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Strictly monotone polynomial with an extra condition
Popescu   11
N 10 minutes ago by Iveela
Source: IMSC 2024 Day 2 Problem 2
Let $\mathbb{R}_{>0}$ be the set of all positive real numbers. Find all strictly monotone (increasing or decreasing) functions $f:\mathbb{R}_{>0} \to \mathbb{R}$ such that there exists a two-variable polynomial $P(x, y)$ with real coefficients satisfying
$$ f(xy)=P(f(x), f(y)) $$for all $x, y\in\mathbb{R}_{>0}$.

Proposed by Navid Safaei, Iran
11 replies
Popescu
Jun 29, 2024
Iveela
10 minutes ago
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Russian Diophantine Equation
LeYohan   1
N 21 minutes ago by Natrium
Source: Moscow, 1963
Find all integer solutions to

$\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3$.
1 reply
LeYohan
Yesterday at 2:59 PM
Natrium
21 minutes ago
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Simple Geometry
AbdulWaheed   6
N 23 minutes ago by Gggvds1
Source: EGMO
Try to avoid Directed angles
Let ABC be an acute triangle inscribed in circle $\Omega$. Let $X$ be the midpoint of the arc $\overarc{BC}$ not containing $A$ and define $Y, Z$ similarly. Show that the orthocenter of $XYZ$ is the incenter $I$ of $ABC$.
6 replies
AbdulWaheed
May 23, 2025
Gggvds1
23 minutes ago
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Bosnia and Herzegovina JBMO TST 2013 Problem 4
gobathegreat   4
N 25 minutes ago by FishkoBiH
Source: Bosnia and Herzegovina Junior Balkan Mathematical Olympiad TST 2013
It is given polygon with $2013$ sides $A_{1}A_{2}...A_{2013}$. His vertices are marked with numbers such that sum of numbers marked by any $9$ consecutive vertices is constant and its value is $300$. If we know that $A_{13}$ is marked with $13$ and $A_{20}$ is marked with $20$, determine with which number is marked $A_{2013}$
4 replies
gobathegreat
Sep 16, 2018
FishkoBiH
25 minutes ago
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A geometry problem
Lttgeometry   2
N 27 minutes ago by Acrylic3491
Triangle $ABC$ has two isogonal conjugate points $P$ and $Q$. The circle $(BPC)$ intersects circle $(AP)$ at $R \neq P$, and the circle $(BQC)$ intersects circle $(AQ)$ at $S\neq Q$. Prove that $R$ and $S$ are isogonal conjugates in triangle $ABC$.
Note: Circle $(AP)$ is the circle with diameter $AP$, Circle $(AQ)$ is the circle with diameter $AQ$.
2 replies
Lttgeometry
Today at 4:03 AM
Acrylic3491
27 minutes ago
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anglechasing , circumcenter wanted
parmenides51   1
N 32 minutes ago by Captainscrubz
Source: Sharygin 2011 Final 9.2
In triangle $ABC, \angle B = 2\angle C$. Points $P$ and $Q$ on the medial perpendicular to $CB$ are such that $\angle CAP = \angle PAQ = \angle QAB = \frac{\angle A}{3}$ . Prove that $Q$ is the circumcenter of triangle $CPB$.
1 reply
parmenides51
Dec 16, 2018
Captainscrubz
32 minutes ago
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Nice FE over R+
doanquangdang   4
N an hour ago by jasperE3
Source: collect
Let $\mathbb{R}^+$ denote the set of positive real numbers. Find all functions $f:\mathbb{R}^+ \to \mathbb{R}^+$ such that
\[x+f(yf(x)+1)=xf(x+y)+yf(yf(x))\]for all $x,y>0.$
4 replies
doanquangdang
Jul 19, 2022
jasperE3
an hour ago
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right triangle, midpoints, two circles, find angle
star-1ord   0
an hour ago
Source: Estonia Final Round 2025 8-3
In the right triangle $ABC$, $M$ is the midpoint of the hypotenuse $AB$. Point $D$ is chosen on the leg $BC$ so that the line segment $DM$ meets $(ACD)$ again at $K$ ($K\neq D$). Let $L$ be the reflection of $K$ in $M$. The circles $(ACD)$ and $(BCL)$ meet again at $N$ ($N\neq C$). Find the measure of $\angle KNL$.
0 replies
star-1ord
an hour ago
0 replies
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interesting functional equation
tabel   3
N an hour ago by waterbottle432
Source: random romanian contest
Determine all functions \( f : (0, \infty) \to (0, \infty) \) that satisfy the functional equation:
\[ f(f(x)(1 + y)) = f(x) + f(xy), \quad \forall x, y > 0. \]
3 replies
tabel
2 hours ago
waterbottle432
an hour ago
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Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   4
N an hour ago by GreenTea2593
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
4 replies
OgnjenTesic
May 22, 2025
GreenTea2593
an hour ago
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pairs (m, n) such that a fractional expression is an integer
cielblue   2
N 2 hours ago by cielblue
Find all pairs $(m,\ n)$ of positive integers such that $\frac{m^3-mn+1}{m^2+mn+2}$ is an integer.
2 replies
cielblue
Yesterday at 8:38 PM
cielblue
2 hours ago
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Sociable set of people
jgnr   23
N 2 hours ago by quantam13
Source: RMM 2012 day 1 problem 1
Given a finite number of boys and girls, a sociable set of boys is a set of boys such that every girl knows at least one boy in that set; and a sociable set of girls is a set of girls such that every boy knows at least one girl in that set. Prove that the number of sociable sets of boys and the number of sociable sets of girls have the same parity. (Acquaintance is assumed to be mutual.)

