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Jane street swag package? USA(J)MO
arfekete   44
N 2 hours ago by pieMax2713
Hey! People are starting to get their swag packages from Jane Street for qualifying for USA(J)MO, and after some initial discussion on what we got, people are getting different things. Out of curiosity, I was wondering how they decide who gets what.
Please enter the following info:

- USAMO or USAJMO
- Grade
- Score
- Award/Medal/HM
- MOP (yes or no, if yes then color)
- List of items you got in your package

I will reply with my info as an example.
44 replies
arfekete
May 7, 2025
pieMax2713
2 hours ago
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[TEST RELEASED] OMMC Year 5
DottedCaculator   102
N 3 hours ago by Pengu14
Test portal: https://ommc-test-portal-2025.vercel.app/

Hello to all creative problem solvers,

Do you want to work on a fun, untimed team math competition with amazing questions by MOPpers and IMO & EGMO medalists? $\phantom{You lost the game.}$
Do you want to have a chance to win thousands in cash and raffle prizes (no matter your skill level)?

Check out the fifth annual iteration of the

Online Monmouth Math Competition!

Online Monmouth Math Competition, or OMMC, is a 501c3 accredited nonprofit organization managed by adults, college students, and high schoolers which aims to give talented high school and middle school students an exciting way to develop their skills in mathematics.

Our website: https://www.ommcofficial.org/

This is not a local competition; any student 18 or younger anywhere in the world can attend. We have changed some elements of our contest format, so read carefully and thoroughly. Join our Discord or monitor this thread for updates and test releases.

How hard is it?

We plan to raffle out a TON of prizes over all competitors regardless of performance. So just submit: a few minutes of your time will give you a great chance to win amazing prizes!

How are the problems?

You can check out our past problems and sample problems here:
https://www.ommcofficial.org/sample
https://www.ommcofficial.org/2022-documents
https://www.ommcofficial.org/2023-documents
https://www.ommcofficial.org/ommc-amc

How will the test be held?/How do I sign up?

Solo teams?

Test Policy

Timeline:
Main Round: May 17th - May 24th
Test Portal Released. The Main Round of the contest is held. The Main Round consists of 25 questions that each have a numerical answer. Teams will have the entire time interval to work on the questions. They can submit any time during the interval. Teams are free to edit their submissions before the period ends, even after they submit.

Final Round: May 26th - May 28th
The top placing teams will qualify for this invitational round (5-10 questions). The final round consists of 5-10 proof questions. Teams again will have the entire time interval to work on these questions and can submit their proofs any time during this interval. Teams are free to edit their submissions before the period ends, even after they submit.

Conclusion of Competition: Early June
Solutions will be released, winners announced, and prizes sent out to winners.

Scoring:

Prizes:

I have more questions. Whom do I ask?

We hope for your participation, and good luck!

OMMC staff

OMMC’S 2025 EVENTS ARE SPONSORED BY:

[list]
[*]Nontrivial Fellowship
[*]Citadel
[*]SPARC
[*]Jane Street
[*]And counting!
[/list]
102 replies
DottedCaculator
Apr 26, 2025
Pengu14
3 hours ago
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Asymptotic Casework
peace09   12
N 5 hours ago by daijobu
Source: 2024 AMC 12A #25
A graph is $\textit{symmetric}$ about a line if the graph remains unchanged after reflection in that line. For how many quadruples of integers $(a,b,c,d)$, where $|a|,|b|,|c|,|d|\le5$ and $c$ and $d$ are not both $0$, is the graph of \[y=\frac{ax+b}{cx+d}\]symmetric about the line $y=x$?

$\textbf{(A) }1282\qquad\textbf{(B) }1292\qquad\textbf{(C) }1310\qquad\textbf{(D) }1320\qquad\textbf{(E) }1330$
12 replies
peace09
Nov 7, 2024
daijobu
5 hours ago
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Do you need to attend mop
averageguy   2
N Yesterday at 6:33 PM by ZMB038
So I got accepted into a summer program and already paid the fee of around $5000 dollars. It's for 8 weeks (my entire summer) and it's in person. I have a few questions
1. If I was to make MOP this year am I forced to attend?
2.If I don't attend the program but still qualify can I still put on my college application that I qualified for MOP or can you only put MOP qualifier if you actually attend the program.
2 replies
averageguy
Mar 5, 2025
ZMB038
Yesterday at 6:33 PM
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MOP Cutoffs Out?
Mathandski   30
N Yesterday at 6:31 PM by ZMB038
MAA has just emailed a press release announcing the formula they will be using this year to come up with the MOP cutoff that applies to you! Here's the process:

1. Multiply your age by $1434$, let $n$ be the result.

2. Calculate $\varphi(n)$, where $\varphi$ is the Euler's totient theorem, which calculates the number of integers less than $n$ relatively prime to $n$.

