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Easy Geometry
pokmui9909   4
N 9 minutes ago by aidenkim119
Source: FKMO 2025 P4
Triangle $ABC$ satisfies $\overline{CA} > \overline{AB}$. Let the incenter of triangle $ABC$ be $\omega$, which touches $BC, CA, AB$ at $D, E, F$, respectively. Let $M$ be the midpoint of $BC$. Let the circle centered at $M$ passing through $D$ intersect $DE, DF$ at $P(\neq D), Q(\neq D)$, respecively. Let line $AP$ meet $BC$ at $N$, line $BP$ meet $CA$ at $L$. Prove that the three lines $EQ, FP, NL$ are concurrent.
4 replies
pokmui9909
Today at 5:18 AM
aidenkim119
9 minutes ago
orz otl fr
Hip1zzzil   6
N 12 minutes ago by space10
Source: FKMO 2025 P3
An acute triangle $\bigtriangleup ABC$ is given which $BC>CA>AB$.
$I$ is the interior and the incircle of $\bigtriangleup ABC$ meets $BC, CA, AB$ at $D,E,F$. $AD$ and $BE$ meet at $P$. Let $l_{1}$ be a tangent from D to the circumcircle of $\bigtriangleup DIP$, and define $l_{2}$ and $l_{3}$ on $E$ and $F$, respectively.
Prove $l_{1},l_{2},l_{3}$ meet at one point.
6 replies
Hip1zzzil
Yesterday at 10:23 AM
space10
12 minutes ago
VERY HARD MATH PROBLEM!
slimshadyyy.3.60   7
N 14 minutes ago by ThresherShark_
Let a ≥b ≥c ≥0 be real numbers such that a^2 +b^2 +c^2 +abc = 4. Prove that
a+b+c+(√a−√c)^2 ≥3.
7 replies
slimshadyyy.3.60
Yesterday at 10:49 PM
ThresherShark_
14 minutes ago
The Tetrahedral Space Partition
jannatiar   1
N 15 minutes ago by alinazarboland
Source: 2025 AlborzMO Day 2 P3
Is it possible to partition three-dimensional space into tetrahedra (not necessarily regular) such that there exists a plane that intersects the edges of each tetrahedron at exactly 4 or 0 points?

Proposed by Arvin Taheri
1 reply
jannatiar
Mar 9, 2025
alinazarboland
15 minutes ago
gcd(a,b) x lcm(b,c), gcd(b,c)x lcm(c,a), ,gcd(c,a) x lcm(a,b)
parmenides51   3
N 19 minutes ago by Nari_Tom
Source: 2024 Czech and Slovak Olympiad III A p1
Let $a, b, c$ be positive integers such that one of the values $$gcd(a,b) \cdot  lcm(b,c), \,\,\,\, gcd(b,c)\cdot  lcm(c,a), \,\,\,\, gcd(c,a)-\cdot lcm(a,b)$$is equal to the product of the remaining two. Prove that one of the numbers $a, b, c$ is a multiple of another of them.
3 replies
parmenides51
May 18, 2024
Nari_Tom
19 minutes ago
Escape from the room
jannatiar   1
N 20 minutes ago by alinazarboland
Source: 2024 AlborzMO P3
A person is locked in a room with a password-protected computer. If they enter the correct password, the door opens and they are freed. However, the password changes every time it is entered incorrectly. The person knows that the password is always a 10-digit number, and they also know that the password change follows a fixed pattern. This means that if the current password is \( b \) and \( a \) is entered, the new password is \( c \), which is determined by \( b \) and \( a \) (naturally, the person does not know \( c \) or \( b \)). Prove that regardless of the characteristics of this computer, the prisoner can free themselves.

