Art of Problem Solving
AoPS Online AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy AoPS Academy
Small live classes for advanced math
and language arts learners in grades 1-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Summer is a great time to explore cool problems to keep your skills sharp!  Schedule a class today!
geometry
B
No tags match your search
M
G
Topic
First Poster
Last Poster
H
Random Points = Problem
kingu   5
N 4 hours ago by happypi31415
Source: Chinese Geometry Handout
Let $ABC$ be a triangle. Let $\omega$ be a circle passing through $B$ intersecting $AB$ at $D$ and $BC$ at $F$. Let $G$ be the intersection of $AF$ and $\omega$. Further, let $M$ and $N$ be the intersections of $FD$ and $DG$ with the tangent to $(ABC)$ at $A$. Now, let $L$ be the second intersection of $MC$ and $(ABC)$. Then, prove that $M$ , $L$ , $D$ , $E$ and $N$ are concyclic.
5 replies
+1 w
kingu
Apr 27, 2024
happypi31415
4 hours ago
J
No more topics!
H
J
H
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 $AH$
G H J
Reveal topic tags
geometry
Spain
L
G H BBookmark kLocked kLocked NReply
Source: Spain MO 2025 P4
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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 $\angle ATX=\angle OTB$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
WLOGQED1729
50 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 $\angle ATX = \angle AMX = \angle OAM =\angle OTM = \angle OTB$ $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
HormigaCebolla
4 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
Z K Y
N Quick Reply
G
H
=
a