(Poland) Marek Cygan
23 replies
jgnr
Mar 3, 2012
quantam13
2 hours ago
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Inspired by learningimprove
sqing   7
N Apr 19, 2025 by sqing
Source: Own
Let $ a,b,c,d\geq0, (a+b)(c+d)=2 . $ Prove that
$$ a^2+b^2+c^2+d^2-ac-bd \geq1 $$Let $ a,b,c,d\geq0, (a+2b)(c+2d)=2 . $ Prove that
$$ a^2+b^2+c^2+d^2-ac-bd \geq\frac{2}{5} $$Let $ a,b,c,d\geq0, (a+2b)(2c+ d)=2 . $ Prove that
$$ a^2+b^2+c^2+d^2-ac-bd \geq\frac{3}{7} $$
7 replies
sqing
Apr 19, 2025
sqing
Apr 19, 2025
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Inspired by learningimprove
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Reveal topic tags
inequalities proposed
algebra
inequalities
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Source: Own
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sqing
42398 posts
sqing
#1 h  Apr 19, 2025, 7:52 AM
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Let $ a,b,c,d\geq0, (a+b)(c+d)=2 . $ Prove that
$$ a^2+b^2+c^2+d^2-ac-bd \geq1 $$Let $ a,b,c,d\geq0, (a+2b)(c+2d)=2 . $ Prove that
$$ a^2+b^2+c^2+d^2-ac-bd \geq\frac{2}{5} $$Let $ a,b,c,d\geq0, (a+2b)(2c+ d)=2 . $ Prove that
$$ a^2+b^2+c^2+d^2-ac-bd \geq\frac{3}{7} $$
This post has been edited 1 time. Last edited by sqing, Apr 19, 2025, 7:54 AM
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GeoMorocco
44 posts
GeoMorocco
#2 h  Apr 19, 2025, 8:08 AM
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sqing wrote:
Let $ a,b,c,d\geq0, (a+b)(c+d)=2 . $ Prove that
$$ a^2+b^2+c^2+d^2-ac-bd \geq1 $$
we have:
$$ 2(a^2+b^2+c^2+d^2-ac-bd-1) =(a-c)^2 + (b-d)^2+(a^2+b^2+c^2+d^2-ac-ad-bc-bd)$$which is true by rearrangement. Equality for $a=b=c=d$.

