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INMO 2022
Flying-Man   40
N 26 minutes ago by endless_abyss
Let $D$ be an interior point on the side $BC$ of an acute-angled triangle $ABC$. Let the circumcircle of triangle $ADB$ intersect $AC$ again at $E(\ne A)$ and the circumcircle of triangle $ADC$ intersect $AB$ again at $F(\ne A)$. Let $AD$, $BE$, and $CF$ intersect the circumcircle of triangle $ABC$ again at $D_1(\ne A)$, $E_1(\ne B)$ and $F_1(\ne C)$, respectively. Let $I$ and $I_1$ be the incentres of triangles $DEF$ and $D_1E_1F_1$, respectively. Prove that $E,F, I, I_1$ are concyclic.
40 replies
Flying-Man
Mar 6, 2022
endless_abyss
26 minutes ago
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circumcenter of BJK lies on line AC, median, right angle, circumcircle related
parmenides51   23
N 38 minutes ago by endless_abyss
Source: 2019 RMM Shortlist G1
Let $BM$ be a median in an acute-angled triangle $ABC$. A point $K$ is chosen on the line through $C$ tangent to the circumcircle of $\vartriangle BMC$ so that $\angle KBC = 90^\circ$. The segments $AK$ and $BM$ meet at $J$. Prove that the circumcenter of $\triangle BJK$ lies on the line $AC$.

Aleksandr Kuznetsov, Russia
23 replies
parmenides51
Jun 18, 2020
endless_abyss
38 minutes ago
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Linear recurrence fits with factorial finitely often
Assassino9931   2
N 43 minutes ago by Assassino9931
Source: Bulgaria Balkan MO TST 2025
Let $k\geq 3$ be an integer. The sequence $(a_n)_{n\geq 1}$ is defined via $a_1 = 1$, $a_2 = k$ and
\[ a_{n+2} = ka_{n+1} + a_n \]for any positive integer $n$. Prove that there are finitely many pairs $(m, \ell)$ of positive integers such that $a_m = \ell!$.
2 replies
Assassino9931
Yesterday at 10:25 PM
Assassino9931
43 minutes ago
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APMO 2017: (ADZ) passes through M
BartSimpsons   76
N an hour ago by Mathgloggers
Source: APMO 2017, problem 2
Let $ABC$ be a triangle with $AB < AC$. Let $D$ be the intersection point of the internal bisector of angle $BAC$ and the circumcircle of $ABC$. Let $Z$ be the intersection point of the perpendicular bisector of $AC$ with the external bisector of angle $\angle{BAC}$. Prove that the midpoint of the segment $AB$ lies on the circumcircle of triangle $ADZ$.

