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

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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Wednesday at 3:18 PM
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Wednesday at 3:18 PM
0 replies
Inspired by JK1603JK
sqing   13
N 4 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 0 $ and $ab+bc+ca=1.$ Prove that$$\frac{abc-2}{abc-1}\ge \frac{4(a^2b+b^2c+c^2a)}{a^3b+b^3c+c^3a+1} $$
13 replies
1 viewing
sqing
Today at 3:31 AM
sqing
4 minutes ago
Problem 1
SlovEcience   0
10 minutes ago
Prove that
\[
C(p-1, k-1) \equiv (-1)^{k-1} \pmod{p}
\]for \( 1 \leq k \leq p-1 \), where \( C(n, m) \) is the binomial coefficient \( n \) choose \( m \).
0 replies
SlovEcience
10 minutes ago
0 replies
A simple power
Rushil   19
N 11 minutes ago by Raj_singh1432
Source: Indian RMO 1993 Problem 2
Prove that the ten's digit of any power of 3 is even.
19 replies
Rushil
Oct 16, 2005
Raj_singh1432
11 minutes ago
Problem 1
blug   3
N 16 minutes ago by blug
Source: Polish Math Olympiad 2025 Finals P1
Find all $(a, b, c, d)\in \mathbb{R}$ satisfying
\[\begin{aligned}
\begin{cases}
    a+b+c+d=0,\\
    a^2+b^2+c^2+d^2=12,\\
    abcd=-3.\\
\end{cases}
\end{aligned}\]
3 replies
blug
3 hours ago
blug
16 minutes ago
No more topics!
Hard geometry
jannatiar   3
N Mar 30, 2025 by alinazarboland
Source: 2024 AlborzMO P4
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
3 replies
jannatiar
Mar 4, 2025
alinazarboland
Mar 30, 2025
Hard geometry
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G H BBookmark kLocked kLocked NReply
Source: 2024 AlborzMO P4
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jannatiar
21 posts
#1 • 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
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alinazarboland
168 posts
#2 • 1 Y
Y by sami1618
Bump
It would be really nice if someone can present a synthetic solution
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sami1618
881 posts
#3 • 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
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alinazarboland
168 posts
#4 • 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
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