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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.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 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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Geometry
MathsII-enjoy   1
N 29 minutes ago by aidenkim119
Given triangle $ABC$ inscribed in $(O)$, $S$ is the midpoint of arc $BAC$ of $(O)$. The perpendicular bisector $BO$ intersects $BS$ at $I$. $(I;IB)$ intersects $AB$ at $U$ different from $B$. $H$ is the orthocenter of triangle $ABC$. Prove that $UH$ = $US$
1 reply
MathsII-enjoy
Yesterday at 3:03 PM
aidenkim119
29 minutes ago
J
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Merlin's castle
gnoka   2
N 30 minutes ago by Anzoteh
Source: 46th International Tournament of Towns, Senior A-Level P6, Fall 2024
Merlin's castle has 100 rooms and 1000 corridors. Each corridor links some two rooms. Each pair of rooms is linked by one corridor at most. Merlin has given out the plan of the castle to the wise men and declared the rules of the challenge. The wise men need to scatter across the rooms in a manner they wish. Each minute Merlin will choose a corridor and one of the wise men will have to pass along it from one of the rooms at its ends to the other one. Merlin wins when in both rooms on the ends of the chosen corridor there are no wise men. Let us call a number $m$ the magic number of the castle if $m$ wise men can pre-agree before the challenge and act in such a way that Merlin never wins, $m$ being the minimal possible number. What are the possible values of the magic number of the castle? (Merlin and all the wise men always know the location of all the wise men).

Timofey Vasilyev
2 replies
gnoka
Nov 15, 2024
Anzoteh
30 minutes ago
J
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A drunk frog jumping ona grid in a weird way
Tintarn   4
N 37 minutes ago by pi_quadrat_sechstel
Source: Baltic Way 2024, Problem 10
A frog is located on a unit square of an infinite grid oriented according to the cardinal directions. The frog makes moves consisting of jumping either one or two squares in the direction it is facing, and then turning according to the following rules:
i) If the frog jumps one square, it then turns $90^\circ$ to the right;
ii) If the frog jumps two squares, it then turns $90^\circ$ to the left.

Is it possible for the frog to reach the square exactly $2024$ squares north of the initial square after some finite number of moves if it is initially facing:
a) North;
b) East?
4 replies
Tintarn
Nov 16, 2024
pi_quadrat_sechstel
37 minutes ago
J
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Interesting inequalities
sqing   6
N 41 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 0 $ and $ ab+bc+ca+abc=4$ . Prove that
$$k(a+b+c) -ab-bc\geq 4\sqrt{k(k+1)}-(k+4)$$Where $ k\geq \frac{16}{9}. $
$$ \frac{16}{9}(a+b+c) -ab-bc\geq \frac{28}{9}$$
6 replies
sqing
Today at 3:36 AM
sqing
41 minutes ago
J
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Inspired by Omerking
sqing   2
N 42 minutes ago by sqing
Source: Own
Let $ a,b,c>0 $ and $ ab+bc+ca\geq \dfrac{1}{3}. $ Prove that
$$ ka+ b+kc\geq \sqrt{\frac{4k-1}{3}}$$Where $ k\geq 1.$$$ 4a+ b+4c\geq \sqrt{5}$$
2 replies
sqing
Today at 5:11 AM
sqing
42 minutes ago
J
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Number Theory Chain!
JetFire008   57
N an hour ago by CHESSR1DER
I will post a question and someone has to answer it. Then they have to post a question and someone else will answer it and so on. We can only post questions related to Number Theory and each problem should be more difficult than the previous. Let's start!

Question 1
57 replies
JetFire008
Apr 7, 2025
CHESSR1DER
an hour ago
J
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Perpendicular bisector meets the circumcircle of another triangle
steppewolf   3
N an hour ago by Omerking
Source: 2023 Junior Macedonian Mathematical Olympiad P4
We are given an acute $\triangle ABC$ with circumcenter $O$ such that $BC<AB$. The bisector of $\angle ACB$ meets the circumcircle of $\triangle ABC$ at a second point $D$. The perpendicular bisector of $AC$ meets the circumcircle of $\triangle BOD$ for the second time at $E$. The line $DE$ meets the circumcircle of $\triangle ABC$ for the second time at $F$. Prove that the lines $CF$, $OE$ and $AB$ are concurrent.

