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Contests & Programs AMC and other contests, summer programs, etc.
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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)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
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MATHCOUNTS/AMC 8 Basics
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WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
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Programming

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Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
May 1, 2025
0 replies
k i Peer-to-Peer Programs Forum
jwelsh   157
N Dec 11, 2023 by cw357
Many of our AoPS Community members share their knowledge with their peers in a variety of ways, ranging from creating mock contests to creating real contests to writing handouts to hosting sessions as part of our partnership with schoolhouse.world.

To facilitate students in these efforts, we have created a new Peer-to-Peer Programs forum. With the creation of this forum, we are starting a new process for those of you who want to advertise your efforts. These advertisements and ensuing discussions have been cluttering up some of the forums that were meant for other purposes, so we’re gathering these topics in one place. This also allows students to find new peer-to-peer learning opportunities without having to poke around all the other forums.

To announce your program, or to invite others to work with you on it, here’s what to do:

1) Post a new topic in the Peer-to-Peer Programs forum. This will be the discussion thread for your program.

2) Post a single brief post in this thread that links the discussion thread of your program in the Peer-to-Peer Programs forum.

Please note that we’ll move or delete any future advertisement posts that are outside the Peer-to-Peer Programs forum, as well as any posts in this topic that are not brief announcements of new opportunities. In particular, this topic should not be used to discuss specific programs; those discussions should occur in topics in the Peer-to-Peer Programs forum.

Your post in this thread should have what you're sharing (class, session, tutoring, handout, math or coding game/other program) and a link to the thread in the Peer-to-Peer Programs forum, which should have more information (like where to find what you're sharing).
157 replies
jwelsh
Mar 15, 2021
cw357
Dec 11, 2023
k i C&P posting recs by mods
v_Enhance   0
Jun 12, 2020
The purpose of this post is to lay out a few suggestions about what kind of posts work well for the C&P forum. Except in a few cases these are mostly meant to be "suggestions based on historical trends" rather than firm hard rules; we may eventually replace this with an actual list of firm rules but that requires admin approval :) That said, if you post something in the "discouraged" category, you should not be totally surprised if it gets locked; they are discouraged exactly because past experience shows they tend to go badly.
-----------------------------
1. Program discussion: Allowed
If you have questions about specific camps or programs (e.g. which classes are good at X camp?), these questions fit well here. Many camps/programs have specific sub-forums too but we understand a lot of them are not active.
-----------------------------
2. Results discussion: Allowed
You can make threads about e.g. how you did on contests (including AMC), though on AMC day when there is a lot of discussion. Moderators and administrators may do a lot of thread-merging / forum-wrangling to keep things in one place.
-----------------------------
3. Reposting solutions or questions to past AMC/AIME/USAMO problems: Allowed
This forum contains a post for nearly every problem from AMC8, AMC10, AMC12, AIME, USAJMO, USAMO (and these links give you an index of all these posts). It is always permitted to post a full solution to any problem in its own thread (linked above), regardless of how old the problem is, and even if this solution is similar to one that has already been posted. We encourage this type of posting because it is helpful for the user to explain their solution in full to an audience, and for future users who want to see multiple approaches to a problem or even just the frequency distribution of common approaches. We do ask for some explanation; if you just post "the answer is (B); ez" then you are not adding anything useful.

You are also encouraged to post questions about a specific problem in the specific thread for that problem, or about previous user's solutions. It's almost always better to use the existing thread than to start a new one, to keep all the discussion in one place easily searchable for future visitors.
-----------------------------
4. Advice posts: Allowed, but read below first
You can use this forum to ask for advice about how to prepare for math competitions in general. But you should be aware that this question has been asked many many times. Before making a post, you are encouraged to look at the following:
[list]
[*] Stop looking for the right training: A generic post about advice that keeps getting stickied :)
[*] There is an enormous list of links on the Wiki of books / problems / etc for all levels.
[/list]
When you do post, we really encourage you to be as specific as possible in your question. Tell us about your background, what you've tried already, etc.

