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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.

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[*]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]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
GCD of terms in a sequence
BBNoDollar   0
an hour ago
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
0 replies
BBNoDollar
an hour ago
0 replies
Number Theory
fasttrust_12-mn   13
N an hour ago by KTYC
Source: Pan African Mathematics Olympiad P1
Find all positive intgers $a,b$ and $c$ such that $\frac{a+b}{a+c}=\frac{b+c}{b+a}$ and $ab+bc+ca$ is a prime number
13 replies
fasttrust_12-mn
Aug 15, 2024
KTYC
an hour ago
GCD of terms in a sequence
BBNoDollar   0
an hour ago
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
0 replies
BBNoDollar
an hour ago
0 replies
Aime type Geo
ehuseyinyigit   3
N an hour ago by sami1618
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
3 replies
ehuseyinyigit
Yesterday at 9:04 PM
sami1618
an hour ago
minimizing sum
gggzul   1
N 2 hours ago by RedFireTruck
Let $x, y, z$ be real numbers such that $x^2+y^2+z^2=1$. Find
$$min\{12x-4y-3z\}.$$
1 reply
gggzul
3 hours ago
RedFireTruck
2 hours ago
Equilateral Triangle inside Equilateral Triangles.
abhisruta03   2
N 2 hours ago by Reacheddreams
Source: ISI 2021 P6
If a given equilateral triangle $\Delta$ of side length $a$ lies in the union of five equilateral triangles of side length $b$, show that there exist four equilateral triangles of side length $b$ whose union contains $\Delta$.
2 replies
abhisruta03
Jul 18, 2021
Reacheddreams
2 hours ago
USAMO 1984 Problem 5 - Polynomial of degree 3n
Binomial-theorem   8
N 2 hours ago by Assassino9931
Source: USAMO 1984 Problem 5
$P(x)$ is a polynomial of degree $3n$ such that

\begin{eqnarray*}
P(0) = P(3) = \cdots &=& P(3n) = 2, \\
P(1) = P(4) = \cdots &=& P(3n-2) = 1, \\
P(2) = P(5) = \cdots &=& P(3n-1) = 0, \quad\text{ and }\\
&& P(3n+1) = 730.\end{eqnarray*}

Determine $n$.
8 replies
Binomial-theorem
Aug 16, 2011
Assassino9931
2 hours ago
Finding positive integers with good divisors
nAalniaOMliO   2
N 2 hours ago by KTYC
Source: Belarusian National Olympiad 2025
For every positive integer $n$ write all its divisors in increasing order: $1=d_1<d_2<\ldots<d_k=n$.
Find all $n$ such that $2025 \cdot n=d_{20} \cdot d_{25}$.
2 replies
1 viewing
nAalniaOMliO
Mar 28, 2025
KTYC
2 hours ago
Balkan MO 2025 p1
Mamadi   1
N 3 hours ago by KevinYang2.71
Source: Balkan MO 2025
An integer \( n > 1 \) is called good if there exists a permutation \( a_1, a_2, \dots, a_n \) of the numbers \( 1, 2, 3, \dots, n \), such that:

\( a_i \) and \( a_{i+1} \) have different parities for every \( 1 \le i \le n - 1 \)

the sum \( a_1 + a_2 + \dots + a_k \) is a quadratic residue modulo \( n \) for every \( 1 \le k \le n \)

Prove that there exist infinitely many good numbers, as well as infinitely many positive integers which are not good.

Remark: Here an integer \( z \) is considered a quadratic residue modulo \( n \) if there exists an integer \( y \) such that \( y^2 \equiv z \pmod{n} \).
1 reply
Mamadi
4 hours ago
KevinYang2.71
3 hours ago
Random Points = Problem
kingu   4
N 3 hours ago by zuat.e
Source: Chinese Geometry Handout
Let $ABC$ be a triangle. Let $\omega$ be a circle passing through $B$ intersecting $AB$ at $D$ and $BC$ at $F$. Let $G$ be the intersection of $AF$ and $\omega$. Further, let $M$ and $N$ be the intersections of $FD$ and $DG$ with the tangent to $(ABC)$ at $A$. Now, let $L$ be the second intersection of $MC$ and $(ABC)$. Then, prove that $M$ , $L$ , $D$ , $E$ and $N$ are concyclic.
4 replies
kingu
Apr 27, 2024
zuat.e
3 hours ago
CooL geo
Pomegranat   2
N 3 hours ago by Curious_Droid
Source: Idk

In triangle \( ABC \), \( D \) is the midpoint of \( BC \). \( E \) is an arbitrary point on \( AC \). Let \( S \) be the intersection of \( AD \) and \( BE \). The line \( CS \) intersects with the circumcircle of \( ACD \), for the second time at \( K \). \( P \) is the circumcenter of triangle \( ABE \). Prove that \( PK \perp CK \).
2 replies
Pomegranat
Yesterday at 5:57 AM
Curious_Droid
3 hours ago
Unique Rational Number Representation
abhisruta03   18
N 3 hours ago by Reacheddreams
Source: ISI 2021 P3
Prove that every positive rational number can be expressed uniquely as a finite sum of the form $$a_1+\frac{a_2}{2!}+\frac{a_3}{3!}+\dots+\frac{a_n}{n!},$$where $a_n$ are integers such that $0 \leq a_n \leq n-1$ for all $n > 1$.
18 replies
abhisruta03
Jul 18, 2021
Reacheddreams
3 hours ago
Math solution
Techno0-8   1
N 3 hours ago by jasperE3
Solution
1 reply
Techno0-8
6 hours ago
jasperE3
3 hours ago
D1027 : Super Schoof
Dattier   1
N 3 hours ago by Dattier
Source: les dattes à Dattier
Let $p>11$ a prime number with $a=\text{card}\{(x,y) \in \mathbb Z/ p \mathbb Z: y^2=x^3+1\}$ and $b=\dfrac 1 {((p-1)/2)! \times ((p-1)/3)! \times ((p-1)/6)!} \mod p$ when $p \mod 3=1$.



Is it true that if $p \mod 3=1$ then $a \in \{b,p-b, \min\{b,p-b\}+p\}$ else $A=p$.
1 reply
Dattier
Today at 5:15 PM
Dattier
3 hours ago
Two permutations
Nima Ahmadi Pour   12
N Apr 23, 2025 by Zhaom
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
12 replies
Nima Ahmadi Pour
Apr 24, 2006
Zhaom
Apr 23, 2025
Two permutations
G H J
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
Z K Y
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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.]
Z K Y
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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?
Z K Y
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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
Z K Y
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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.
Z K Y
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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!
Z K Y
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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 :)
Z K Y
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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...).
Z K Y
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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
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The case with $n$ prime is also resolved in this paper
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Bataw
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any other solutions ?
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Zhaom
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: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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