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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
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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!
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J
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Vietnam NMO 2004_2
April   1
N Nov 2, 2008 by sunken rock
In a triangle $ ABC$, the bisector of $ \angle ACB$ cuts the side $ AB$ at $ D$. An arbitrary circle $ (O)$ passing through $ C$ and $ D$ meets the lines $ BC$ and $ AC$ at $ M$ and $ N$ (different from $ C$), respectively.
(a) Prove that there is a circle $ (S)$ touching $ DM$ at $ M$ and $ DN$ at $ N$.
(b) If circle $ (S)$ intersects the lines $ BC$ and $ CA$ again at $ P$ and $ Q$ respectively, prove that the lengths of the segments $ MP$ and $ NQ$ are constant as $ (O)$ varies.
1 reply
April
Oct 26, 2008
sunken rock
Nov 2, 2008
J
Vietnam NMO 2004_2
G H J
Reveal topic tags
geometry
angle bisector
geometry proposed
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G H BBookmark kLocked kLocked NReply
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April
1270 posts
April
#1 h  Oct 26, 2008, 1:43 AM • 2 Y
Y by Adventure10, Mango247
In a triangle $ ABC$, the bisector of $ \angle ACB$ cuts the side $ AB$ at $ D$. An arbitrary circle $ (O)$ passing through $ C$ and $ D$ meets the lines $ BC$ and $ AC$ at $ M$ and $ N$ (different from $ C$), respectively.
(a) Prove that there is a circle $ (S)$ touching $ DM$ at $ M$ and $ DN$ at $ N$.
(b) If circle $ (S)$ intersects the lines $ BC$ and $ CA$ again at $ P$ and $ Q$ respectively, prove that the lengths of the segments $ MP$ and $ NQ$ are constant as $ (O)$ varies.
Z K Y
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sunken rock
4384 posts
sunken rock
#2 h  Nov 2, 2008, 6:43 AM • 3 Y
Y by AlastorMoody, Adventure10, Mango247
As CD is the angle bisector of <ACB, it follows that DM = DN, which proves that there exists the circle (S), as required; its centre is T, the midpoint of the arc MN of the circle (O), the arc containing C.
T belongs to the external angle bisector of <ACB, which gives CP = CN and CQ = CM, therefore MP = NQ = CM + CN, so we have to prove this sum is constant.
If R and S are the orthogonal projections of D onto BC and AC respectively, the triangles DMR and DNS are equal, therefore MR = NS and, consequently,
CM + CN = CR + CS = ct.

Best regards,
sunken rock
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