Olimpiada de Matemáticas, Nicaragua
76 replies
BartSimpsons
May 14, 2017
Mathgloggers
an hour ago
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Interesting inequalities
sqing   4
N an hour ago by sqing
Source: Own
Let $ a,b $ be real numbers . Prove that
$$-\frac{11+8\sqrt 2}{7}\leq \frac{ab+a+b-1}{(a^2+a+1)(b^2+b+1)}\leq \frac{2}{9} $$$$-\frac{37+13\sqrt{13}}{414}\leq \frac{ab+a+b-2}{(a^2+a+4)(b^2+b+4)}\leq \frac{3}{50} $$$$-\frac{5\sqrt 5+9}{22}\leq \frac{ab+a+b-2}{(a^2+a+2)(b^2+b+2)}\leq \frac{5\sqrt 5-9}{22}$$
4 replies
sqing
Yesterday at 3:06 AM
sqing
an hour ago
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Inspired by Ruji2018252
sqing   2
N an hour ago by sqing
Source: Own
Let $ a,b,c $ be reals such that $ a^2+b^2+c^2-2a-4b-4c=7. $ Prove that
$$ -4\leq 2a+b+2c\leq 20$$$$5-4\sqrt 3\leq a+b+c\leq 5+4\sqrt 3$$$$ 11-4\sqrt {14}\leq a+2b+3c\leq 11+4\sqrt {14}$$
2 replies
sqing
Today at 2:00 AM
sqing
an hour ago
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Some Identity that I need help
ItzsleepyXD   2
N an hour ago by Tkn
Given $\triangle ABC$ with orthocenter , circumcenter and incenter $H,O,I$ , circum-radius $R$ , in-radius $r$.
Prove that $OH^2 = 2 HI^2 - 4r^2 + R^2$ .
2 replies
ItzsleepyXD
Dec 28, 2024
Tkn
an hour ago
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lcm and gcd FE
jasperE3   9
N 2 hours ago by AshAuktober
Source: IMOC 2018 N2
Find all functions $f:\mathbb N\to\mathbb N$ satisfying
$$\operatorname{lcm}(f(x),y)\gcd(f(x),f(y))=f(x)f(f(y))$$for all $x,y\in\mathbb N$.
9 replies
jasperE3
Aug 17, 2021
AshAuktober
2 hours ago
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A three-variable functional inequality on non-negative reals
Tintarn   9
N 2 hours ago by ErTeeEs06
Source: Dutch TST 2024, 1.2
Find all functions $f:\mathbb{R}_{\ge 0} \to \mathbb{R}$ with
\[2x^3zf(z)+yf(y) \ge 3yz^2f(x)\]for all $x,y,z \in \mathbb{R}_{\ge 0}$.
9 replies
Tintarn
Jun 28, 2024
ErTeeEs06
2 hours ago
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*pop* Noice FE
EmilXM   3
N 2 hours ago by jasperE3
Source: Own
Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$, such that,
$$f((x+y)f(xy)-f(x))= x^2y+xy^2-f(x)$$for all $x,y \in \mathbb{R}$.
3 replies
EmilXM
Jul 4, 2021
jasperE3
2 hours ago
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Troll number theory
kokcio   0
2 hours ago
Let $p,q,r>2$ are primes, which satisfy:
$q^4+(q+4)(r^2-q^2)=p$
$q|p+5$
$\sqrt{\frac{q^4-1}{r^2-1}}$ is an integer
Find all possible values of $p$.
0 replies
kokcio
2 hours ago
0 replies
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A metrical relation in a trapezoid.
Virgil Nicula   4
N Jan 22, 2021 by tiny123tt
PP4. A trapezoid $ABCD\ ,\ AD\parallel BC$ with $I\in AC\cap BD\ ;\ P\in (AB)\ ,\ R\in (CD)$ so that $I\in PR\ ;\ S\in AR\cap BD$ . Prove that $CS\parallel AB\iff \frac {PB}{PA}+\frac {BC}{AD}=1$ .
4 replies
Virgil Nicula
Jun 15, 2015
tiny123tt
Jan 22, 2021
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A metrical relation in a trapezoid.
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Reveal topic tags
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Virgil Nicula
7054 posts
Virgil Nicula
#1 h  Jun 15, 2015, 4:59 PM • 2 Y
Y by Adventure10, Mango247
PP4. A trapezoid $ABCD\ ,\ AD\parallel BC$ with $I\in AC\cap BD\ ;\ P\in (AB)\ ,\ R\in (CD)$ so that $I\in PR\ ;\ S\in AR\cap BD$ . Prove that $CS\parallel AB\iff \frac {PB}{PA}+\frac {BC}{AD}=1$ .
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Luis González
4146 posts
Luis González
#2 h  Jun 15, 2015, 6:02 PM • 2 Y
Y by Adventure10, Mango247
Assume that $CS \parallel AB.$ $CS$ cuts $AD$ at $E$ and $BE$ cuts $AC,PR$ at $M,G,$ resp. $AECB$ is parallelogram with diagonal intersection $M$ $\Longrightarrow$ $M$ is midpoint of $\overline{AC}$ and from the complete $CRSI,$ it follows that $(M,B,G,E)=I(C,S,R,E)=-1$ $\Longrightarrow$ $\overline{GM}:\overline{GB}=-\overline{EM}:\overline{EB}=-1:2$ $\Longrightarrow$ $G$ is centroid of $\triangle ABC$ $\Longrightarrow$ $\tfrac{PB}{PA}+\tfrac{IC}{IA}=1$ (Cristea's theorem) $\Longrightarrow$ $\tfrac{PB}{PA}+\tfrac{BC}{AD}=1.$

The converse is taken for granted by the uniqueness of $S \in BD$ and $P \in AB.$
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drmzjoseph
445 posts
drmzjoseph
#3 h  Jun 15, 2015, 6:22 PM • 2 Y
Y by Adventure10, Mango247
$X \equiv AB \cap CD, CS \parallel AB \Longleftrightarrow (\infty,B,A,X)=(S,B,I,D)=R(A,B,P,X)$
$\Longleftrightarrow \frac{BX}{AB}=\frac{AP}{BP} \times \frac{BX}{AX} \Longleftrightarrow \frac{BP}{AP}=\frac{AB}{AX}=1-\frac{BC}{AD}$
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Virgil Nicula
7054 posts
Virgil Nicula
#4 h  Jun 16, 2015, 4:01 PM • 2 Y
Y by Adventure10, Mango247
See PP4 (with my proof) from here.
This post has been edited 1 time. Last edited by Virgil Nicula, Jun 16, 2015, 4:03 PM
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tiny123tt
5 posts
tiny123tt
#5 h  Jan 22, 2021, 2:28 PM
Y by
Luis González wrote:
Assume that $CS \parallel AB.$ $CS$ cuts $AD$ at $E$ and $BE$ cuts $AC,PR$ at $M,G,$ resp. $AECB$ is parallelogram with diagonal intersection $M$ $\Longrightarrow$ $M$ is midpoint of $\overline{AC}$ and from the complete $CRSI,$ it follows that $(M,B,G,E)=I(C,S,R,E)=-1$ $\Longrightarrow$ $\overline{GM}:\overline{GB}=-\overline{EM}:\overline{EB}=-1:2$ $\Longrightarrow$ $G$ is centroid of $\triangle ABC$ $\Longrightarrow$ $\tfrac{PB}{PA}+\tfrac{IC}{IA}=1$ (Cristea's theorem) $\Longrightarrow$ $\tfrac{PB}{PA}+\tfrac{BC}{AD}=1.$

The converse is taken for granted by the uniqueness of $S \in BD$ and $P \in AB.$

Cristea's theorem?
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