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Geometry
smartvong   0
2 minutes ago
Source: UM Mathematical Olympiad 2024
Let $P$ be a point inside a triangle $ABC$. Let $AP$ meet $BC$ at $A_1$, let $BP$ meet $CA$ at $B_1$, and let $CP$ meet $AB$ at $C_1$. Let $A_2$ be the point such that $A_1$ is the midpoint of $PA_2$, let $B_2$ be the point such that $B_1$ is the midpoint of $PB_2$, and let $C_2$ be the point such that $C_1$ is the midpoint of $PC_2$. Prove that points $A_2, B_2, C_2$ cannot all lie strictly inside the circumcircle of triangle $ABC$.
0 replies
smartvong
2 minutes ago
0 replies
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angles in triangle
AndrewTom   34
N 19 minutes ago by happypi31415
Source: BrMO 2012/13 Round 2
The point $P$ lies inside triangle $ABC$ so that $\angle ABP = \angle PCA$. The point $Q$ is such that $PBQC$ is a parallelogram. Prove that $\angle QAB = \angle CAP$.
34 replies
AndrewTom
Feb 1, 2013
happypi31415
19 minutes ago
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Hard to approach it !
BogG   130
N 2 hours ago by Ilikeminecraft
Source: Swiss Imo Selection 2006
Let $\triangle ABC$ be an acute-angled triangle with $AB \not= AC$. Let $H$ be the orthocenter of triangle $ABC$, and let $M$ be the midpoint of the side $BC$. Let $D$ be a point on the side $AB$ and $E$ a point on the side $AC$ such that $AE=AD$ and the points $D$, $H$, $E$ are on the same line. Prove that the line $HM$ is perpendicular to the common chord of the circumscribed circles of triangle $\triangle ABC$ and triangle $\triangle ADE$.
130 replies
BogG
May 25, 2006
Ilikeminecraft
2 hours ago
J
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A game with balls and boxes
egxa   5
N 2 hours ago by compoly2010
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?
5 replies
egxa
Apr 30, 2023
compoly2010
2 hours ago
J
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Two lines meet on semicircle
va2010   87
N 2 hours ago by Ilikeminecraft
Source: 2015 ISL G3
Let $ABC$ be a triangle with $\angle{C} = 90^{\circ}$, and let $H$ be the foot of the altitude from $C$. A point $D$ is chosen inside the triangle $CBH$ so that $CH$ bisects $AD$. Let $P$ be the intersection point of the lines $BD$ and $CH$. Let $\omega$ be the semicircle with diameter $BD$ that meets the segment $CB$ at an interior point. A line through $P$ is tangent to $\omega$ at $Q$. Prove that the lines $CQ$ and $AD$ meet on $\omega$.
87 replies
va2010
Jul 7, 2016
Ilikeminecraft
2 hours ago
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something...
SunnyEvan   2
N 2 hours ago by SunnyEvan
Source: unknown
Try to prove : $$ \sum csc^{20} \frac{2^{i} \pi}{7} csc^{23} \frac{2^{j}\pi }{7} csc^{2023} \frac{2^{k} \pi}{7} $$is a rational number.
Where $ (i,j,k)=(1,2,3) $ and other permutations.
2 replies
SunnyEvan
May 5, 2025
SunnyEvan
2 hours ago
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USAMO 2002 Problem 2
MithsApprentice   34
N 3 hours ago by Giant_PT
Let $ABC$ be a triangle such that
\[ \left( \cot \dfrac{A}{2} \right)^2 + \left( 2\cot \dfrac{B}{2} \right)^2 + \left( 3\cot \dfrac{C}{2} \right)^2 = \left( \dfrac{6s}{7r} \right)^2, \]
where $s$ and $r$ denote its semiperimeter and its inradius, respectively. Prove that triangle $ABC$ is similar to a triangle $T$ whose side lengths are all positive integers with no common divisors and determine these integers.
34 replies
MithsApprentice
Sep 30, 2005
Giant_PT
3 hours ago
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Another config geo with concurrent lines
a_507_bc   17
N 3 hours ago by Rayvhs
Source: BMO SL 2023 G5
Let $ABC$ be a triangle with circumcenter $O$. Point $X$ is the intersection of the parallel line from $O$ to $AB$ with the perpendicular line to $AC$ from $C$. Let $Y$ be the point where the external bisector of $\angle BXC$ intersects with $AC$. Let $K$ be the projection of $X$ onto $BY$. Prove that the lines $AK, XO, BC$ have a common point.
17 replies
a_507_bc
May 3, 2024
Rayvhs
3 hours ago
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Nice sequence problem.
mathlover1231   1
N 3 hours ago by vgtcross
Source: Own
Scientists found a new species of bird called “N-coloured rainbow”. They also found out 3 interesting facts about the bird’s life: 1) every day, N-coloured rainbow is coloured in one of N colors.
2) every day, the color is different from yesterday (not every previous day, just yesterday).
3) there are no four days i, j, k, l in the bird’s life such that i<j<k<l with colours a, b, c, d respectively for which a=c ≠ b=d.
Find the greatest possible age (in days) of this bird as a function of N.
1 reply
mathlover1231
Apr 10, 2025
vgtcross
3 hours ago
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Three circles are concurrent
Twoisaprime   23
N 4 hours ago by Curious_Droid
Source: RMM 2025 P5
Let triangle $ABC$ be an acute triangle with $AB<AC$ and let $H$ and $O$ be its orthocenter and circumcenter, respectively. Let $\Gamma$ be the circle $BOC$. The line $AO$ and the circle of radius $AO$ centered at $A$ cross $\Gamma$ at $A’$ and $F$, respectively. Prove that $\Gamma$ , the circle on diameter $AA’$ and circle $AFH$ are concurrent.
Proposed by Romania, Radu-Andrew Lecoiu
23 replies
Twoisaprime
Feb 13, 2025
Curious_Droid
4 hours ago
J
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|a_i/a_j - a_k/a_l| <= C
mathwizard888   32
N 4 hours ago by ezpotd
Source: 2016 IMO Shortlist A2
Find the smallest constant $C > 0$ for which the following statement holds: among any five positive real numbers $a_1,a_2,a_3,a_4,a_5$ (not necessarily distinct), one can always choose distinct subscripts $i,j,k,l$ such that
\[ \left| \frac{a_i}{a_j} - \frac {a_k}{a_l} \right| \le C. \]
32 replies
mathwizard888
Jul 19, 2017
ezpotd
4 hours ago
J
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Two lines meeting on circumcircle
Zhero   54
N 4 hours ago by Ilikeminecraft
Source: ELMO Shortlist 2010, G4; also ELMO #6
Let $ABC$ be a triangle with circumcircle $\omega$, incenter $I$, and $A$-excenter $I_A$. Let the incircle and the $A$-excircle hit $BC$ at $D$ and $E$, respectively, and let $M$ be the midpoint of arc $BC$ without $A$. Consider the circle tangent to $BC$ at $D$ and arc $BAC$ at $T$. If $TI$ intersects $\omega$ again at $S$, prove that $SI_A$ and $ME$ meet on $\omega$.

