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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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
Apr 2, 2025
0 replies
J
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My problem that I could not find(NT)
Nuran2010   1
N 2 minutes ago by Nuran2010
Source: Own
While I was thinking on some other geometry problem, a NT problem came to my mind. Despite some tries(which were mostly order), I could not find a way to solve the problem. As I searched, this problem has never been posted before. Here is the problem.

Find all positive integers $a,b$ such that:
$a+b|2^{ab}+1$

Moreover, I wonder if there is a way to solve the question in this variant:

Find all positive integers $a,b,n$ such that:
$a+b|n^{ab}+1$
1 reply
Nuran2010
Apr 24, 2025
Nuran2010
2 minutes ago
J
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FE over R+
jasperE3   10
N 27 minutes ago by jasperE3
Source: Slovenia TST 2005 Test 1 Problem 2
Find all functions $f:\mathbb R^+\to\mathbb R^+$ such that for any $x,y>0$,
$$x^2\left(f(x)+f(y)\right)=(x+y)f\left(f(x)y\right).$$
10 replies
jasperE3
Apr 5, 2021
jasperE3
27 minutes ago
J
H
FE over R
jasperE3   7
N an hour ago by jasperE3
Source: Evan Chen
Find all functions $f:\mathbb R\to\mathbb R$ such that $f(x+f(y))+f(xy)=f(x+1)f(y+1)-1\forall x,y\in\mathbb R$.
7 replies
1 viewing
jasperE3
Apr 27, 2021
jasperE3
an hour ago
J
H
IMO ShortList 2002, geometry problem 7
orl   109
N an hour ago by Ilikeminecraft
Source: IMO ShortList 2002, geometry problem 7
The incircle $ \Omega$ of the acute-angled triangle $ ABC$ is tangent to its side $ BC$ at a point $ K$. Let $ AD$ be an altitude of triangle $ ABC$, and let $ M$ be the midpoint of the segment $ AD$. If $ N$ is the common point of the circle $ \Omega$ and the line $ KM$ (distinct from $ K$), then prove that the incircle $ \Omega$ and the circumcircle of triangle $ BCN$ are tangent to each other at the point $ N$.
109 replies
orl
Sep 28, 2004
Ilikeminecraft
an hour ago
J
No more topics!
H
J
H
Actually, D could be any point on the plane
Sadigly   2
N Apr 25, 2025 by Ianis
An arbitary point $D$ is selected on arc $BC$ not containing $A$ on $(ABC)$. $P$ and $Q$ are the reflections of point $B$ and $C$ with respect to $AD$, respectively. Circumcircles of $ABQ$ and $ACP$ intersect at $E\neq A$. Prove that $A;D;E$ is colinear
2 replies
Sadigly
Apr 13, 2025
Ianis
Apr 25, 2025
J
Actually, D could be any point on the plane
G H J
Reveal topic tags
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G H BBookmark kLocked kLocked NReply
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Sadigly
154 posts
Sadigly
#1 h  Apr 13, 2025, 6:40 PM
Y by
An arbitary point $D$ is selected on arc $BC$ not containing $A$ on $(ABC)$. $P$ and $Q$ are the reflections of point $B$ and $C$ with respect to $AD$, respectively. Circumcircles of $ABQ$ and $ACP$ intersect at $E\neq A$. Prove that $A;D;E$ is colinear
This post has been edited 1 time. Last edited by Sadigly, Apr 14, 2025, 2:58 PM
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Turkish_sniper
3 posts
Turkish_sniper
#2 h  Apr 25, 2025, 2:32 PM
Y by
Let $E$ is intersection of $(ABQ)$ and $[AD]$.
We have $\angle BAE = \angle PAE$.
So $A,E,P,C$ is circular $\Leftrightarrow \angle PCE=\angle PAE=\angle BAE=\angle BQE$.
If $ \triangle QBE \cong \triangle CPE$, we are done. Moreover, we know that $BE = PE$ and $QE = CE$
In that case, we should prove that $QB = CP$.
This is not too hard, just see that $Q,B,P,C$ is circular and $BP\parallel PC$.
Thus, proof is completed. $\blacksquare$

.
This post has been edited 3 times. Last edited by Turkish_sniper, Apr 25, 2025, 2:34 PM
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Ianis
404 posts
Ianis
#3 h  Apr 25, 2025, 3:05 PM • 1 Y
Y by Turkish_sniper
The circles are symmetric wrt $AD$. Done.
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