GMO P3 2024

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

GMO P3 2024

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Geometry Mains Olympiad (GMO) 2024 P3

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

64 posts

#1 h Oct 20, 2024, 10:42 PM • 1 Y

Y by Rounak_iitr

Let $ABC$ be a scalene triangle. $D$ is the foot of the altitude from $A$ to $BC$ . $H$ is the orthocenter, $M$ is the midpoint of $BC$ , and $N$ is the midpoint of $AH$ . Tangents from $B$ and $C$ to the circumcircle of $ABC$ intersect at $T$ . Prove that line $TH$ passes through one of the intersections of circles $ADT$ with $MHN$ .

Author:Haris Shaqiri (Kosovo)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1546 posts

#2 h Oct 21, 2024, 3:30 AM • 1 Y

Y by MihaiT

Curious how exactly this is p3 difficulty. (EDIT: oh nonstandard format)

Invert about $H$ with radius $-\sqrt{AT \cdot AD}$ . Then $(ADT)$ is fixed under this so $TH \cap (ADT)$ is the inverse of $T$ . Inverting, we get the following equivalent problem.

Quote:

Let $ABC$ be a scalene triangle. $D$ is the foot of the altitude from $A$ to $BC$ . $H$ is the orthocenter, $M$ is the midpoint of $BC$ . $H_1$ is the reflection of $H$ over $D$ . Tangents from $B$ and $C$ to the circumcircle of $ABC$ intersect at $T$ . Prove that line $TH_1 \cap HM$ lies on the circumcircle.

Let $H_2$ be the reflection of $H$ over $M$ . Let $K = MH \cap (ABC), K \ne H_2$ . Then we want to show that $\triangle BH_1CK$ is harmonic. We can eqiuvalently show that $\measuredangle BH_1M = \measuredangle KH_1C$ , or that $\measuredangle MHB = \measuredangle KBC$ , or that $(BKH)$ is tangent to $BM$ . Except this follows since
$BM^2 = BM \cdot CM = MH_2 \cdot MK = MH \cdot MK$

Attachments:

This post has been edited 2 times. Last edited by YaoAOPS, Oct 21, 2024, 3:41 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

737 posts

#3 h Oct 21, 2024, 5:36 AM • 2 Y

Y by Collatz09, MihaiT

Let $AD\cap (ABC)=E$ and $Q$ be $A-$ queue point. Note that $M,H,Q$ are collinear. Also $-1=(AQ,AD;AB,AC)=(Q,E;B,C)$ hence $T,E,Q$ are collinear. Let $TH\cap (ADT)=S$ . Since $HM.HQ=HA.HD=HN.HE$ , we get that $Q,N,M,E$ are concyclic.
$HS.HT=HA.HD=HN.HE=HQ.HM$ Thus,
$\measuredangle QMS=\measuredangle QTS=\measuredangle ETS=\measuredangle ENS=\measuredangle HNS$ Which implies $H,M,S,N$ are concyclic as desired. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

204 posts

Eka01

#4 h Oct 22, 2024, 5:16 PM

Y by

We prove that $(MHN)$ passes through $TH \cap (ADT)=G$ . By power of point, $GH.HT=AH.HD$ . Now consider the $\sqrt{-HA.HD}$ inversion. By the above result from power of point, we get that $G$ is mapped to $T$ . $N$ is mapped to the reflection of $H$ about $D$ which we call $D'$ and $M$ is mapped to the $A$ queue point $Q$ . Now it suffices to prove that $\overline {Q-D'-T}$ are collinear.
Now if $X$ is the $A$ expoint then $(XD;BC) \stackrel{A}{=} (Q,D';BC)=-1$ and since $T$ is pole of $BC$ , it follows that $\overline{Q-D'-T}$ are collinear.

This post has been edited 4 times. Last edited by Eka01, Oct 23, 2024, 8:32 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

SomeonesPenguin

129 posts

SomeonesPenguin

#5 h Oct 26, 2024, 8:02 AM • 1 Y

Y by zzSpartan

Let $HM$ intersect $(ABC)$ at $S$ , different from the $A$ antipode and let $E$ be the reflection of $H$ over $D$ . Now we invert at $H$ with radius $\sqrt{HA\cdot HD}$ . It's easy to see that $(ADT)$ is invariant and $(HNM)$ maps to $SE$ so it suffices to prove that $\overline{S-E-T}$ .

This is equivalent to $(SE;BC)=-1$ . If $AS$ meets $BC$ at $P$ , it is known that $(PD;BC)=-1$ and the conclusion follows by projecting through $A$ . $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

245 posts

#6 h Apr 17, 2025, 5:59 AM

Y by

Troll aah problem. Let $H'$ , $Q_A$ , $X_A$ denote $2D-H$ , $A$ -Queue, $A$ -Ex point respectively. Now invert about $H$ with radius $-\sqrt{HA \cdot HD}$ then $\overline{TH}$ , $(ADT)$ are fixed; $M$ goes to $Q_A$ and $N$ goes to $H'$ . Now since $(Q_A,H';B,C) \overset A= (X_A,D;B,C)=-1$ and so $T \in \overline{Q_AH'}$ and so the concurrency point in this picture is just $T$ lol.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a