Proposed by Reza Tahernejad Karizi
1 reply
jannatiar
Mar 4, 2025
alinazarboland
20 minutes ago
A number theory about divisors which no one fully solved at the contest
nAalniaOMliO   23
N 26 minutes ago by AL1296
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
23 replies
nAalniaOMliO
Jul 24, 2024
AL1296
26 minutes ago
Thanks u!
Ruji2018252   0
34 minutes ago
Let $x,y,z,t\in\mathbb{R}$ and $\begin{cases}x^2+y^2=4\\z^2+t^2=9\\xt+yz\geqslant 6\end{cases}$.
$1,$ Prove $xz=yt$
$2,$ Find maximum $P=x+z$
0 replies
Ruji2018252
34 minutes ago
0 replies
a^{2m}+a^{n}+1 is perfect square
kmh1   2
N 43 minutes ago by cuden
Source: own
Find all positive integer triplets $(a,m,n)$ such that $2m>n$ and $a^{2m}+a^{n}+1$ is a perfect square.
2 replies
kmh1
Mar 20, 2025
cuden
43 minutes ago
IMO 2016 Problem 1
quangminhltv99   150
N an hour ago by ehuseyinyigit
Source: IMO 2016
Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$. Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ are concurrent.
150 replies
quangminhltv99
Jul 11, 2016
ehuseyinyigit
an hour ago
OA=OB if <PAD = <ADP=< CBP =< PCB =< CPD
parmenides51   2
N an hour ago by Nari_Tom
Source: 2024 Czech and Slovak Olympiad III A p2
Let the interior point $P$ of the convex quadrilateral $ABCD$ be such that $$|\angle PAD| = |\angle ADP| = |\angle CBP| = |\angle PCB| = |\angle CPD|.$$Let $O$ be the center of the circumcircle of the triangle $CPD$. Prove that $|OA| = |OB|$.
2 replies
parmenides51
May 18, 2024
Nari_Tom
an hour ago
Classic Inequality
KHOMNYO2   2
N an hour ago by GreekIdiot
Given positive real numbers $x,y,z$ such that $x^2 + y^2 + z^2 = 3$. Prove the following holds

$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 2x + 2y + 2z \geq 9$$
2 replies
1 viewing
KHOMNYO2
Mar 28, 2025
GreekIdiot
an hour ago
Belgium 2004.
Virgil Nicula   6
N Aug 17, 2016 by lebathanh
PP (NMO Belgium 2004). Let $\triangle ABC$ with $c>b\ ,$ $D\in (AC$ so that $BD=CD$ and

$E\in (BC)\ ,$ $F\in AB$ so that $EF\parallel BD$ and $G\in AE\cap BD\ .$ Prove that $\widehat{BCF}\equiv\widehat{BCG}\ .$
6 replies
Virgil Nicula
Aug 15, 2016
lebathanh
Aug 17, 2016
Belgium 2004.
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Virgil Nicula
7054 posts
#1 • 1 Y
Y by Adventure10
PP (NMO Belgium 2004). Let $\triangle ABC$ with $c>b\ ,$ $D\in (AC$ so that $BD=CD$ and

$E\in (BC)\ ,$ $F\in AB$ so that $EF\parallel BD$ and $G\in AE\cap BD\ .$ Prove that $\widehat{BCF}\equiv\widehat{BCG}\ .$
This post has been edited 4 times. Last edited by Virgil Nicula, Aug 15, 2016, 11:02 PM
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vanstraelen
8944 posts
#2 • 2 Y
Y by Adventure10, Mango247
$D \in [AC]$ extended?
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lebathanh
464 posts
#3 • 1 Y
Y by Adventure10
any solution please
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Arab
612 posts
#4 • 2 Y
Y by Adventure10, Mango247
Let $H$ be a point on $AD$ such that $GH\parallel BC$.

Now $FE\parallel BD\implies \frac{EF}{GB}=\frac{AE}{AG}=\frac{EC}{GH}\implies \triangle CEF\sim \triangle HGB$.

$BCHG$ is an isosceles trapezoid since $BD=CD$. Hence we have that $\angle BCF=\angle BHG=\angle BCG$.
This post has been edited 1 time. Last edited by Arab, Aug 16, 2016, 5:20 PM
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Stens
49 posts
#5 • 1 Y
Y by Adventure10
Arab's solution is very nice. Below is a sketch of a not so clever solution:

Let $CF$ meet $GD$ at $F'$. By some angle chasing we observe that we are done if $DC$ is tangent to the circumcircle of $\triangle GCF'$, which is equivalent to $DB^2=DF'\cdot DG$. Finding the lengths $DF',DB$ and $DG$ in terms of $AB,BC,CA$ is not too hard, and then we just verify that we do in fact have equality above.
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Virgil Nicula
7054 posts
#6 • 2 Y
Y by Adventure10, Mango247
Thank you for your proofs. See here my proof which is more intricate (with harmonic division).
This post has been edited 2 times. Last edited by Virgil Nicula, Aug 17, 2016, 2:09 AM
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lebathanh
464 posts
#7 • 2 Y
Y by Adventure10, Mango247
Arab wrote:
Let $H$ be a point on $AD$ such that $GH\parallel BC$.

Now $FE\parallel BD\implies \frac{EF}{GB}=\frac{AE}{AG}=\frac{EC}{GH}\implies \triangle CEF\sim \triangle HGB$.

$BCHG$ is an isosceles trapezoid since $BD=CD$. Hence we have that $\angle BCF=\angle BHG=\angle BCG$.

the proof is very nice ,thanks
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