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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.

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Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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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
parallel wanted, right triangle, circumcircle, angle bisector related
parmenides51   6
N 7 minutes ago by Ianis
Source: Norwegian Mathematical Olympiad 2020 - Abel Competition p4b
The triangle $ABC$ has a right angle at $A$. The centre of the circumcircle is called $O$, and the base point of the normal from $O$ to $AC$ is called $D$. The point $E$ lies on $AO$ with $AE = AD$. The angle bisector of $\angle CAO$ meets $CE$ in $Q$. The lines $BE$ and $OQ$ intersect in $F$. Show that the lines $CF$ and $OE$ are parallel.
6 replies
parmenides51
Apr 26, 2020
Ianis
7 minutes ago
IMO ShortList 2008, Number Theory problem 5
April   25
N an hour ago by awesomeming327.
Source: IMO ShortList 2008, Number Theory problem 5, German TST 6, P2, 2009
For every $ n\in\mathbb{N}$ let $ d(n)$ denote the number of (positive) divisors of $ n$. Find all functions $ f: \mathbb{N}\to\mathbb{N}$ with the following properties: [list][*] $ d\left(f(x)\right) = x$ for all $ x\in\mathbb{N}$.
[*] $ f(xy)$ divides $ (x - 1)y^{xy - 1}f(x)$ for all $ x$, $ y\in\mathbb{N}$.[/list]

Proposed by Bruno Le Floch, France
25 replies
April
Jul 9, 2009
awesomeming327.
an hour ago
IMO Shortlist 2014 N2
hajimbrak   32
N an hour ago by ezpotd
Determine all pairs $(x, y)$ of positive integers such that \[\sqrt[3]{7x^2-13xy+7y^2}=|x-y|+1.\]
Proposed by Titu Andreescu, USA
32 replies
hajimbrak
Jul 11, 2015
ezpotd
an hour ago
An easy geometry problem in NEHS Mock APMO
chengbilly   2
N 2 hours ago by MathLuis
Source: own
Let $ABC$ be a triangle with circumcenter $O$ and orthocenter $H$. $AD,BE,CF$ the altitudes of $\triangle ABC$. A point $T$ lies on line $EF$ such that $DT \perp EF$. A point $X$ lies on the circumcircle of $\triangle ABC$ such that $AX,EF,DO$ are concurrent. $DT$ meets $AX$ at $R$. Prove that $H,T,R,X$ are concyclic.
2 replies
chengbilly
May 23, 2021
MathLuis
2 hours ago
The reflection of AD intersect (ABC) lies on (AEF)
alifenix-   62
N 2 hours ago by Rayvhs
Source: USA TST for EGMO 2020, Problem 4
Let $ABC$ be a triangle. Distinct points $D$, $E$, $F$ lie on sides $BC$, $AC$, and $AB$, respectively, such that quadrilaterals $ABDE$ and $ACDF$ are cyclic. Line $AD$ meets the circumcircle of $\triangle ABC$ again at $P$. Let $Q$ denote the reflection of $P$ across $BC$. Show that $Q$ lies on the circumcircle of $\triangle AEF$.

Proposed by Ankan Bhattacharya
62 replies
alifenix-
Jan 27, 2020
Rayvhs
2 hours ago
PAMO 2022 Problem 1 - Line Tangent to Circle Through Orthocenter
DylanN   5
N 2 hours ago by Y77
Source: 2022 Pan-African Mathematics Olympiad Problem 1
Let $ABC$ be a triangle with $\angle ABC \neq 90^\circ$, and $AB$ its shortest side. Let $H$ be the orthocenter of $ABC$. Let $\Gamma$ be the circle with center $B$ and radius $BA$. Let $D$ be the second point where the line $CA$ meets $\Gamma$. Let $E$ be the second point where $\Gamma$ meets the circumcircle of the triangle $BCD$. Let $F$ be the intersection point of the lines $DE$ and $BH$.