Amol Aggarwal.
54 replies
Zhero
Jul 5, 2012
Ilikeminecraft
4 hours ago
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Help me this problem. Thank you
illybest   3
N 5 hours ago by jasperE3
Find f: R->R such that
f( xy + f(z) ) = (( xf(y) + yf(x) )/2) + z
3 replies
illybest
Yesterday at 11:05 AM
jasperE3
5 hours ago
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J
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Equal angles with midpoint of $AH$
Stuttgarden   2
N Apr 24, 2025 by HormigaCebolla
Source: Spain MO 2025 P4
Let $ABC$ be an acute triangle with circumcenter $O$ and orthocenter $H$, satisfying $AB<AC$. The tangent line at $A$ to the circumcicle of $ABC$ intersects $BC$ in $T$. Let $X$ be the midpoint of $AH$. Prove that $\angle ATX=\angle OTB$.
2 replies
Stuttgarden
Mar 31, 2025
HormigaCebolla
Apr 24, 2025
J
Equal angles with midpoint of $AH$
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G H BBookmark kLocked kLocked NReply
Source: Spain MO 2025 P4
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Stuttgarden
34 posts
Stuttgarden
#1 h  Mar 31, 2025, 1:10 PM • 1 Y
Y by Rounak_iitr
Let $ABC$ be an acute triangle with circumcenter $O$ and orthocenter $H$, satisfying $AB<AC$. The tangent line at $A$ to the circumcicle of $ABC$ intersects $BC$ in $T$. Let $X$ be the midpoint of $AH$. Prove that $\angle ATX=\angle OTB$.
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WLOGQED1729
47 posts
WLOGQED1729
#2 h  Mar 31, 2025, 2:47 PM • 1 Y
Y by ehuseyinyigit
Nice property! Here is my solution.

Let $M$ be midpoint of side $BC$. It is well-known that $AXMO$ is parallelogram.
Claim $X$ is orthocenter of triangle $ATM$
Proof We have $AX \perp TM$. Since $OA \perp AT$ and $XM \parallel AO$, we deduce that $MX \perp AT$. $\square$

The rest is just angle chasing.
Note that $\angle ATX = \angle AMX = \angle OAM =\angle OTM = \angle OTB$ $\blacksquare$
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HormigaCebolla
3 posts
HormigaCebolla
#3 h  Apr 24, 2025, 9:11 AM • 1 Y
Y by Steve12345
My solution during the exam: parallelogram isogonality lemma applied to parallelogram $AXMO$ and triangle $TAM$, where $M$ is the midpoint of side $BC$.
This post has been edited 1 time. Last edited by HormigaCebolla, Apr 25, 2025, 1:46 PM
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