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Do you need to attend mop
averageguy   5
N Yesterday at 5:55 PM by babyzombievillager
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.
5 replies
averageguy
Mar 5, 2025
babyzombievillager
Yesterday at 5:55 PM
2014 amc 10 a problem 23
Rook567   3
N Yesterday at 4:13 PM by Rook567
Why do solutions assume 30 60 90 triangles?
If you assume 45 45 90 you get 5/6 as answer, don’t you?
3 replies
Rook567
Monday at 7:31 PM
Rook567
Yesterday at 4:13 PM
Inequality with a^2+b^2+c^2+abc=4
cn2_71828182846   72
N Yesterday at 4:12 PM by endless_abyss
Source: USAMO 2001 #3
Let $a, b, c \geq 0$ and satisfy \[ a^2+b^2+c^2 +abc = 4 . \] Show that \[ 0 \le ab + bc + ca - abc \leq 2. \]
72 replies
cn2_71828182846
Jun 27, 2004
endless_abyss
Yesterday at 4:12 PM
Special Points on $BC$
tenniskidperson3   40
N Yesterday at 3:43 PM by dipinsubedi
Source: 2013 USAMO Problem 6
Let $ABC$ be a triangle. Find all points $P$ on segment $BC$ satisfying the following property: If $X$ and $Y$ are the intersections of line $PA$ with the common external tangent lines of the circumcircles of triangles $PAB$ and $PAC$, then \[\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.\]
40 replies
tenniskidperson3
May 1, 2013
dipinsubedi
Yesterday at 3:43 PM
Alcumus vs books
UnbeatableJJ   12
N Yesterday at 2:59 PM by pingpongmerrily
If I am aiming for AIME, then JMO afterwards, is Alcumus adequate, or I still need to do the problems on AoPS books?

I got AMC 23 this year, and never took amc 10 before. If I master the alcumus of intermediate algebra (making all of the bars blue). How likely I can qualify for AIME 2026?
12 replies
UnbeatableJJ
Apr 23, 2025
pingpongmerrily
Yesterday at 2:59 PM
Zsigmondy's theorem
V0305   17
N Yesterday at 7:22 AM by whwlqkd
Is Zsigmondy's theorem allowed on the IMO, and is it allowed on the AMC series of proof competitions (e.g. USAJMO, USA TSTST)?
17 replies
V0305
May 24, 2025
whwlqkd
Yesterday at 7:22 AM
USAJMO problem 3: Inequality
BOGTRO   106
N Monday at 11:12 PM by Learning11
Let $a,b,c$ be positive real numbers. Prove that $\frac{a^3+3b^3}{5a+b}+\frac{b^3+3c^3}{5b+c}+\frac{c^3+3a^3}{5c+a} \geq \frac{2}{3}(a^2+b^2+c^2)$.
106 replies
BOGTRO
Apr 24, 2012
Learning11
Monday at 11:12 PM
what the yap
KevinYang2.71   31
N Monday at 3:26 PM by Blast_S1
Source: USAMO 2025/3
Alice the architect and Bob the builder play a game. First, Alice chooses two points $P$ and $Q$ in the plane and a subset $\mathcal{S}$ of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: for each pair $A,\,B$ of cities, they are connected with a road along the line segment $AB$ if and only if the following condition holds:
[center]For every city $C$ distinct from $A$ and $B$, there exists $R\in\mathcal{S}$ such[/center]
[center]that $\triangle PQR$ is directly similar to either $\triangle ABC$ or $\triangle BAC$.[/center]
Alice wins the game if (i) the resulting roads allow for travel between any pair of cities via a finite sequence of roads and (ii) no two roads cross. Otherwise, Bob wins. Determine, with proof, which player has a winning strategy.

Note: $\triangle UVW$ is directly similar to $\triangle XYZ$ if there exists a sequence of rotations, translations, and dilations sending $U$ to $X$, $V$ to $Y$, and $W$ to $Z$.
31 replies
KevinYang2.71
Mar 20, 2025
Blast_S1
Monday at 3:26 PM
Mustang Math Recruitment is Open!
MustangMathTournament   3
N Monday at 3:21 PM by Puzzlebooks206
The Interest Form for joining Mustang Math is open!

Hello all!

We're Mustang Math, and we are currently recruiting for the 2025-2026 year! If you are a high school or college student and are passionate about promoting an interest in competition math to younger students, you should strongly consider filling out the following form: https://link.mustangmath.com/join. Every member in MM truly has the potential to make a huge impact, no matter your experience!