Prove that the line $BD$ is tangent to the circumcircle of the triangle $DFH$.
5 replies
DylanN
Jun 25, 2022
Y77
2 hours ago
Conditional geo with centroid
a_507_bc   6
N 3 hours ago by LeYohan
Source: Singapore Open MO Round 2 2023 P1
In a scalene triangle $ABC$ with centroid $G$ and circumcircle $\omega$ centred at $O$, the extension of $AG$ meets $\omega$ at $M$; lines $AB$ and $CM$ intersect at $P$; and lines $AC$ and $BM$ intersect at $Q$. Suppose the circumcentre $S$ of the triangle $APQ$ lies on $\omega$ and $A, O, S$ are collinear. Prove that $\angle AGO = 90^{o}$.
6 replies
a_507_bc
Jul 1, 2023
LeYohan
3 hours ago
Channel name changed
Plane_geometry_youtuber   0
3 hours ago
Hi,

Due to the search handle issue in youtube. My channel is renamed to Olympiad Geometry Club. And the new link is as following:

https://www.youtube.com/@OlympiadGeometryClub

Recently I introduced the concept of harmonic bundle. I will move on to the conjugate median soon. In the future, I will discuss more than a thousand theorems on plane geometry and hopefully it can help to the students preparing for the Olympiad competition.

Please share this to the people may need it.

Thank you!
0 replies
Plane_geometry_youtuber
3 hours ago
0 replies
IMO Shortlist 2010 - Problem G1
Amir Hossein   134
N 3 hours ago by happypi31415
Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$

Proposed by Christopher Bradley, United Kingdom
134 replies
Amir Hossein
Jul 17, 2011
happypi31415
3 hours ago
interesting incenter/tangent circle config
LeYohan   0
4 hours ago
Source: 2022 St. Mary's Canossian College F4 Final Exam Mathematics Paper 1, Q 18d of 18 (modified)
$BC$ is tangent to the circle $AFDE$ at $D$. $AB$ and $AC$ cut the circle at $F$ and $E$ respectively. $I$ is the in-centre of $\triangle ABC$, and $D$ is on the line $AI$. $CI$ and $DE$ intersect at $G$, while $BI$ and $FD$ intersect at $P$. Prove that the points $P, F, G, E$ lie on a circle.
0 replies
LeYohan
4 hours ago
0 replies
interesting geo config (2/3)
Royal_mhyasd   5
N 4 hours ago by Royal_mhyasd
Source: own
Let $\triangle ABC$ be an acute triangle and $H$ its orthocenter. Let $P$ be a point on the parallel through $A$ to $BC$ such that $\angle APH = |\angle ABC-\angle ACB|$. Define $Q$ and $R$ as points on the parallels through $B$ to $AC$ and through $C$ to $AB$ similarly. If $P,Q,R$ are positioned around the sides of $\triangle ABC$ as in the given configuration, prove that $P,Q,R$ are collinear.
5 replies
Royal_mhyasd
Saturday at 11:36 PM
Royal_mhyasd
4 hours ago
interesting geometry config (3/3)
Royal_mhyasd   2
N 4 hours ago by Royal_mhyasd
Let $\triangle ABC$ be an acute triangle, $H$ its orthocenter and $E$ the center of its nine point circle. Let $P$ be a point on the parallel through $C$ to $AB$ such that $\angle CPH = |\angle BAC-\angle ABC|$ and $P$ and $A$ are on different sides of $BC$ and $Q$ a point on the parallel through $B$ to $AC$ such that $\angle BQH = |\angle BAC - \angle ACB|$ and $C$ and $Q$ are on different sides of $AB$. If $B'$ and $C'$ are the reflections of $H$ over $AC$ and $AB$ respectively, $S$ and $T$ are the intersections of $B'Q$ and $C'P$ respectively with the circumcircle of $\triangle ABC$, prove that the intersection of lines $CT$ and $BS$ lies on $HE$.

