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P2 Geo that most of contestants died
AlephG_64   2
N 8 minutes ago by Tsikaloudakis
Source: 2025 Finals Portuguese Mathematical Olympiad P2
Let $ABCD$ be a quadrilateral such that $\angle A$ and $\angle D$ are acute and $\overline{AB} = \overline{BC} = \overline{CD}$. Suppose that $\angle BDA = 30^\circ$, prove that $\angle DAC= 30^\circ$.
2 replies
AlephG_64
Yesterday at 1:23 PM
Tsikaloudakis
8 minutes ago
Beautiful problem
luutrongphuc   3
N 8 minutes ago by aidenkim119
Let triangle $ABC$ be circumscribed about circle $(I)$, and let $H$ be the orthocenter of $\triangle ABC$. The circle $(I)$ touches line $BC$ at $D$. The tangent to the circle $(BHC)$ at $H$ meets $BC$ at $S$. Let $J$ be the midpoint of $HI$, and let the line $DJ$ meet $(I)$ again at $X$. The tangent to $(I)$ parallel to $BC$ meets the line $AX$ at $T$. Prove that $ST$ is tangent to $(I)$.
3 replies
luutrongphuc
Apr 4, 2025
aidenkim119
8 minutes ago
Collinearity with orthocenter
Retemoeg   6
N 10 minutes ago by aidenkim119
Source: Own?
Given scalene triangle $ABC$ with circumcenter $(O)$. Let $H$ be a point on $(BOC)$ such that $\angle AOH = 90^{\circ}$. Denote $N$ the point on $(O)$ satisfying $AN \parallel BC$. If $L$ is the projection of $H$ onto $BC$, show that $LN$ passes through the orthocenter of $\triangle ABC$.
6 replies
Retemoeg
Mar 30, 2025
aidenkim119
10 minutes ago
Geometry
youochange   0
10 minutes ago
m:}
Let $\triangle ABC$ be a triangle inscribed in a circle, where the tangents to the circle at points $B$ and $C$ intersect at the point $P$. Let $M$ be a point on the arc $AC$ (not containing $B$) such that $M \neq A$ and $M \neq C$. Let the lines $BC$ and $AM$ intersect at point $K$. Let $P'$ be the reflection of $P$ with respect to the line $AM$. The lines $AP'$ and $PM$ intersect at point $Q$, and $PM$ intersects the circumcircle of $\triangle ABC$ again at point $N$.

Prove that the point $Q$ lies on the circumcircle of $\triangle ANK$.
0 replies
youochange
10 minutes ago
0 replies
comp. geo starting with a 90-75-15 triangle. <APB =<CPQ, <BQA =<CQP.
parmenides51   1
N 20 minutes ago by Mathzeus1024
Source: 2013 Cuba 2.9
Let ABC be a triangle with $\angle A = 90^o$, $\angle B = 75^o$, and $AB = 2$. Points $P$ and $Q$ of the sides $AC$ and $BC$ respectively, are such that $\angle APB =  \angle CPQ$ and $\angle BQA = \angle CQP$. Calculate the lenght of $QA$.
1 reply
parmenides51
Sep 20, 2024
Mathzeus1024
20 minutes ago
Fridolin just can't get enough from jumping on the number line
Tintarn   2
N 27 minutes ago by Sadigly
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 1
Fridolin the frog jumps on the number line: He starts at $0$, then jumps in some order on each of the numbers $1,2,\dots,9$ exactly once and finally returns with his last jump to $0$. Can the total distance he travelled with these $10$ jumps be a) $20$, b) $25$?
2 replies
Tintarn
Mar 17, 2025
Sadigly
27 minutes ago
Geometry
Captainscrubz   2
N 35 minutes ago by MrdiuryPeter
Source: Own
Let $D$ be any point on side $BC$ of $\triangle ABC$ .Let $E$ and $F$ be points on $AB$ and $AC$ such that $EB=ED$ and $FD=FC$ respectively. Prove that the locus of circumcenter of $(DEF)$ is a line.
Prove without using moving points :D
2 replies
Captainscrubz
2 hours ago
MrdiuryPeter
35 minutes ago
inequality ( 4 var
SunnyEvan   4
N 37 minutes ago by SunnyEvan
Let $ a,b,c,d \in R $ , such that $ a+b+c+d=4 . $ Prove that :
$$ a^4+b^4+c^4+d^4+3 \geq \frac{7}{4}(a^3+b^3+c^3+d^3) $$$$ a^4+b^4+c^4+d^4+ \frac{252}{25} \geq \frac{88}{25}(a^3+b^3+c^3+d^3) $$equality cases : ?
4 replies
SunnyEvan
Apr 4, 2025
SunnyEvan
37 minutes ago
Find the constant
JK1603JK   1
N an hour ago by Quantum-Phantom
Source: unknown
Find all $k$ such that $$\left(a^{3}+b^{3}+c^{3}-3abc\right)^{2}-\left[a^{3}+b^{3}+c^{3}+3abc-ab(a+b)-bc(b+c)-ca(c+a)\right]^{2}\ge 2k\cdot(a-b)^{2}(b-c)^{2}(c-a)^{2}$$forall $a,b,c\ge 0.$
1 reply
JK1603JK
4 hours ago
Quantum-Phantom
an hour ago
2025 - Turkmenistan National Math Olympiad
A_E_R   4
N an hour ago by NODIRKHON_UZ
Source: Turkmenistan Math Olympiad - 2025
Let k,m,n>=2 positive integers and GCD(m,n)=1, Prove that the equation has infinitely many solutions in distict positive integers: x_1^m+x_2^m+⋯x_k^m=x_(k+1)^n
4 replies
A_E_R
2 hours ago
NODIRKHON_UZ
an hour ago
hard problem
Cobedangiu   15
N an hour ago by Nguyenhuyen_AG
problem
15 replies
Cobedangiu
Mar 27, 2025
Nguyenhuyen_AG
an hour ago
9x9 board
oneplusone   8
N an hour ago by lightsynth123
Source: Singapore MO 2011 open round 2 Q2
If 46 squares are colored red in a $9\times 9$ board, show that there is a $2\times 2$ block on the board in which at least 3 of the squares are colored red.
8 replies
oneplusone
Jul 2, 2011
lightsynth123
an hour ago
Triangles with equal areas
socrates   11
N 2 hours ago by Nari_Tom
Source: Baltic Way 2014, Problem 13
Let $ABCD$ be a square inscribed in a circle $\omega$ and let $P$ be a point on the shorter arc $AB$ of $\omega$. Let $CP\cap BD = R$ and $DP \cap  AC = S.$
Show that triangles $ARB$ and $DSR$ have equal areas.
11 replies
socrates
Nov 11, 2014
Nari_Tom
2 hours ago
prove that $\angle Q L A=\angle M L A$
NJAX   3
N Apr 2, 2025 by Baimukh
Source: 2nd Al-Khwarizmi International Junior Mathematical Olympiad 2024, Day2, Problem7
Two circles with centers $O_{1}$ and $O_{2}$ intersect at $P$ and $Q$. Let $\omega$ be the circumcircle of the triangle $P O_{1} O_{2}$; the circle $\omega$ intersect the circles centered at $O_{1}$ and $O_{2}$ at points $A$ and $B$, respectively. The point $Q$ is inside triangle $P A B$ and $P Q$ intersects $\omega$ at $M$. The point $E$ on $\omega$ is such that $P Q=Q E$. Let $M E$ and $A B$ meet at $L$, prove that $\angle Q L A=\angle M L A$.

