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D1031 : A general result on polynomial 1
Dattier   1
N 25 minutes ago by Dattier
Source: les dattes à Dattier
Let $P(x,y) \in \mathbb Q(x,y)$ with $\forall (a,b) \in \mathbb Z^2, P(a,b) \in \mathbb Z  $.

Is it true that $P(x,y) \in \mathbb Q[x,y]$?
1 reply
Dattier
4 hours ago
Dattier
25 minutes ago
Asymmetric FE
sman96   18
N 26 minutes ago by jasperE3
Source: BdMO 2025 Higher Secondary P8
Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that$$f(xf(y)-y) + f(xy-x) + f(x+y) = 2xy$$for all $x, y \in \mathbb{R}$.
18 replies
sman96
Feb 8, 2025
jasperE3
26 minutes ago
Easy Geometry
pokmui9909   6
N 38 minutes ago by reni_wee
Source: FKMO 2025 P4
Triangle $ABC$ satisfies $\overline{CA} > \overline{AB}$. Let the incenter of triangle $ABC$ be $\omega$, which touches $BC, CA, AB$ at $D, E, F$, respectively. Let $M$ be the midpoint of $BC$. Let the circle centered at $M$ passing through $D$ intersect $DE, DF$ at $P(\neq D), Q(\neq D)$, respecively. Let line $AP$ meet $BC$ at $N$, line $BP$ meet $CA$ at $L$. Prove that the three lines $EQ, FP, NL$ are concurrent.
6 replies
pokmui9909
Mar 30, 2025
reni_wee
38 minutes ago
Old hard problem
ItzsleepyXD   3
N an hour ago by Funcshun840
Source: IDK
Let $ABC$ be a triangle and let $O$ be its circumcenter and $I$ its incenter.
Let $P$ be the radical center of its three mixtilinears and let $Q$ be the isogonal conjugate of $P$.
Let $G$ be the Gergonne point of the triangle $ABC$.
Prove that line $QG$ is parallel with line $OI$ .
3 replies
ItzsleepyXD
Apr 25, 2025
Funcshun840
an hour ago
Beautiful Angle Sum Property in Hexagon with Incenter
Raufrahim68   0
2 hours ago
Hello everyone! I discovered an interesting geometric property and would like to share it with the community. I'm curious if this is a known result and whether it can be generalized.

Problem Statement:
Let
A
B
C
D
E
K
ABCDEK be a convex hexagon with an incircle centered at
O
O. Prove that:


A
O
B
+

C
O
D
+

E
O
K
=
180

∠AOB+∠COD+∠EOK=180
0 replies
Raufrahim68
2 hours ago
0 replies
IMO 2009 P2, but in space
Miquel-point   1
N 2 hours ago by Miquel-point
Source: KoMaL A. 485
Let $ABCD$ be a tetrahedron with circumcenter $O$. Suppose that the points $P, Q$ and $R$ are interior points of the edges $AB, AC$ and $AD$, respectively. Let $K, L, M$ and $N$ be the centroids of the triangles $PQD$, $PRC,$ $QRB$ and $PQR$, respectively. Prove that if the plane $PQR$ is tangent to the sphere $KLMN$ then $OP=OQ=OR.$

1 reply
Miquel-point
2 hours ago
Miquel-point
2 hours ago
Imtersecting two regular pentagons
Miquel-point   0
2 hours ago
Source: KoMaL B. 5093
The intersection of two congruent regular pentagons is a decagon with sides of $a_1,a_2,\ldots ,a_{10}$ in this order. Prove that
\[a_1a_3+a_3a_5+a_5a_7+a_7a_9+a_9a_1=a_2a_4+a_4a_6+a_6a_8+a_8a_{10}+a_{10}a_2.\]
0 replies
Miquel-point
2 hours ago
0 replies
Dissecting regular heptagon in similar isosceles trapezoids
Miquel-point   0
2 hours ago
Source: KoMaL B. 5085
Show that a regular heptagon can be dissected into a finite number of symmetrical trapezoids, all similar to each other.

Proposed by M. Laczkovich, Budapest
0 replies
Miquel-point
2 hours ago
0 replies
Amazing projective stereometry
Miquel-point   0
3 hours ago
Source: KoMaL B 5060
In the plane $\Sigma$, given a circle $k$ and a point $P$ in its interior, not coinciding with the center of $k$. Call a point $O$ of space, not lying on $\Sigma$, a proper projection center if there exists a plane $\Sigma'$, not passing through $O$, such that, by projecting the points of $\Sigma$ from $O$ to $\Sigma'$, the projection of $k$ is also a circle, and its center is the projection of $P$. Show that the proper projection centers lie on a circle.
0 replies
Miquel-point
3 hours ago
0 replies
Proving radical axis through orthocenter
azzam2912   2
N 3 hours ago by Miquel-point
In acute triangle $ABC$ let $D, E$ and $F$ denote the feet of the altitudes from $A, B$ and $C$, respectively. Let line $DE$ intersect circumcircle $ABC$ at points $G, H$. Similarly, let line $DF$ intersect circumcircle $ABC$ at points $I, J$. Prove that the radical axis of circles $EIJ$ and $FGH$ passes through the orthocenter of triangle $ABC$
2 replies
azzam2912
Today at 12:02 PM
Miquel-point
3 hours ago
Hard Inequality
Asilbek777   2
N 3 hours ago by Ritwin
Waits for Solution
2 replies
Asilbek777
6 hours ago
Ritwin
3 hours ago
A geometry problem from the TOT
Invert_DOG_about_centre_O   11
N 3 hours ago by seriousPossibilist
Source: Tournament of towns Spring 2018 A-level P4
Let O be the center of the circumscribed circle of the triangle ABC. Let AH be the altitude in this triangle, and let P be the base of the perpendicular drawn from point A to the line CO. Prove that the line HP passes through the midpoint of the side AB. (6 points)

Egor Bakaev
11 replies
Invert_DOG_about_centre_O
Mar 10, 2020
seriousPossibilist
3 hours ago
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
An easy FE
G H J
G H BBookmark kLocked kLocked NReply
Source: Romania EGMO TST 2017 Day 1 P3
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oVlad
1746 posts
#1
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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
23514 posts
#2 • 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
#3
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It is also 2015 Argentina TST P3
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jasperE3
11337 posts
#4
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https://artofproblemsolving.com/community/c6h2426810p20007653
https://artofproblemsolving.com/community/c6h2990785p26842638
https://artofproblemsolving.com/community/c6h3469044p33552067
https://artofproblemsolving.com/community/c6h2115304p15348032
https://artofproblemsolving.com/community/c6h1309256p7009219
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