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

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

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.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/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

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Contest Preparation: Grades 6-12

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


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0 replies
jlacosta
Mar 2, 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
average FE
KevinYang2.71   76
N 23 minutes ago by Maximilian113
Source: USAJMO 2024/5
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy
\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]for all $x,y\in\mathbb{R}$.

Proposed by Carl Schildkraut
76 replies
+1 w
KevinYang2.71
Mar 21, 2024
Maximilian113
23 minutes ago
[Registration Open] Mustang Math Tournament 2025
MustangMathTournament   30
N 31 minutes ago by sadas123
Mustang Math is excited to announce that registration for our annual tournament, MMT 2025, is open! This year, we are bringing our tournament to 9 in-person locations, as well as online!

Locations include: Colorado, Norcal, Socal, Georgia, Illinois, Massachusetts, New Jersey, Nevada, Washington, and online. For registration and more information, check out https://mustangmath.com/competitions/mmt-2025.

MMT 2025 is a math tournament run by a group of 150+ mathematically experienced high school and college students who are dedicated to providing a high-quality and enjoyable contest for middle school students. Our tournament centers around teamwork and collaboration, incentivizing students to work with their teams not only to navigate the challenging and interesting problems of the tournament but also to develop strategies to master the unique rounds. This includes a logic puzzle round, a strategy-filled hexes round, a race-like gallop round, and our trademark ‘Mystery Mare’ round!

Awards:
[list]
[*] Medals for the top teams
[*] Shirts, pins, stickers and certificates for all participants
[*] Additional awards provided by our wonderful sponsors!
[/list]

We are also holding a free MMT prep seminar from 3/15-3/16 to help students prepare for the upcoming tournament. Join the Google Classroom! https://classroom.google.com/c/NzQ5NDUyNDY2NjM1?cjc=7sogth4
30 replies
MustangMathTournament
Mar 8, 2025
sadas123
31 minutes ago
JSMCR Results
FuturePanda   14
N an hour ago by Pengu14
Hi everyone,

Did anyone get their JSMCR decisions back yet? They were supposed to release on 2/28

Thanks!
14 replies
FuturePanda
Mar 1, 2025
Pengu14
an hour ago
AMC 10.........
BAM10   8
N 2 hours ago by ChickensEatGrass
I'm in 8th grade and have never taken the AMC 10. I am currently in alg2. I have scored 20 on AMC 8 this year and 34 on the chapter math counts last year. Can I qualify for AIME. Also what should I practice AMC 10 next year?
8 replies
BAM10
Mar 2, 2025
ChickensEatGrass
2 hours ago
Tennessee Math Tournament (TMT) Online 2025
TennesseeMathTournament   31
N 2 hours ago by NashvilleSC
Hello everyone! We are excited to announce a new competition, the Tennessee Math Tournament, created by the Tennessee Math Coalition! Anyone can participate in the virtual competition for free.

The testing window is from March 22nd to April 5th, 2025. Virtual competitors may participate in the competition at any time during that window.

The virtual competition consists of three rounds: Individual, Bullet, and Team. The Individual Round is 60 minutes long and consists of 30 questions (AMC 10 level). The Bullet Round is 20 minutes long and consists of 80 questions (Mathcounts Chapter level). The Team Round is 30 minutes long and consists of 16 questions (AMC 12 level). Virtual competitors may compete in teams of four, or choose to not participate in the team round.

To register and see more information, click here!

If you have any questions, please email connect@tnmathcoalition.org or reply to this thread!
31 replies
TennesseeMathTournament
Mar 9, 2025
NashvilleSC
2 hours ago
hcssim application question
enya_yurself   2
N 2 hours ago by Magnetoninja
do they send the Interesting Test to everyone who applied or do they read the friendly letter first and only send to the kids they like?
2 replies
enya_yurself
5 hours ago
Magnetoninja
2 hours ago
The return of American geo
brianzjk   77
N 3 hours ago by Ilikeminecraft
Source: USAJMO 2023/6
Isosceles triangle $ABC$, with $AB=AC$, is inscribed in circle $\omega$. Let $D$ be an arbitrary point inside $BC$ such that $BD\neq DC$. Ray $AD$ intersects $\omega$ again at $E$ (other than $A$). Point $F$ (other than $E$) is chosen on $\omega$ such that $\angle DFE = 90^\circ$. Line $FE$ intersects rays $AB$ and $AC$ at points $X$ and $Y$, respectively. Prove that $\angle XDE = \angle EDY$.

Proposed by Anton Trygub
77 replies
brianzjk
Mar 23, 2023
Ilikeminecraft
3 hours ago
Apply for Team USA at the International Math Competition (IMC)!
peace09   53
N 3 hours ago by stjwyl
The International Math Competition (IMC) is essentially the elementary and middle school equivalent of the IMO, with individual and team rounds featuring both short-answer and proof-based problems. See past problems here.