Proposed by Amir Parsa Hoseini Nayeri, Iran
3 replies
NJAX
May 31, 2024
Baimukh
Apr 2, 2025
prove that $\angle Q L A=\angle M L A$
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G H BBookmark kLocked kLocked NReply
Source: 2nd Al-Khwarizmi International Junior Mathematical Olympiad 2024, Day2, Problem7
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NJAX
29 posts
#1 • 2 Y
Y by GeoKing, Rounak_iitr
Two circles with centers $O_{1}$ and $O_{2}$ intersect at $P$ and $Q$. Let $\omega$ be the circumcircle of the triangle $P O_{1} O_{2}$; the circle $\omega$ intersect the circles centered at $O_{1}$ and $O_{2}$ at points $A$ and $B$, respectively. The point $Q$ is inside triangle $P A B$ and $P Q$ intersects $\omega$ at $M$. The point $E$ on $\omega$ is such that $P Q=Q E$. Let $M E$ and $A B$ meet at $L$, prove that $\angle Q L A=\angle M L A$.

Proposed by Amir Parsa Hoseini Nayeri, Iran
This post has been edited 1 time. Last edited by NJAX, May 31, 2024, 12:34 PM
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v4913
1650 posts
#2 • 1 Y
Y by GeoKing
Since $MP \perp O_1O_2$, $Q$ is the orthocenter of $\triangle{MO_1O_2}$, so $MA = MQ = MB$ and $MQ^2 = ML \cdot ME \implies (LQE)$ is tangent to $MQ$. So, $\angle{QLM} = 2\angle{QPE} = 2\angle{MLA}$.
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sami1618
883 posts
#3 • 1 Y
Y by NJAX
Let $O$ be the circumcenter of $PAB$.

Claim: $Q$ is the incenter of $PAB$
The following equalities are sufficient $$\angle AQP=180^{\circ}-\frac{1}{2}\angle AO_1P=90^{\circ}+\frac{1}{2}\angle ABP$$$$\angle BQP=180^{\circ}-\frac{1}{2}\angle BO_2P=90^{\circ}+\frac{1}{2}\angle BAP$$Claim: $EQOM$ is concyclic
$$\angle QOE=\frac{1}{2}\angle POE=\angle PME$$Claim: $MQ$ is tangent to the circumcircle of $ELQ$
Notice that $MA$ is tangent to the circumcircle of $ELA$ as $\angle MAL=\angle AEM$. Thus $MQ^2=MA^2=ME\cdot ML$.
Claim: $\angle QLA=\angle MLA$
$$2\angle MLB=\angle MOE=\angle MQE=\angle MLQ$$
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Baimukh
7 posts
#4
Y by
Let $Q'$ be the symmetric point of $Q$ relative to $AB$, and $M'$ be the symmetric point of $M$ relative to $AB$. Then $MM' \parallel QQ'$ $\angle QLQ'=\angle MLM'$ $LQ=LQ';$ $LM=LM';$ $ \Longrightarrow \angle LMM'=\angle LM'M=\angle LQQ'=\angle LQ'Q \Longrightarrow \angle QLA=\angle ALQ'=\angle ALM$
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