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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Sequences problem
BBNoDollar   3
N an hour ago by BBNoDollar
Source: Mathematical Gazette Contest
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
3 replies
BBNoDollar
Yesterday at 5:53 PM
BBNoDollar
an hour ago
Arbitrary point on BC and its relation with orthocenter
falantrng   33
N an hour ago by Thapakazi
Source: Balkan MO 2025 P2
In an acute-angled triangle \(ABC\), \(H\) be the orthocenter of it and \(D\) be any point on the side \(BC\). The points \(E, F\) are on the segments \(AB, AC\), respectively, such that the points \(A, B, D, F\) and \(A, C, D, E\) are cyclic. The segments \(BF\) and \(CE\) intersect at \(P.\) \(L\) is a point on \(HA\) such that \(LC\) is tangent to the circumcircle of triangle \(PBC\) at \(C.\) \(BH\) and \(CP\) intersect at \(X\). Prove that the points \(D, X, \) and \(L\) lie on the same line.

Proposed by Theoklitos Parayiou, Cyprus
33 replies
falantrng
Apr 27, 2025
Thapakazi
an hour ago
Inequality
lgx57   4
N an hour ago by GeoMorocco
Source: Own
$a,b,c>0,ab+bc+ca=1$. Prove that

$$\sum \sqrt{8ab+1} \ge 5$$
(I don't know whether the equality holds)
4 replies
lgx57
Yesterday at 3:14 PM
GeoMorocco
an hour ago
Rubber bands
v_Enhance   5
N an hour ago by lpieleanu
Source: OTIS Mock AIME 2024 #12
Let $\mathcal G_n$ denote a triangular grid of side length $n$ consisting of $\frac{(n+1)(n+2)}{2}$ pegs. Charles the Otter wishes to place some rubber bands along the pegs of $\mathcal G_n$ such that every edge of the grid is covered by exactly one rubber band (and no rubber band traverses an edge twice). He considers two placements to be different if the sets of edges covered by the rubber bands are different or if any rubber band traverses its edges in a different order. The ordering of which bands are over and under does not matter.
For example, Charles finds there are exactly $10$ different ways to cover $\mathcal G_2$ using exactly two rubber bands; the full list is shown below, with one rubber band in orange and the other in blue.
IMAGE
Let $N$ denote the total number of ways to cover $\mathcal G_4$ with any number of rubber bands. Compute the remainder when $N$ is divided by $1000$.

