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Nordic 2025 P3
anirbanbz   8
N 19 minutes ago by Primeniyazidayi
Source: Nordic 2025
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. Let $E$ and $F$ be points on the line segments $AC$ and $AB$ respectively such that $AEHF$ is a parallelogram. Prove that $\vert OE \vert = \vert OF \vert$.
8 replies
anirbanbz
Mar 25, 2025
Primeniyazidayi
19 minutes ago
A number theory about divisors which no one fully solved at the contest
nAalniaOMliO   21
N an hour ago by nAalniaOMliO
Source: Belarusian national olympiad 2024
Let's call a pair of positive integers $(k,n)$ interesting if $n$ is composite and for every divisor $d<n$ of $n$ at least one of $d-k$ and $d+k$ is also a divisor of $n$
Find the number of interesting pairs $(k,n)$ with $k \leq 100$
M. Karpuk
21 replies
nAalniaOMliO
Jul 24, 2024
nAalniaOMliO
an hour ago
2025 Caucasus MO Seniors P2
BR1F1SZ   3
N an hour ago by X.Luser
Source: Caucasus MO
Let $ABC$ be a triangle, and let $B_1$ and $B_2$ be points on segment $AC$ symmetric with respect to the midpoint of $AC$. Let $\gamma_A$ denote the circle passing through $B_1$ and tangent to line $AB$ at $A$. Similarly, let $\gamma_C$ denote the circle passing through $B_1$ and tangent to line $BC$ at $C$. Let the circles $\gamma_A$ and $\gamma_C$ intersect again at point $B'$ ($B' \neq B_1$). Prove that $\angle ABB' = \angle CBB_2$.
3 replies
BR1F1SZ
Mar 26, 2025
X.Luser
an hour ago
IMO Shortlist 2010 - Problem G1
Amir Hossein   130
N 2 hours ago by LeYohan
Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$