Team USA is looking for 6th graders and below with AIME qualification or AMC 8 DHR (or equivalent), and for 9th graders and below with JMO or Mathcounts Nationals qualification. If you think you meet said criteria, fill out the initial form here.

Here are a couple quick links for further information:
[list=disc]
[*] Dr. Tao Hong's website, which contains a detailed recap of the 2024 competition (and previous years'), as well as Team USA's historical results. (You may recognize a couple names... @channing421 @vrondoS et al.: back me up here :P)
[*] My journal, which gives an insider's perspective on the camp :ninja:
[/list]
53 replies
peace09
Aug 13, 2024
stjwyl
3 hours ago
OTIS Mock AIME 2025 airs Dec 19th
v_Enhance   39
N 3 hours ago by MonkeyLuffy
Source: https://web.evanchen.cc/mockaime.html
Satisfactory. Keep cooking.
IMAGE

Problems are posted at https://web.evanchen.cc/mockaime.html#current now!

Like last year, we're running the OTIS Mock AIME 2025 again, except this time there will actually be both a I and a II because we had enough problems to pull it off. However, the two versions will feel quite different from each other:

[list]
[*] The OTIS Mock AIME I is going to be tough. It will definitely be harder than the actual AIME, by perhaps 2 to 4 problems. But more tangibly, it will also have significant artistic license. Problems will freely assume IMO-style background throughout the test, and intentionally stretch the boundary of what constitutes an “AIME problem”.
[*] The OTIS Mock AIME II is meant to be more practically useful. It will adhere more closely to the difficulty and style of the real AIME. There will inevitably still be some more IMO-flavored problems, but they’ll appear later in the ordering.
[/list]
Like last time, all 30 problems are set by current and past OTIS students.

Details are written out at https://web.evanchen.cc/mockaime.html, but to highlight important info:
[list]
[*] Free, obviously. Anyone can participate.
[*]Both tests will release sometime Dec 19th. You can do either/both.
[*]If you'd like to submit for scoring, you should do so by January 20th at 23:59 Pacific time (same deadline for both). Please hold off on public spoilers before then.
[*]Solutions, statistics, and maybe some high scores will be published shortly after that.
[/list]
Feel free to post questions, hype comments, etc. in this thread.
39 replies
v_Enhance
Dec 6, 2024
MonkeyLuffy
3 hours ago
Gunn Math Competition
the_math_prodigy   12
N 3 hours ago by the_math_prodigy
Gunn Math Circle is excited to host the fourth annual Gunn Math Competition (GMC)! GMC will take place at Gunn High School in Palo Alto, California on Sunday, March 30th. Gather a team of up to four and compete for over $7,500 in prizes! The contest features three rounds: Individual, Guts, and Team. We welcome participants of all skill levels, with separate Beginner and Advanced divisions for all students.

Registration is free and now open at compete.gunnmathcircle.org. The deadline to sign up is March 27th.

Special Guest Speaker: Po-Shen Loh!!!
We are honored to welcome Po-Shen Loh, a world-renowned mathematician, Carnegie Mellon professor, and former coach of the USA International Math Olympiad team. He will deliver a 30-minute talk to both students and parents, offering deep insights into mathematical thinking and problem-solving in the age of AI!

View competition manual, schedule, prize pool at compete.gunnmathcircle.org . Stay updated by joining our Discord discord.gg/fqcxukv3Dq server. For any questions, reach out at ghsmathcircle@gmail.com or ask in Discord.
12 replies
the_math_prodigy
Mar 8, 2025
the_math_prodigy
3 hours ago
They mixed up USAJMO and AIME I guess
Math4Life7   54
N 3 hours ago by littlefox_amc
Source: USAJMO 2024/1
Let $ABCD$ be a cyclic quadrilateral with $AB = 7$ and $CD = 8$. Point $P$ and $Q$ are selected on segment $AB$ such that $AP = BQ = 3$. Points $R$ and $S$ are selected on segment $CD$ such that $CR = DS = 2$. Prove that $PQRS$ is a cyclic quadrilateral.

Proposed by Evan O'Dorney
54 replies
Math4Life7
Mar 20, 2024
littlefox_amc
3 hours ago
USAMO vs USAJMO Prestige
elasticwealth   47
N 3 hours ago by axusus
Just curious, what does everyone think about the prestige of a USAMO qualification vs a USAJMO qual? Obv USAMO > USAJMO but how much?

And while we’re at it, what about amc 10 dhr vs amc 12 dhr? Thoughts?