final problem for this "points on parallels forming strange angles with the orthocenter" config, for now. personally i think its pretty cool :D
2 replies
Royal_mhyasd
Yesterday at 7:06 AM
Royal_mhyasd
4 hours ago
Convex Quadrilateral with Bisector Diagonal
matinyousefi   8
N 5 hours ago by lpieleanu
Source: Germany TST 2017
In a convex quadrilateral $ABCD$, $BD$ is the angle bisector of $\angle{ABC}$. The circumcircle of $ABC$ intersects $CD,AD$ in $P,Q$ respectively and the line through $D$ parallel to $AC$ cuts $AB,AC$ in $R,S$ respectively. Prove that point $P,Q,R,S$ lie on a circle.
8 replies
matinyousefi
Apr 11, 2020
lpieleanu
5 hours ago
collinear wanted, toucpoints of incircle related
parmenides51   2
N 5 hours ago by Tamam
Source: 2018 Thailand October Camp 1.2
Let $\Omega$ be the inscribed circle of a triangle $\vartriangle ABC$. Let $D, E$ and $F$ be the tangency points of $\Omega$ and the sides $BC, CA$ and $AB$, respectively, and let $AD, BE$ and $CF$ intersect $\Omega$ at $K, L$ and $M$, respectively, such that $D, E, F, K, L$ and $M$ are all distinct. The tangent line of $\Omega$ at $K$ intersects $EF$ at $X$, the tangent line of $\Omega$ at $L$ intersects $DE$ at $Y$ , and the tangent line of $\Omega$ at M intersects $DF$ at $Z$. Prove that $X,Y$ and $Z$ are collinear.
2 replies
parmenides51
Oct 15, 2020
Tamam
5 hours ago
Bishops and permutations
Assassino9931   8
N Mar 30, 2025 by awesomeming327.
Source: RMM 2024 Problem 1
Let $n$ be a positive integer. Initially, a bishop is placed in each square of the top row of a $2^n \times 2^n$
chessboard; those bishops are numbered from $1$ to $2^n$ from left to right. A jump is a simultaneous move made by all bishops such that each bishop moves diagonally, in a straight line, some number of squares, and at the end of the jump, the bishops all stand in different squares of the same row.

Find the total number of permutations $\sigma$ of the numbers $1, 2, \ldots, 2^n$ with the following property: There exists a sequence of jumps such that all bishops end up on the bottom row arranged in the order $\sigma(1), \sigma(2), \ldots, \sigma(2^n)$, from left to right.

Israel
8 replies
Assassino9931
Feb 29, 2024
awesomeming327.
Mar 30, 2025
Bishops and permutations
G H J
Source: RMM 2024 Problem 1
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Assassino9931
1383 posts
#1 • 2 Y
Y by Phorphyrion, Zfn.nom-_nom
Let $n$ be a positive integer. Initially, a bishop is placed in each square of the top row of a $2^n \times 2^n$
chessboard; those bishops are numbered from $1$ to $2^n$ from left to right. A jump is a simultaneous move made by all bishops such that each bishop moves diagonally, in a straight line, some number of squares, and at the end of the jump, the bishops all stand in different squares of the same row.

Find the total number of permutations $\sigma$ of the numbers $1, 2, \ldots, 2^n$ with the following property: There exists a sequence of jumps such that all bishops end up on the bottom row arranged in the order $\sigma(1), \sigma(2), \ldots, \sigma(2^n)$, from left to right.

Israel
This post has been edited 1 time. Last edited by Assassino9931, Mar 4, 2024, 10:59 AM
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DottedCaculator
7357 posts
#2 • 1 Y
Y by khina
Reindex to $0$ through $2^n-1$. Notice that the possible jumps map $k$ to $k\oplus2^m$ and move $2^m$ rows, so $\sigma(k)=k\oplus z$ for all odd $0\leq z<2^n$ by parity, giving a total of $\boxed{2^{n-1}}$ possible $\sigma$.
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ihatemath123
3449 posts
#3 • 1 Y
Y by ATGY
Number the columns $0$ thru $2^n-1$. It's easy to see by induction or whatnot that the "bishop moving" process is equivalent to picking a place value $i$ from $0$ to $n-1$, and toggling the $i$th binary digit of each bishop's column number.

