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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.

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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
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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
J
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AIME qual outside US?
daijobu   8
N an hour ago by qwedsazxc
Can students outside the US take the AIME if they earn a qualifying score?
8 replies
1 viewing
daijobu
Yesterday at 7:10 PM
qwedsazxc
an hour ago
J
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[$10K+ IN PRIZES] Poolesville Math Tournament (PVMT) 2025
qwerty123456asdfgzxcvb   20
N an hour ago by panda2018
Hi everyone!

After the resounding success of the first three years of PVMT, the Poolesville High School Math Team is excited to announce the fourth annual Poolesville High School Math Tournament (PVMT)! The PVMT team includes a MOPper and multiple USA(J)MO and AIME qualifiers!

PVMT is open to all 6th-9th graders in the country (including rising 10th graders). Students will compete in teams of up to 4 people, and each participant will take three subject tests as well as the team round. The contest is completely free, and will be held virtually on June 7, 2025, from 10:00 AM to 4:00 PM (EST).

Additionally, thanks to our sponsors, we will be awarding approximately $10K+ worth of prizes (including gift cards, Citadel merch, AoPS coupons, Wolfram licenses) to top teams and individuals. More details regarding the actual prizes will be released as we get closer to the competition date.

Further, newly for this year we might run some interesting mini-events, which we will announce closer to the competition date, such as potentially a puzzle hunt and integration bee!

If you would like to register for the competition, the registration form can be found at https://pvmt.org/register.html or https://tinyurl.com/PVMT25.

Additionally, more information about PVMT can be found at https://pvmt.org

If you have any questions not answered in the below FAQ, feel free to ask in this thread or email us at falconsdomath@gmail.com!

We look forward to your participation!

FAQ
20 replies
qwerty123456asdfgzxcvb
Apr 5, 2025
panda2018
an hour ago
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<BAC = 2 <ABC wanted, AC + AI = BC given , incenter I
parmenides51   3
N 2 hours ago by LeYohan
Source: 2020 Dutch IMO TST 1.1
In acute-angled triangle $ABC, I$ is the center of the inscribed circle and holds $| AC | + | AI | = | BC |$. Prove that $\angle BAC = 2 \angle ABC$.
3 replies
parmenides51
Nov 21, 2020
LeYohan
2 hours ago
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China South East Mathematical Olympiad 2014 Q3B
sqing   5
N 2 hours ago by MathLuis
Source: China Zhejiang Fuyang , 27 Jul 2014
Let $p$ be a primes ,$x,y,z $ be positive integers such that $x<y<z<p$ and $\{\frac{x^3}{p}\}=\{\frac{y^3}{p}\}=\{\frac{z^3}{p}\}$.
Prove that $(x+y+z)|(x^5+y^5+z^5).$
5 replies
sqing
Aug 17, 2014
MathLuis
2 hours ago
J
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Parallelograms and concyclicity
Lukaluce   32
N 2 hours ago by v_Enhance
Source: EGMO 2025 P4
Let $ABC$ be an acute triangle with incentre $I$ and $AB \neq AC$. Let lines $BI$ and $CI$ intersect the circumcircle of $ABC$ at $P \neq B$ and $Q \neq C$, respectively. Consider points $R$ and $S$ such that $AQRB$ and $ACSP$ are parallelograms (with $AQ \parallel RB, AB \parallel QR, AC \parallel SP$, and $AP \parallel CS$). Let $T$ be the point of intersection of lines $RB$ and $SC$. Prove that points $R, S, T$, and $I$ are concyclic.
32 replies
Lukaluce
Apr 14, 2025
v_Enhance
2 hours ago
J
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Expression is a Cube
nosaj   38
N 2 hours ago by NicoN9
Source: 2015 AIME I Problem 3
There is a prime number $p$ such that $16p+1$ is the cube of a positive integer. Find $p$.
38 replies
nosaj
Mar 20, 2015
NicoN9
2 hours ago
J
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Gcd of N and its coprime pair sum
EeEeRUT   18
N 2 hours ago by lksb
Source: EGMO 2025 P1
For a positive integer $N$, let $c_1 < c_2 < \cdots < c_m$ be all positive integers smaller than $N$ that are coprime to $N$. Find all $N \geqslant 3$ such that $$\gcd( N, c_i + c_{i+1}) \neq 1$$for all $1 \leqslant i \leqslant m-1$

Here $\gcd(a, b)$ is the largest positive integer that divides both $a$ and $b$. Integers $a$ and $b$ are coprime if $\gcd(a, b) = 1$.