About Mustang Math

Mustang Math is a nonprofit organization of high school and college volunteers that is dedicated to providing middle schoolers access to challenging, interesting, fun, and collaborative math competitions and resources. Having reached over 4000 U.S. competitors and 1150 international competitors in our first six years, we are excited to expand our team to offer our events to even more mathematically inclined students.

PROJECTS
We have worked on various math-related projects. Our annual team math competition, Mustang Math Tournament (MMT) recently ran. We hosted 8 in-person competitions based in Washington, NorCal, SoCal, Illinois, Georgia, Massachusetts, Nevada and New Jersey, as well as an online competition run nationally. In total, we had almost 900 competitors, and the students had glowing reviews of the event. MMT International will once again be running later in August, and with it, we anticipate our contest to reach over a thousand students.

In our classes, we teach students math in fun and engaging math lessons and help them discover the beauty of mathematics. Our aspiring tech team is working on a variety of unique projects like our website and custom test platform. We also have a newsletter, which, combined with our social media presence, helps to keep the mathematics community engaged with cool puzzles, tidbits, and information about the math world! Our design team ensures all our merch and material is aesthetically pleasing.

Some highlights of this past year include 1000+ students in our classes, AMC10 mock with 150+ participants, our monthly newsletter to a subscriber base of 6000+, creating 8 designs for 800 pieces of physical merchandise, as well as improving our custom website (mustangmath.com, 20k visits) and test-taking platform (comp.mt, 6500+ users).

Why Join Mustang Math?

As a non-profit organization on the rise, there are numerous opportunities for volunteers to share ideas and suggest projects that they are interested in. Through our organizational structure, members who are committed have the opportunity to become a part of the leadership team. Overall, working in the Mustang Math team is both a fun and fulfilling experience where volunteers are able to pursue their passion all while learning how to take initiative and work with peers. We welcome everyone interested in joining!

More Information

To learn more, visit https://link.mustangmath.com/RecruitmentInfo. If you have any questions or concerns, please email us at contact@mustangmath.com.

https://link.mustangmath.com/join
3 replies
MustangMathTournament
May 24, 2025
Puzzlebooks206
Monday at 3:21 PM
[TEST RELEASED] OMMC Year 5
DottedCaculator   173
N Monday at 3:01 PM by drhong
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]
173 replies
DottedCaculator
Apr 26, 2025
drhong
Monday at 3:01 PM
Two permutations
Nima Ahmadi Pour   13
N Yesterday at 7:50 PM by awesomeming327.
Source: Iran prepration exam
Suppose that $ a_1$, $ a_2$, $ \ldots$, $ a_n$ are integers such that $ n\mid a_1 + a_2 + \ldots + a_n$.
Prove that there exist two permutations $ \left(b_1,b_2,\ldots,b_n\right)$ and $ \left(c_1,c_2,\ldots,c_n\right)$ of $ \left(1,2,\ldots,n\right)$ such that for each integer $ i$ with $ 1\leq i\leq n$, we have
\[ n\mid a_i - b_i - c_i
\]

Proposed by Ricky Liu & Zuming Feng, USA
13 replies
Nima Ahmadi Pour
Apr 24, 2006
awesomeming327.
Yesterday at 7:50 PM
Two permutations
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Source: Iran prepration exam
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Nima Ahmadi Pour
160 posts
#1 • 2 Y
Y by Adventure10 and 1 other user
Suppose that $ a_1$, $ a_2$, $ \ldots$, $ a_n$ are integers such that $ n\mid a_1 + a_2 + \ldots + a_n$.
Prove that there exist two permutations $ \left(b_1,b_2,\ldots,b_n\right)$ and $ \left(c_1,c_2,\ldots,c_n\right)$ of $ \left(1,2,\ldots,n\right)$ such that for each integer $ i$ with $ 1\leq i\leq n$, we have
\[ n\mid a_i - b_i - c_i
\]

Proposed by Ricky Liu & Zuming Feng, USA
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ZetaX
7579 posts
#2 • 3 Y
Y by Adventure10, Mango247, and 1 other user
This problem and the following generalisation appeared 1979 in Ars Combinatoria (thanks to Darij who found it):

Let $ (G, + )$ be a finite abelian group of order $ n$.
Let also $ a_1,a_2,...,a_{n - 1} \in G$ be arbitrary.
Then there exist pairwise distinct $ b_1,b_2,...,b_{n - 1} \in G$ and pairwise distinct $ c_1,c_2,...,c_{n - 1} \in G$ such that $ a_k = b_k + c_k$ for $ k = 1,2,...,n - 1$.

