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angle wanted, right ABC, AM=CB , CN=MB
parmenides51   2
N 14 minutes ago by Tsikaloudakis
Source: 2022 European Math Tournament - Senior First + Grand League - Math Battle 1.3
In a right-angled triangle $ABC$, points $M$ and $N$ are taken on the legs $AB$ and $BC$, respectively, so that $AM=CB$ and $CN=MB$. Find the acute angle between line segments $AN$ and $CM$.
2 replies
parmenides51
Dec 19, 2022
Tsikaloudakis
14 minutes ago
J
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Colors over colors in a grid
FAA2533   1
N 36 minutes ago by Rohit-2006
Source: BdMO 2025 Higher Secondary P3
Two player are playing in an $100 \times 100$ grid. Initially the whole board is black. On $A$'s move, he selects $4 \times 4$ subgrid and color it white. On $B$'s move, he selects a $3 \times 3$ subgrid and colors it black. $A$ wants to make the whole board white. Can he do it?

Proposed by S M A Nahian
1 reply
FAA2533
Feb 8, 2025
Rohit-2006
36 minutes ago
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AD and BC meet MN at P and Q
WakeUp   6
N 37 minutes ago by Nari_Tom
Source: Baltic Way 2005
Let $ABCD$ be a convex quadrilateral such that $BC=AD$. Let $M$ and $N$ be the midpoints of $AB$ and $CD$, respectively. The lines $AD$ and $BC$ meet the line $MN$ at $P$ and $Q$, respectively. Prove that $CQ=DP$.
6 replies
WakeUp
Dec 28, 2010
Nari_Tom
37 minutes ago
J
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thank you
Piwbo   0
42 minutes ago
Let $p_n$ be the n-th prime number in increasing order for $n\geq 1$. Prove that there exists a sequence of distinct prime numbers $q_n$ satisfying $q_1+q_2+...+q_n=p_n$ for all $n\geq 1 $
0 replies
Piwbo
42 minutes ago
0 replies
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IMO ShortList 2002, geometry problem 4
orl   24
N 44 minutes ago by Avron
Source: IMO ShortList 2002, geometry problem 4
Circles $S_1$ and $S_2$ intersect at points $P$ and $Q$. Distinct points $A_1$ and $B_1$ (not at $P$ or $Q$) are selected on $S_1$. The lines $A_1P$ and $B_1P$ meet $S_2$ again at $A_2$ and $B_2$ respectively, and the lines $A_1B_1$ and $A_2B_2$ meet at $C$. Prove that, as $A_1$ and $B_1$ vary, the circumcentres of triangles $A_1A_2C$ all lie on one fixed circle.
24 replies
orl
Sep 28, 2004
Avron
44 minutes ago
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comb and nt mixed
MR.1   1
N an hour ago by MR.1
Source: own
in antarctica there is $n\geq3$ penguin and each one is numbered using numbers $1,2,\dots n$. penguin $i$ is called $mirza$ if $\binom{\underline i}{p_i}=1$ where $p_i$ is $i_{th}$ prime divisor of $n$ ($p_{kt+i}=p_{i}$(where k is number of $n$'s prime divisors). what is maximum ratio of $\frac{M}{n}$ where $M$ is number of $mirza$ in antarctica?($p_1\neq 1$)
1 reply
MR.1
Yesterday at 2:34 PM
MR.1
an hour ago
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The best computer problems of the year
NT_G   9
N an hour ago by bin_sherlo
Source: https://t.me/NeuroGeometry
1. I is incenter of triangle $ABC$. $K$ is the midpoint of arc $BC$ not containing $A$. $L$ is the midpoint of arc $(BAC)$ $M$ is the midpoint of $IK$ and $N$ is the midpoint of $LM$.  $D$ is projection of $I$ onto $BC$. $S$ is the second intersection of $KD$ and $(ABC)$
Prove that $(ISD)$ is tangent to $(AMN)$.

2. Circle $w$ with center $I$ is incircle of triangle $ABC$. $D$ is projection of $I$ onto $BC$ Points $E, F$ on segments $BI$, $CI$ are following condition: $IE = IF$. $M$ is the midpoint of $EF$. $P$ is the second intersection of $(IMD)$ and $w$. $Q$ is the second intersection of $AP$ and $(IMD)$.
Prove, that $I,E,F,Q$ are concyclic.

