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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
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0 replies
jlacosta
Mar 2, 2025
0 replies
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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
J
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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
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funny title placeholder
pikapika007   46
N 5 minutes ago by aliz
Source: USAJMO 2025/6
Let $S$ be a set of integers with the following properties:
[list]
[*] $\{ 1, 2, \dots, 2025 \} \subseteq S$.
[*] If $a, b \in S$ and $\gcd(a, b) = 1$, then $ab \in S$.
[*] If for some $s \in S$, $s + 1$ is composite, then all positive divisors of $s + 1$ are in $S$.
[/list]
Prove that $S$ contains all positive integers.
46 replies
pikapika007
Yesterday at 12:10 PM
aliz
5 minutes ago
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Scary Binomial Coefficient Sum
EpicBird08   31
N 18 minutes ago by john0512
Source: USAMO 2025/5
Determine, with proof, all positive integers $k$ such that $$\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k$$is an integer for every positive integer $n.$
31 replies
EpicBird08
Yesterday at 11:59 AM
john0512
18 minutes ago
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sequence and number theory
srnjbr   0
an hour ago
Let a1 be a member of the integers and an+1=an^2-an-1. Show that (an+1,2n+1)=1
0 replies
srnjbr
an hour ago
0 replies
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2022 Junior Balkan MO, Problem 1
sarjinius   25
N an hour ago by anudeep
Source: 2022 JBMO Problem 1
Find all pairs of positive integers $(a, b)$ such that $$11ab \le a^3 - b^3 \le 12ab.$$
25 replies
sarjinius
Jun 30, 2022
anudeep
an hour ago
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Mathroots
Ruegerbyrd   0
an hour ago
Has anyone gotten acceptances from MIT's Mathroots yet? Did they ever say they wouldn't send letters to anyone unless accepted?
0 replies
Ruegerbyrd
an hour ago
0 replies
J
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Nice function question
srnjbr   0
an hour ago
Find all functions f:R+--R+ such that for all a,b>0, f(af(b)+a)(f(bf(a))+a)=1
0 replies
srnjbr
an hour ago
0 replies
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[TEST RELEASED] Mock Geometry Test for College Competitions
Bluesoul   16
N an hour ago by lord_of_the_rook
Hi AOPSers,

I have finished writing a mock geometry test for fun and practice for the real college competitions like HMMT/PUMaC/CMIMC... There would be 10 questions and you should finish the test in 60 minutes, the test would be close to the actual test (hopefully). You could sign up under this thread, PM me your answers!. The submission would close on March 31st at 11:59PM PST.

I would create a private discussion forum so everyone could discuss after finishing the test. This is the first mock I've written, please sign up and enjoy geometry!!

~Bluesoul

Leaderboard
16 replies
Bluesoul
Feb 24, 2025
lord_of_the_rook
an hour ago
J
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Find min
hunghd8   6
N an hour ago by imnotgoodatmathsorry
Let $a,b,c$ be nonnegative real numbers such that $ a+b+c\geq 2+abc $. Find min
$$P=a^2+b^2+c^2.$$
6 replies
hunghd8
Yesterday at 12:10 PM
imnotgoodatmathsorry
an hour ago
J
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Interesting inequality
sqing   1
N an hour ago by ionbursuc
Source: Own
Let $ a,b> 0$ and $ a+b=1 . $ Prove that
$$ \frac{1}{a}+\frac{1}{b}\geq \frac{2k}{1+k^2 a^2b^2}$$Where $ 5\leq k\in N^+.$
1 reply
sqing
2 hours ago
ionbursuc
an hour ago
J
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Integral with dt
RenheMiResembleRice   0
2 hours ago
Source: Yanxue Lu
Solve the attached:
0 replies
RenheMiResembleRice
2 hours ago
0 replies
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Inequality
srnjbr   1
N 2 hours ago by sqing
a^2+b^2+c^2+x^2+y^2=1. Find the maximum value of the expression (ax+by)^2+(bx+cy)^2
1 reply
1 viewing
srnjbr
Yesterday at 4:32 PM
sqing
2 hours ago
J
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9 Three concurrent chords
v_Enhance   3
N 3 hours ago by ohiorizzler1434
Three distinct circles $\Omega_1$, $\Omega_2$, $\Omega_3$ cut three common chords concurrent at $X$. Consider two distinct circles $\Gamma_1$, $\Gamma_2$ which are internally tangent to all $\Omega_i$. Determine, with proof, which of the following two statements is true.