Authored by Petar Filipovski
3 replies
steppewolf
Jun 10, 2023
Omerking
an hour ago
J
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Quad formed by orthocenters has same area (all 7's!)
v_Enhance   35
N an hour ago by Wictro
Source: USA January TST for the 55th IMO 2014
Let $ABCD$ be a cyclic quadrilateral, and let $E$, $F$, $G$, and $H$ be the midpoints of $AB$, $BC$, $CD$, and $DA$ respectively. Let $W$, $X$, $Y$ and $Z$ be the orthocenters of triangles $AHE$, $BEF$, $CFG$ and $DGH$, respectively. Prove that the quadrilaterals $ABCD$ and $WXYZ$ have the same area.
35 replies
v_Enhance
Apr 28, 2014
Wictro
an hour ago
J
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Two sets
steven_zhang123   5
N an hour ago by Filipjack
Given \(0 < b < a\), let
\[ A = \left\{ r \, \middle| \, r = \frac{a}{3}\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) + b\sqrt[3]{xyz}, \quad x, y, z \in \left[1, \frac{a}{b}\right] \right\}, \]and
\[ B = \left[2\sqrt{ab}, a + b\right]. \]
Prove that \(A = B\).
5 replies
steven_zhang123
5 hours ago
Filipjack
an hour ago
J
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A Segment Bisection Problem
buratinogigle   2
N an hour ago by aidenkim119
Source: VN Math Olympiad For High School Students P9 - 2025
In triangle $ABC$, let the incircle $\omega$ touch sides $BC, CA, AB$ at $D, E, F$, respectively. Let $P$ lie on the line through $D$ perpendicular to $BC$. Let $Q, R$ be the intersections of $PC, PB$ with $EF$, respectively. Let $K, L$ be the projections of $R, Q$ onto line $BC$. Let $M, N$ be the second intersections of $DQ, DR$ with the incircle $\omega$. Let $S$ be the intersection of $KM$ and $LN$. Prove that the line $DS$ bisects segment $QR$.
2 replies
buratinogigle
Today at 1:36 AM
aidenkim119
an hour ago
J
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Predicted AMC 8 Scores
megahertz13   154
N an hour ago by Aaronjudgeisgoat
$\begin{tabular}{c|c|c|c}Username & Grade & AMC8 Score \\ \hline megahertz13 & 5 & 23 \\ \end{tabular}$
154 replies
1 viewing
megahertz13
Jan 25, 2024
Aaronjudgeisgoat
an hour ago
J
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Another Cubic Curve!
v_Enhance   164
N 4 hours ago by IndexLibrorumProhibitorum
Source: USAMO 2015 Problem 1, JMO Problem 2
Solve in integers the equation
\[ x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3. \]
164 replies
v_Enhance
Apr 28, 2015
IndexLibrorumProhibitorum
4 hours ago
J
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How to get good at comp math
fossasor   24
N Today at 6:40 AM by Cha0s
I'm a rising ninth grader who wasn't in the school math league this year, and basically put aside comp math for a year. Unfortunately, that means that now that I'm in high school and having the epiphany about how important comp math actually is, and how much it would help my chances of getting involved in other math-related programs. In addition, I do enjoy math in general, and suspect that things like the AMCs are probably going to be some of the best practice I can get. What this all means is that I'm trying to go from mediocre to orz, 2 years after I probably should have started if I wanted to be any good.

So my question is: how do I get good at comp math?

