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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
28 minutes ago
aidan0626
22 minutes ago
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Geo to make por people happy
AlephG_64   1
N 35 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_
35 minutes ago
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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
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No more topics!
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Trapezoid - Austria 2011
Martin N.   4
N Sep 11, 2020 by A-Thought-Of-God
Let $k$ be a circle centered at $M$ and let $t$ be a tangentline to $k$ through some point $T\in k$. Let $P$ be a point on $t$ and let $g\neq t$ be a line through $P$ intersecting $k$ at $U$ and $V$. Let $S$ be the point on $k$ bisecting the arc $UV$ not containing $T$ and let $Q$ be the the image of $P$ under a reflection over $ST$.
Prove that $Q$, $T$, $U$ and $V$ are vertices of a trapezoid.
4 replies
Martin N.
Apr 14, 2011
A-Thought-Of-God
Sep 11, 2020
J
Trapezoid - Austria 2011
G H J
Reveal topic tags
geometry
trapezoid
geometric transformation
reflection
LaTeX
geometry proposed
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Martin N.
434 posts
Martin N.
#1 h  Apr 14, 2011, 10:56 AM • 3 Y
Y by nguyendangkhoa17112003, Adventure10, Mango247
Let $k$ be a circle centered at $M$ and let $t$ be a tangentline to $k$ through some point $T\in k$. Let $P$ be a point on $t$ and let $g\neq t$ be a line through $P$ intersecting $k$ at $U$ and $V$. Let $S$ be the point on $k$ bisecting the arc $UV$ not containing $T$ and let $Q$ be the the image of $P$ under a reflection over $ST$.
Prove that $Q$, $T$, $U$ and $V$ are vertices of a trapezoid.
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exiri
11 posts
exiri
#2 h  Apr 14, 2011, 12:32 PM • 2 Y
Y by Adventure10, Mango247
too easy!
we must prove QT is parallel to UV
assume U is between P and V:
angles QTS=STP because Q is reflection of P
and angle STP=arc SUT/2 because t is tangent to k
assume UV and ST intersect at N.then angle TNP=(arc UT+arc SK)/2
and arc SK=arc SU so TNP=arc SUT/2
then angles STP=TNP and QT is parallel to UV
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Martin N.
434 posts
Martin N.
#3 h  Apr 14, 2011, 5:00 PM • 2 Y
Y by Adventure10, Mango247
Noone said it would be difficult. It is Austria's first round (out of three, basically) by the way, so the problems are really not supposed to be too hard.

You are new here, but you could at least try to use $\LaTeX$, though. ;)
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aops29
452 posts
aops29
#4 h  Dec 27, 2018, 12:00 PM • 2 Y
Y by Adventure10, Mango247
solution

I would really appreciate it if someone would help me improve the LaTeX in the iff part.
This post has been edited 1 time. Last edited by aops29, Dec 27, 2018, 12:01 PM
Reason: ;aslkdjfpasiodhf
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A-Thought-Of-God
454 posts
A-Thought-Of-God
#5 h  Sep 11, 2020, 8:25 AM
Y by
:wallbash: Darn I wasted my 20 min, misread it as Isosceles Trapezoid Solution
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