Edit: also how much is a perfect score on 10/12 worth? Asking for a friend….
47 replies
elasticwealth
Feb 17, 2025
axusus
3 hours ago
My opinions on this years AIMES problems+cutoff range??
isache   35
N 4 hours ago by axusus
1: Little bit harder than most Nr 1's, but you can just bash this out either way.
2. Kinda annoying, but once u break it down it isnt that bad. Average nr2.
3. Cool problem, just break it down into the 3 cases and it isnt too bad. Tiny bit easy for a n3.
4. Dividing by xy, x^2, or y^2 makes this problem a lot easier. You can also factor. avg no 4.
5. Easily sillyable again, kinda annoying for a nr 5. Honestly I would switch no 5 and 6.
6. Pretty simple if you use pithot. If not it can be difficult. Avg no 6.
7. Very hard for a nr7. The whole problem itself is not bad, just it is extremely sillyable
8. Very bashy, and not super easy to solve, but putting this one on a cartesian plane makes it easier. Harder than last years nr8 by a mile.
9. Easy if you see the trick, impossible if not. Without the trick, this problem becomes super bashy, but probably avg nr9. I spent like 1hr on this to absolutely no avail. I got into an equation with degree 6 bc I didnt see the trick.
10. Not very easy, as you have to break it down well for the solution to flow nicely. Quite hard for a nr10 imo
11. Difficult to understand, but once you have the hang of it down it is not horrible. Despite this, many struggled to understand it in the first place. Fairly hard for a nr 11.
12. Lotta people struggle to graph inequalities in 3d planes (So do I). little hard for a nr 12.
13. Very confusing for a bunch of people (including me). Avg for a nr 13 tho.
14. Super hard problem, but extremely elegant. Fermats point is a cool concept here. Hard for a nr 14 tho
15. LTE helps a lot. Honestly I would switch 14 and 15.

Overall, I think problems 1-6 were avg for an aime, but after that the problem got significantly harder. Much harder than last yrs imo. Tough to say what cutoffs will be given all that has gone last year. Tbf I predict sub 220 for both 10a and 10b, but I could be wrong. We kinda just have to wait and see.
35 replies
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isache
Feb 11, 2025
axusus
4 hours ago
Convolution of order f(n)
trumpeter   74
N 5 hours ago by EpicBird08
Source: 2019 USAMO Problem 1
Let $\mathbb{N}$ be the set of positive integers. A function $f:\mathbb{N}\to\mathbb{N}$ satisfies the equation \[\underbrace{f(f(\ldots f}_{f(n)\text{ times}}(n)\ldots))=\frac{n^2}{f(f(n))}\]for all positive integers $n$. Given this information, determine all possible values of $f(1000)$.

Proposed by Evan Chen
74 replies
trumpeter
Apr 17, 2019
EpicBird08
5 hours ago
Degree Six Polynomial's Roots
ksun48   42
N Yesterday at 2:20 AM by eg4334
Source: 2014 AIME I Problem 14
Let $m$ be the largest real solution to the equation \[\frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}= x^2-11x-4.\] There are positive integers $a,b,c$ such that $m = a + \sqrt{b+\sqrt{c}}$. Find $a+b+c$.
42 replies
ksun48
Mar 14, 2014
eg4334
Yesterday at 2:20 AM
Degree Six Polynomial's Roots
G H J
Source: 2014 AIME I Problem 14
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ksun48
1514 posts
#1 • 4 Y
Y by ahmedosama, tdeng, Casetoo, Adventure10
Let $m$ be the largest real solution to the equation \[\frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}= x^2-11x-4.\] There are positive integers $a,b,c$ such that $m = a + \sqrt{b+\sqrt{c}}$. Find $a+b+c$.
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ABCDE
1963 posts
#2 • 13 Y
Y by niraekjs, NumberGiant, LaQuesha_BakerJones, mentalgenius, Not_a_Username, Mathcat1234, bowenying24, ThisUsernameIsTaken, rayfish, Adventure10, think4l, mrtheory, and 1 other user
Solution
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sat113
430 posts
#3 • 1 Y
Y by Adventure10
I didn't have enough time to solve this during the test, but after working the problem at home, I got 263.
Sniped
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borntobeweild
331 posts
#4 • 2 Y
Y by Adventure10, Math4Life7
Particularly brutal to careless this one... :wallbash:
(Edit: Okay, apparently I'm in the same league as Kent Merryfield here, so maybe I shouldn't feel so bad after all)

Add $4$ to both sides to get \[\frac{x}{x-3}+\frac{x}{x-5}+\frac{x}{x-17}+\frac{x}{x-19}=x^2-11x\] Divide through x to get \[\frac{1}{x-3}+\frac{1}{x-5}+\frac{1}{x-17}+\frac{1}{x-19}=x-11\] Adding the two outside and two inside terms, the left hand side is \[\frac{2x-22}{x^2-22x+57}+\frac{2x-22}{x^2-22x+85}\] Therefore the equation becomes \[\frac{1}{x^2-22x+57}+\frac{1}{x^2-22x+85}=\frac{1}{2}\] Let $y=x^2-22x+57$, so this is \[\frac{1}{y}+\frac{1}{y+28}=\frac{1}{2}\] This clears to \[y^2+24y-56=0\] which solves to be $y=-12\pm\sqrt{200}$ (but I somehow got negative that on the AIME). Then $x^2-22x+69\pm\sqrt{200}$ gives $x=11\pm\sqrt{52\pm\sqrt{200}}$ (Pedantic and irrelevant edit: by this, I mean the four possible combinations of $+$ and $-$, no just the two possibilities of $+, +$ and $-, -$), so taking the obvious maximum we get $11+52+200=\boxed{263}$
This post has been edited 2 times. Last edited by borntobeweild, Mar 14, 2014, 8:35 PM
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bobthesmartypants
4337 posts
#5 • 2 Y
Y by Adventure10, Mango247
I solved this, but then I subtracted $484-276$ wrong. :wallbash:
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Kent Merryfield
18574 posts
#6 • 2 Y
Y by Adventure10, Mango247
I was working the problems while proctoring. For this one, I started by setting $x=y+11,$ and then pairing terms on the left to get denominators of $y^2-64$ and $y^2-36.$ That immediately led to factoring out a $y,$ and then I also discovered that I could factor out $2y+11$. That left a quadratic for $y^2$ to solve by the quadratic formula.