Proposed by Paulius Aleknavičius, Lithuania
18 replies
EeEeRUT
Apr 16, 2025
lksb
2 hours ago
J
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Easy right-angled triangle problem
gghx   7
N 2 hours ago by LeYohan
Source: SMO open 2024 Q1
In triangle $ABC$, $\angle B=90^\circ$, $AB>BC$, and $P$ is the point such that $BP=BC$ and $\angle APB=90^\circ$, where $P$ and $C$ lie on the same side of $AB$. Let $Q$ be the point on $AB$ such that $AP=AQ$, and let $M$ be the midpoint of $QC$. Prove that the line through $M$ parallel to $AP$ passes through the midpoint of $AB$.
7 replies
gghx
Aug 3, 2024
LeYohan
2 hours ago
J
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EGMO (geo) Radical Center Question
gulab_jamun   9
N 3 hours ago by MathRook7817
For this theorem, Evan says that the power of point $P$ with respect to $\omega_1$ is greater than 0 if $P$ lies between $A$ and $B$. (I've underlined it). But, I'm a little confused as I thought the power was $OP^2 - r^2$ and since $P$ is inside the circle, wouldn't the power be negative since $OP < r$?
9 replies
gulab_jamun
May 25, 2025
MathRook7817
3 hours ago
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ai+aj is the multiple of n
Jackson0423   2
N 3 hours ago by Jackson0423
Consider an strictly increasing sequence of integers \( a_n \).
For every positive integer \( n \), there exist indices \( 1 \leq i < j \leq n \) such that \( a_i + a_j \) is divisible by \( n \).
Given that \( a_1 \geq 1 \), find the minimum possible value of \( a_{100} \).
2 replies
Jackson0423
Yesterday at 12:41 AM
Jackson0423
3 hours ago
J
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Triple mixtilinear then sum of segments
Noob_at_math_69_level   5
N 3 hours ago by awesomeming327.
Source: DGO 2023 Team P4
Let $\triangle{ABC}$ be an acute triangle with the $A,B,C-$mixtilinear incircles are $\Omega_A,\Omega_B,\Omega_C$ respectively. $\Omega_A$ is tangent to the circumcircle of $\triangle{ABC}$ at $X$. $O_2,O_3$ are the centers of circles $\Omega_B,\Omega_C$ respectively. Suppose the reflection of line $BX$ over $BO_3$ intersects the reflection of line $CX$ over $CO_2$ at point $S.$ Prove that: $BS+BX=CS+CX.$

Proposed by many authors
5 replies
Noob_at_math_69_level
Dec 18, 2023
awesomeming327.
3 hours ago
J
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Inspired by a cool result
DoThinh2001   0
3 hours ago
Source: Old?
Let three real numbers $a,b,c\geq 0$, no two of which are $0$. Prove that:
$$\sqrt{\frac{a^2+bc}{b^2+c^2}}+\sqrt{\frac{b^2+ca}{c^2+a^2}}+\sqrt{\frac{c^2+ab}{a^2+b^2}}\geq 2+\sqrt{\frac{ab+bc+ca}{a^2+b^2+c^2}}.$$
Inspiration
0 replies
DoThinh2001
3 hours ago
0 replies
J
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Basic ideas in junior diophantine equations
Maths_VC   4
N 4 hours ago by TopGbulliedU
Source: Serbia JBMO TST 2025, Problem 3
Determine all positive integers $a, b$ and $c$ such that
$2$ $\cdot$ $10^a + 5^b = 2025^c$
4 replies
Maths_VC
May 27, 2025
TopGbulliedU
4 hours ago
J
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Iran TST Starter
M11100111001Y1R   8
N 4 hours ago by flower417477
Source: Iran TST 2025 Test 1 Problem 1
Let \( a_n \) be a sequence of positive real numbers such that for every \( n > 2025 \), we have:
\[ a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i} \]Prove that there exists a natural number \( M \) such that for all \( n > M \), the following holds:
\[ a_n < \frac{1}{1404} \]
8 replies
M11100111001Y1R
May 27, 2025
flower417477
4 hours ago
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J
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five digit multiplication?
fruitmonster97   48
N Apr 4, 2025 by Apple_maths60
Source: 2024 AMC 10A #1/AMC 12A #1
What is the value of $9901\cdot101-99\cdot10101?$

