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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.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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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
Find (AB * CD) / (AC * BD) & prove orthogonality of circles
Maverick   15
N 41 minutes ago by Ilikeminecraft
Source: IMO 1993, Day 1, Problem 2
Let $A$, $B$, $C$, $D$ be four points in the plane, with $C$ and $D$ on the same side of the line $AB$, such that $AC \cdot BD = AD \cdot BC$ and $\angle ADB = 90^{\circ}+\angle ACB$. Find the ratio
\[\frac{AB \cdot CD}{AC \cdot BD}, \]
and prove that the circumcircles of the triangles $ACD$ and $BCD$ are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicuar. Thus, proving that the circumcircles of the triangles $ACD$ and $BCD$ are orthogonal is equivalent to proving that the tangents to the circumcircles of the triangles $ACD$ and $BCD$ at the point $C$ are perpendicular.)
15 replies
Maverick
Jul 13, 2004
Ilikeminecraft
41 minutes ago
f(x+f(x)+f(y))=x+f(x+y)
dangerousliri   10
N an hour ago by jasperE3
Source: FEOO, Shortlist A5
Find all functions $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ such that for any positive real numbers $x$ and $y$,
$$f(x+f(x)+f(y))=x+f(x+y)$$Proposed by Athanasios Kontogeorgis, Grecce, and Dorlir Ahmeti, Kosovo
10 replies
dangerousliri
May 31, 2020
jasperE3
an hour ago
n-variable inequality
ABCDE   66
N an hour ago by ND_
Source: 2015 IMO Shortlist A1, Original 2015 IMO #5
Suppose that a sequence $a_1,a_2,\ldots$ of positive real numbers satisfies \[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]for every positive integer $k$. Prove that $a_1+a_2+\ldots+a_n\geq n$ for every $n\geq2$.
66 replies
1 viewing
ABCDE
Jul 7, 2016
ND_
an hour ago
Euler Line Madness
raxu   75
N 2 hours ago by lakshya2009
Source: TSTST 2015 Problem 2
Let ABC be a scalene triangle. Let $K_a$, $L_a$ and $M_a$ be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of $AK_aL_a$ intersects $AM_a$ a second time at point $X_a$ different from A. Define $X_b$ and $X_c$ analogously. Prove that the circumcenter of $X_aX_bX_c$ lies on the Euler line of ABC.
(The Euler line of ABC is the line passing through the circumcenter, centroid, and orthocenter of ABC.)