I got $11+\sqrt{52+\sqrt{230}}.$ So I must have made an arithmetic error very near the final step.
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msinghal
725 posts
#7 • 2 Y
Y by Adventure10, Mango247
Solution
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howie2000
413 posts
#9 • 2 Y
Y by Adventure10, Mango247
Solution 14
EDIT: dang looks like i was late
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sat113
430 posts
#10 • 1 Y
Y by Adventure10
I was so mad at myself for ignoring this problem after I solved it in <7,8 minutes at my house.
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tanuagg13
137 posts
#11 • 2 Y
Y by Adventure10, Mango247
How did you guys come up with this clever manipulation? Was it something you say before or did you just immediately come up with it?
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v_Enhance
6858 posts
#12 • 161 Y
Y by DVA6102, SMOJ, happiface, yugrey, AMN300, mathocean97, Binomial-theorem, MSTang, AkshajK, tanuagg13, forthegreatergood, droid347, sindennisz, etude, Einstein314, pedronr, MathematicsOfPi, NextEinstein, dantx5, Hydroxide, vlchen888, yuxiao, Tan, AnonymousBunny, banonymous322, howie2000, r31415, AndrewKwon97, sat113, lucylai, niraekjs, ProblemPro, geoishard, chezbgone, pinetree1, NumberGiant, LaQuesha_BakerJones, mentalgenius, blasterboy, Tuxianeer, turkeybob777, cnnwy1282, ABCDE, addictedtomath, crastybow, dyang, bacca2002, snapdragon, abishek99, nbute, bcp123, MuseOfMath, sicilianfan, mathtastic, HYP135peppers, Boomer, ptes77, hamup1, ahaanomegas, Not_a_Username, Royalreter1, DrMath, hexagram, va2010, TheMaskedMagician, TheCrafter, mathman523, WalkerTesla, rjiang16, TheStrangeCharm, 15Pandabears, 62861, epiclucario, mathmagic12, JNEW, MathStudent2002, donot, rkm0959, Alnitak, wu2481632, Shaddoll, Th3Numb3rThr33, coffee_bean, acegikmoqsuwy2000, maverick8, shiningsunnyday, budu, trumpeter, champion999, mathwiz0803, liberator, BIGBUBBLE, CurryinaHurry, mathisawesome2169, bluephoenix, yrnsmurf, Mudkipswims42, GeronimoStilton, speulers_theorem, Ultroid999OCPN, sa2001, mathfun1, GeoMetrix, Greenleaf5002, Juno, fidgetboss_4000, mathleticguyyy, CALCMAN, mathapple101, HamstPan38825, MathJams, jacoporizzo, ike.chen, tigerzhang, Zorger74, celestialphoenix3768, OlympusHero, AOPqghj, megarnie, rayfish, FalconMaster, centslordm, eagles2018, IMUKAT, tenebrine, smileapple, peelybonehead, rg_ryse, michaelwenquan, Sedro, eibc, aidan0626, awesomehuman, Mogmog8, TheUltimate123, ChromeRaptor777, ex-center, Adventure10, ihatemath123, iamhungry, vrondoS, Jiwan, roribaki, think4l, akliu, Funcshun840, megahertz13, MarioLuigi8972, and 13 other users
tanuagg13 wrote:
How did you guys come up with this clever manipulation? Was it something you saw before or did you just immediately come up with it?

So here's the (dramatized) story of me and this problem. Hopefully it provides some insight on how problems get solved. If not, hopefully it's kind of funny. :P
EDIT: Added embellishments

Prologue
It was about 25 minutes into the test and I had knocked out 9, 12, 15, 8, 3 (in that order). This problem had caught my eye at the beginning because it was pure algebra, but I hadn't jumped at it because I didn't have any immediate ideas how to attack it. Now is the hour, I decided, and I began my siege. Even if I didn't solve it right away, I could have it in the back of my head while I knocked out easy problems.

Act I
Right away I noticed that $3+19=5+17 = 2 \cdot 11$. That HAD to be important. It probably meant that $a=11$, for example. There's no other possible value for $a$ that would even make sense. But I wasn't sure what was up with the right hand side. The left-hand side looked natural, but the $x(x-11)-4$ seemed completely haphazardly thrown. The problem was hiding something, but I couldn't see what.