The other inequalities can be proved in a similar way.
This post has been edited 1 time. Last edited by GeoMorocco, Apr 19, 2025, 8:08 AM
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sqing
42398 posts
sqing
#3 h  Apr 19, 2025, 8:09 AM
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Nice.Thanks.
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sqing
42398 posts
sqing
#4 h  Apr 19, 2025, 8:14 AM
Y by
Let $ a,b,c $ be reals such that $ a+2b+c=0 $ and $ a^2+b^2+c^2=3. $ Prove that
$$abc\leq 1$$Let $ a,b,c $ be reals such that $ a+2b+c=0 $ and $ a^2+2b^2+c^2=3. $ Prove that
$$abc\leq\frac{3\sqrt{3}}{8}$$Let $ a,b,c $ be reals such that $ a+4b+c=0 $ and $ a^2+2b^2+c^2=3. $ Prove that
$$abc\leq\frac{3 }{5}\sqrt{\frac{6}{5}}$$
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SunnyEvan
130 posts
SunnyEvan
#5 h  Apr 19, 2025, 10:38 AM
Y by
$$ a+2b+c=0 \iff b= -\frac{a+c}{2}$$$$ a^2+b^2+c^2=3 \iff 5b^2-2ac=3 $$$$ abc \leq 1 \iff 5b^3-3b-2 \leq 0 \iff b \leq 1 \iff |a+c| \leq 2 $$because : $$ \frac{5}{4}(a+c)^2 \leq 3+ \frac{1}{2}(a+c)^2 \iff |a+c| \leq 2 $$The other inequalities can be proved in a similar way. :-D
equality case : $ (a,b,c)=(-1,1,-1)$
This post has been edited 2 times. Last edited by SunnyEvan, Apr 19, 2025, 10:45 AM
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GeoMorocco
44 posts
GeoMorocco
#6 h  Apr 19, 2025, 11:05 AM
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SunnyEvan wrote:
$$ a+2b+c=0 \iff b= -\frac{a+c}{2}$$$$ a^2+b^2+c^2=3 \iff 5b^2-2ac=3 $$$$ abc \leq 1 \iff 5b^3-3b-2 \leq 0 \iff b \leq 1 \iff |a+c| \leq 2 $$because : $$ \frac{5}{4}(a+c)^2 \leq 3+ \frac{1}{2}(a+c)^2 \iff |a+c| \leq 2 $$The other inequalities can be proved in a similar way. :-D
equality case : $ (a,b,c)=(-1,1,-1)$
The first one is trivial by Am-Gm. In fact, we have:
$$3 = a^2+b^2+c^2 \geq 3\sqrt[3]{a^2b^2c^2}$$Therefore
$$abc \leq |abc| \leq 1$$
This post has been edited 1 time. Last edited by GeoMorocco, Apr 19, 2025, 11:05 AM
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SunnyEvan
130 posts
SunnyEvan
#7 h  Apr 19, 2025, 11:41 AM
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GeoMorocco wrote:
SunnyEvan wrote:
$$ a+2b+c=0 \iff b= -\frac{a+c}{2}$$$$ a^2+b^2+c^2=3 \iff 5b^2-2ac=3 $$$$ abc \leq 1 \iff 5b^3-3b-2 \leq 0 \iff b \leq 1 \iff |a+c| \leq 2 $$because : $$ \frac{5}{4}(a+c)^2 \leq 3+ \frac{1}{2}(a+c)^2 \iff |a+c| \leq 2 $$The other inequalities can be proved in a similar way. :-D
equality case : $ (a,b,c)=(-1,1,-1)$
The first one is trivial by Am-Gm. In fact, we have:
$$3 = a^2+b^2+c^2 \geq 3\sqrt[3]{a^2b^2c^2}$$Therefore
$$abc \leq |abc| \leq 1$$

Yeah . :-D
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sqing
42398 posts
sqing
#8 h  Apr 19, 2025, 12:23 PM
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Thank you.
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