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polynomial problem from russia 2002
vineet   8
N 13 minutes ago by LawofCosine
Source: 2002 All-Russian MO, Grade 10, Problem 1
The polynomials $P$, $Q$, $R$ with real coefficients, one of which is degree $2$ and two of degree $3$, satisfy the equality $P^2+Q^2=R^2$. Prove that one of the polynomials of degree $3$ has three real roots.
8 replies
+1 w
vineet
Feb 25, 2003
LawofCosine
13 minutes ago
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IMO Shortlist 2010 - Problem G7
Amir Hossein   21
N an hour ago by MathLuis
Three circular arcs $\gamma_1, \gamma_2,$ and $\gamma_3$ connect the points $A$ and $C.$ These arcs lie in the same half-plane defined by line $AC$ in such a way that arc $\gamma_2$ lies between the arcs $\gamma_1$ and $\gamma_3.$ Point $B$ lies on the segment $AC.$ Let $h_1, h_2$, and $h_3$ be three rays starting at $B,$ lying in the same half-plane, $h_2$ being between $h_1$ and $h_3.$ For $i, j = 1, 2, 3,$ denote by $V_{ij}$ the point of intersection of $h_i$ and $\gamma_j$ (see the Figure below). Denote by $\widehat{V_{ij}V_{kj}}\widehat{V_{kl}V_{il}}$ the curved quadrilateral, whose sides are the segments $V_{ij}V_{il},$ $V_{kj}V_{kl}$ and arcs $V_{ij}V_{kj}$ and $V_{il}V_{kl}.$ We say that this quadrilateral is $circumscribed$ if there exists a circle touching these two segments and two arcs. Prove that if the curved quadrilaterals $\widehat{V_{11}V_{21}}\widehat{V_{22}V_{12}}, \widehat{V_{12}V_{22}}\widehat{V_{23}V_{13}},\widehat{V_{21}V_{31}}\widehat{V_{32}V_{22}}$ are circumscribed, then the curved quadrilateral $\widehat{V_{22}V_{32}}\widehat{V_{33}V_{23}}$ is circumscribed, too.

Proposed by Géza Kós, Hungary

IMAGE
21 replies
Amir Hossein
Jul 17, 2011
MathLuis
an hour ago
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Interesting inequality
sqing   5
N an hour ago by sqing
Source: Own
Let $a,b\geq 0, 2a+2b+ab=5.$ Prove that
$$a+b^3+a^3b+\frac{101}{8}ab\leq\frac{125}{8}$$
5 replies
sqing
3 hours ago
sqing
an hour ago
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Number of functions satisfying sum inequality
CyclicISLscelesTrapezoid   19
N an hour ago by john0512
Source: ISL 2022 C5
Let $m,n \geqslant 2$ be integers, let $X$ be a set with $n$ elements, and let $X_1,X_2,\ldots,X_m$ be pairwise distinct non-empty, not necessary disjoint subset of $X$. A function $f \colon X \to \{1,2,\ldots,n+1\}$ is called nice if there exists an index $k$ such that \[\sum_{x \in X_k} f(x)>\sum_{x \in X_i} f(x) \quad \text{for all } i \ne k.\]Prove that the number of nice functions is at least $n^n$.
19 replies
CyclicISLscelesTrapezoid
Jul 9, 2023
john0512
an hour ago
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Inequality em981
oldbeginner   21
N an hour ago by sqing
Source: Own
Let $a, b, c>0, a+b+c=3$. Prove that
\[\sqrt{a+\frac{9}{b+2c}}+\sqrt{b+\frac{9}{c+2a}}+\sqrt{c+\frac{9}{a+2b}}+\frac{2(ab+bc+ca)}{9}\ge\frac{20}{3}\]
21 replies
oldbeginner
Sep 22, 2016
sqing
an hour ago
J
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Interesting inequality
sqing   3
N an hour ago by sqing
Source: Own
Let $ (a+b)^2+(a-b)^2=1. $ Prove that
$$0\geq (a+b-1)(a-b+1)\geq -\frac{3}{2}-\sqrt 2$$$$ -\frac{9}{2}+2\sqrt 2\geq (a+b-2)(a-b+2)\geq -\frac{9}{2}-2\sqrt 2$$
3 replies
sqing
2 hours ago
sqing
an hour ago
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Simple triangle geometry [a fixed point]
darij grinberg   50
N an hour ago by ezpotd
Source: German TST 2004, IMO ShortList 2003, geometry problem 2
Three distinct points $A$, $B$, and $C$ are fixed on a line in this order. Let $\Gamma$ be a circle passing through $A$ and $C$ whose center does not lie on the line $AC$. Denote by $P$ the intersection of the tangents to $\Gamma$ at $A$ and $C$. Suppose $\Gamma$ meets the segment $PB$ at $Q$. Prove that the intersection of the bisector of $\angle AQC$ and the line $AC$ does not depend on the choice of $\Gamma$.
50 replies
darij grinberg
May 18, 2004
ezpotd
an hour ago
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Inspired by RMO 2006
sqing   6
N an hour ago by sqing
Source: Own
Let $ a,b >0 . $ Prove that
$$ \frac {a^{2}+1}{b+k}+\frac { b^{2}+1}{ka+1}+\frac {2}{a+kb} \geq \frac {6}{k+1} $$Where $k\geq 0.03 $
$$ \frac {a^{2}+1}{b+1}+\frac { b^{2}+1}{a+1}+\frac {2}{a+b} \geq 3 $$
6 replies
sqing
Saturday at 3:24 PM
sqing
an hour ago
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IMO 2009, Problem 2
orl   143
N an hour ago by ezpotd
Source: IMO 2009, Problem 2
Let $ ABC$ be a triangle with circumcentre $ O$. The points $ P$ and $ Q$ are interior points of the sides $ CA$ and $ AB$ respectively. Let $ K,L$ and $ M$ be the midpoints of the segments $ BP,CQ$ and $ PQ$. respectively, and let $ \Gamma$ be the circle passing through $ K,L$ and $ M$. Suppose that the line $ PQ$ is tangent to the circle $ \Gamma$. Prove that $ OP = OQ.$