(1) $X$ is the insimilicenter of $\Gamma_1$ and $\Gamma_2$
(2) $X$ is the exsimilicenter of $\Gamma_1$ and $\Gamma_2$.
3 replies
v_Enhance
Yesterday at 8:45 PM
ohiorizzler1434
3 hours ago
J
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Mathhhhh
mathbetter   9
N 3 hours ago by ohiorizzler1434
Three turtles are crawling along a straight road heading in the same
direction. "Two other turtles are behind me," says the first turtle. "One turtle is
behind me and one other is ahead," says the second. "Two turtles are ahead of me
and one other is behind," says the third turtle. How can this be possible?
9 replies
mathbetter
Thursday at 11:21 AM
ohiorizzler1434
3 hours ago
J
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An inequality about xy+yz+zx+2xyz=1
jokehim   0
4 hours ago
Source: my problem
Problem. Let $x,y,z>0: xy+yz+zx+2xyz=1.$ Prove that$$\frac{1}{6x+1}+\frac{1}{6y+1}+\frac{1}{6z+1}\ge \frac{9(xy+yz+zx)-3}{5}.$$Proposed by Phan Ngoc Chau
0 replies
jokehim
4 hours ago
0 replies
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2025 AMC 8 Problem
Kexinshi   8
N Mar 15, 2025 by CJB19
Source: 2025 AMC 8 Problem #15
Kei draws a $6$-by-$6$ grid. He colors $13$ of the unit squares silver and the remaining squares gold. Kei then folds the grid in half vertically, forming pairs of overlapping unit squares. Let $m$ and $M$ equal equal the least and greatest possible number of gold-on-gold pairs, respectively. What is the value of $m+M$?

$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 16\qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 20$

Had fun doing this one!
8 replies
Kexinshi
Jan 31, 2025
CJB19
Mar 15, 2025
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2025 AMC 8 Problem
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Reveal topic tags
AMC 8
2025 AMC 8
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G H BBookmark kLocked kLocked NReply
Source: 2025 AMC 8 Problem #15
The post below has been deleted. Click to close.
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Kexinshi
172 posts
Kexinshi
#1 h  Jan 31, 2025, 3:29 AM
Y by
Kei draws a $6$-by-$6$ grid. He colors $13$ of the unit squares silver and the remaining squares gold. Kei then folds the grid in half vertically, forming pairs of overlapping unit squares. Let $m$ and $M$ equal equal the least and greatest possible number of gold-on-gold pairs, respectively. What is the value of $m+M$?

$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 16\qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 20$

Had fun doing this one!
Z K Y
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PaperMath
958 posts
PaperMath
#2 h  Jan 31, 2025, 3:36 AM
Y by
I got this wrong
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RedFireTruck
4220 posts
RedFireTruck
#3 h  Jan 31, 2025, 3:42 AM
Y by
aight so we have $18$ pairs

these pairs are one of gold-gold, gold-silver, silver-silver

let $a$ be gold-gold, $b$ be gold-silver, $c$ be silver-silver

$a+b+c=18$ and $b+2c=13$

subtracting both equations gives $a-c=5$, so $a\ge 5$

we substitute to get $2a+b=23$ so $a\le 11$

our answer is $5+11=\boxed{16}$

the specific constructions are $(a,b,c)=(5,13,0)$ and $(a,b,c)=(11,1,6)$
Z K Y
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Gavin_Deng
755 posts
Gavin_Deng
#4 h  Jan 31, 2025, 4:14 AM
Y by
Wait I might have put 11+3. Shoot….
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KMSONI
143 posts
KMSONI
#5 h  Feb 1, 2025, 12:43 AM
Y by
Thank god, i put 16. WHat question number was this again?
Z K Y
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mithu542
1557 posts
mithu542
#6 h  Feb 1, 2025, 1:10 AM
Y by
15; it says at the top where it says the source
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Kexinshi
172 posts
Kexinshi
#7 h  Feb 1, 2025, 2:03 AM
Y by
@RedFireTruck
No offence, but you should probably hide your solutions so people don't get spoiled.
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sadas123
1066 posts
sadas123
#8 h  Feb 1, 2025, 2:16 AM
Y by
I got Answer for this question
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CJB19
3 posts
CJB19
#9 h  Mar 15, 2025, 3:00 AM
Y by
I remember this one! Don't read this if you want to solve it yourself (Idk how to hide stuff)! So basically, the most overlapping gold occurs when we maximize silver overlap. This can be don't by having 6 pairs of silver-on-silver and 1 pair of silver-on-gold. this leaves 18-(6+1)=11 pairs as our maximum. To minimize gold-on-gold, we want to maximize silver-on-gold. This gives 13-silver-on-gold, and 5 gold-on-gold. Therefore, our answer is 5+11=16
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