Hard geometry

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

Hard geometry

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: 2024 AlborzMO P4

The post below has been deleted. Click to close.

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

21 posts

#1 h Mar 4, 2025, 6:59 AM • 1 Y

Y by sami1618

In triangle $ABC$ , let $I$ be the $A$ -excenter. Points $X$ and $Y$ are placed on line $BC$ such that $B$ is between $X$ and $C$ , and $C$ is between $Y$ and $B$ . Moreover, $B$ and $C$ are the contact points of $BC$ with the $A$ -excircle of triangles $BAY$ and $AXC$ , respectively. Let $J$ be the $A$ -excenter of triangle $AXY$ , and let $H'$ be the reflection of the orthocenter of triangle $ABC$ with respect to its circumcenter. Prove that $I$ , $J$ , and $H'$ are collinear.

Proposed by Ali Nazarboland

This post has been edited 2 times. Last edited by jannatiar, Mar 4, 2025, 7:58 AM

Z K Y

The post below has been deleted. Click to close.

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

168 posts

#2 h Mar 29, 2025, 10:57 PM • 1 Y

Y by sami1618

Bump
It would be really nice if someone can present a synthetic solution

Z K Y

The post below has been deleted. Click to close.

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

881 posts

#3 h Mar 30, 2025, 2:52 AM • 1 Y

Y by jannatiar

Very interesting problem. Not sure if the following solution is considered synthetic, but it was the best I could come up with

Let $\mathcal{E}_b$ be the ellipse passing through $B$ with foci $A$ and $C$ and let $\mathcal{E}_c$ be the ellipse passing through $C$ with foci $A$ and $B$ .

Part I: $\mathcal{E}_b$ and $\mathcal{E}_c$ intersect at exactly two points $P$ and $Q$ proof

Part II: $H'$ lies on $PQ$ proof

Part III: $I$ lies on $PQ$ proof

Part IV: $J$ lies on $PQ$ proof

Attachments:

This post has been edited 1 time. Last edited by sami1618, Mar 30, 2025, 2:53 AM

Z K Y

The post below has been deleted. Click to close.

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

168 posts

#4 h Mar 30, 2025, 9:03 AM • 1 Y

Y by sami1618

@above Wonderful solution. Althought, it's not synthetic at all...
I added the ellipses too, and I proved the fact that $I \in PQ$ using the Bogdanov's theorem as well. The fact that $H' \in PQ$ was just a fact I heard somewhere and it also appeared in Sharygin 2023 first round. I didn't prove it myself but there exist a lot of different methods to prove that like coordinate bash or even taking the configuration into the 3D space. The fact that $J \in PQ$ could be easily proven by 2 point-DIT (having $I \in P$ ) but your method was new for me and I enjoyed it.
But still, I believe your approach, isn't a synthetic one

This post has been edited 1 time. Last edited by alinazarboland, Mar 30, 2025, 9:03 AM

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a