Proposed by Ivan Borsenco
75 replies
raxu
Jun 26, 2015
lakshya2009
2 hours ago
Own made functional equation
Primeniyazidayi   8
N 2 hours ago by MathsII-enjoy
Source: own(probably)
Find all functions $f:R \rightarrow R$ such that $xf(x^2+2f(y)-yf(x))=f(x)^3-f(y)(f(x^2)-2f(x))$ for all $x,y \in \mathbb{R}$
8 replies
Primeniyazidayi
May 26, 2025
MathsII-enjoy
2 hours ago
IMO ShortList 2002, geometry problem 7
orl   110
N 3 hours ago by SimplisticFormulas
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$.
110 replies
orl
Sep 28, 2004
SimplisticFormulas
3 hours ago
Cute NT Problem
M11100111001Y1R   6
N 3 hours ago by X.Allaberdiyev
Source: Iran TST 2025 Test 4 Problem 1
A number \( n \) is called lucky if it has at least two distinct prime divisors and can be written in the form:
\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]where \( p_1, \dots, p_k \) are distinct prime numbers that divide \( n \). (Note: it is possible that \( n \) has other prime divisors not among \( p_1, \dots, p_k \).) Prove that for every prime number \( p \), there exists a lucky number \( n \) such that \( p \mid n \).
6 replies
M11100111001Y1R
May 27, 2025
X.Allaberdiyev
3 hours ago
China MO 2021 P6
NTssu   23
N 3 hours ago by bin_sherlo
Source: CMO 2021 P6
Find $f: \mathbb{Z}_+ \rightarrow \mathbb{Z}_+$, such that for any $x,y \in \mathbb{Z}_+$, $$f(f(x)+y)\mid x+f(y).$$
23 replies
NTssu
Nov 25, 2020
bin_sherlo
3 hours ago
Cyclic Sum
P162008   1
N 3 hours ago by aaravdodhia
If $\sum_{cyc} \alpha = 0$ and $\sum_{cyc} \frac{\alpha^4}{2\alpha^2 + \beta\gamma} = 1$ then find the greatest possible value of $\sum_{cyc}\alpha^4.$
1 reply
P162008
May 26, 2025
aaravdodhia
3 hours ago
Prove that the circumcentres of the triangles are collinear
orl   19
N 3 hours ago by Ilikeminecraft
Source: IMO Shortlist 1997, Q9
Let $ A_1A_2A_3$ be a non-isosceles triangle with incenter $ I.$ Let $ C_i,$ $ i = 1, 2, 3,$ be the smaller circle through $ I$ tangent to $ A_iA_{i+1}$ and $ A_iA_{i+2}$ (the addition of indices being mod 3). Let $ B_i, i = 1, 2, 3,$ be the second point of intersection of $ C_{i+1}$ and $ C_{i+2}.$ Prove that the circumcentres of the triangles $ A_1 B_1I,A_2B_2I,A_3B_3I$ are collinear.
19 replies
orl
Aug 10, 2008
Ilikeminecraft
3 hours ago
c^a + a = 2^b
Havu   9
N 3 hours ago by Havu
Find $a, b, c\in\mathbb{Z}^+$ such that $a,b,c$ coprime, $a + b = 2c$ and $c^a + a = 2^b$.
9 replies
Havu
May 10, 2025
Havu
3 hours ago
Original Problem
wonderboy807   1
N 4 hours ago by jasperE3
f(0)=f(1)=1. \frac{f(n)f(n-m+1)}{f(n-m)} + \frac{f(n+1)f(n-m)}{f(m-n)} = \frac{f(n+2)f(n-m)f(m-n)}{f(n-m+1)f(m-n+1)}. Find f(10).

Answer: Click to reveal hidden text

Solution: Click to reveal hidden text
1 reply
wonderboy807
Today at 11:36 AM
jasperE3
4 hours ago
Find x^2 + y^2
Darealzolt   3
N 4 hours ago by jasperE3
Let \(x,y\) be positive real numbers that fulfill
\[
\frac{x^2}{y^2}+\frac{4x^2-3xy-4y^2}{2xy-5y^2}=2
\]Hence find the value of \(x^2+y^2\)
3 replies
Darealzolt
Today at 11:45 AM
jasperE3
4 hours ago
Own Problem
BinariouslyRandom   1
N 4 hours ago by BinariouslyRandom
LuAn is a renowned treasure hunter from Muntinlupa City. One day while patrolling the jungles of Mindanao, he found a safe with the following problem written on it:
[quote]Convert \(50392420515\) (base-10) into base-\(n\), where \(n\) is the smallest integer that has 8 factors. The answer will be the code to the safe.[/quote]
What is the secret password?

Note: If n > 10, A = 11, B = 12, and so on.
1 reply
BinariouslyRandom
Yesterday at 11:21 AM
BinariouslyRandom
4 hours ago
Rubik's cube problem
ilikejam   20
N Mar 30, 2025 by jasperE3
If I have a solved Rubik's cube, and I make a finite sequence of (legal) moves repeatedly, prove that I will eventually resolve the puzzle.

(this wording is kinda goofy but i hope its sorta intuitive)
20 replies
ilikejam
Mar 28, 2025
jasperE3
Mar 30, 2025
Rubik's cube problem
G H J
G H BBookmark kLocked kLocked NReply
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ilikejam
34 posts
#1
Y by
If I have a solved Rubik's cube, and I make a finite sequence of (legal) moves repeatedly, prove that I will eventually resolve the puzzle.

