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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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0 replies
jlacosta
Apr 2, 2025
0 replies
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Factorial: n!|a^n+1
Nima Ahmadi Pour   66
N 3 minutes ago by cursed_tangent1434
Source: IMO Shortlist 2005 N4, Iran preparation exam
Find all positive integers $ n$ such that there exists a unique integer $ a$ such that $ 0\leq a < n!$ with the following property:
\[ n!\mid a^n + 1 \]

Proposed by Carlos Caicedo, Colombia
66 replies
Nima Ahmadi Pour
Apr 24, 2006
cursed_tangent1434
3 minutes ago
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Problems for v_p(n)
xytunghoanh   15
N an hour ago by AshAuktober
Hello everyone. I need some easy problems for $v_p(n)$ (use in a problem for junior) to practice. Can anyone share to me?
Thanks :>
15 replies
+1 w
xytunghoanh
3 hours ago
AshAuktober
an hour ago
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BMO2 1997 Q4
rstather   1
N an hour ago by aidan0626
Source: BMO2 1997 Q4
Problem Statement:
The set S= {1/r : r = 1, 2, 3,...} of reciprocals of the positive integers contains arithmetic progressions of various lengths. For instance, 1/20, 1/8, 1/5 is such a progression, of length 3 (and common difference 3/40). Moreover, this is a maximal progression in S of length 3 since it cannot be extended to the left or right within S (−1/40 and 11/40 not
being members of S).
(i) Find a maximal progression in
S of length 1996.
(ii) Is there a maximal progression in
S of length 1997?


For part (i) I constructed the sequence: 1/1996!, 2/1996!, 3/1996!, ..., 1996/1996!. Which has common difference of 1/1996! and because of the nature of the factorial function, each of them are elements in S, and continuing to the left we get 0, which is not in S, and continuing to the right we get 1997/1996! which is not in S because 1997 is prime. Therefore we have found a maximal progression in S of length 1996 as desired.

For part (ii), let A = (4000!)/(2003!). Now consider the sequence: 2004/A, 2005/A, 2006/A, ... , 4000/A. Each is an element of S, by the nature of the factorial function and the common difference is 1/A. Continuing to the left gives 2003/A, 2003 is prime which is not a factor of A, so 2003/A is not an element of S. Continuing to the right gives 4001/A, 4001 is prime which is not a factor of A, so 4001/A is not an element of S. Therefore we have found a maximal progression in S of length 1997, and so the answer is Yes.


Is this proof sound?
1 reply
rstather
an hour ago
aidan0626
an hour ago
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Geo to make por people happy
AlephG_64   1
N an hour ago by ND_
Source: 2025 Finals Portuguese Mathematical Olympiad P4
Let $ABCD$ be a square with $2cm$ side length and with center $T$. A rhombus $ARTE$ is drawn where point $E$ lies on line $DC$. What is the area of $ARTE$?
1 reply
AlephG_64
3 hours ago
ND_
an hour ago
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No more topics!
H
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angle chasing with the incircle
parmenides51   4
N Oct 25, 2018 by omriya200
Source: Spanish Mathematical Olympiad 1991 P4
The incircle of $ABC$ touches the sides $BC,CA,AB$ at $A' ,B' ,C'$ respectively. The line $A' C'$ meets the angle bisector of $\angle A$ at $D$. Find $\angle ADC$.
4 replies
parmenides51
Aug 1, 2018
omriya200
Oct 25, 2018
J
angle chasing with the incircle
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Reveal topic tags
geometry
incircle
angle
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G H BBookmark kLocked kLocked NReply
Source: Spanish Mathematical Olympiad 1991 P4
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parmenides51
30629 posts
parmenides51
#1 h  Aug 1, 2018, 4:10 PM • 2 Y
Y by Adventure10, Mango247
The incircle of $ABC$ touches the sides $BC,CA,AB$ at $A' ,B' ,C'$ respectively. The line $A' C'$ meets the angle bisector of $\angle A$ at $D$. Find $\angle ADC$.
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jayme
9775 posts
jayme
#2 h  Aug 2, 2018, 8:52 AM • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=479013
or
http://jl.ayme.pagesperso-orange.fr/Docs/An%20unlikely%20concurrence.pdf for more...

Sincerely
Jean-Louis
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AlastorMoody
2125 posts
AlastorMoody
#3 h  Oct 25, 2018, 8:38 AM • 1 Y
Y by Adventure10
We know that $BC'IA'$ is cyclic $\implies \angle C'A'I = \frac {\angle B }{2} $
Infact, it also easy to notice that $$\angle DA'C =90^{\circ} +\frac {\angle B}{2} ; \angle BID=\frac{1}{2} (\angle A+\angle B) ; \angle BIC = 90^{\circ} +\frac {\angle A}{2} $$
Now, $$\angle DIC =\angle BIC - \angle BID = 90^{\circ} -\frac {\angle B}{2} \implies A'DIC \text { is cyclic}$$Hence, $\boxed {\angle ADC = 90^{\circ} } $
This post has been edited 1 time. Last edited by AlastorMoody, Oct 25, 2018, 8:39 AM
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MathematicalPhysicist
179 posts
MathematicalPhysicist
#4 h  Oct 25, 2018, 9:57 AM • 2 Y
Y by Adventure10, Mango247
Right angles on incircle chord :D
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omriya200
317 posts
omriya200
#5 h  Oct 25, 2018, 10:52 AM • 2 Y
Y by Adventure10, Mango247
Angle chasing
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