Actually, the absolute best way to get a helpful response is to take a few examples of problems that you tried to solve but couldn't, and explain what you tried on them / why you couldn't solve them. Here is a great example of a specific question.
-----------------------------
5. Publicity: use P2P forum instead
See https://artofproblemsolving.com/community/c5h2489297_peertopeer_programs_forum.
Some exceptions have been allowed in the past, but these require approval from administrators. (I am not totally sure what the criteria is. I am not an administrator.)
-----------------------------
6. Mock contests: use Mock Contests forum instead
Mock contests should be posted in the dedicated forum instead:
https://artofproblemsolving.com/community/c594864_aops_mock_contests
-----------------------------
7. AMC procedural questions: suggest to contact the AMC HQ instead
If you have a question like "how do I submit a change of venue form for the AIME" or "why is my name not on the qualifiers list even though I have a 300 index", you would be better off calling or emailing the AMC program to ask, they are the ones who can help you :)
-----------------------------
8. Discussion of random math problems: suggest to use MSM/HSM/HSO instead
If you are discussing a specific math problem that isn't from the AMC/AIME/USAMO, it's better to post these in Middle School Math, High School Math, High School Olympiads instead.
-----------------------------
9. Politics: suggest to use Round Table instead
There are important conversations to be had about things like gender diversity in math contests, etc., for sure. However, from experience we think that C&P is historically not a good place to have these conversations, as they go off the rails very quickly. We encourage you to use the Round Table instead, where it is much more clear that all posts need to be serious.
-----------------------------
10. MAA complaints: discouraged
We don't want to pretend that the MAA is perfect or that we agree with everything they do. However, we chose to discourage this sort of behavior because in practice most of the comments we see are not useful and some are frankly offensive.
[list] [*] If you just want to blow off steam, do it on your blog instead.
[*] When you have criticism, it should be reasoned, well-thought and constructive. What we mean by this is, for example, when the AOIME was announced, there was great outrage about potential cheating. Well, do you really think that this is something the organizers didn't think about too? Simply posting that "people will cheat and steal my USAMOO qualification, the MAA are idiots!" is not helpful as it is not bringing any new information to the table.
[*] Even if you do have reasoned, well-thought, constructive criticism, we think it is actually better to email it the MAA instead, rather than post it here. Experience shows that even polite, well-meaning suggestions posted in C&P are often derailed by less mature users who insist on complaining about everything.
[/list]
-----------------------------
11. Memes and joke posts: discouraged
It's fine to make jokes or lighthearted posts every so often. But it should be done with discretion. Ideally, jokes should be done within a longer post that has other content. For example, in my response to one user's question about olympiad combinatorics, I used a silly picture of Sogiita Gunha, but it was done within a context of a much longer post where it was meant to actually make a point.

On the other hand, there are many threads which consist largely of posts whose only content is an attached meme with the word "MAA" in it. When done in excess like this, the jokes reflect poorly on the community, so we explicitly discourage them.
-----------------------------
12. Questions that no one can answer: discouraged
Examples of this: "will MIT ask for AOIME scores?", "what will the AIME 2021 cutoffs be (asked in 2020)", etc. Basically, if you ask a question on this forum, it's better if the question is something that a user can plausibly answer :)
-----------------------------
13. Blind speculation: discouraged
Along these lines, if you do see a question that you don't have an answer to, we discourage "blindly guessing" as it leads to spreading of baseless rumors. For example, if you see some user posting "why are there fewer qualifiers than usual this year?", you should not reply "the MAA must have been worried about online cheating so they took fewer people!!". Was sich überhaupt sagen lässt, lässt sich klar sagen; und wovon man nicht reden kann, darüber muss man schweigen.
-----------------------------
14. Discussion of cheating: strongly discouraged
If you have evidence or reasonable suspicion of cheating, please report this to your Competition Manager or to the AMC HQ; these forums cannot help you.
Otherwise, please avoid public discussion of cheating. That is: no discussion of methods of cheating, no speculation about how cheating affects cutoffs, and so on --- it is not helpful to anyone, and it creates a sour atmosphere. A longer explanation is given in Seriously, please stop discussing how to cheat.
-----------------------------
15. Cutoff jokes: never allowed
Whenever the cutoffs for any major contest are released, it is very obvious when they are official. In the past, this has been achieved by the numbers being posted on the official AMC website (here) or through a post from the AMCDirector account.

You must never post fake cutoffs, even as a joke. You should also refrain from posting cutoffs that you've heard of via email, etc., because it is better to wait for the obvious official announcement. A longer explanation is given in A Treatise on Cutoff Trolling.
-----------------------------
16. Meanness: never allowed
Being mean is worse than being immature and unproductive. If another user does something which you think is inappropriate, use the Report button to bring the post to moderator attention, or if you really must reply, do so in a way that is tactful and constructive rather than inflammatory.
-----------------------------

Finally, we remind you all to sit back and enjoy the problems. :D

-----------------------------
(EDIT 2024-09-13: AoPS has asked to me to add the following item.)