[Moderator edit: The Ars Combinatoria paper is:
F. Salzborn, G. Szekeres, A problem in Combinatorial Group Theory, Ars Combinatoria 7 (1979), pp. 3-5.]
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epitomy01
240 posts
#3 • 3 Y
Y by Dan37kosothangnao, Adventure10, Mango247
so could someone post a proof of either the problem or its generalization?
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spdf
136 posts
#4 • 1 Y
Y by Adventure10
you can find the proof in file shortlist 2005 which has been posted by orl
The main idea is given two sequence $a_{1}...a_{n}$ and $b_{1}...b_{n}$ s.t $\sum a_{i}\equiv 0(mod n)$ and $\sum b_{i}\equiv 0(mod n)$ and there are exactly two i;j s.t $a_{i}\neq\ b_{i}(modn)$ and $a_{j}\neq\ b_{j}(modn)$.Then if we know two permutation good for the sequence (a_1...a_n) then we can build two permuttionm good for (b_1...b_n)
i will come back with detail if you need
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ZetaX
7579 posts
#5 • 1 Y
Y by Adventure10
Well, I will post the solution from Ars Combinatoria if a re-find that two sheets of paper...
It's a bit different from the ISL one.
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keira_khtn
485 posts
#6 • 2 Y
Y by Adventure10, Mango247
I think you didnt keep promise,Zetax :lol: Please post it here and now!
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bodom
123 posts
#7 • 2 Y
Y by Adventure10, Mango247
to spdf: that was also my idea when i first saw the problem but i can't find a good way to contruct those 2 permutations for $ b_j$.you said you can post details.please do so :)
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ZetaX
7579 posts
#8 • 2 Y
Y by Adventure10, Mango247
Sorry for not repsonding (I merely forgot...). But I just saw that problem again: In a slightly different manner (but being equivalent to this one here) it is solved in "The Mathematics of Juggling", called the "Converse of the Average Theorem".

Main ideas:
You show that this property (being a sum of two permutations) is invariant under the operations $ a_{i,j,d}$ that add $ d$ to $ a_i$ and subtract $ d$ from $ a_j$.
For this, you need to do it algorithmically (but describing it is a bit hard without that graphics...).
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arandomperson123
430 posts
#9 • 2 Y
Y by Adventure10, Mango247
ZetaX wrote:
Sorry for not repsonding (I merely forgot...). But I just saw that problem again: In a slightly different manner (but being equivalent to this one here) it is solved in "The Mathematics of Juggling", called the "Converse of the Average Theorem".

Main ideas:
You show that this property (being a sum of two permutations) is invariant under the operations $ a_{i,j,d}$ that add $ d$ to $ a_i$ and subtract $ d$ from $ a_j$.
For this, you need to do it algorithmically (but describing it is a bit hard without that graphics...).

that is what I tried to do, but I can not prove that we can do it for the general case... can someone please help?
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ThE-dArK-lOrD
4071 posts
#10 • 16 Y
Y by Lam.DL.01, Mosquitall, nmd27082001, Arc_archer, MathbugAOPS, iceillusion, Aryan-23, magicarrow, k12byda5h, gabrupro, Mop2018, Adventure10, Mango247, bhan2025, CyclicISLscelesTrapezoid, winniep008hfi
Since it's almost twelve years without complete solution, here's the official solution:

Suppose there exists permutations $\sigma$ and $\tau$ of $[n]$ for some sequence $\{ a_i\}_{i\in [n]}$ so that $a_i\equiv_n \sigma (i)+\tau (i)$ for all $i\in [n]$.
Given a sequence $\{ b_i\}_{i\in [n]}$ with sum divisible by $n$ that differ, in modulo $n$, from $\{ a_i\}_{i\in [n]}$ in only two positions, say $i_1$ and $i_2$.
We want to construct permutations $\sigma'$ and $\tau'$ of $[n]$ so that $b_i\equiv_n \sigma' (i) +\tau' (i)$ for all $i\in [n]$.
Recall that $b_i\equiv a_i\pmod{n}$ for all $i\in [n]$ that $i\neq i_1,i_2$.
Construct a sequence $i_1,i_2,i_3,...$ by, for each integer $k\geq 2$, define $i_{k+1}\in [n]$ to be the unique integer satisfy $\sigma (i_{k-1})+\tau (i_{k+1})\equiv_n b_{i_k}$.
Let (clearly exists) $p<q$ are the indices that $i_p=i_q$ with minimal $p$, and then minimal $q$.