Proposed by Christopher Bradley, United Kingdom
130 replies
Amir Hossein
Jul 17, 2011
LeYohan
2 hours ago
CGMO6: Airline companies and cities
v_Enhance   13
N 2 hours ago by Marcus_Zhang
Source: 2012 China Girl's Mathematical Olympiad
There are $n$ cities, $2$ airline companies in a country. Between any two cities, there is exactly one $2$-way flight connecting them which is operated by one of the two companies. A female mathematician plans a travel route, so that it starts and ends at the same city, passes through at least two other cities, and each city in the route is visited once. She finds out that wherever she starts and whatever route she chooses, she must take flights of both companies. Find the maximum value of $n$.
13 replies
v_Enhance
Aug 13, 2012
Marcus_Zhang
2 hours ago
nice problem
hanzo.ei   0
2 hours ago
Source: I forgot
Let triangle $ABC$ be inscribed in the circumcircle $(O)$ and circumscribed about the incircle $(I)$, with $AB < AC$. The incircle $(I)$ touches the sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. A line through $I$, perpendicular to $AI$, intersects $BC$, $CA$, and $AB$ at $X$, $Y$, and $Z$, respectively. The line $AI$ meets $(O)$ at $M$ (distinct from $A$). The circumcircle of triangle $AYZ$ intersects $(O)$ at $N$ (distinct from $A$). Let $P$ be the midpoint of the arc $BAC$ of $(O)$. The line $AI$ cuts segments $DF$ and $DE$ at $K$ and $L$, respectively, and the tangents to the circle $(DKL)$ at $K$ and $L$ intersect at $T$. Prove that $AT \perp BC$.
0 replies
hanzo.ei
2 hours ago
0 replies
Find a given number of divisors of ab
proglote   9
N 2 hours ago by zuat.e
Source: Brazil MO 2013, problem #2
Arnaldo and Bernaldo play the following game: given a fixed finite set of positive integers $A$ known by both players, Arnaldo picks a number $a \in A$ but doesn't tell it to anyone. Bernaldo thens pick an arbitrary positive integer $b$ (not necessarily in $A$). Then Arnaldo tells the number of divisors of $ab$. Show that Bernaldo can choose $b$ in a way that he can find out the number $a$ chosen by Arnaldo.
9 replies
proglote
Oct 24, 2013
zuat.e
2 hours ago
2025 TST 22
EthanWYX2009   1
N 3 hours ago by hukilau17
Source: 2025 TST 22
Let \( A \) be a set of 2025 positive real numbers. For a subset \( T \subseteq A \), define \( M_T \) as the median of \( T \) when all elements of \( T \) are arranged in increasing order, with the convention that \( M_\emptyset = 0 \). Define
\[
P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T.
\]Find the smallest real number \( C \) such that for any set \( A \) of 2025 positive real numbers, the following inequality holds:
\[
P(A) - Q(A) \leq C \cdot \max(A),
\]where \(\max(A)\) denotes the largest element in \( A \).
1 reply
EthanWYX2009
5 hours ago
hukilau17
3 hours ago
Deriving Van der Waerden Theorem
Didier2   0
3 hours ago
Source: Khamovniki 2023-2024 (group 10-1)
Suppose we have already proved that for any coloring of $\Large \mathbb{N}$ in $r$ colors, there exists an arithmetic progression of size $k$. How can we derive Van der Waerden's theorem for $W(r, k)$ from this?
0 replies
Didier2
3 hours ago
0 replies
Not so classic orthocenter problem
m4thbl3nd3r   6
N 3 hours ago by maths_enthusiast_0001
Source: own?
Let $O$ be circumcenter of a non-isosceles triangle $ABC$ and $H$ be a point in the interior of $\triangle ABC$. Let $E,F$ be foots of perpendicular lines from $H$ to $AC,AB$. Suppose that $BCEF$ is cyclic and $M$ is the circumcenter of $BCEF$, $HM\cap AB=K,AO\cap BE=T$. Prove that $KT$ bisects $EF$
6 replies
m4thbl3nd3r
Yesterday at 4:59 PM
maths_enthusiast_0001
3 hours ago
Functional equations
hanzo.ei   1
N 3 hours ago by GreekIdiot
Source: Greekldiot
Find all $f: \mathbb R_+ \rightarrow \mathbb R_+$ such that $f(xf(y)+f(x))=yf(x+yf(x)) \: \forall \: x,y \in \mathbb R_+$
1 reply
hanzo.ei
4 hours ago
GreekIdiot
3 hours ago
Symmetric orthotransversal
buratinogigle   5
N Jun 3, 2015 by jayme
Source: Own
Prove that orthotransversal of incenter wrt reference triangle and wrt incentral triangle are symmetric through such this incenter.
5 replies
buratinogigle
May 18, 2015
jayme
Jun 3, 2015
Symmetric orthotransversal
G H J
G H BBookmark kLocked kLocked NReply
Source: Own
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buratinogigle
2317 posts
#1 • 2 Y
Y by Adventure10, Mango247
Prove that orthotransversal of incenter wrt reference triangle and wrt incentral triangle are symmetric through such this incenter.
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Luis González
4145 posts
#2 • 3 Y
Y by buratinogigle, Adventure10, Mango247
$I$ is incenter of $\triangle ABC$ and $AI,BI,CI$ cut $BC,CA,AB$ at $D,E,F.$ Perpendicular to $AI$ at $I$ cuts $BC,EF$ at $X,X^*,$ respectively. Points $\{Y,Y^*\}$ and $\{Z,Z^*\}$ are defined cyclically. $\overline{XYZ}$ and $\overline{X^*Y^*Z^*}$ are the orthotransversals of $I$ WRT $\triangle ABC$ and $\triangle DEF.$

