Stay ahead of learning milestones! Enroll in a class over the summer!

G
Topic
First Poster
Last Poster
Sintetic geometry problem
ICE_CNME_4   5
N an hour ago by Ianis
Source: Math Gazette Contest 2025
Let there be the triangle ABC and the points E ∈ (AC), F ∈ (AB), such that BE and CF are concurrent in O.
If {L} = AO ∩ EF and K ∈ BC, such that LK ⊥ BC, show that EKL = FKL.
5 replies
ICE_CNME_4
5 hours ago
Ianis
an hour ago
Solution needed ASAP
UglyScientist   9
N an hour ago by MathsII-enjoy
$ABC$ is acute triangle. $H$ is orthocenter, $M$ is the midpoint of $BC$, $L$ is the midpoint of smaller arc $BC$. Point $K$ is on $AH$ such that, $MK$ is perpendicular to $AL$. Prove that: $HMLK$ is paralelogram(Synthetic sol needed).
9 replies
UglyScientist
Yesterday at 1:18 PM
MathsII-enjoy
an hour ago
AC, BF, DE concurrent
a1267ab   75
N an hour ago by NicoN9
Source: APMO 2020 Problem 1
Let $\Gamma$ be the circumcircle of $\triangle ABC$. Let $D$ be a point on the side $BC$. The tangent to $\Gamma$ at $A$ intersects the parallel line to $BA$ through $D$ at point $E$. The segment $CE$ intersects $\Gamma$ again at $F$. Suppose $B$, $D$, $F$, $E$ are concyclic. Prove that $AC$, $BF$, $DE$ are concurrent.
75 replies
a1267ab
Jun 9, 2020
NicoN9
an hour ago
FE on R+
AshAuktober   7
N an hour ago by GingerMan
Source: 2007 MOP
(Note I couldn't find a post w/ this from AoPS search so I'm posting, please do tell if there exists a post.)

Solve over positive real numbers the functional equation
\[ f\left( f(x) y + \frac xy \right) = xyf(x^2+y^2). \]
7 replies
AshAuktober
Sep 2, 2024
GingerMan
an hour ago
2-adic Valuation Unbounded
tigerzhang   14
N an hour ago by GingerMan
Source: Own
For any nonzero integer, define $\nu_2(n)$ as the largest integer $k$ such that $2^k \mid n$. Find all integers $n$ that are not powers of $3$ such that the sequence $\nu_2\left(3^0-n\right),\nu_2\left(3^1-n\right),\nu_2\left(3^2-n\right),\ldots$ is unbounded.
14 replies
tigerzhang
Nov 5, 2021
GingerMan
an hour ago
Combinatorial Sum
P162008   2
N an hour ago by cazanova19921
Evaluate $\sum_{n=0}^{\infty} \frac{2^n + 1}{(2n + 1) \binom{2n}{n}}$
2 replies
P162008
Apr 24, 2025
cazanova19921
an hour ago
IMO 90/3 and IMO 00/5 cross-up
v_Enhance   60
N an hour ago by GingerMan
Source: USA TSTST 2018 Problem 8
For which positive integers $b > 2$ do there exist infinitely many positive integers $n$ such that $n^2$ divides $b^n+1$?

Evan Chen and Ankan Bhattacharya
60 replies
v_Enhance
Jun 26, 2018
GingerMan
an hour ago
square root problem
kjhgyuio   4
N an hour ago by aidan0626
........
4 replies
kjhgyuio
Yesterday at 4:48 AM
aidan0626
an hour ago
gcd (a^n+b,b^n+a) is constant
EthanWYX2009   81
N 2 hours ago by GingerMan
Source: 2024 IMO P2
Determine all pairs $(a,b)$ of positive integers for which there exist positive integers $g$ and $N$ such that
$$\gcd (a^n+b,b^n+a)=g$$holds for all integers $n\geqslant N.$ (Note that $\gcd(x, y)$ denotes the greatest common divisor of integers $x$ and $y.$)

Proposed by Valentio Iverson, Indonesia
81 replies
EthanWYX2009
Jul 16, 2024
GingerMan
2 hours ago
Orthocenter
jayme   5
N 2 hours ago by Ianis
Dear Mathlinkers,

1. ABC an acuatangle triangle
2. H the orthcenter of ABC
3. DEF the orthic triangle of ABC
4. A* the midpoint of AH
5. X the point of intersection of AH and EF.

Prove : X is the orthocenter of A*BC.

Sincerely
Jean-Louis
5 replies
jayme
Mar 25, 2015
Ianis
2 hours ago
Geometry inmo problem
Enes040612   2
N Apr 25, 2025 by Red_gear5246
Source: INMO 2008
Can anyone help with this problem? the book also starts the solution by saying IA1=IB1=IC1=2r which I'm not understanding why. The problem is at the attachments.
2 replies
Enes040612
Feb 13, 2025
Red_gear5246
Apr 25, 2025
Geometry inmo problem
G H J
G H BBookmark kLocked kLocked NReply
Source: INMO 2008
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Enes040612
4 posts
#1
Y by
Can anyone help with this problem? the book also starts the solution by saying IA1=IB1=IC1=2r which I'm not understanding why. The problem is at the attachments.
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
CHESSR1DER
57 posts
#2
Y by
Let $IA1 \cap BC = H$. Then since $IH = r$ and $IH = A1H$, then $IA1= 2r$. Other are similar.
This post has been edited 2 times. Last edited by CHESSR1DER, Feb 13, 2025, 9:21 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Red_gear5246
1 post
#3
Y by
@Enes040612 which book are u solving?
Z K Y
N Quick Reply
G
H
=
a