This year, my scores on AMC 10 (and these are the highest I've ever gotten) were a 73.5 and an 82.5 (AMC 8 was 21/25, but that doesn't matter much). This is not good enough to qualify for AIME, and I probably need to raise my performance on each by at least 10 points. I've been decently good in the past at Number Theory, but I need to work on Geo and Combinatorics, and I'm trying to find the best resources to do that. My biggest flaw is probably not knowing many algorithms like Stars and Bars, and the path is clear here (learn them) but I'm still not sure which ones I need to know.

I'm aware that some of this advice is going to be something like "Practice 5 hours a day and start hardgrinding" or something along those lines. Unfortunately, I have other extracurriculars I need to balance, and for me, time is a limiting resource. My parents are somewhat frowning upon me doing a lot of comp math, which limits my time as well. I have neither the time nor motivation to do more than an hour a day, and in practice, I don't think I can be doing that consistently. As such, I would need to make that time count.

I know this is a very general question, and that aops is chock-full of detailed advice for math competitions. However, I'd appreciate it if anyone here could help me out, or show me the best resources I should use to get started. What mocks are any good, or what textbooks should I use? Where do I get the best practice with the shortest time? Is there some place I can find a list of useful formulas that have appeared in math comps before?

All advice is welcome!

24 replies
fossasor
Apr 10, 2025
Cha0s
Today at 6:40 AM
J
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2025 Math and AI 4 Girls Competition: Win Up To $1,000!!!
audio-on   49
N Today at 5:46 AM by EvaLin
Join the 2025 Math and AI 4 Girls Competition for a chance to win up to $1,000!

Hey Everyone, I'm pleased to announce the dates for the 2025 MA4G Competition are set!
Applications will open on March 22nd, 2025, and they will close on April 26th, 2025 (@ 11:59pm PST).

Applicants will have one month to fill out an application with prizes for the top 50 contestants & cash prizes for the top 20 contestants (including $1,000 for the winner!). More details below!

Eligibility:
The competition is free to enter, and open to middle school female students living in the US (5th-8th grade).
Award recipients are selected based on their aptitude, activities and aspirations in STEM.

Event dates:
Applications will open on March 22nd, 2025, and they will close on April 26th, 2025 (by 11:59pm PST)
Winners will be announced on June 28, 2025 during an online award ceremony.

Application requirements:
Complete a 12 question problem set on math and computer science/AI related topics
Write 2 short essays

Prizes:
1st place: $1,000 Cash prize
2nd place: $500 Cash prize
3rd place: $300 Cash prize
4th-10th: $100 Cash prize each
11th-20th: $50 Cash prize each
Top 50 contestants: Over $50 worth of gadgets and stationary

Many thanks to our current and past sponsors and partners: Hudson River Trading, MATHCOUNTS, Hewlett Packard Enterprise, Automation Anywhere, JP Morgan Chase, D.E. Shaw, and AI4ALL.

Math and AI 4 Girls is a nonprofit organization aiming to encourage young girls to develop an interest in math and AI by taking part in STEM competitions and activities at an early age. The organization will be hosting an inaugural Math and AI 4 Girls competition to identify talent and encourage long-term planning of academic and career goals in STEM.

Contact:
mathandAI4girls@yahoo.com

For more information on the competition:
https://www.mathandai4girls.org/math-and-ai-4-girls-competition

More information on how to register will be posted on the website. If you have any questions, please ask here!


49 replies
audio-on
Jan 26, 2025
EvaLin
Today at 5:46 AM
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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?
G H J
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
770 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
7551 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.
Z K Y
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wangzrpi
158 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
Z K Y
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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
Z K Y
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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
Z K Y
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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
208 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)
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Catcumber
162 posts
Catcumber
#45 h  Nov 8, 2024, 1:27 AM
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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
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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
108 posts
navier3072
#47 h  Nov 9, 2024, 8:26 AM
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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
1098 posts
Challengees24
#48 h  Nov 9, 2024, 7:23 PM
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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
496 posts
Mr.Sharkman
#49 h  Dec 11, 2024, 12:43 PM
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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
24 posts
Apple_maths60
#51 h  Apr 4, 2025, 3:15 PM
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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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