Advertising paid program or service: never allowed

Per the AoPS Terms of Service (rule 5h), general advertisements are not allowed.

While we do allow advertisements of official contests (at the MAA and MATHCOUNTS level) and those run by college students with at least one successful year, any and all advertisements of a paid service or program is not allowed and will be deleted.
0 replies
v_Enhance
Jun 12, 2020
0 replies
k i Stop looking for the "right" training
v_Enhance   50
N Oct 16, 2017 by blawho12
Source: Contest advice
EDIT 2019-02-01: https://blog.evanchen.cc/2019/01/31/math-contest-platitudes-v3/ is the updated version of this.

EDIT 2021-06-09: see also https://web.evanchen.cc/faq-contest.html.

Original 2013 post
50 replies
v_Enhance
Feb 15, 2013
blawho12
Oct 16, 2017
HCSSiM results
SurvivingInEnglish   63
N 33 minutes ago by KevinChen_Yay
Anyone already got results for HCSSiM? Are there any point in sending additional work if I applied on March 19?
63 replies
SurvivingInEnglish
Apr 5, 2024
KevinChen_Yay
33 minutes ago
Sad Algebra
tastymath75025   46
N an hour ago by Ilikeminecraft
Source: 2019 USAMO 6, by Titu Andreescu and Gabriel Dospinescu
Find all polynomials $P$ with real coefficients such that $$\frac{P(x)}{yz}+\frac{P(y)}{zx}+\frac{P(z)}{xy}=P(x-y)+P(y-z)+P(z-x)$$holds for all nonzero real numbers $x,y,z$ satisfying $2xyz=x+y+z$.

Proposed by Titu Andreescu and Gabriel Dospinescu
46 replies
tastymath75025
Apr 18, 2019
Ilikeminecraft
an hour ago
USAMO Medals
YauYauFilter   50
N an hour ago by ConfidentKoala4
YauYauFilter
Apr 24, 2025
ConfidentKoala4
an hour ago
Will I fail again
hashbrown2009   8
N an hour ago by anticodon
so this year I got 34 on JMO 772 774 and got docked 1 point from top honors + mop

I just got info that I pretty much cannot do math for the rest of summer due to family reasons, and the only time I have is winter break

do you guys think it's enough time to practice/grind to qualify mop through USAMO, or should I tell my parents to reschedule the stuff because I really want to make mop

(Note: I'm aiming for like 25+ on USAMO so at least silver but I'm not sure that's realistic given the circumstances i'm in)
8 replies
hashbrown2009
Yesterday at 1:54 PM
anticodon
an hour ago
Jane street swag package? USA(J)MO
arfekete   15
N 2 hours ago by llddmmtt1
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.
15 replies
arfekete
Wednesday at 4:34 PM
llddmmtt1
2 hours ago
Aime ll 2022 problem 5
Rook567   1
N 3 hours ago by clarkculus
I don’t understand the solution. I got 220 as answer. Why does it insist, for example two primes must add to the third, when you can take 2,19,19 or 2,7,11 which for drawing purposes is equivalent to 1,1,2 and 2,7,9?
1 reply
Rook567
3 hours ago
clarkculus
3 hours ago
MathILy 2025 Decisions Thread
mysterynotfound   41
N 3 hours ago by cappucher
Discuss your decisions here!
also share any relevant details about your decisions if you want
41 replies
mysterynotfound
Apr 21, 2025
cappucher
3 hours ago
9 ARML Location
deduck   37
N 4 hours ago by imbadatmath1233
UNR -> Nevada
St Anselm -> New Hampshire
PSU -> Pennsylvania
WCU -> North Carolina


Put your USERNAME in the list ONLY IF YOU WANT TO!!!! !!!!!

I'm going to UNR if anyone wants to meetup!!! :D

Current List:
Iowa
UNR
PSU
St Anselm
WCU
37 replies
deduck
May 6, 2025
imbadatmath1233
4 hours ago
Can I make the IMO team next year?
aopslover08   15
N 4 hours ago by Alex-131
Hi everyone,

I am a current 11th grader living in Orange, Texas. I recently started doing competition math and I think I am pretty good at it. Recently I did a mock AMC8 and achieved a score of 21/25, which falls in the top 1% DHR. I also talked to my math teacher and she says I am an above average student.