(this wording is kinda goofy but i hope its sorta intuitive)
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jkim0656
1077 posts
#2
Y by
if u say "eventually," doesn't that mean an infinite number of moves?
I mean all rubik's cubes can be solved from any position in 20 moves, so with luck and infinite time you will get those 20 or less moves all correct.
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aidan0626
1964 posts
#3
Y by
oh i've thought about this before
imagine all possible rubik's cube positions as vertices, and put them on a directed graph, with an edge from one position to another if the sequence turns the first position into the second
note that every vertex has an indegree of 1 and outdegree of 1
now assume you can't resolve the puzzle, that means you get stuck in a cycle not containing the original position
but to get into such a cycle without the original position, there must be a vertex with an indegree greater than 1 (not sure how to rigorously show this, but inuitively makes sense), which is a contradiction
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kred9
1022 posts
#4
Y by
By the Pigeonhole Principle, there must be some state of the cube that is achieved twice after applying this algorithm arbitrarily many times. Therefore, repeating the algorithm some number of times brings a state back to itself, and hence it will bring the solved state back to itself.
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jkim0656
1077 posts
#5
Y by
kred9 wrote:
By the Pigeonhole Principle, there must be some state of the cube that is achieved twice after applying this algorithm arbitrarily many times. Therefore, repeating the algorithm some number of times brings a state back to itself, and hence it will bring the solved state back to itself.

dang well said
that's smart
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jasperE3
11394 posts
#6
Y by
Theorem 4.2.2 provides an upper limit with only face turns (standard HTM) of $1260$ moves.
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fruitmonster97
2506 posts
#7
Y by
kred9 wrote:
By the Pigeonhole Principle, there must be some state of the cube that is achieved twice after applying this algorithm arbitrarily many times. Therefore, repeating the algorithm some number of times brings a state back to itself, and hence it will bring the solved state back to itself.

I must be missing something, but what if(theoretically) every position except the solved state is in a loop? Yes, the first unsolved position will repeat, but that yields no insight on why the solved face should repeat?
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Coolmanppap3
1 post
#8
Y by
fruitmonster97 wrote:
kred9 wrote:
By the Pigeonhole Principle, there must be some state of the cube that is achieved twice after applying this algorithm arbitrarily many times. Therefore, repeating the algorithm some number of times brings a state back to itself, and hence it will bring the solved state back to itself.

I must be missing something, but what if(theoretically) every position except the solved state is in a loop? Yes, the first unsolved position will repeat, but that yields no insight on why the solved face should repeat?

In theory, that solution would take a long time, but it will happen, right? Imagine it like a circle. Every time you bend the circle to make another, the start still remains, even though there is a detour. In a sense, it would loop back, after many loops. (I have no idea btw)
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greenturtle3141
3562 posts
#9
Y by
The generalization of this is phrased in terms of group theory: "in a finite group, every element has a finite order". Here the group is the permutations you can make on a rubiks cube through legal moves. the order of a certain combination of moves is how many times it needs to be repeated to go back to the "nothing happened" outcome (called the "group identity").
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mpcnotnpc
55 posts
#10
Y by
kred9 wrote:
By the Pigeonhole Principle, there must be some state of the cube that is achieved twice after applying this algorithm arbitrarily many times. Therefore, repeating the algorithm some number of times brings a state back to itself, and hence it will bring the solved state back to itself.

Wait is this true? This only implies that some state might repeat, doesn't mean that the initial state will.
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vincentwant
1446 posts
#11
Y by
mpcnotnpc wrote:
kred9 wrote:
By the Pigeonhole Principle, there must be some state of the cube that is achieved twice after applying this algorithm arbitrarily many times. Therefore, repeating the algorithm some number of times brings a state back to itself, and hence it will bring the solved state back to itself.