If $p>2$. This means $i_j\not\in \{ i_1,i_2\} \implies \sigma (i_j) +\tau (i_j) \equiv_n b_{i_j}$ for all $j\in \{ p,p+1,...,q\}$.
Summing the equation $\sigma (i_{k-1})+\tau (i_{k+1})\equiv_n b_{i_k}$ for $k\in \{ p,p+1,...,q-1\}$ gives us
$$\sum_{j=p-1}^{q-2}{\sigma (i_j) } +\sum_{j=p+1}^{q}{\tau (i_j)} \equiv_n\sum_{j=p}^{q-1}{b_{i_j}} \implies \sigma (i_{p-1}) +\sigma (i_p) +\tau (i_{q-1}) +\tau (i_q) \equiv_nb_{i_p}+b_{i_{q-1}}.$$Plugging $i_p=i_q$ and use $\sigma (i_p) +\tau (i_p)\equiv_n b_{i_p}$ gives us $\sigma (i_{p-1}) +\tau (i_{q-1})\equiv_n b_{i_{q-1}} \equiv_n \sigma (i_{q-1})+\tau (i_{q-1})$.
Hence, $\sigma (i_{p-1}) \equiv_n \sigma (i_{q-1})\implies i_{p-1}=i_{q-1}$, contradiction to the definition of $p,q$.

So, we've $p\in \{ 1,2\}$. Let $p'=3-p$. Define the desired permutations $\sigma'$ and $\tau'$ as follows:
$$\sigma' (i_l)=\begin{cases} 
\sigma (i_{l-1}), & \text{ if } l\in \{ 2,3,...,q-1\} \\
\sigma (i_{q-1}), & \text{ if } l=1
\end{cases} ,\tau' (i_l)= \begin{cases} 
\tau (i_{l+1}), & \text{ if } l\in \{ 2,3,...,q-1\} \\
\tau (i_{p'}), & \text{ if } l=1
\end{cases}  $$and $\sigma' (i) =\sigma (i),\tau' (i)=\tau (i)$ for the rest $i\in [n]$ that $i\not\in \{ i_1,i_2,...,i_{q-1}\}$.
Note that the reason we choose $\tau (i_{p'})$ is just to not use $\tau (i_p)=\tau (i_{(q-1)+1})$ more than one time.
This construction gives us $\sigma' (i)+\tau' (i)\equiv_n b_i$ for all $i\in [n]$ except when $i=i_1$.
But since both $\sigma'$ and $\tau'$ are permutations of $[n]$, we've $\sum_{i\in [n]}{(\sigma' (i)+\tau' (i))} \equiv_n 2\times \frac{n(n-1)}{2}\equiv_n 0\equiv_n \sum_{i\in [n]}{b_i}$.
This guarantee that $\sigma' (i) +\tau' (i)\equiv_n b_i$ when $i=i_1$ too. This prove the validity of permutations we constructed.
This post has been edited 3 times. Last edited by ThE-dArK-lOrD, Jan 16, 2018, 3:07 PM
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mathleticguyyy
3217 posts
#11
Y by
The case with $n$ prime is also resolved in this paper
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Bataw
43 posts
#13
Y by
any other solutions ?
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Zhaom
5124 posts
#16
Y by
:star_struck:

Plot $\left(b_k,c_k\right)$ for $k=1,2,\cdots,n$ on the coordinate plane with coordinates modulo $n$. We want to show that it is possible to choose the $n$ points plotted to have $\left\{b_1+c_1,b_2+c_2,\cdots,b_n+c_n\right\}$ be any multiset of $n$ elements $\pmod{n}$ such that sum of the elements is $0\pmod{n}$. Note that the $n$ points plotted can be any $n$ points not in the same row or column.

Instead, we will prove that we can choose $n-1$ points not in the same row or column such that the multiset $S$ of $x+y$ for all of the points $(x,y)$ can be any multiset of $n-1$ elements $\pmod{n}$. Then, we can choose the unique point not in the same row or column as the $n-1$ points to obtain $n$ points for the original statement, as the sum of the coordinates of all $n$ points is necessarily $2\cdot\left(0+1+\cdots+(n-1)\right)\equiv(n-1)n\equiv0\pmod{n}$.