Define the value of a move as $i$, if it toggles the $i$th binary digit in the column number of each bishop.

Moves of the same value that appear twice will cancel each other out, so the final permutation of our bishops is just determined by whether we make an even or odd number of moves of each value. We must make an odd number of moves of value $1$, since the bishops traverse an odd number of rows, but otherwise, the other $n-1$ parities can be anything. It follows that the answer is $2^{n-1}$.
This post has been edited 2 times. Last edited by ihatemath123, Mar 1, 2024, 2:50 AM
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VicKmath7
1391 posts
#4
Y by
Solution
This post has been edited 15 times. Last edited by VicKmath7, Mar 2, 2024, 3:17 PM
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vsamc
3789 posts
#5
Y by
Solution
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Leo.Euler
577 posts
#7 • 3 Y
Y by TheBlackPuzzle913, Radmandookheh, MATH-TITAN
First, suppose that the bishops can displace by $k$ rows down. Then, it follows that the first $k$ bishops have to displace by $+k$ horizontally, and the next $k$ are consequently forced to displace by $-k$. This gives us a ``connected block" of $2k$ bishops: an example with $k=3$ is shown below.
https://cdn.discordapp.com/attachments/1069403895304040531/1214294379540385852/Screen_Shot_2024-03-04_at_11.32.18_AM.png?ex=65f896c4&is=65e621c4&hm=997671dfdfdd13deb23c574978cdd1d5f5a3e1a69bf11e49bda9322e586b1943& Repeating this on the remaining bishops, we find that $2k \mid 2^n$, or $k \mid 2^{n-1}$. The construction for such $k$ recycles the connected blocks idea.

Using the construction for $k \mid 2^{n-1}$, it is easy to see that the move of displacing by $k$ rows down is equivalent to toggling the $(1 + \log_2 k)$th digit from the right in the binary representation of the $x$-coordinate of a bishop, where we set the $x$-coordinates to be from $0$ to $2^n-1$.

Thus, we realize that the end permutation corresponds to toggling some of the digits of the binary representations of the numbers. In particular, each end permutation corresponds to some subset of $S$ of $\{2^0, 2^1, \dots, 2^{n-1}\}$ such that there exists a linear combination of the elements of $S$ with odd coefficients that sums to $2^n-1$. We can construct a bijection with such $S$ and the binary representations of all odd positive integers less than or equal to $2^n - 1$, which is $\boxed{2^{n-1}}$. Done.
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lnzhonglp
120 posts
#8
Y by
Let the size of a jump be the number of rows down it moves the bishops. It's easy to see that we can only make jumps of size $2^k$ for $k < n$. A jump of size $2^k$ swaps adjacent groups of $2^k$ bishops, so the order of the jumps doesn't matter. Two jumps of the same size cancel each other out, so we can replace any jump of size $2^k$ with two jumps of size $2^{k-1}$, effectively deleting the jump. $2^n = 2^{n-1} + \dots + 1$, so we can choose whether to include a jump of size $2^k$ for each of $1 \leq k \leq n-1$. Therefore, the answer is $2^{n-1}$.
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MathLuis
1559 posts
#10 • 1 Y
Y by dolphinday
Thought process is extremely important for this question, i like it.
Fitst of all why do we have $2^n$ here?, try to perform a move from a row to a row that is a distance that contains an odd prime factor of row's ahead ;) . You will notice that you really can't do this from a forcing algorithm to perform such move (i.e. looking where bishop $1$ goes, then $2$ then $3$ and so on) it must satisfy to cover the entire row so in fact the distance tells the movement of the bishops in "chunks", and since it has an odd prime factor it can't divide $2^n$ so you can only move $2^m$ rows ahead for some non-negative $m \le n-1$ (you can't achieve $m=n$ directly because of the bishops in the middle).
The following observation that you can permute the process, is key and this is true because you can check by the "chunks" where each bishop is assigned so you can swap two processes that are next to each other and thus do that to swap any two. And also note that performing a move an even number of times is basically not performing a move at all (from this observation).
The reason why this is so important is because then by a chessboard coloring a bishop keeps it's square color invariant so it only has $2^{n-1}$ possibilities in the otherside, but when we pick any of them the rest of the bishop movements are fixed because the sequence of moves can be permuted and that will change nothing in the end result!, so we can put the ones with the bigger moves at the start and check using the "chunks" for each move to realice that in the back-tracking there was only one way (after performing an optimization of repeating a move the least possible number of times (by transferring the repetitions to moves that move "the farest away" possible)) to get a bishop from a certain square to another, therefore there exists only $2^{n-1}$ possible permutations that can be achieved thus we are done :cool:.
This post has been edited 2 times. Last edited by MathLuis, Sep 5, 2024, 3:45 PM
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awesomeming327.
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We can represent each jump by a bijective function
\[f:\{1,2,\dots,2^n\}\to\{1,2,\dots,2^n\}\]such that the bishop that is on the $i$th column goes to the $f(i)$th column. A bishop move satisfies: $|f(i)-i|$ is the same as number of rows that the bishop traveled, so there exists a positive integer $c\le 2^n-1$ such that $|f(i)-i|=c$.

