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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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Problems for v_p(n)
xytunghoanh   15
N 20 minutes 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
xytunghoanh
3 hours ago
AshAuktober
20 minutes ago
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BMO2 1997 Q4
rstather   1
N 22 minutes 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
27 minutes ago
aidan0626
22 minutes ago
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Geo to make por people happy
AlephG_64   1
N 34 minutes 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_
34 minutes ago
J
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Modular Matching Pairs
steven_zhang123   3
N an hour ago by flower417477
Source: China TST 2025 P20
Let \( n \) be an odd integer, \( m = \frac{n+1}{2} \). Consider \( 2m \) integers \( a_1, a_2, \ldots, a_m, b_1, b_2, \ldots, b_m \) such that for any \( 1 \leq i < j \leq m \), \( a_i \not\equiv a_j \pmod{n} \) and \( b_i \not\equiv b_j \pmod{n} \). Prove that the number of \( k \in \{0, 1, \ldots, n-1\} \) for which satisfy \( a_i + b_j \equiv k \pmod{n} \) for some \( i \neq j \), $i, j \in \left \{ 1,2,\cdots,m \right \} $ is greater than \( n - \sqrt{n} - \frac{1}{2} \).
3 replies
steven_zhang123
Mar 29, 2025
flower417477
an hour ago
J
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midpoints wanted, 4 equal inradii in an equilateral triangle
parmenides51   0
Jul 30, 2021
Source: 1995 Bulgaria NMO, Round 4, p4
Points $A_1,B_1,C_1$ are selected on the sides $BC$,$CA$,$AB$ respectively of an equilateral triangle $ABC$ in such a way that the inradii of the triangles $C_1AB_1$, $A_1BC_1$, $B_1CA_1$ and $A_1B_1C_1$ are equal. Prove that $A_1,B_1,C_1$ are the midpoints of the corresponding sides.
0 replies
parmenides51
Jul 30, 2021
0 replies
J
midpoints wanted, 4 equal inradii in an equilateral triangle
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Reveal topic tags
geometry
midpoint
equal circles
inradii
Equilateral
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Source: 1995 Bulgaria NMO, Round 4, p4
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parmenides51
30629 posts
parmenides51
#1 h  Jul 30, 2021, 5:16 PM
Y by
Points $A_1,B_1,C_1$ are selected on the sides $BC$,$CA$,$AB$ respectively of an equilateral triangle $ABC$ in such a way that the inradii of the triangles $C_1AB_1$, $A_1BC_1$, $B_1CA_1$ and $A_1B_1C_1$ are equal. Prove that $A_1,B_1,C_1$ are the midpoints of the corresponding sides.
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