If $XX^*$ cuts $AC,AB$ at $U,V,$ then by Desargues involution theorem for $BFEC,$ it follows that $X \mapsto X^*,$ $U \mapsto V$ is an involution where $I$ is double and $U,V$ are symmetric WRT $I$ $\Longrightarrow$ it coincides with the central symmetry WRT $I$ $\Longrightarrow$ $X,X^*$ are symmetric WRT $I$ and similarly $Y,Y^*$ and $Z,Z^*$ are symmetric WRT $I$ $\Longrightarrow$ $\overline{XYZ}$ and $\overline{X^*Y^*Z^*}$ are symmetryc WRT $I.$
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TelvCohl
2312 posts
#3 • 3 Y
Y by buratinogigle, enhanced, Adventure10
My solution :

Let $ I $ be the incenter of $ \triangle ABC $ and $ \triangle DEF $ be its cevian triangle WRT $ \triangle ABC $ .
Let the orthotransversal of $ I $ WRT $ \triangle ABC $ cuts $ BC, CA, AB $ at $ X, Y, Z $, respectively .
Let the orthotransversal of $ I $ WRT $ \triangle DEF $ cuts $ EF, FD, DE $ at $ X^*, Y^*, Z^* $, respectively .
Let $ T $ be the intersection of $ BC $ with A-external bisctor of $ \triangle ABC $ ( it's well-known $ T \in EF $ ) .

From $ T(A,I;X^*,X)=-1 $ and $ XX^* \parallel TA $ ( both $ \perp AI $ ) $ \Longrightarrow XI=X^*I $ .
Similarly, we can prove $ YI=Y^*I $ and $ ZI=Z^*I \Longrightarrow \overline{XYZ} $ and $ \overline{X^*Y^*Z^*} $ are symmetry WRT $ I $ .

Q.E.D
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buratinogigle
2317 posts
#4 • 2 Y
Y by Adventure10, Mango247
Thank you so much, actually I have found solution the same as Telv base on idea of Luis but Telv posted. I only note that. Then line passes through I cuts $BC, EF$ at $G,H$ then $AG,AH$ are isogonal conjugate wrt $\angle BAC$.
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tranquanghuy7198
253 posts
#5 • 3 Y
Y by buratinogigle, Adventure10, Mango247
Mr.Buratino wrote:
Prove that orthotransversal of incenter wrt reference triangle and wrt incentral triangle are symmetric through such this incenter (post #1).
My solution:

Lemma 1. Let $X, Y$ be 2 points on the side $BC$ of $\triangle{ABC}$. We have: $\odot{(AXY)}$ is tangent to $\odot{(ABC)} \Longleftrightarrow AX, AY$ are isogonal in $\angle{BAC}$
Proof. Simple angle chasing.

Lemma 2. Given $\triangle{ABC}, E = R_{AC}(B), F = R_{AB}(C)$. The line passing through $A$ which is tangent to $\odot{(AEF)}$ intersects $BC, EF$ at $X, Y$. Prove that: $AX = AY$

Back to our main problem:
Consider $\triangle{ABC}$ and the bisectors $AD, BE, CF$ concur at $I$
The orthotransversal of $I$ WRT $\triangle{ABC}$ intersects $BC, CA, AB$ at $X, Y, Z$
The orthotransversal of $I$ WRT $\triangle{DEF}$ intersects $EF, FD, DE$ at $M, N, P$
In order to prove $\overline{X, Y, Z} = S_I(\overline{M, N, P})$, we only need to indicate that $IX = IM$
$K = R_{IF}(E), L = R_{IE}(F)$ $\Rightarrow K, L\in{BC}$
Notice that $\angle{CIK} = \angle{CIE} = \angle{BIF} = \angle{BIL}$
$\Rightarrow (IKL), (IBC)$ are tangent to each other at $I$ (lemma 1)
$\Rightarrow (IKL)$ is tangent to $\overline{X, I, M}$
Now apply the lemma 2 for $\triangle{IEF}$ we receive: $IX = IM$ and our proof is completed!
Q.E.D
This post has been edited 1 time. Last edited by tranquanghuy7198, Jun 2, 2015, 4:50 PM
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jayme
9772 posts
#6 • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,

a proof based on anharmonic pencils can be very nice...

Sincerely
Jean-Louis
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