Claim 1: The function $f$ exists if and only if $c\in \{1,2,4,8,\dots, 2^{n-1}\}$, and this function is unique when it exists.
Note that $|f(c)-c|=c$ means that $f(c)=2c$. In order for $2c$ to be in the range we must have $c\le 2^{n-1}$.

When $i\le c$, $f(i)$ can equal $i-c$ or $i+c$. Since $i-c$ is not in the range, $f(i)=i+c$. The same can be said for the inverse: $f^{-1}(i)=i+c$. This implies that \[f(\{1,2,\dots,2c\})=\{1,2,\dots,2c\}\]Therefore, when we remove the values from $1$ to $2c$, $f$ is still a bijection. We can therefore apply the same logic inductively to get that we must be able to partition the domain of $f$ in to sets of size $2c$, which means $2c\mid 2^n$, or $c$ is a power of two. In each of these sets, the behavior of $f$ is determined. The claim is proved.
Let the functions that result from the previous claim be $f_i$ for $i=0$, $1$, $\dots$, $n-1$, where $f_i$ has $c=2^i$. Each of these functions, when considering disjoint cycle form of the permutation, is in reality $2^{n-1}$ different cycles of size $2$.

Claim 2: For any two functions $f_i$ and $f_j$ with $i\neq j$, we have $f_i\circ f_j=f_j\circ f_i$.
We will consider the behavior of each individual element. Suppose $i<j$. Let $x$ and $x+2^i$ be mapped to each other. Then $f_j(x)\equiv x\pmod {2^{i+1}}$ so the behavior of $f_j(x)$ and $x$ should be the same under $f_i(x)$, as should $f_j(x+2^i)$ and $x+2^i$. That is,
\[f_i(f_j(x+2^i))=f_j(x)=f_j(f_i(x+2^i))\]and less obviously,
\[f_i(f_j(x))=f_j(x)+2^i=f_j(x+2^i)=f_j(f_i(x))\]where the second equality of the second line is from the fact that $x$ and $x+2^i$ are in the same half of the $2c$-set of $f_j$, so their behavior should be the same.

Since after all our bishop jumps, we must end up an odd number of rows away, so $f_0$ is used an odd number of times. Since $f_i\circ f_i=\text{id}$ for all $i$, and since composition is commutative, we can view $\sigma$ as the composition of zero or one of the $f_i$s for positive integer $i$, and then $f_0$.

Note that there are $2^{n-1}$ of these functions. To prove that they are all different, simply note that the first bishop is in a different location each time because if we order the functions in increasing order of index, we note that the first bishop is always in the first half of the first $2c$-set, so it simply adds $2^i$ for each used $f_i$.
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