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Very odd geo
Royal_mhyasd   1
N 2 hours ago by Royal_mhyasd
Source: own (i think)
Let $\triangle ABC$ be an acute triangle with $AC>AB>BC$ and let $H$ be its orthocenter. Let $P$ be a point on the perpendicular bisector of $AH$ such that $\angle APH=2(\angle ABC - \angle ACB)$ and $P$ and $C$ are on different sides of $AB$, $Q$ a point on the perpendicular bisector of $BH$ such that $\angle BQH = 2(\angle ACB-\angle BAC)$ and $R$ a point on the perpendicular bisector of $CH$ such that $\angle CRH=2(\angle ABC - \angle BAC)$ and $Q,R$ lie on the opposite side of $BC$ w.r.t $A$. Prove that $P,Q$ and $R$ are collinear.
1 reply
Royal_mhyasd
2 hours ago
Royal_mhyasd
2 hours ago
Calculating sum of the numbers
Sadigly   5
N 2 hours ago by aokmh3n2i2rt
Source: Azerbaijan Junior MO 2025 P4
A $3\times3$ square is filled with numbers $1;2;3...;9$.The numbers inside four $2\times2$ squares is summed,and arranged in an increasing order. Is it possible to obtain the following sequences as a result of this operation?

$\text{a)}$ $24,24,25,25$

$\text{b)}$ $20,23,26,29$
5 replies
Sadigly
May 9, 2025
aokmh3n2i2rt
2 hours ago
Swap to the symmedian
Noob_at_math_69_level   7
N 2 hours ago by awesomeming327.
Source: DGO 2023 Team P1
Let $\triangle{ABC}$ be a triangle with points $U,V$ lie on the perpendicular bisector of $BC$ such that $B,U,V,C$ lie on a circle. Suppose $UD,UE,UF$ are perpendicular to sides $BC,AC,AB$ at points $D,E,F.$ The tangent lines from points $E,F$ to the circumcircle of $\triangle{DEF}$ intersects at point $S.$ Prove that: $AV,DS$ are parallel.

Proposed by Paramizo Dicrominique
7 replies
Noob_at_math_69_level
Dec 18, 2023
awesomeming327.
2 hours ago
Find (AB * CD) / (AC * BD) & prove orthogonality of circles
Maverick   15
N 2 hours ago by Ilikeminecraft
Source: IMO 1993, Day 1, Problem 2
Let $A$, $B$, $C$, $D$ be four points in the plane, with $C$ and $D$ on the same side of the line $AB$, such that $AC \cdot BD = AD \cdot BC$ and $\angle ADB = 90^{\circ}+\angle ACB$. Find the ratio
\[\frac{AB \cdot CD}{AC \cdot BD}, \]
and prove that the circumcircles of the triangles $ACD$ and $BCD$ are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicuar. Thus, proving that the circumcircles of the triangles $ACD$ and $BCD$ are orthogonal is equivalent to proving that the tangents to the circumcircles of the triangles $ACD$ and $BCD$ at the point $C$ are perpendicular.)
15 replies
Maverick
Jul 13, 2004
Ilikeminecraft
2 hours ago
f(x+f(x)+f(y))=x+f(x+y)
dangerousliri   10
N 3 hours ago by jasperE3
Source: FEOO, Shortlist A5
Find all functions $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ such that for any positive real numbers $x$ and $y$,
$$f(x+f(x)+f(y))=x+f(x+y)$$Proposed by Athanasios Kontogeorgis, Grecce, and Dorlir Ahmeti, Kosovo
10 replies
dangerousliri
May 31, 2020
jasperE3
3 hours ago
n-variable inequality
ABCDE   66
N 3 hours ago by ND_
Source: 2015 IMO Shortlist A1, Original 2015 IMO #5
Suppose that a sequence $a_1,a_2,\ldots$ of positive real numbers satisfies \[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]for every positive integer $k$. Prove that $a_1+a_2+\ldots+a_n\geq n$ for every $n\geq2$.
66 replies
ABCDE
Jul 7, 2016
ND_
3 hours ago
Euler Line Madness
raxu   75
N 4 hours ago by lakshya2009
Source: TSTST 2015 Problem 2
Let ABC be a scalene triangle. Let $K_a$, $L_a$ and $M_a$ be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of $AK_aL_a$ intersects $AM_a$ a second time at point $X_a$ different from A. Define $X_b$ and $X_c$ analogously. Prove that the circumcenter of $X_aX_bX_c$ lies on the Euler line of ABC.
(The Euler line of ABC is the line passing through the circumcenter, centroid, and orthocenter of ABC.)