Given my natural talent and the fact that I am willing to work ~3.5 hours a week studying competition math, do you think I will be able to make IMO next year? I am aware of the difficulty of this task but my mom says that I can achieve whatever I put my mind to, as long as I work hard.

Here is my plan for the next few months:

month 1-2: finish studying pre-algebra and learn geometry
month 3-4: learn pre-calculus
month 5-6: start doing IMO shortlist problems
month 7+: keep doing ISL/IMO problems.

Is this a feasible task? I am a girl btw.
15 replies
aopslover08
5 hours ago
Alex-131
4 hours ago
The answer of 2022 AIME II #5 is incorrect
minz32   5
N 4 hours ago by Rook567
Source: 2022 AIME II problem 5
The answer posted on the AOPS page for the 2022 AIME II problem 5 has some flaw in it and the final answer is incorrect.

The answer is trying to use the symmetric property for the a, b, c. However the numbers are not exactly cyclically symmetric in the problem.

For example, the solution works well with a = 20, it derives 4 different pairs of b and c for p2 is in (3, 5, 11, 17) as stated. However, for a = 19, only 3 p2 in the set works, since if the p2 = 17, we will have c = 0, which is not a possible vertex label. So only three pairs of b and c work for a= 19, which are (19, 17, 14), (19, 17, 12), (19, 17, 6).

Same for a = 18.

However, for a = 17, one more possibility joined. c can be bigger than a. So the total triple satisfying the problem is back to 4.

There are two more cases that the total number of working pairs of b, c is 3, when a = 5, and a = 4.

So the final answer for this problem is not 072. It should be 068.



5 replies
minz32
Nov 20, 2022
Rook567
4 hours ago
Past USAMO Medals
sdpandit   0
5 hours ago
Does anyone know where to find lists of USAMO medalists from past years? I can find the 2025 list on their website, but they don't seem to keep lists from previous years and I can't find it anywhere else. Thanks!
0 replies
sdpandit
5 hours ago
0 replies
Deciding between Ross and HCSSiM
akliu   28
N 5 hours ago by Vivaandax
Hey! I got accepted into Ross Indiana, and I think I'll probably also get accepted into HCSSiM. I've been looking between the two camps, and I'm trying to decide which one to go to -- both seem like really fun options.

Instead of trying to explain my personal preferences and thought processes, I thought it might be a good idea to ask the community for their personal opinions on these camps. What are some things that you like or dislike about both camps? (Whether it be through personal experience or by word-of-mouth, but please specify if it's just something you've heard)

This will probably help me be more informed on making a final decision, so I'd appreciate any advice. Thanks in advance!
28 replies
akliu
Apr 18, 2025
Vivaandax
5 hours ago
Mathcounts state
happymoose666   40
N 5 hours ago by ZMB038
Hi everyone,
I just have a question. I live in PA and I sadly didn't make it to nationals this year. Is PA a competitive state? I'm new into mathcounts and not sure
40 replies
happymoose666
Mar 24, 2025
ZMB038
5 hours ago
How to PHO Qual
itsjeyanth   1
N Yesterday at 4:06 PM by BelowAverageAsian
Hi guys, I wanted to try out physics, this year I got a 14 on the f=ma and missed the usahpo cutoff by 1 point. But I am pretty bad at physics so I wanted to do a class to get better, I was wondering like what books or classes I should take, do you guys think physwoot would be a good fit? Would I struggle too much?
1 reply
itsjeyanth
Yesterday at 3:29 PM
BelowAverageAsian
Yesterday at 4:06 PM
sussy baka stop intersecting in my lattice points
Spectator   24
N Apr 29, 2025 by ilikemath247365
Source: 2022 AMC 10A #25
Let $R$, $S$, and $T$ be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the x-axis. The left edge of $R$ and the right edge of $S$ are on the $y$-axis, and $R$ contains $\frac{9}{4}$ as many lattice points as does $S$. The top two vertices of $T$ are in $R \cup S$, and $T$ contains $\frac{1}{4}$ of the lattice points contained in $R \cup S$. See the figure (not drawn to scale).

IMAGE

The fraction of lattice points in $S$ that are in $S \cap T$ is 27 times the fraction of lattice points in $R$ that are in $R \cap T$. What is the minimum possible value of the edge length of $R$ plus the edge length of $S$ plus the edge length of $T$?