Ethan Lee
5 replies
v_Enhance
Jan 16, 2024
lpieleanu
an hour ago
Geometry with orthocenter config
thdnder   6
N 2 hours ago by ohhh
Source: Own
Let $ABC$ be a triangle, and let $AD, BE, CF$ be its altitudes. Let $H$ be its orthocenter, and let $O_B$ and $O_C$ be the circumcenters of triangles $AHC$ and $AHB$. Let $G$ be the second intersection of the circumcircles of triangles $FDO_B$ and $EDO_C$. Prove that the lines $DG$, $EF$, and $A$-median of $\triangle ABC$ are concurrent.
6 replies
1 viewing
thdnder
Apr 29, 2025
ohhh
2 hours ago
Strange Inequality
anantmudgal09   40
N 2 hours ago by starchan
Source: INMO 2020 P4
Let $n \geqslant 2$ be an integer and let $1<a_1 \le a_2 \le \dots \le a_n$ be $n$ real numbers such that $a_1+a_2+\dots+a_n=2n$. Prove that$$a_1a_2\dots a_{n-1}+a_1a_2\dots a_{n-2}+\dots+a_1a_2+a_1+2 \leqslant a_1a_2\dots a_n.$$
Proposed by Kapil Pause
40 replies
anantmudgal09
Jan 19, 2020
starchan
2 hours ago
Finding Solutions
MathStudent2002   22
N 2 hours ago by ihategeo_1969
Source: Shortlist 2016, Number Theory 5
Let $a$ be a positive integer which is not a perfect square, and consider the equation \[k = \frac{x^2-a}{x^2-y^2}.\]Let $A$ be the set of positive integers $k$ for which the equation admits a solution in $\mathbb Z^2$ with $x>\sqrt{a}$, and let $B$ be the set of positive integers for which the equation admits a solution in $\mathbb Z^2$ with $0\leq x<\sqrt{a}$. Show that $A=B$.
22 replies
MathStudent2002
Jul 19, 2017
ihategeo_1969
2 hours ago
USAMO 2000 Problem 3
MithsApprentice   10
N 2 hours ago by HamstPan38825
A game of solitaire is played with $R$ red cards, $W$ white cards, and $B$ blue cards. A player plays all the cards one at a time. With each play he accumulates a penalty. If he plays a blue card, then he is charged a penalty which is the number of white cards still in his hand. If he plays a white card, then he is charged a penalty which is twice the number of red cards still in his hand. If he plays a red card, then he is charged a penalty which is three times the number of blue cards still in his hand. Find, as a function of $R, W,$ and $B,$ the minimal total penalty a player can amass and all the ways in which this minimum can be achieved.
10 replies
MithsApprentice
Oct 1, 2005
HamstPan38825
2 hours ago
Hard limits
Snoop76   7
N 2 hours ago by MihaiT
$a_n$ and $b_n$ satisfies the following recursion formulas: $a_{0}=1, $ $b_{0}=1$, $ a_{n+1}=a_{n}+b_{n}$$ $ and $ $$ b_{n+1}=(2n+3)b_{n}+a_{n}$. Find $ \lim_{n \to \infty} \frac{a_n}{(2n-1)!!}$ $ $ and $ $ $\lim_{n \to \infty} \frac{b_n}{(2n+1)!!}.$
7 replies
Snoop76
Mar 25, 2025
MihaiT
2 hours ago
Additive combinatorics (re Cauchy-Davenport)
mavropnevma   3
N 2 hours ago by Orzify
Source: Romania TST 3 2010, Problem 4
Let $X$ and $Y$ be two finite subsets of the half-open interval $[0, 1)$ such that $0 \in X \cap Y$ and $x + y = 1$ for no $x \in X$ and no $y \in Y$. Prove that the set $\{x + y - \lfloor x + y \rfloor : x \in X \textrm{ and } y \in Y\}$ has at least $|X| + |Y| - 1$ elements.

***
3 replies
mavropnevma
Aug 25, 2012
Orzify
2 hours ago
Ducks can play games now apparently
MortemEtInteritum   34
N 3 hours ago by HamstPan38825
Source: USA TST(ST) 2020 #1
Let $a$, $b$, $c$ be fixed positive integers. There are $a+b+c$ ducks sitting in a
circle, one behind the other. Each duck picks either rock, paper, or scissors, with $a$ ducks
picking rock, $b$ ducks picking paper, and $c$ ducks picking scissors.
A move consists of an operation of one of the following three forms:

[list]
[*] If a duck picking rock sits behind a duck picking scissors, they switch places.
[*] If a duck picking paper sits behind a duck picking rock, they switch places.
[*] If a duck picking scissors sits behind a duck picking paper, they switch places.
[/list]
Determine, in terms of $a$, $b$, and $c$, the maximum number of moves which could take
place, over all possible initial configurations.
34 replies
MortemEtInteritum
Nov 16, 2020
HamstPan38825
3 hours ago
Floor sequence
va2010   87
N 3 hours ago by Mathgloggers
Source: 2015 ISL N1
Determine all positive integers $M$ such that the sequence $a_0, a_1, a_2, \cdots$ defined by \[ a_0 = M + \frac{1}{2}   \qquad  \textrm{and} \qquad    a_{k+1} = a_k\lfloor a_k \rfloor   \quad \textrm{for} \, k = 0, 1, 2, \cdots \]contains at least one integer term.
87 replies
va2010
Jul 7, 2016
Mathgloggers
3 hours ago
INMO 2019 P3
div5252   45
N 3 hours ago by anudeep
Let $m,n$ be distinct positive integers. Prove that
$$gcd(m,n) + gcd(m+1,n+1) + gcd(m+2,n+2) \le 2|m-n| + 1. $$Further, determine when equality holds.
45 replies
div5252
Jan 20, 2019
anudeep
3 hours ago
My unsolved problem
ZeltaQN2008   1
N 3 hours ago by Adywastaken
Source: Belarus 2017
Find all funcition $f:(0,\infty)\rightarrow (0,\infty)$ such that for all any $x,y\in (0,\infty)$ :
$f(x+f(xy))=xf(1+f(y))$
1 reply
ZeltaQN2008
4 hours ago
Adywastaken
3 hours ago
hard problem
Cobedangiu   18
N Apr 15, 2025 by giangtruong13
problem
18 replies
Cobedangiu
Mar 27, 2025
giangtruong13
Apr 15, 2025
hard problem
G H J
G H BBookmark kLocked kLocked NReply
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Cobedangiu
66 posts
#1 • 1 Y
Y by PikaPika999
problem
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Cobedangiu
66 posts
#2 • 1 Y
Y by PikaPika999
no one? .
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Alex-131
5378 posts
#3 • 1 Y
Y by PikaPika999
Wrong solution in-english class, will check out later
This post has been edited 5 times. Last edited by Alex-131, Mar 27, 2025, 4:54 PM
Reason: r
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ND_
51 posts
#4 • 1 Y
Y by PikaPika999
When is equality achieved? a=b=c doesn't satisfy the initial condition. I think minimum is 10.
This post has been edited 2 times. Last edited by ND_, Mar 27, 2025, 4:50 PM
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Nuran2010
82 posts
#7 • 2 Y
Y by PikaPika999, TunarHasanzade
https://artofproblemsolving.com/community/c6h1834400p23600351
Also posted here
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Cobedangiu
66 posts
#8 • 1 Y
Y by PikaPika999
ND_ wrote:
When is equality achieved? a=b=c doesn't satisfy the initial condition. I think minimum is 10.

with equality occurring when $a=2, b=1,c=1$
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ND_
51 posts
#10 • 1 Y
Y by PikaPika999
\( b+c \ge 2\sqrt{bc} \iff \frac{b+c}{bc} \ge \frac{4}{b+c} \iff \) $\frac{1}{b} + \frac{1}{c} \ge \frac{2}{b+c} + \frac{2}{b+c}$
\( \Rightarrow (a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \ge \left(a+\frac{b+c}{2}+\frac{b+c}{2}\right)\left(\frac{1}{a} + \frac{2}{b+c} + \frac{2}{b+c}\right) \)

Hence, minimum occurs at \( b=c \). Plugging onto original equation, we get \( 2 + \left(\frac{b}{a}\right)^3=5\left(\frac{b}{a}\right) \), or \( b=2a \). So, minimum of 10 occurs at \( (2b, b, b) \).
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EaZ_Shadow
1265 posts
#11 • 1 Y
Y by PikaPika999
I used Cauchy and got $(a+b+c)(\frac1a+\frac1b+\frac1c)\geq9$ so i think 10 is the min, and I dont think 9 is achievable.
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xytunghoanh
33 posts
#12 • 1 Y
Y by PikaPika999
Cobedangiu wrote:
problem

cmath?
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Cobedangiu
66 posts
#13 • 2 Y
Y by xytunghoanh, PikaPika999
xytunghoanh wrote:
Cobedangiu wrote:
problem

cmath?
yes, but this inequality is from someone else, just took it back from that person.
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xytunghoanh
33 posts
#14 • 1 Y
Y by PikaPika999
I can help you to solve this. Check pm for Vietnamese solution.
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Cobedangiu
66 posts
#15 • 2 Y
Y by xytunghoanh, PikaPika999
xytunghoanh wrote:
I can help you to solve this. Check pm for Vietnamese solution.
I know how to do it
This post has been edited 1 time. Last edited by Cobedangiu, Mar 27, 2025, 5:27 PM
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sqing
41989 posts
#16 • 1 Y
Y by PikaPika999
Cobedangiu wrote:
problem
Let $a,b,c>0 $ and $a^3+b^3+c^3=5abc.$ Prove that$$10 \leq (a+b+c)\left(\frac {1}{a}+\frac {1}{b}+\frac {1}{c}\right)\leq5+4\sqrt{2}$$https://artofproblemsolving.com/community/c6h3455063p33365029
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InftyByond
205 posts
#17 • 2 Y
Y by Maximilian113, PikaPika999
k this is SUPER hard
why is inequality not satisfied by a=b=c?
suggests something funny like calculus