Proposed by Ivan Borsenco
75 replies
raxu
Jun 26, 2015
lakshya2009
4 hours ago
Own made functional equation
Primeniyazidayi   8
N 4 hours ago by MathsII-enjoy
Source: own(probably)
Find all functions $f:R \rightarrow R$ such that $xf(x^2+2f(y)-yf(x))=f(x)^3-f(y)(f(x^2)-2f(x))$ for all $x,y \in \mathbb{R}$
8 replies
Primeniyazidayi
May 26, 2025
MathsII-enjoy
4 hours ago
IMO ShortList 2002, geometry problem 7
orl   110
N 4 hours ago by SimplisticFormulas
Source: IMO ShortList 2002, geometry problem 7
The incircle $ \Omega$ of the acute-angled triangle $ ABC$ is tangent to its side $ BC$ at a point $ K$. Let $ AD$ be an altitude of triangle $ ABC$, and let $ M$ be the midpoint of the segment $ AD$. If $ N$ is the common point of the circle $ \Omega$ and the line $ KM$ (distinct from $ K$), then prove that the incircle $ \Omega$ and the circumcircle of triangle $ BCN$ are tangent to each other at the point $ N$.
110 replies
orl
Sep 28, 2004
SimplisticFormulas
4 hours ago
Cute NT Problem
M11100111001Y1R   6
N 4 hours ago by X.Allaberdiyev
Source: Iran TST 2025 Test 4 Problem 1
A number \( n \) is called lucky if it has at least two distinct prime divisors and can be written in the form:
\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]where \( p_1, \dots, p_k \) are distinct prime numbers that divide \( n \). (Note: it is possible that \( n \) has other prime divisors not among \( p_1, \dots, p_k \).) Prove that for every prime number \( p \), there exists a lucky number \( n \) such that \( p \mid n \).
6 replies
M11100111001Y1R
May 27, 2025
X.Allaberdiyev
4 hours ago
China MO 2021 P6
NTssu   23
N 5 hours ago by bin_sherlo
Source: CMO 2021 P6
Find $f: \mathbb{Z}_+ \rightarrow \mathbb{Z}_+$, such that for any $x,y \in \mathbb{Z}_+$, $$f(f(x)+y)\mid x+f(y).$$
23 replies
NTssu
Nov 25, 2020
bin_sherlo
5 hours ago
incenter of ABC lies on (ACQ) , AN = NM = MC
parmenides51   1
N Jul 10, 2022 by StarLex1
Source: 2006 Rusanovsky Lyceum Olympiad p139
On the sides $AB$ and $BC$ of the triangle $ABC$, points $N$ and $M$ are marked, respectively, such that $AN = NM = MC$. Let $Q$ be the intersection point of the segments $AM$ and $CN$. Prove that the center of the inscribed circle of triangle $ABC$ lies on the circumcircle of triangle $ACQ$.
1 reply
parmenides51
Jun 29, 2022
StarLex1
Jul 10, 2022
incenter of ABC lies on (ACQ) , AN = NM = MC
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Source: 2006 Rusanovsky Lyceum Olympiad p139
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parmenides51
30653 posts
#1
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On the sides $AB$ and $BC$ of the triangle $ABC$, points $N$ and $M$ are marked, respectively, such that $AN = NM = MC$. Let $Q$ be the intersection point of the segments $AM$ and $CN$. Prove that the center of the inscribed circle of triangle $ABC$ lies on the circumcircle of triangle $ACQ$.
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StarLex1
816 posts
#2
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$\angle{NAM} = x,\angle{NMA} = x,\angle{MNC}=a, \angle{NCM} = a$
By LoS
$\frac{MC}{\sin(x+a)} = \frac{QM}{\sin(a)}$
$\frac{NM}{\sin(x+a)} = \frac{QM}{\sin(x)}$
$\sin(a) = \sin(x)$
$a = x $ or $180-a = x$
if $180-a=x$ then $\angle{ANQ} = $ $180-x-a-a= -a$ which is absurd
$a= x$
by thales since $\frac{BN}{NA} = \frac{BM}{MC}$ then implicating it is parallel
let we extend BQ and meet AC at K then $AK=CK$ by ceva
$\frac{BA}{AK}=\frac{BC}{CK}$ implicating BK is also bisector thus Q is incenter
This post has been edited 1 time. Last edited by StarLex1, Jul 10, 2022, 5:09 PM
Reason: typo
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