I am grateful to Savva Chuev for making the diagrams.
9 replies
NT_G
Dec 31, 2023
bin_sherlo
an hour ago
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geometry problem
Medjl   4
N an hour ago by tigerBoss101
Source: Netherlands TST for IMO 2017 day 3 problem 1
A circle $\omega$ with diameter $AK$ is given. The point $M$ lies in the interior of the circle, but not on $AK$. The line $AM$ intersects $\omega$ in $A$ and $Q$. The tangent to $\omega$ at $Q$ intersects the line through $M$ perpendicular to $AK$, at $P$. The point $L$ lies on $\omega$, and is such that $PL$ is tangent to $\omega$ and $L\neq Q$.
Show that $K, L$, and $M$ are collinear.
4 replies
Medjl
Feb 1, 2018
tigerBoss101
an hour ago
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IMO ShortList 2002, geometry problem 3
orl   71
N an hour ago by Avron
Source: IMO ShortList 2002, geometry problem 3
The circle $S$ has centre $O$, and $BC$ is a diameter of $S$. Let $A$ be a point of $S$ such that $\angle AOB<120{{}^\circ}$. Let $D$ be the midpoint of the arc $AB$ which does not contain $C$. The line through $O$ parallel to $DA$ meets the line $AC$ at $I$. The perpendicular bisector of $OA$ meets $S$ at $E$ and at $F$. Prove that $I$ is the incentre of the triangle $CEF.$
71 replies
orl
Sep 28, 2004
Avron
an hour ago
J
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Perpendicularity
April   31
N an hour ago by Tsikaloudakis
Source: CGMO 2007 P5
Point $D$ lies inside triangle $ABC$ such that $\angle DAC = \angle DCA = 30^{\circ}$ and $\angle DBA = 60^{\circ}$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF = 2FC$. Prove that $DE \perp EF$.
31 replies
April
Dec 28, 2008
Tsikaloudakis
an hour ago
J
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China Mathematical Olympiad 1986 problem3
jred   4
N an hour ago by alexanderhamilton124
Source: China Mathematical Olympiad 1986 problem3
Let $Z_1,Z_2,\cdots ,Z_n$ be complex numbers satisfying $|Z_1|+|Z_2|+\cdots +|Z_n|=1$. Show that there exist some among the $n$ complex numbers such that the modulus of the sum of these complex numbers is not less than $1/6$.
4 replies
jred
Jan 17, 2014
alexanderhamilton124
an hour ago
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A remarkable tangent
jayme   3
N Feb 4, 2019 by ayamgabut
Dear Mathlinkers,

1. ABC a triangle,
2. P, I, Be the foot of the A-altitude, the incenter, the Bevan’s point of ABC.

Prove : ABe is the tangent to the circle (API) at A.

Sincerely
Jean-Louis
3 replies
jayme
Jul 12, 2018
ayamgabut
Feb 4, 2019
J
A remarkable tangent
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jayme
9775 posts
jayme
#1 h  Jul 12, 2018, 12:26 PM • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,

1. ABC a triangle,
2. P, I, Be the foot of the A-altitude, the incenter, the Bevan’s point of ABC.

Prove : ABe is the tangent to the circle (API) at A.

Sincerely
Jean-Louis
Z K Y
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WizardMath
2487 posts
WizardMath
#2 h  Jul 12, 2018, 1:23 PM • 3 Y
Y by amar_04, Adventure10, Mango247
Firstly note that the intersection of the perpendicular from $I$ to $AI$ meets $BC$ on $(API)$, say at $X$. From here, if $DEF$ is the intouch triangle, and the Miquel point of $BCEF$ is $K_A$, then the antipode of $A$ in $(ABC)$, $I$, $K_A$ are collinear, and $AK_A$ meets $BC$ on $(API)$. So due to the right angled triangles $AK_AI$ and $AIX$, we just need $AB_e \parallel K_AI$, which is obvious due to reflection in $O$.
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Luis González
4146 posts
Luis González
#3 h  Jul 13, 2018, 3:56 AM • 2 Y
Y by aops29, Adventure10
Let $\triangle I_aI_bI_c$ be the excentral triangle of $\triangle ABC$ with circumcenter $Be$ (well-known). Let $D$ be the midpoint of $\overline{II_a}$ (circumcenter of $\triangle I_aBC$), $J \equiv AI \cap BC$ and let $X$ be the projection of $I_a$ on $BC.$ We have

$(I,A,J,I_a)=-1 \Longrightarrow \frac{AP}{I_aX}=\frac{JA}{JI_a}=2 \cdot \frac{IA}{II_a}=\frac{IA}{DI_a} \Longrightarrow DX \parallel IP.$

Now since $\triangle I_aBC \cup \{D,X\} \sim \triangle I_aI_bI_c \cup \{Be,P\}$ $\Longrightarrow$ $\angle API=\angle DXI_a=\angle BeAI_a$ $\Longrightarrow$ $ABe$ is tangent of $\odot(AIP),$ as desired.
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ayamgabut
10 posts
ayamgabut
#4 h  Feb 4, 2019, 2:13 AM • 1 Y
Y by Adventure10
Suppose $\Delta I_{a}I_{b}I_{c}$ is the excentral triangle of $\Delta ABC$, the circle with center I and radius AI intersect BC at $X_{b},X_{x}$, and D,E,F be the touch points of the incircle to BC,AC,AB.

Notice that: $$\Delta IDX_{b} \cong \Delta IEA \implies \Delta EID \sim \Delta AIX_{b}$$
Now, $\angle AX_{b}X_{c}=\angle IED + \angle IEF = \angle I_{a}I_{b}I_{c}$.
Analogously, $\angle AX_{c}X_{b} = \angle I_{a}I_{c}I_{b}$ which implies $\Delta AX_{b}X_{c} \sim \Delta I_{a}I_{b}I_{c}$.
Since P,A are both the foot of altitude from $A,I_{a}$ and $I,B_{e}$ are both the circumcenter's of the two triangles, then $\Delta API \sim \Delta I_{a}AB_{e}$; which implies $\angle IAB_{e} = \angle API$ as desired.
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