$\textbf{(A) }336\qquad\textbf{(B) }337\qquad\textbf{(C) }338\qquad\textbf{(D) }339\qquad\textbf{(E) }340$
24 replies
Spectator
Nov 11, 2022
ilikemath247365
Apr 29, 2025
sussy baka stop intersecting in my lattice points
G H J
Source: 2022 AMC 10A #25
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Spectator
657 posts
#1 • 1 Y
Y by megarnie
Let $R$, $S$, and $T$ be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the x-axis. The left edge of $R$ and the right edge of $S$ are on the $y$-axis, and $R$ contains $\frac{9}{4}$ as many lattice points as does $S$. The top two vertices of $T$ are in $R \cup S$, and $T$ contains $\frac{1}{4}$ of the lattice points contained in $R \cup S$. See the figure (not drawn to scale).

[asy]
//kaaaaaaaaaante314
size(8cm);
import olympiad;
label(scale(.8)*"$y$", (0,60), N);
label(scale(.8)*"$x$", (60,0), E);
filldraw((0,0)--(55,0)--(55,55)--(0,55)--cycle, yellow+orange+white+white);
label(scale(1.3)*"$R$", (55/2,55/2));
filldraw((0,0)--(0,28)--(-28,28)--(-28,0)--cycle, green+white+white);
label(scale(1.3)*"$S$",(-14,14));
filldraw((-10,0)--(15,0)--(15,25)--(-10,25)--cycle, red+white+white);
label(scale(1.3)*"$T$",(3.5,25/2));
draw((0,-10)--(0,60),EndArrow(TeXHead));
draw((-34,0)--(60,0),EndArrow(TeXHead));[/asy]

The fraction of lattice points in $S$ that are in $S \cap T$ is 27 times the fraction of lattice points in $R$ that are in $R \cap T$. What is the minimum possible value of the edge length of $R$ plus the edge length of $S$ plus the edge length of $T$?

$\textbf{(A) }336\qquad\textbf{(B) }337\qquad\textbf{(C) }338\qquad\textbf{(D) }339\qquad\textbf{(E) }340$
This post has been edited 3 times. Last edited by Spectator, Dec 24, 2022, 12:54 PM
Reason: Asy
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peelybonehead
6291 posts
#3
Y by
You guys had time to solve P25?!!?
Z K Y
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iamhungry
149 posts
#4 • 22 Y
Y by Paul10, balllightning37, bobthebuilder1234, andrewwang2623, bestzack66, michaelwenquan, Ritwin, ivyshine13, eibc, aidan0626, Lamboreghini, plang2008, mahaler, mathboy100, megarnie, mathmax12, spiritshine1234, the_mathmagician, EpicBird08, aidensharp, akliu, Sedro
I looked at this problem, saw colored squares and the word "lattice points" and immediately went back to the previous problems
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Snowyowl2005
57 posts
#5 • 8 Y
Y by Spectator, wamofan, iamhungry, Mogmog8, Lamboreghini, cosmicgenius, Aopamy, centslordm
Solution at the end of https://randommath.app.box.com/s/fnajq15ew2gvx0bmt8ucri8karksulyo.