We will construct the $n-1$ points in the following way.

We start with $n-1$ arbitrary points not in the same row or column. We will change the $n-1$ points in such a way that we can replace any element of $S$ by any residue $\pmod{n}$. Suppose we change the element in $S$ corresponding to a point $P$. Then, we will shift $P$ horizontally until the sum of the coordinates of $P$ is the desired value $\pmod{n}$. Now, we might have two points in the same column. We will repeatedly perform the following operation.

Note that there are exactly $n-1$ distinct rows taken up by the $n-1$ points. If the last point moved was $P$ and it is in the same column as $Q$, we will move $Q$ on the line through $Q$ with slope $-1$ until $Q$ occupies the previously empty row.

It suffices that this cannot last forever.

Claim. This operation is a reflection over a fixed line.

Proof. Suppose that this operation moves $P$ to a point $P'$, then on the next step it moves $Q$ to a point $Q'$. We see that $\overline{PP'}$ and $\overline{QQ'}$ both have slope $-1$. Now, note that $P'$ and $Q$ must be in the same row. Furthermore, after the operation moves $P$ to $P'$, the row containing $P$ must be empty, meaning that $Q$ must be moved to that row. This implies that $P$ and $Q'$ are in the same row. Therefore, we see that $PP'QQ'$ is a cyclic isosceles trapezoid and in particular the perpendicular bisectors of $\overline{PP'}$ and $\overline{QQ'}$ are the same, proving the claim.

Now, assume for the sake of contradiction that the operation lasts forever. We will label the $n-1$ points to distinguish them. Now, there are finitely many states of the $n-1$ points, so eventually the operation must loop through a cycle of states. In this cycle, start at an arbitrary state. Now, since at least $1$ point must be moved and then return back to its original position, consider the point $P$ which returns to its original position the earliest. When $P$ was moved, the next point $Q$ which was moved must have occupied the original row of $P$. However, if $P$ was moved back to its original position, then the original row of $P$ must be vacant, so $Q$ must have been moved. However, this implies that $Q$ was moved back to its original position before $P$ by the claim, a contradiction, so we are done.
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awesomeming327.
1736 posts
#17
Y by
Call a sequence $a_1$, $a_2$, $a_3$, $\dots$, $a_n$ good if the two permutations $b$ and $c$ exist such that $b_i+c_i=a_i$ for all $i$. Now consider another sequence that differs from $a$ in only two spots, at $a_{i_1}$ and $a_{i_2}$. Let the modified sequence be $a'_1,a'_2,\dots,a'_n$.

Let $a'_{i_2}=b_{i_1}+c_{i_3}$ for some $i_3$. Let $a'_{i_3}=b_{i_2}+c_{i_4}$ for some $i_4$. Let this process continue until we see a repeat of indices. That is, let $y$ be the smallest index such that there exists $x<y$ for which $i_x=i_y$ following this pattern. We have:
\[b_{i_{x-1}}+\dots+b_{i_{y-2}}+c_{i_{x+2}}+\dots+c_{i_y}=a'_{i_x}+\dots+a'_{i_{y-1}}\]Claim 1: $x=1$ or $2$.
If $i_y$ is not $i_1$ or $i_2$ then all of the ones from $i_x$ to $i_{y-1}$ are not $i_1$ or $i_2$, so $a'$ and $a$ sequences are identical. Thus,
\begin{align*}
b_{i_{x-1}}+\dots+b_{i_{y-2}}+c_{i_{x+1}}+\dots+c_{i_y} &= a_{i_x}+\dots+a_{i_{y-1}} \\
&= b_{i_{x}}+\dots+b_{i_{y-1}}+c_{i_{x}}+\dots+c_{i_{y-1}}
\end{align*}which implies $b_{i_{x-1}}-b_{i_{y-1}}=c_{i_{x}}-c_{i_y}=0$ so $i_{x-1}=i_{y-1}$, contradiction.
If $a_{i_y}=b_{i_y-1}+c_{i_2}$ then $a_{i_1}=b_{i_y}+c_{i_1}$ and vice versa, which works because the sum of the terms match, so matching $n-1$ of the terms will automatically match the last one. Leave all the rest of the terms unchanged.

We proved that any sequence that differs by a good sequence by two values is also good, so we can use this method to reach any sequence satisfying the property.
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