-x \implies 0 \leq f(x)+f(-x) \leq f(x) -x^3 \implies x^3 \leq f(x) \implies f(x)=x^3$$for all $x,$ which clearly works.
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Assassino9931
1286 posts
Assassino9931
#7 h  Apr 6, 2025, 10:18 PM • 1 Y
Y by cubres
jasperE3 wrote:
Let $g(x)=f(x)-x^3+x$, it reduces to this old problem:
https://artofproblemsolving.com/community/c6h257315p1403063

Nice catch haha
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John_Mgr
67 posts
John_Mgr
#8 h  Apr 7, 2025, 3:59 AM • 1 Y
Y by cubres
jasperE3 wrote:
Let $g(x)=f(x)-x^3+x$, it reduces to this old problem:
https://artofproblemsolving.com/community/c6h257315p1403063

Good one!!
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