Suppose that the square $R$ has $r^2$ lattice points (i.e. has coordinates ranging from $(0, 0)$ to $(r - 1, r - 1)$). Similarly, suppose that the square $S$ has $s^2$ lattice point (i.e. has coordinates ranging from $(0, 0)$ to $(1 - s, s - 1)$).
Then $\frac{r^2}{s^2} = \frac{9}{4}$, and so $r = 3x$ and $s = 2x$ for some positive integer $x$.
Furthermore, the union of $R$ and $S$ contains $r^2 + s^2 - s$ lattice points since they share $s$ lattice points of the form $(0, 0), (0, 1), (0, 2), \ldots, (0, s - 1)$. Therefore:
\[t^2 = \frac{r^2 + s^2 - s}{4} = \frac{13x^2 - 2x}{4}\]and so:
\[13x^2 - 2x = 4t^2\]Since the right hand side is even, $13x^2$ is even, and so we can write $x = 2y$ for some integer $y$. Thus:
\[52y^2 - 4y = 4t^2\]\[13y^2 - y = y(13y - 1) = t^2\]But by the Euclidean algorithm, $y$ and $13y - 1$ are relatively prime, and so both $y$ and $13y - 1$ must be perfect squares. Write:
\[y = m^2\]\[13y - 1 = 13m^2 - 1 = n^2\]Suppose that the top left corner of $T$ is given by $(1 - k, 0)$. Then the top right corner must be given by $(t - k, 0)$. We have:
\[\frac{|S \cap T|}{|S|} = 27 \cdot \frac{|R \cap T|}{|R|}\]where $|A|$ denotes the number of lattice points in a region $A$. Plugging in the relevant values, we get:
\[\frac{k \cdot t}{s^2} = 27 \cdot \frac{(t - k + 1) \cdot t}{r^2}\]Cross-multiplying, we get:
\[k \cdot t \cdot r^2 = 27 \cdot (t - k + 1) \cdot t \cdot s^2\]Simplifying and using $r^2 = \frac{9}{4}s^2$, we get:
\[9 \cdot k = (t - k + 1) \cdot 4 \cdot 27\]\[12 \cdot (t - k + 1) = k\]\[12 \cdot (t + 1) = 13k\]Therefore, for any $t \equiv 12 \pmod{13}$, we have a solution for $k$. But:
\[y(13y - 1) \equiv -y \equiv t^2 \equiv 1 \pmod{13}\]and so $y \equiv 12 \pmod{13}$. Since $y = m^2$, we have:
\[m^2 + 1 \equiv 0 \pmod{13}\]and that $13m^2 - 1$ is a perfect square. The smallest $m$ satisfying both conditions is $m = 5$. Plugging this in, we get:
\[y = m^2 = 25\]\[x = 2y = 50\]and so $r = 3x = 150$, $s = 2x = 100$, and $t^2 = y(13y - 1) = 25 \cdot 324$, so $t = 5 \cdot 18 = 90$, and:
\[(r - 1) + (s - 1) + (t - 1) = 150 + 100 + 90 - 3 = 340 - 3 = 337\]so the answer is B.
This post has been edited 2 times. Last edited by Snowyowl2005, Nov 11, 2022, 6:09 PM
Reason: The problem is asking for the sum of the edge lengths, which are $r - 1, s - 1, t - 1$, instead of $r + s + t$. It has been fixed above.
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mathboy100
675 posts
#6
Y by
Can we use Pick's theorem on this?
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bryanguo
1032 posts
#7
Y by
Answer E is incorrect, they did sum of coordinates which happened to give an answer $3$ larger than what it's supposed to be.

Rough sketch is to label side length of $S$ as $x,$ then $(x+1)^2$ lattice points in $S$ and $\tfrac{9}{4}(x+1)^2$ lattice points in $R,$ so side length of $R$ is $\tfrac{3x+1}{2},$ from there just use the other two given conditions are you get modular congruences to give side lengths $(89,99,149) \implies 337.$
This post has been edited 1 time. Last edited by bryanguo, Nov 11, 2022, 5:35 PM
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ihatemath123
3446 posts
#8
Y by
Uhhh what would you need picks for here
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Spectator
657 posts
#9
Y by
can someone include asy code for this
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Snowyowl2005
57 posts
#10 • 1 Y
Y by bryanguo
bryanguo wrote:
Answer E is incorrect, they did sum of coordinates which happened to give an answer $3$ larger than what it's supposed to be.

Rough sketch is to label side length of $S$ as $x,$ then $(x+1)^2$ lattice points in $S$ and $\tfrac{9}{4}(x+1)^2$ lattice points in $R,$ so side length of $R$ is $\tfrac{3x+1}{2},$ from there just use the other two given conditions are you get modular congruences to give side lengths $(89,99,149) \implies 337.$

The only error seems to be the answer extraction - it has been fixed above.
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HamstPan38825
8860 posts
#11
Y by
This problem is essentially how quickly you can do number theory.

Let the side length of $R$ be $x$, so the number of lattice points in $R$ is $(x+1)^2$, the number of lattice points in $S$ is $\frac 49(x+1)^2$. It follows that the side length of $S$ is $\frac 23 x - \frac 13$.

Now, let $a$ and $b$ be the lengths of the portions of the sides of $T$ in $S$ and $R$, respectively. We have $$\frac{(a+1)(k+1)}{\frac 49(x+1)^2} = 27 \cdot \frac{(b+1)(k+1)}{(x+1)^2}$$by the third condition, so $a = 12b+11$, and the side length $k$ of $T$ is of the form $13b+11$ for some $b$.