edit: ok im back with a fake solve!!!!!!!
Solve:
Expand $$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$$to get $$\sum (1+\frac{a}{b}+\frac{b}{a})$$into which we now substitute $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$.
this now gives us $$\sum (1+x+\frac{1}{x})$$which taking the double derivative gives that this function is convex.
so we can apply Karamata (smoothing kind of) to give us that equality case is where two of $x, y, z$ are equal and consequently two of $a, b, c$ equal.
note that scaling doesn't change anything in the inequality so you homogenize by setting idk $a+2c=4$ ;) and then going back into the condition
we should get $$a=2, b=1, c=1$$must be the minimum and consequently the minimum value is achieved at 10????????????
This post has been edited 1 time. Last edited by InftyByond, Apr 6, 2025, 5:34 AM
Reason: fake proof issues die hard
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Cobedangiu
66 posts
#18 • 1 Y
Y by PikaPika999
InftyByond wrote:
k this is SUPER hard
why is inequality not satisfied by a=b=c?
suggests something funny like calculus

edit: ok im back with a fake solve!!!!!!!
Solve:
Expand $$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$$to get $$\sum (1+\frac{a}{b}+\frac{b}{a})$$into which we now substitute $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$.
this now gives us $$\sum (1+x+\frac{1}{x})$$which taking the double derivative gives that this function is convex.
so we can apply Karamata (smoothing kind of) to give us that equality case is where two of $x, y, z$ are equal and consequently two of $a, b, c$ equal.
note that scaling doesn't change anything in the inequality so you homogenize by setting idk $a+2c=4$ ;) and then going back into the condition
we should get $$a=2, b=1, c=1$$must be the minimum and consequently the minimum value is achieved at 10????????????

The problem has been solved ;) , but the solution will not be posted here :P
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Nguyenhuyen_AG
3320 posts
#19 • 3 Y
Y by PikaPika999, anduran, truongphatt2668
Cobedangiu wrote:
problem
See here: https://nguyenhuyenag.wordpress.com/2025/04/06/inequality-43/
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InftyByond
205 posts
#20 • 1 Y
Y by PikaPika999
Cobedangiu wrote:
InftyByond wrote:
k this is SUPER hard
why is inequality not satisfied by a=b=c?
suggests something funny like calculus

edit: ok im back with a fake solve!!!!!!!
Solve:
Expand $$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$$to get $$\sum (1+\frac{a}{b}+\frac{b}{a})$$into which we now substitute $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$.
this now gives us $$\sum (1+x+\frac{1}{x})$$which taking the double derivative gives that this function is convex.
so we can apply Karamata (smoothing kind of) to give us that equality case is where two of $x, y, z$ are equal and consequently two of $a, b, c$ equal.
note that scaling doesn't change anything in the inequality so you homogenize by setting idk $a+2c=4$ ;) and then going back into the condition
we should get $$a=2, b=1, c=1$$must be the minimum and consequently the minimum value is achieved at 10????????????

The problem has been solved ;) , but the solution will not be posted here :P

K so i looked at the solve
Its pretty nice and slick with no calculus
Is my alternate solve wrong then? not surprised
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DKI
32 posts
#21
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we have
\[
(a+b-c)(a+b-2c+2\sqrt2c)(a+b-2c-2\sqrt2c)=4(a^3+b^3+c^3-5abc)-(a-b)^2(3a+3b+5c)
\]hence $a+b\ge c$, and similarly $a+c\ge b$, $b+c\ge a$,
\[
T=(a+b+c)(\frac1a+\frac1b+\frac1c)=10+\frac{a^3+b^3+c^3-5abc+(a+b-c)(a+c-b)(b+c-a)}{abc}\ge10
\]
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giangtruong13
141 posts
#22
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See this https://artofproblemsolving.com/community/c6h1834400p23632179 for more information
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