Wait is this true? This only implies that some state might repeat, doesn't mean that the initial state will.

correct me if im wrong but if its not the initial state then its not a group (im rusty at group theory tho so)
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joeym2011
501 posts
#12
Y by
Redacted, misinterpreted problem
This post has been edited 1 time. Last edited by joeym2011, Mar 28, 2025, 11:18 PM
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kred9
1022 posts
#13
Y by
fruitmonster97 wrote:
kred9 wrote:
By the Pigeonhole Principle, there must be some state of the cube that is achieved twice after applying this algorithm arbitrarily many times. Therefore, repeating the algorithm some number of times brings a state back to itself, and hence it will bring the solved state back to itself.

I must be missing something, but what if(theoretically) every position except the solved state is in a loop? Yes, the first unsolved position will repeat, but that yields no insight on why the solved face should repeat?
mpcnotnpc wrote:
kred9 wrote:
By the Pigeonhole Principle, there must be some state of the cube that is achieved twice after applying this algorithm arbitrarily many times. Therefore, repeating the algorithm some number of times brings a state back to itself, and hence it will bring the solved state back to itself.

Wait is this true? This only implies that some state might repeat, doesn't mean that the initial state will.

I'll address these two concerns with my solution. Write the algorithm as $g$. Then suppose $g^a$ and $g^b$ correspond to the same state (some distinct values of $a$ and $b$ must satisfy this by PHP). Therefore, applying the algorithm $b-a$ times to a state returns that state back to itself. This is true no matter what state we are currently at, since every piece is unique on the cube. This means that after $b-a$ algorithms at the beginning, we will return to the solved state.
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ilikejam
34 posts
#14
Y by
my solution was that every square (the small ones, there are 9 on each side) moves to a certain position, and the square that was in that position moves to a new one, etc. this will eventually become a cycle, and even though there may be more than one such cycle, given enough repetitions, all the cycles will eventually loop back to the starting, and thus solved, state.
This post has been edited 1 time. Last edited by ilikejam, Mar 28, 2025, 11:46 PM
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ilikejam
34 posts
#15
Y by
jkim0656 wrote:
if u say "eventually," doesn't that mean an infinite number of moves?
I mean all rubik's cubes can be solved from any position in 20 moves, so with luck and infinite time you will get those 20 or less moves all correct.

no i mean i have a certain sequence of moves, maybe like ULU that i repeat until it solves itself
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mpcnotnpc
55 posts
#16
Y by
oop im not familiar with rubik's cube rules, but "legal" moves don't incorporate like turning a side back and forth right; cuz that's like a finite sequence?
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jasperE3
11394 posts
#17
Y by
If you turn a side back and forth by $90^\circ$ and then $-90^\circ$ then trivially the puzzle gets resolved after applying this finite sequence of legal moves.
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wuwang2002
1215 posts
#18
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jasperE3 wrote:
Theorem 4.2.2 provides an upper limit with only face turns (standard HTM) of $1260$ moves.

when i was small i counted all 1260 moves and surprisingly didn’t mess up (so much dedication)
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mpcnotnpc
55 posts
#19 • 1 Y
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jasperE3 wrote:
If you turn a side back and forth by $90^\circ$ and then $-90^\circ$ then trivially the puzzle gets resolved after applying this finite sequence of legal moves.

:skull: mb
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ilikejam
34 posts
#20
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some clarifications:
mpcnotnpc wrote:
oop im not familiar with rubik's cube rules, but "legal" moves don't incorporate like turning a side back and forth right; cuz that's like a finite sequence?

no i mean like no breaking the cube, only twisting the sides, back and forth is ok.
jasperE3 wrote:
If you turn a side back and forth by $90^\circ$ and then $-90^\circ$ then trivially the puzzle gets resolved after applying this finite sequence of legal moves.

yes, however the problem is to prove that every sequence eventually solves the cube, not to find just one.
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jasperE3
11394 posts
#21
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I was responding to #16 not the original post.
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