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Fond all functions in M with a) f(1)=5/2, b) f(1)=√3
Amir Hossein   5
N 2 hours ago by jasperE3
Source: IMO LongList 1982 - P34
Let $M$ be the set of all functions $f$ with the following properties:

(i) $f$ is defined for all real numbers and takes only real values.

(ii) For all $x, y \in \mathbb R$ the following equality holds: $f(x)f(y) = f(x + y) + f(x - y).$

(iii) $f(0) \neq 0.$

Determine all functions $f \in M$ such that

(a) $f(1)=\frac 52$,

(b) $f(1)= \sqrt 3$.
5 replies
Amir Hossein
Mar 18, 2011
jasperE3
2 hours ago
help me please
thuanz123   6
N 3 hours ago by pavel kozlov
find all $a,b \in \mathbb{Z}$ such that:
a) $3a^2-2b^2=1$
b) $a^2-6b^2=1$
6 replies
thuanz123
Jan 17, 2016
pavel kozlov
3 hours ago
Problem 5 (Second Day)
darij grinberg   78
N 3 hours ago by cj13609517288
Source: IMO 2004 Athens
In a convex quadrilateral $ABCD$, the diagonal $BD$ bisects neither the angle $ABC$ nor the angle $CDA$. The point $P$ lies inside $ABCD$ and satisfies \[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\] Prove that $ABCD$ is a cyclic quadrilateral if and only if $AP=CP$.
78 replies
darij grinberg
Jul 13, 2004
cj13609517288
3 hours ago
concyclic wanted, PQ = BP, cyclic quadrilateral and 2 parallelograms related
parmenides51   2
N 3 hours ago by SuperBarsh
Source: 2011 Italy TST 2.2
Let $ABCD$ be a cyclic quadrilateral in which the lines $BC$ and $AD$ meet at a point $P$. Let $Q$ be the point of the line $BP$, different from $B$, such that $PQ = BP$. We construct the parallelograms $CAQR$ and $DBCS$. Prove that the points $C, Q, R, S$ lie on the same circle.
2 replies
parmenides51
Sep 25, 2020
SuperBarsh
3 hours ago
Integer FE Again
popcorn1   43
N 3 hours ago by DeathIsAwe
Source: ISL 2020 N5
Determine all functions $f$ defined on the set of all positive integers and taking non-negative integer values, satisfying the three conditions:
[list]
[*] $(i)$ $f(n) \neq 0$ for at least one $n$;
[*] $(ii)$ $f(x y)=f(x)+f(y)$ for every positive integers $x$ and $y$;
[*] $(iii)$ there are infinitely many positive integers $n$ such that $f(k)=f(n-k)$ for all $k<n$.
[/list]
43 replies
popcorn1
Jul 20, 2021
DeathIsAwe
3 hours ago
Long and wacky inequality
Royal_mhyasd   2
N 4 hours ago by Royal_mhyasd
Source: Me
Let $x, y, z$ be positive real numbers such that $x^2 + y^2 + z^2 = 12$. Find the minimum value of the following sum :
$$\sum_{cyc}\frac{(x^3+2y)^3}{3x^2yz - 16z - 8yz + 6x^2z}$$knowing that the denominators are positive real numbers.
2 replies
Royal_mhyasd
May 12, 2025
Royal_mhyasd
4 hours ago
Perpendicular passes from the intersection of diagonals, \angle AEB = \angle CED
NO_SQUARES   1
N 4 hours ago by mathuz
Source: 239 MO 2025 10-11 p3
Inside of convex quadrilateral $ABCD$ point $E$ was chosen such that $\angle DAE = \angle CAB$ and $\angle ADE = \angle CDB$. Prove that if perpendicular from $E$ to $AD$ passes from the intersection of diagonals of $ABCD$, then $\angle AEB = \angle CED$.
1 reply
NO_SQUARES
May 5, 2025
mathuz
4 hours ago
A game with balls and boxes
egxa   6
N 4 hours ago by Sh309had
Source: Turkey JBMO TST 2023 Day 1 P4
Initially, Aslı distributes $1000$ balls to $30$ boxes as she wishes. After that, Aslı and Zehra make alternated moves which consists of taking a ball in any wanted box starting with Aslı. One who takes the last ball from any box takes that box to herself. What is the maximum number of boxes can Aslı guarantee to take herself regardless of Zehra's moves?
6 replies
egxa
Apr 30, 2023
Sh309had
4 hours ago
Angle Relationships in Triangles
steven_zhang123   2
N 4 hours ago by Captainscrubz
In $\triangle ABC$, $AB > AC$. The internal angle bisector of $\angle BAC$ and the external angle bisector of $\angle BAC$ intersect the ray $BC$ at points $D$ and $E$, respectively. Given that $CE - CD = 2AC$, prove that $\angle ACB = 2\angle ABC$.
2 replies
steven_zhang123
Yesterday at 11:09 PM
Captainscrubz
4 hours ago
Easy functional equation
fattypiggy123   14
N 4 hours ago by Fly_into_the_sky
Source: Singapore Mathematical Olympiad 2014 Problem 2
Find all functions from the reals to the reals satisfying
\[f(xf(y) + x) = xy + f(x)\]
14 replies
fattypiggy123
Jul 5, 2014
Fly_into_the_sky
4 hours ago
sharky devil or no sharky devil
rafaello   1
N Jan 15, 2023 by Zena_B
Source: MODSMO 2019 September Advanced Contest P3
$ABC$ is a triangle with incentre $I$. The feet of the altitudes from $I$ to $BC, AC, AB$ are $D, E, F$ respectively, and the line through $D$ parallel to $AI$ intersects \(AB\) and \(AC\) at \(X\) and \(Y\) respectively. Prove that the circles with diameters $XF$ and $YE$ have a common point on the circumcircle of $ABC.$
1 reply
rafaello
Oct 27, 2021
Zena_B
Jan 15, 2023
sharky devil or no sharky devil
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G H BBookmark kLocked kLocked NReply
Source: MODSMO 2019 September Advanced Contest P3
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rafaello
1079 posts
#1
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$ABC$ is a triangle with incentre $I$. The feet of the altitudes from $I$ to $BC, AC, AB$ are $D, E, F$ respectively, and the line through $D$ parallel to $AI$ intersects \(AB\) and \(AC\) at \(X\) and \(Y\) respectively. Prove that the circles with diameters $XF$ and $YE$ have a common point on the circumcircle of $ABC.$
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Zena_B
3 posts
#2
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Notice that $EF\perp AI, EF\perp DX $ Define $P=DX\cap EF $ and $ S=IP\cap (AIF) $ See that $P$ and $S$ swap under incircle inversion. Also, $(ABC)$ swaps with the nine-point circle of $DEF$. Hence, $S\in (ABC)$ (sharky-devil)
Now let $F'$ be the $F$-Antipode in $(AIF)$. Applying Pascal's Theorem on $SF'EFAI$ we get $X\in SF'$, since $EF'\perp EF, EF'\parallel AI\parallel XP$. Thus, we have $\angle XSF=90, S\in (XF) $. Similarly, $S\in (YE) $ so we are done.
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