With no better idea, I got a new sheet of paper and attempted brute force. Knowing that $a=11$ was almost certain, I set $x = r+11$ and began crunching. Pairing, rearranging, whatever I could do to try to knock this out.

Unfortunately all I got was messes. (I had expected $r$, $r^3$, $r^5$ to cancel, but no such luck). After some minutes of this I accepted a temporary defeat and decided to come back later. I proceeded to squash the other eight problems I hadn't attempted yet within an hour, then returned to the sole survivor.

Act II
By this time I had about 90 minutes left so I was feeling pretty good about my chances of solving it. I looked at the equation again. The extra time had helped, because I noticed something I didn't before -- the equation was degree six (if you expand), but the roots of the form $a \pm \sqrt{b \pm \sqrt{c}}$ only covered four roots. So there had to be other roots left. I wondered if a quadratic factored out of the thing somehow, with some strange irrational roots.

Or maybe there were some more obvious roots...

Act III
It was at this point I decided to try plugging in $x=0$ to see what the constant term might be under a full expansion. This gives $(-1) + (-1) + (-1) + (-1) = -4$... which gives precisely the right-hand side. Oops! I swore audibly in the testing room. At least I knew what the $-4$ there was for now.

So if $x=0$ is a root, that means I could annihilate a factor of $x$ from both sides. That caused me to add the $4$ to both sides, and then kill the $x$'s. This left \[ \frac{1}{x-3} + \frac{1}{x-5} + \frac{1}{x-17} + \frac{1}{x-19} = x-11. \] Now I did the same $r$ substitution as before, and at this point the fractions paired off nicely to wipe out the $r$ from both sides. Indeed, you just get \[ \frac{2r}{r^2-64} + \frac{2r}{r^2-36} = r \] which immediately drops to a quadratic in $r^2$. And that was straightforward!

Epilogue
This was it! I had solved all 15 problems with over an hour to boot. Maybe this would finally be the year when I didn't screw up and got that 15. All that stood in the way was a simple quadratic, and I had all the time in the world. With deep satisfaction and a sense of triumph, I knew I had solved all 15 problems.

I then proceeded to put 663 because of an arithmetic error. The end.
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mathocean97
606 posts
#13 • 30 Y
Y by v_Enhance, 62861, bluephoenix, mathleticguyyy, Aryan-23, HamstPan38825, OlympusHero, megarnie, rayfish, ConfidentKoala4, centslordm, michaelwenquan, IMUKAT, Adventure10, Mango247, vrondoS, and 14 other users
@above: Well, from what I heard, #14 was not the only thing standing between you and a 15...
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v_Enhance
6858 posts
#14 • 25 Y
Y by 62861, bluephoenix, mathleticguyyy, HamstPan38825, OlympusHero, megarnie, rayfish, centslordm, Zorger74, michaelwenquan, IMUKAT, Adventure10, Mango247, vrondoS, Sedro, and 10 other users
Haha yeah this is pretty true. Certainly felt that way during the test though :P
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bobthesmartypants
4337 posts
#15 • 3 Y
Y by dantx5, Chimphechunu, Adventure10
my motivation
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briantix
138 posts
#16 • 4 Y
Y by Einstein314, Adventure10, and 2 other users
Done without substitution (unless you consider algebraic shorthand substitution)
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raptorw
1012 posts
#17 • 2 Y
Y by Adventure10, Mango247
...I got it in the form of $11+\sqrt{52+10\sqrt{2}}$... got 075 as my answer... :( Why couldn't have the form of the answer be different...
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bobthesmartypants
4337 posts
#18 • 5 Y
Y by aidan0626, Adventure10, Mango247, and 2 other users
Well $10\sqrt{2}=\sqrt{200}$ and you didn't notice?
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raptorw
1012 posts
#19 • 2 Y
Y by Adventure10, Mango247
I didn't see what the answer form was supposed to be.
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mathtastic
3258 posts
#20 • 2 Y
Y by Adventure10, Mango247
very similar to this problem i just found