3. Multiply your result by $1434$ again because why not, let the result be $m$.

4. Define the Fibonacci sequence $F_0 = 1, F_1 = 1, F_n = F_{n-1} + F_{n-2}$ for $n \ge 2$. Let $r$ be the remainder $F_m$ leaves when you divide it by $69$.

5. Let $x$ be your predicted USA(J)MO score.

6. You will be invited if your score is at least $\lfloor \frac{x + \sqrt[r]{r^2} + r \ln(r)}{r} \rfloor$.

7. Note that there may be additional age restrictions for non-high schoolers.

See here for MAA's original news message.

.

.

.


Edit (4/2/2025): This was an April Fool's post.
Here's the punchline
30 replies
Mathandski
Apr 1, 2025
ZMB038
Yesterday at 6:31 PM
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Essentially, how to get good at olympiad math?
gulab_jamun   10
N Yesterday at 6:22 PM by ZMB038
Ok, so I'm posting this as an anynonymous user cuz I don't want to get flamed by anyone I know for my goals but I really do want to improve on my math skill.

Basically, I'm alright at computational math (10 AIME, dhr stanford math meet twice) and I hope I can get good enough at olympiad math over the summer to make MOP next year (I will be entering 10th as after next year, it becomes much harder :( )) Essentially, I just want to get good at olympiad math. If someone could, please tell me how to study, like what books (currently thinking of doing EGMO) but I don't know how to get better at the other topics. Also, how would I prepare? Like would I study both proof geometry and proof number theory concurrently or just study each topic one by one?? Would I do mock jmo/amo or js prioritize olympiad problems in each topic. I have the whole summer ahead of me, and intend to dedicate it to olympiad math, so any advice would be really appreciated. Thank you!
10 replies
gulab_jamun
May 18, 2025
ZMB038
Yesterday at 6:22 PM
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OTIS or MathWOOT 2
math_on_top   15
N Yesterday at 5:33 PM by JeffreyTao
Hey AoPS community I took MathWOOT 1 this year and scored an 8 on AIME (last year I got a 6). My goal is to make it to MOP next year through USAMO. It's gonna be a lot of work, but do you think that I should do MathWOOT 2 or OTIS? Personally, I felt that MathWOOT 1 taught me a lot but was more focused on computational - not sure how to split computation vs olympiad prep. So, for those who can address this question:

(1) How much compuational vs olympiad
(2) OTIS or MathWOOT 2 and why
15 replies
math_on_top
May 18, 2025
JeffreyTao
Yesterday at 5:33 PM
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Two Sequences
worthawholebean   11
N Yesterday at 10:09 AM by P162008
Source: AIME 2008II Problem 6
The sequence $ \{a_n\}$ is defined by
\[ a_0 = 1,a_1 = 1, \text{ and } a_n = a_{n - 1} + \frac {a_{n - 1}^2}{a_{n - 2}}\text{ for }n\ge2. \]The sequence $ \{b_n\}$ is defined by
\[ b_0 = 1,b_1 = 3, \text{ and } b_n = b_{n - 1} + \frac {b_{n - 1}^2}{b_{n - 2}}\text{ for }n\ge2. \]Find $ \frac {b_{32}}{a_{32}}$.
11 replies
worthawholebean
Apr 3, 2008
P162008
Yesterday at 10:09 AM
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Goals for 2025-2026
Airbus320-214   158
N Yesterday at 7:05 AM by pieMax2713
Please write down your goal/goals for competitions here for 2025-2026.
158 replies
Airbus320-214
May 11, 2025
pieMax2713
Yesterday at 7:05 AM
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looks like roots of unity filter!
math31415926535   37
N Tuesday at 9:06 PM by xHypotenuse
Source: 2022 AIME II Problem 13
There is a polynomial $P(x)$ with integer coefficients such that $$P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}$$holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$
37 replies
math31415926535
Feb 17, 2022
xHypotenuse
Tuesday at 9:06 PM
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Outcome related combinatorics problem
egxa   1
N Apr 29, 2025 by iliya8788
Source: All Russian 2025 10.7
A competition consists of $25$ sports, each awarding one gold medal to a winner. $25$ athletes participate, each in all $25$ sports. There are also $25$ experts, each of whom must predict the number of gold medals each athlete will win. In each prediction, the medal counts must be non-negative integers summing to $25$. An expert is called competent if they correctly guess the number of gold medals for at least one athlete. What is the maximum number \( k \) such that the experts can make their predictions so that at least \( k \) of them are guaranteed to be competent regardless of the outcome?
1 reply
egxa
Apr 18, 2025
iliya8788
Apr 29, 2025
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Outcome related combinatorics problem
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Source: All Russian 2025 10.7
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egxa
211 posts
egxa
#1 h  Apr 18, 2025, 5:12 PM
Y by
A competition consists of $25$ sports, each awarding one gold medal to a winner. $25$ athletes participate, each in all $25$ sports. There are also $25$ experts, each of whom must predict the number of gold medals each athlete will win. In each prediction, the medal counts must be non-negative integers summing to $25$. An expert is called competent if they correctly guess the number of gold medals for at least one athlete. What is the maximum number \( k \) such that the experts can make their predictions so that at least \( k \) of them are guaranteed to be competent regardless of the outcome?
This post has been edited 1 time. Last edited by egxa, Apr 18, 2025, 5:21 PM
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iliya8788
8 posts
iliya8788
#2 h  Apr 29, 2025, 10:46 AM
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We claim that the maximum $k$ is equal to $24$.
We correspond each athlete to a number from $1$ to $25$.
Define Podium as the $n$-tuple $(a_{1},a_{2},...,a_{25})$ where $a_{i}$ is the amount of medals won by athlete $i$.
First notice that none of the experts can predict the $n$-tuple $(1,1,...,1)$ because then if the podium is equal to $(25,0,0,...,0)$ then the most possible amount of judges predicting correctly is equal to $24$. From the previous statement it follows that one of the judges predicts a $n$-tuple which has a $0$. So we proceed with the following algorithm. Pick one of the prediction $n$-tuples and convert it with the following algorithm to a new $n$-tuple: turn anything that isn't equal to $0$ to $0$ and then spread the 25 medals among the zero's.
Now if the podium $n$-tuple is the same as this $n$-tuple then at least one of the experts isn't competent and so maximum $k$ is equal to $24$.
Now we will prove that no matter what the podium looks like we can achieve $k=24$.
So we propose the following predictions: $(24,1,0,0,...,0),(24,0,1,0,0,...,0),(24,0,0,1,0,0,...0),...,(24,0,0,...,1),(1,1,...,1,1,1)$.
Define the count of zeros of the podium as $C$.
Now 3 cases might happen:
$C>2$: then it's easy to notice that every expert except the last one is competent.
$C=2$: an expert isn't competent if and only if both zeros are in the same place as the $1$ and $24$. notice that no $2$ of the predictions have these 2 numbers in the same spot so worst case possible $24$ of the experts are competent.
$C=1$: notice that the podium $n$-tuple must have $23$, $1$'s and a single $2$ and a single $0$. an expert isn't competent if and only if one of the following 2 cases occur:
First case: the $1$ in the prediction is in the same place as the $0$ of the podium.
Second case: the $24$ in the prediction is in the same place as the $0$ of the podium and the $1$ in the prediction is in the same place as the $2$ of the podium.
notice that the position of the $1$ differs among the predictions so for a fixed podium only one expert can be not competent because of the first case. The same applies for the second case as well. Since the union of the positions of $1$ and $24$ differs among the predictions there can't be 2 experts that are incompetent because of the second case. Also both cases can't make experts incompetent simultaneously since in case of both cases happening the $1$ and $24$ must basically switch places but in each of the first $24$ predictions $24$ is not in the first spot and all of the $1$'s are in the first spot. So only one of these $24$ experts can be incompetent and the last expert is competent so at least $24$ of the experts are competent. $\blacksquare$
This post has been edited 5 times. Last edited by iliya8788, Apr 29, 2025, 12:11 PM
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