Proposed by Sergei Berlov, Russia
143 replies
orl
Jul 15, 2009
ezpotd
an hour ago
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A sharp one with 3 var (2)
mihaig   0
2 hours ago
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$a+b+c+\sqrt{abc}\geq4.$$
0 replies
mihaig
2 hours ago
0 replies
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f(1)f(2)...f(n) has at most n prime factors
MarkBcc168   39
N 2 hours ago by cursed_tangent1434
Source: 2020 Cyberspace Mathematical Competition P2
Let $f(x) = 3x^2 + 1$. Prove that for any given positive integer $n$, the product
$$f(1)\cdot f(2)\cdot\dots\cdot f(n)$$has at most $n$ distinct prime divisors.

Proposed by Géza Kós
39 replies
MarkBcc168
Jul 15, 2020
cursed_tangent1434
2 hours ago
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An easy FE
oVlad   3
N Apr 21, 2025 by jasperE3
Source: Romania EGMO TST 2017 Day 1 P3
Determine all functions $f:\mathbb R\to\mathbb R$ such that \[f(xy-1)+f(x)f(y)=2xy-1,\]for any real numbers $x{}$ and $y{}.$
3 replies
oVlad
Apr 21, 2025
jasperE3
Apr 21, 2025
J
An easy FE
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Source: Romania EGMO TST 2017 Day 1 P3
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oVlad
1746 posts
oVlad
#1 h  Apr 21, 2025, 1:36 PM
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Determine all functions $f:\mathbb R\to\mathbb R$ such that \[f(xy-1)+f(x)f(y)=2xy-1,\]for any real numbers $x{}$ and $y{}.$
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pco
23515 posts
pco
#2 h  Apr 21, 2025, 3:54 PM • 1 Y
Y by ATM_
oVlad wrote:
Determine all functions $f:\mathbb R\to\mathbb R$ such that \[f(xy-1)+f(x)f(y)=2xy-1,\]for any real numbers $x{}$ and $y{}.$
Let $P(x,y)$ be the assertion $f(xy-1)+f(x)f(y)=2xy-1$
Let $c=f(1)$

If $f(0)\ne 0$, $P(x,0)$ $\implies$ $f(x)$ constant, which is never a solution. So $f(0)=0$

$P(0,0)$ $\implies$ $f(-1)=-1$
$P(1,1)$ $\implies$ $c=\pm 1$
Subtracting $P(x,1)$ from $P(-x,-1)$, we get $f(-x)=-cf(x)$

Subtracting $P(x,y)$ from $P(xy,1)$, we get new assertion $Q(x,y)$ : $f(x)f(y)=cf(xy)$
If $f(u)=0$ for some $u\ne 0$, $Q(x,u)$ implies $f(ux)=0$ $\forall x$ and so $f\equiv 0$, which is not a solution.
So $f(x)=0$ $\iff$ $x=0$

$Q(x,x)$ implies $\frac{f(x)}c$ is multiplicative and positive $\forall x>0$ and so $g(x)=\ln \frac{f(e^x)}c$ is additive

If $g(x)$ is not linear, its graph is dense in $\mathbb R^2$ and so graph of $f(x)$ is :
Either dense in $\mathbb R_{>0}\times \mathbb R_{>0}$ if $c=1$
Either dense in $\mathbb R_{>0}\times \mathbb R_{<0}$ if $c=-1$

But $P(x,x)$ $\implies$ $f(x^2-1)\le 2x^2-1$ and so contradiction with both cases
So $g(x)$ is linear and $f(x)=cx^a$ $\forall x>0$ for some real $a$
Then $P(2,1)$ implies $c+2^a=3$ and so :

If $c=1$ : $a=1$ and $f(x)=x$ $\forall x\ge 0$ and $f(-x)=-cf(x)=-f(x)$ imply $\boxed{\text{S1 : }f(x)=x\quad\forall x}$, which indeed fits

If $c=-1$ : $a=2$ and $f(x)=-x^2$ $\forall x\ge 0$ and $f(-x)=-cf(x)=f(x)$ imply $\boxed{\text{S2 : }f(x)=-x^2\quad\forall x}$, which indeed fits
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BR1F1SZ
578 posts
BR1F1SZ
#3 h  Apr 21, 2025, 6:20 PM
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It is also 2015 Argentina TST P3
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jasperE3
11384 posts
jasperE3
#4 h  Apr 21, 2025, 10:44 PM
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