$\textbf{(A) }2\qquad\textbf{(B) }20\qquad\textbf{(C) }21\qquad\textbf{(D) }200\qquad\textbf{(E) }2020$
48 replies
fruitmonster97
Nov 7, 2024
Apple_maths60
Apr 4, 2025
J
five digit multiplication?
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Reveal topic tags
AMC
AMC 10
2024 AMC 10A
2024 AMC 12A
AMC 12
2024 AMC
L
G H BBookmark kLocked kLocked NReply
Source: 2024 AMC 10A #1/AMC 12A #1
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xHypotenuse
788 posts
xHypotenuse
#36 h  Nov 7, 2024, 6:20 PM
Y by
SomeonecoolLovesMaths wrote:
A confirmed

I put 2 but why was my answer choice E for that one
Z K Y
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scrabbler94
7555 posts
scrabbler94
#37 h  Nov 7, 2024, 6:27 PM
Y by
Modulo 10 kills this problem, since $9901 \cdot 101 - 99 \cdot 10101 \equiv 1 - 9 \equiv 2 \pmod{10}$. Only (A) has units digit 2.
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wangzrpi
159 posts
wangzrpi
#38 h  Nov 7, 2024, 6:35 PM
Y by
ayush_agarwal wrote:
Yeah you can just look at the numbers mod 10 and just see that the last digit must be 2

Nice solution. I rounded the products and estimated
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TetraFish
1088 posts
TetraFish
#39 h  Nov 7, 2024, 7:11 PM
Y by
Pretty satisfying p1. $9901(101) - 99(9901 + 200) = 2(9901) - 99(200) = 2(9901) - 2(9900) = 2(1) = \boxed{2}$, so A.
Z K Y
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SkatingKitty
223 posts
SkatingKitty
#40 h  Nov 7, 2024, 7:52 PM
Y by
One problem I got right
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akqbstr
1 post
akqbstr
#41 h  Nov 7, 2024, 7:53 PM
Y by
Just look at the last digit.
9901*101, last digit is 1
99*10101, last digit 9
subtract them, last digit has to be 2
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YunboZ
2 posts
YunboZ
#42 h  Nov 7, 2024, 10:24 PM
Y by
A, I bashed it too
Z K Y
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gicyuraok2
1059 posts
gicyuraok2
#43 h  Nov 7, 2024, 11:17 PM
Y by
ooh new amc drop i guess i'll try all the problems

anyway free problem just compute and your good for A
Z K Y
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Existing_Human1
214 posts
Existing_Human1
#44 h  Nov 7, 2024, 11:45 PM
Y by
POV: You somehow fail to learn how to subtract and multiply, so you have to skip this problem. Come back to it, still forget how to multiply and subtract, and finally get it halfway through the test (I'm not actually that bad, just silly a bunch)
Z K Y
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Catcumber
166 posts
Catcumber
#45 h  Nov 8, 2024, 1:27 AM
Y by
why was p1 lowkey time consuming...
i almost got -899998 cuz i somehow copied down 1000001 as 100001 lol
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brainfertilzer
1831 posts
brainfertilzer
#46 h  Nov 8, 2024, 9:15 PM
Y by
Unironically kind of hard becuase i didn't see mod 10 oops. $990100 + 9901 - 1010100 + 10101 = 20002 - 20000 =\boxed{2}$.
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navier3072
121 posts
navier3072
#47 h  Nov 9, 2024, 8:26 AM
Y by
Since $9900 \cdot 101 = 99 \cdot 10100 = 99 \cdot 10 \cdot 101$, we get $101-99=2$ or $\textbf{(A) }$
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Challengees24
1092 posts
Challengees24
#48 h  Nov 9, 2024, 7:23 PM
Y by
Countmath1 wrote:
Write $9901= 10,000 - 99$ and $10101 = 10000 + 101$, expand, simplify: $\textbf{(A)\ 2}.$

what i did

though just bashing was prob as easy
This post has been edited 1 time. Last edited by Challengees24, Nov 9, 2024, 7:23 PM
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Mr.Sharkman
501 posts
Mr.Sharkman
#49 h  Dec 11, 2024, 12:43 PM
Y by
We have, if $x= 10,$
$$(x^{4}-x^{2}+1)(x^{2}+1)-(x^{4}+x^{2}+1)(x^{2}-1) = x^{6}+1-(x^{6}-1) = 2.$$
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Apple_maths60
27 posts
Apple_maths60
#51 h  Apr 4, 2025, 3:15 PM
Y by
Extremely trivial
Answer is 2
This post has been edited 1 time. Last edited by Apple_maths60, Apr 4, 2025, 4:37 PM
Reason: .
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