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Integer-Valued FE comes again
lminsl   207
N 31 minutes ago by NerdyNashville
Source: IMO 2019 Problem 1
Let $\mathbb{Z}$ be the set of integers. Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that, for all integers $a$ and $b$, $$f(2a)+2f(b)=f(f(a+b)).$$Proposed by Liam Baker, South Africa
207 replies
lminsl
Jul 16, 2019
NerdyNashville
31 minutes ago
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Problem 6 (Second Day)
darij grinberg   42
N 33 minutes ago by OronSH
Source: IMO 2004 Athens
We call a positive integer alternating if every two consecutive digits in its decimal representation are of different parity.

Find all positive integers $n$ such that $n$ has a multiple which is alternating.
42 replies
darij grinberg
Jul 13, 2004
OronSH
33 minutes ago
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Italian WinterCamps test07 Problem4
mattilgale   88
N an hour ago by OronSH
Source: ISL 2006, G3, VAIMO 2007/5
Let $ ABCDE$ be a convex pentagon such that
\[ \angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE. \]The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$.

Proposed by Zuming Feng, USA
88 replies
mattilgale
Jan 29, 2007
OronSH
an hour ago
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Hard to approach it !
BogG   129
N an hour ago by OronSH
Source: Swiss Imo Selection 2006
Let $\triangle ABC$ be an acute-angled triangle with $AB \not= AC$. Let $H$ be the orthocenter of triangle $ABC$, and let $M$ be the midpoint of the side $BC$. Let $D$ be a point on the side $AB$ and $E$ a point on the side $AC$ such that $AE=AD$ and the points $D$, $H$, $E$ are on the same line. Prove that the line $HM$ is perpendicular to the common chord of the circumscribed circles of triangle $\triangle ABC$ and triangle $\triangle ADE$.
129 replies
BogG
May 25, 2006
OronSH
an hour ago
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Simple triangle geometry [a fixed point]
darij grinberg   48
N an hour ago by OronSH
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 $\Gamma$.
48 replies
darij grinberg
May 18, 2004
OronSH
an hour ago
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IMO ShortList 1998, geometry problem 4
orl   14
N an hour ago by OronSH
Source: IMO ShortList 1998, geometry problem 4
Let $ M$ and $ N$ be two points inside triangle $ ABC$ such that
\[ \angle MAB = \angle NAC\quad \mbox{and}\quad \angle MBA = \angle NBC. \]
Prove that
\[ \frac {AM \cdot AN}{AB \cdot AC} + \frac {BM \cdot BN}{BA \cdot BC} + \frac {CM \cdot CN}{CA \cdot CB} = 1. \]
14 replies
orl
Oct 22, 2004
OronSH
an hour ago
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Functional equation
Nima Ahmadi Pour   98
N an hour ago by ezpotd
Source: ISl 2005, A2, Iran prepration exam
We denote by $\mathbb{R}^+$ the set of all positive real numbers.

Find all functions $f: \mathbb R^ + \rightarrow\mathbb R^ +$ which have the property:
\[f(x)f(y)=2f(x+yf(x))\]
for all positive real numbers $x$ and $y$.

Proposed by Nikolai Nikolov, Bulgaria
98 replies
Nima Ahmadi Pour
Apr 24, 2006
ezpotd
an hour ago
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Geometry
noneofyou34   0
an hour ago
Please can someone help me prove that orthocenter of a triangle exists by using Menelau's Theorem!
0 replies
noneofyou34
an hour ago
0 replies
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Inequality with mathematical means
StefanSebez   12
N an hour ago by Sh309had
Source: Serbia JBMO TST 2022 P1
Prove that for all positive real numbers $a$, $b$ the following inequality holds:
\begin{align*} \sqrt{\frac{a^2+b^2}{2}}+\frac{2ab}{a+b}\ge \frac{a+b}{2}+ \sqrt{ab} \end{align*}When does equality hold?
12 replies
StefanSebez
Jun 1, 2022
Sh309had
an hour ago
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Really fun geometry problem
Sadigly   4
N 2 hours ago by Double07
Source: Azerbaijan Senior MO 2025 P6
In the acute triangle $ABC$ with $AB<AC$, the foot of altitudes from $A,B,C$ to the sides $BC,CA,AB$ are $D,E,F$, respectively. $H$ is the orthocenter. $M$ is the midpoint of segment $BC$. Lines $MH$ and $EF$ intersect at $K$. Let the tangents drawn to circumcircle $(ABC)$ from $B$ and $C$ intersect at $T$. Prove that $T;D;K$ are colinear
4 replies
Sadigly
3 hours ago
Double07
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
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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
23511 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
570 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
11308 posts
jasperE3
#4 h  Apr 21, 2025, 10:44 PM
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