Now, using the second condition, $$\frac{(k+1)^2}{\frac {13}9(x+1)^2 - \frac 23(x+1)} = \frac 14.$$Note Then $$36(13b+12)^2 = (x+1)(13x+7).$$Notice that $\gcd(x+1, 13x+7) = \gcd(x+1, 6) = 6$, so actually
\begin{align*}
x+1 &= 6r_1^2 \\
13x+ 7 &= 6r_2^2
\end{align*}for some positive integers $r_1, r_2$. These $r_1, r_2$ satisfy $$13r_1^2 - r_2^2 = 1.$$At this point, there isn't really a better way than to test out values for $r_1$, as we know by the answer choices that the answers cannot be too big. Indeed, $(r_1, r_2) = (5, 18)$ is a solution. Then, $x = 149$, so $$(k+1)^2 = \frac{150 \cdot 1944}{36} = 324 \cdot 25,$$which yields $k = 89$. Finally, the last side length is $99$, so the answer is $89+99+149 = \boxed{337}$.
This post has been edited 2 times. Last edited by HamstPan38825, Nov 11, 2022, 7:32 PM
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Somersett
71 posts
#12
Y by
peelybonehead wrote:
You guys had time to solve P25?!!?

Bruh this test went by so fast i didn’t even get to look at this problem
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Math_Shisa
158 posts
#13
Y by
All the solutions look so long I don’t think I would have time to even answer this one.
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saturnrocket
1306 posts
#14
Y by
I didn't even get to 15
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Dino76
707 posts
#15 • 1 Y
Y by Mango247
That's a lot better than me. I didn't even bother to take it...
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qwerty123456asdfgzxcvb
1086 posts
#16
Y by
me when not 340
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mannshah1211
651 posts
#17
Y by
if There is one more Baka in AMC Titles, i will Riot.
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Bluesoul
894 posts
#19
Y by
Not necessary to solve exact R,S,T, just compute R+S+T mod 13

Denote the side lengths of $R,S,T$ as $r,s,t$ respectively. According to the first condition, we have $\frac{9}{4}(s+1)^2=(r+1)^2, 3s+1=2r$

Second condition yields $4(t+1)^2=(s+1)^2+(r+1)^2-(s+1)$. Plug $r=\frac{3s+1}{2}$, we get $16(t+1)^2=13(s+1)^2-4(s+1)=(s+1)(13s+9)$

Let the length of $T$ in negative x-axis as $x$, the third condition implies $\frac{(x+1)(t+1)}{(s+1)^2}=27\frac{(t+1)(t-x+1)}{(r+1)^2}$ simplify this to $x+1=12(t-x+1), 12t+11=13x$. From here, we can attain $t\equiv -2\pmod{13}$. Consider equation $16(t+1)^2=(s+1)(13s+9)$, we have $(s+1)(13s+9)\equiv 9(s+1)\equiv 3\pmod{13}$

Thus, we have $s\equiv 8\pmod{13}$

Finally, we have $r=\frac{3s+1}{2}$ which yields $r\equiv 6\pmod{13}$. Consequently, $r+s+t\equiv 8+6-2\equiv 12\pmod{13}$ while only $\boxed{B}$ satisfies.
This post has been edited 1 time. Last edited by Bluesoul, Oct 25, 2023, 12:05 AM
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MathyMathMan
250 posts
#20
Y by
mannshah1211 wrote:
if There is one more Baka in AMC Titles, i will Riot.

Bouta say, but what is that title? :huuh:
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Amkan2022
2015 posts
#21
Y by
.........
This post has been edited 1 time. Last edited by Amkan2022, Apr 29, 2025, 4:07 AM
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IbrahimNadeem
888 posts
#22
Y by
aops being wild today
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IbrahimNadeem
888 posts
#23
Y by
why am I getting an irrational side for T?
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ESAOPS
262 posts
#26
Y by
oops
sol
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xHypotenuse
778 posts
#27
Y by
This feels harder than average aime #12s and I lwk have 0 clue on how to approach it
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MC_ADe
182 posts
#28
Y by
write out each condition and turn it into different forms, since you know each is an integer you can use divisibility,modulo, euclidean algorithm and other methods to find the values
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ilikemath247365
253 posts
#29
Y by
I was actually able to solve this problem......except it took me like 30 minutes. :rotfl:
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