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2597249&sid=971159d0f08274af07475064d61a4b47#p2597249
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Nikpour
1748 posts
#21 • 1 Y
Y by Adventure10
\begin{align}
  & \frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}={{x}^{2}}-11x-4 \\ 
 & x-11=u\Rightarrow x=u+11 \\ 
 & \Rightarrow \frac{3}{u+8}+\frac{5}{u+6}+\frac{17}{u-6}+\frac{19}{u-8}=(u+11)u-4 \\ 
 & \Rightarrow \frac{3(u-8)+19(u+8)}{{{u}^{2}}-64}+\frac{5(u-6)+17(u+6)}{{{u}^{2}}-36}={{u}^{2}}+11u-4 \\ 
 & \Rightarrow \frac{22u+128}{{{u}^{2}}-64}+\frac{22u+72}{{{u}^{2}}-36}={{u}^{2}}+11u-4 \\ 
 & \Rightarrow \frac{2({{u}^{2}}+11u)}{{{u}^{2}}-64}+\frac{2({{u}^{2}}+11u)}{{{u}^{2}}-36}={{u}^{2}}+11u \\ 
 & \Rightarrow \left\{ \begin{matrix}
   {{u}^{2}}+11u=0\Rightarrow u=0,-11  \\
   \frac{2}{{{u}^{2}}-64}+\frac{2}{{{u}^{2}}-36}=1\Rightarrow ({{u}^{2}}-36)({{u}^{2}}-64)=4{{u}^{2}}-200  \\
\end{matrix} \right. \\ 
 & {{u}^{4}}-104{{u}^{2}}+2504=0\Rightarrow {{u}^{2}}=52\pm 10\sqrt{2} \\ 
 & \Rightarrow u=\pm \sqrt{52\pm 10\sqrt{2}}\Rightarrow x=11\pm \sqrt{52\pm 10\sqrt{2}} 
\end{align}
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cobbler
2180 posts
#22 • 7 Y
Y by shiningsunnyday, Blast_S1, Adventure10, Mango247, and 3 other users
Seeing terms of the form $\frac{a}{x-a}$ reminds one of the fact that $\frac{1}{1-a}=1+a+a^2+\cdots$ (only valid for $|a|<1$, but we can draw a veil over that for now :P ). Indeed, since $\frac{3}{3a-3}$ $=\frac{1}{a-1}$ $=-(1+a+a^2+\cdots)$, then letting $x:=3a$ yields $\frac{3}{x-3}$ $=-\left(1+\frac{x}{3}+\left(\frac{x}{3}\right)^2+\cdots\right)$, and similarly for the rest. Therefore, the equation becomes \[-\left(1+\frac{x}{3}+\left(\frac{x}{3}\right)^2+\cdots\right)-\left(1+\frac{x}{5}+\left(\frac{x}{5}\right)^2+\cdots\right)\]
\[-\left(1+\frac{x}{17}+\left(\frac{x}{17}\right)^2+\cdots \right)-\left(1+\frac{x}{19}+\left(\frac{x}{19}\right)^2+\cdots\right)\]
\[=x^2-11x-4.\] It is now clear to add $4$ to both sides (by evenly distributing it among the terms on the LHS), because they all have a constant term of $-1$, and $-1-1-1-1=-4$, which is the constant term on the RHS. Therefore, adding $4$ to both sides we get \[-\left(\frac{x}{3}+\left(\frac{x}{3}\right)^2+\cdots\right)-\left(\frac{x}{5}+\left(\frac{x}{5}\right)^2+\cdots\right)\]
\[-\left(\frac{x}{17}+\left(\frac{x}{17}\right)^2+\cdots \right)-\left(\frac{x}{19}+\left(\frac{x}{19}\right)^2+\cdots\right)\]
\[=x^2-11x.\] The rest is easy. (Divide through by $x$, switch back to closed-forms, transform, etc.)
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targo___
493 posts
#23 • 4 Y
Y by maXplanK, Yagami1728, Adventure10, Mango247
Can we say that this is a polynomial ? (i am nt sure ) we are getting a polynomial ..but that polynomial and the given equation are not same ..equation is not continuous at $x=3,5,17,19$
Can we say that an equation is a polynomial also when it is not ccontinuous at some points of it's domain ?
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targo___
493 posts
#24 • 2 Y
Y by Adventure10, Mango247
anyone ?
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KenV
1198 posts
#25 • 3 Y
Y by CurryinaHurry, targo___, Adventure10
It's a polynomial with holes
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iiRishabii
1155 posts
#26
Y by
Adding $1$ to each fraction yields

$$\frac{3}{x-3}+1+\frac{5}{x-5}+1+\frac{17}{x-17}+1+\frac{19}{x-19}+1= x^2-11x-4+4$$$$\frac{x}{x-3}+\frac{x}{x-5}+\frac{x}{x-17}+\frac{x}{x-19}= x^2-11x.$$
Now we multiply by $\frac{1}{x}$ to get rid of the $x$ terms in the numerators of the RHS;

$$\frac{1}{x-3}+\frac{1}{x-5}+\frac{1}{x-17}+\frac{1}{x-19}=x-11.$$
Notice that combining the first and last terms yields $\frac{2x-22}{x^2-22x+57}$ and combining the second and third terms yields $\frac{2x-22}{x^2-22x+85}.$ This is helpful because it allows us to substitute later on.

$$\frac{2x-22}{x^2-22x+85}+\frac{2x-22}{x^2-22x+57}=x-11.$$
Multiplying both sides by $\frac{1}{2x-22}$ (similar step done previously) to clean up the numerators, we have

$$\frac{1}{x^2-22x+85}+\frac{1}{x^2-22x+57}=\frac{1}{2}.$$
Now, substituting $y=x^2-22x+71,$ we have

$$\frac{1}{y+14}+\frac{1}{y-14}=\frac{1}{2}$$$$\frac{2y}{y^2-196}=\frac{1}{2}$$$$4y=y^2-196$$$$y^2-4y-196=0.$$
Using the quadratic formula yields $\frac{4\pm\sqrt{800}}{2}$ or $2\pm20\sqrt{2}.$ Since the problem asks for the largest root, we can ignore $2-2\sqrt{20}.$ Substituting $y=2+2\sqrt{20}$ into $y=x^2-22x+71,$ we have

$$2+2\sqrt{20}=x^2-22x+71$$$$x^2-22x+69-2\sqrt{200}=0.$$
Using the quadratic formula on this again, we have $x=11\pm\sqrt{52+\sqrt{200}},$ and since the problem asks for the largest root, it is obvious that $m=11+\sqrt{52+\sqrt{200}},$ so $a+b+c=\boxed{263}.$
This post has been edited 5 times. Last edited by iiRishabii, Apr 20, 2021, 4:14 AM
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OlympusHero
17018 posts
#27
Y by
What's the motivation for adding $4$ to both sides? Was scrolling through thread and caught that it was first step, but how would I figure that out on the actual test?

Solution
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Geometry285
902 posts
#28
Y by
@above the RHS has $-4$, so you instinctively add. The trick is that you notice how it distributes to all the fractions in the first place.

How is this a #14????
This post has been edited 1 time. Last edited by Geometry285, Jun 3, 2021, 8:21 PM
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HamstPan38825
8853 posts
#29 • 2 Y
Y by fuzimiao2013, Sedro
@above I think this is adequate for #14; this is probably one of the best alg manip problems I've ever seen that are "synthetic algebra". I think moving the 4 is intuitive once one has enough experience with manipulations because it leaves a nice $x(x-11)$ on the RHS, which is compatible with our initial substitution.

Let $y = x-11$, and notice that
\begin{align*}
\frac {11-8}{y+8} + \frac{11-6}{y+6} + \frac{11+6}{y-6} + \frac{11+8}{y-8} + 4 &= y(y+11) \\
\frac {11-8+y+8}{y+8} + \frac{11-6+y+6}{y+6} + \frac{11+6+y-6}{y-6} + \frac{11+8+y-8}{y-8} &= y(y+11) \\
\frac 1{y+8} + \frac 1{y+6} + \frac 1{y-6} + \frac 1{y-8} = y \\
\frac{2y}{y^2-36} + \frac{2y}{y^2-64} = y \\
\frac 1{y^2-36}+\frac 1{y^2-64} = \frac 12.
\end{align*}Now let $z=y^2$, such that $$z^2-104z+2504=0 \implies z = 52 \pm 10\sqrt 2.$$Now we want the largest solution, which is $$x = \boxed{11+\sqrt{52 + 10 \sqrt 2}}.$$We can extract the answer of $\boxed{263}$.
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rayfish
1121 posts
#30
Y by
This is one of my favorite AIME problems.
Solution (same as #2)
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Oyler
20 posts
#31
Y by
I don't feel the need to add another solution, I just wanted to add modified the above solution by @rayfish by instead letting $y = x - 11$ to get $y^4 - 104y^2 + 2504 = 0 \implies y^2 = 52 \pm \sqrt{200} \implies x = 11 + \sqrt{52 + \sqrt{200}} \implies \boxed{263}$ is the answer.
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eagles2018
2733 posts
#32
Y by
I unfortunately got the "add 4 to both sides" by randomly scrolling through this thread. Otherwise I got the solution. Very nice manipulations :) Unfortunately, I got it wrong 'cause I can't multiply.
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jasperE3
11093 posts
#33
Y by
Note that $0$ and $11$ are solutions, but suppose momentarily that $x\notin\{0,11\}$.
Adding $4$ to both sides, we have:
$$\frac x{x-3}+\frac x{x-5}+\frac x{x-17}+\frac x{x-19}=\frac3{x-3}+1+\frac5{x-5}+1+\frac{17}{x-17}+1+\frac{19}{x-19}=x^2-11x$$$$\Rightarrow\frac1{x-3}+\frac1{x-5}+\frac1{x-17}+\frac1{x-19}=x-11.$$Let $y=x-11$. Then:
$$y=\frac1{y+6}+\frac1{y-6}+\frac1{y+8}+\frac1{y-8}=\frac{2y}{y^2-36}+\frac{2y}{y^2-64}$$$$\Rightarrow\frac12=\frac1{y^2-36}+\frac1{y^2-64}.$$Let $z=y^2-50$. Then:
$$\frac12=\frac1{z+14}+\frac1{z-14}=\frac{2z}{z^2-196}$$$$\Rightarrow z^2-4z-196=0\Rightarrow z=2\pm\sqrt{200}.$$Then $y=\pm\sqrt{37\pm\sqrt{200}}$ and $x=11\pm\sqrt{37\pm\sqrt{200}}$. The largest such real solution is $11+\sqrt{37+\sqrt{200}}$, which is greater than $0$ and $11$, so our answer is $11+37+200=\boxed{248}$.
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samrocksnature
8791 posts
#34
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Find the largest real number $x$ satisfying
\[ \frac{3}{x-3} + \frac{5}{x-5} + \frac{17}{x-17}
+ \frac{19}{x-19} = x^2-11x-4.  \]
Adding $4$ to both sides, we obtain
\begin{align*}
    x(x-11) &= \frac{3}{x-3} + \frac{5}{x-5} + \frac{17}{x-17}
+ \frac{19}{x-19}+4 \\
&= \frac{3+(x-3)}{x-3}+\frac{5+(x-5)}{x-5}+\frac{17+(x-17)}{x-17}+\frac{19+(x-19)}{x-19} \\
&=\frac{x}{x-3}+\frac{x}{x-5}+\frac{x}{x-17}+\frac{x}{x-19} \\
\end{align*}and assuming $x \neq 0$ (we will find a larger solution), dividing by $x$ yields
\begin{align*}
x-11 &= \frac{1}{x-3}+\frac{1}{x-5}+\frac{1}{x-17}+\frac{1}{x-19} \\
&= \left(\frac{1}{x-3}+\frac{1}{x-19}\right) \left(\frac{1}{x-5}+\frac{1}{x-17}\right)\\
&= \frac{2x-22}{(x-3)(x-19)}+\frac{2x-22}{(x-5)(x-17)}\\
&= 2(x-11)\left(\frac{1}{x^2-22x+57}+\frac{1}{x^2-22x+85}\right)\\
\end{align*}upon which diving by $2(x-11)$ (assuming $x \neq 11$) and substituting $y=x^2-22x+71$ yields
$$\frac{1}{2}=\frac{1}{y-14}+\frac{1}{y+14}=\frac{2y}{y^2-196}$$$$y^2-196=4y\implies y=2 \pm 10\sqrt{2}.$$Since we want the largest real number $x,$ we take the positive root and finish: $$2+10\sqrt{2}=y=x^2-22x+71 \implies x=11 \pm \sqrt{52+10\sqrt{2}}.$$Thus, the largest value of $x$ is $11 + \sqrt{52+10\sqrt{2}}.$ Extract $\boxed{263}.$
This post has been edited 1 time. Last edited by samrocksnature, Aug 27, 2022, 6:49 AM
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ihatemath123
3429 posts
#35
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The solution to this problem is magic
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euler12345
488 posts
#36
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Kinda
The symmetry just pops out at you when you see the problem
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channing421
1352 posts
#37
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in the otis unit this didn't have an answer extraction and i'm too lazy to change all of my $10\sqrt{2}$s into $\sqrt{200}$s so yeah

solution
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gracemoon124
872 posts
#38
Y by
solution
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lolipepit
394 posts
#39
Y by
mysol
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cinnamon_e
703 posts
#40
Y by
solution
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huashiliao2020
1292 posts
#42
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It's easily deduced by moving over the four that $$\frac{1}{x-3}+\frac{1}{x-5}+\frac{1}{x-17}+\frac{1}{x-19}=x-11\iff \frac{2}{(x-3)(x-19)}+\frac{2}{(x-5)(x-17)}=1.$$This is well motivated because it's very common when there are fractions with pairwise sums to be the same that we combine denominators of some to manipulate. From here, just substitute $y=x^2-22x+c$ (you can choose any c you want, it'll work) and solving for y solves the problem.
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joshualiu315
2513 posts
#43
Y by
sol
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dolphinday
1310 posts
#44
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Add $4$ to both sides and divide by $x$ to get
$\newline$
\[\frac{1}{x - 3} + \frac{1}{x - 5} + \frac{1}{x-17} + \frac{1}{x-19} = x - 11\]and then cross multiplying and dividing by $x - 11$ gives
\[\frac{2}{(x-3)(x-19)} + \frac{2}{(x-5)(x-17)} = 1\]Then let $y = x + 11$, so we have
\[\frac{2}{(y^2 - 64)} + \frac{2}{(y^2 - 36)} = 1 \implies \]\[4y^2 - 200 = (y^2 - 64)(y^2 - 36)\]Now, let $z = y^2$.
\[4z - 200 = (z - 64)(z - 36) \implies z^2 - 104z + 2504 = 0 \implies z = 52 + 10\sqrt{2}\]
So $x = 11 + \sqrt{52 + 10\sqrt{2}} \implies 263$ is our answer.
This post has been edited 1 time. Last edited by dolphinday, Jan 30, 2024, 12:17 AM
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eg4334
601 posts
#45
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get rid of the 4 on the rhs and add one to each term on the left to get $\sum_{i=3, 5, 17, 19} \frac{1}{x-i} = x-11$. On the rhs pair up the terms with sum $22$ to get $\frac{1}{(x-3)(x-19)} + \frac{1}{(x-5)(x-17)} = \frac12$. Then just notice both are symmetric around $x-11$, so let $y=x-11$ and from here it is a very straightfroward classical problem.
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