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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.

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0 replies
jlacosta
Mar 2, 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
is this really supposed to be #13???
hgmium   4
N 4 minutes ago by hgmium
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_13

I managed to do all the other geo problems for that year besides this one
misplaced?
4 replies
hgmium
2 hours ago
hgmium
4 minutes ago
[Registration Open] Mustang Math Tournament 2025
MustangMathTournament   23
N 9 minutes ago by Gavin_Deng
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
23 replies
MustangMathTournament
Mar 8, 2025
Gavin_Deng
9 minutes ago
Perfect Squares, Infinite Integers and Integers
steven_zhang123   4
N 38 minutes ago by ohiorizzler1434
Source: China TST 2001 Quiz 5 P1
For which integer \( h \), are there infinitely many positive integers \( n \) such that \( \lfloor \sqrt{h^2 + 1} \cdot n \rfloor \) is a perfect square? (Here \( \lfloor x \rfloor \) denotes the integer part of the real number \( x \)?
4 replies
steven_zhang123
Yesterday at 12:06 PM
ohiorizzler1434
38 minutes ago
another geometry problem with sharky-devil point
anyone__42   11
N 40 minutes ago by zhenghua
Source: The francophone mathematical olympiads P1
Let $ABC$ be an acute triangle with $AC>AB$, Let $DEF$ be the intouch triangle with $D \in (BC)$,$E \in (AC)$,$F \in (AB)$,, let $G$ be the intersecttion of the perpendicular from $D$ to $EF$ with $AB$, and $X=(ABC)\cap (AEF)$.
Prove that $B,D,G$ and $X$ are concylic
11 replies
anyone__42
Jun 27, 2020
zhenghua
40 minutes ago
AMC- IMO preparation
asyaela.   11
N 44 minutes ago by hashbrown2009
I'm a ninth grader, and I recently attempted the AMC 12, getting 18 questions correct and leaving 7 empty. I started working on Olympiad math in November and currently dedicate about two hours per day to preparation. I'm feeling a bit demotivated, but if it's possible for me to reach IMO level, I'd be willing to put in more time. How realistic is it for me to get there, and how much study would it typically take?
11 replies
1 viewing
asyaela.
6 hours ago
hashbrown2009
44 minutes ago
IMO Shortlist 2011, Algebra 5
orl   18
N an hour ago by mathfun07
Source: IMO Shortlist 2011, Algebra 5
Prove that for every positive integer $n,$ the set $\{2,3,4,\ldots,3n+1\}$ can be partitioned into $n$ triples in such a way that the numbers from each triple are the lengths of the sides of some obtuse triangle.

Proposed by Canada
18 replies
orl
Jul 11, 2012
mathfun07
an hour ago
2025 ROSS Program
scls140511   15
N an hour ago by akliu
Since the application has ended, are we now free to discuss the problems and stats? How do you think this year's problems are?
15 replies
scls140511
Yesterday at 2:36 AM
akliu
an hour ago
Line Perpendicular to Euler Line
tastymath75025   55
N an hour ago by ohiorizzler1434
Source: USA TSTST 2017 Problem 1, by Ray Li
Let $ABC$ be a triangle with circumcircle $\Gamma$, circumcenter $O$, and orthocenter $H$. Assume that $AB\neq AC$ and that $\angle A \neq 90^{\circ}$. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$ in $\triangle ABC$, respectively. Let $P$ be the intersection of line $MN$ with the tangent line to $\Gamma$ at $A$. Let $Q$ be the intersection point, other than $A$, of $\Gamma$ with the circumcircle of $\triangle AEF$. Let $R$ be the intersection of lines $AQ$ and $EF$. Prove that $PR\perp OH$.

Proposed by Ray Li
55 replies
tastymath75025
Jun 29, 2017
ohiorizzler1434
an hour ago
Foot from vertex to Euler line
cjquines0   31
N an hour ago by pUssydestroyer777
Source: 2016 IMO Shortlist G5
Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $ABC$. A circle $\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $AB$ and $AC$ at $X$ and $Y$ respectively. Let $P$ be the foot of altitude from $A$ to $BC$, and let $M$ be the midpoint of $BC$. Prove that the circumcentre of triangle $XSY$ is equidistant from $P$ and $M$.
31 replies
1 viewing
cjquines0
Jul 19, 2017
pUssydestroyer777
an hour ago
Inequality => square
Rushil   12
N 2 hours ago by ohiorizzler1434
Source: INMO 1998 Problem 4
Suppose $ABCD$ is a cyclic quadrilateral inscribed in a circle of radius one unit. If $AB \cdot BC \cdot CD \cdot DA \geq 4$, prove that $ABCD$ is a square.
12 replies
Rushil
Oct 7, 2005
ohiorizzler1434
2 hours ago
p + q + r + s = 9 and p^2 + q^2 + r^2 + s^2 = 21
who   28
N 2 hours ago by asdf334
Source: IMO Shortlist 2005 problem A3
Four real numbers $ p$, $ q$, $ r$, $ s$ satisfy $ p+q+r+s = 9$ and $ p^{2}+q^{2}+r^{2}+s^{2}= 21$. Prove that there exists a permutation $ \left(a,b,c,d\right)$ of $ \left(p,q,r,s\right)$ such that $ ab-cd \geq 2$.
28 replies
who
Jul 8, 2006
asdf334
2 hours ago
H not needed
dchenmathcounts   44
N 3 hours ago by Ilikeminecraft
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
44 replies
dchenmathcounts
May 23, 2020
Ilikeminecraft
3 hours ago
IZHO 2017 Functional equations
user01   51
N 3 hours ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
user01
Jan 14, 2017
lksb
3 hours ago
Inequality with wx + xy + yz + zw = 1
Fermat -Euler   23
N 3 hours ago by hgomamogh
Source: IMO ShortList 1990, Problem 24 (THA 2)
Let $ w, x, y, z$ are non-negative reals such that $ wx + xy + yz + zw = 1$.
Show that $ \frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$.
23 replies
Fermat -Euler
Nov 2, 2005
hgomamogh
3 hours ago
rows are DERANGED and a SOCOURGE to usajmo .
GrantStar   26
N Saturday at 6:00 AM by joshualiu315
Source: USAJMO 2024/4
Let $n \geq 3$ be an integer. Rowan and Colin play a game on an $n \times n$ grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid and Colin is allowed to permute the columns. A grid coloring is orderly if: [list] [*]no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and [*]no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring. [/list] In terms of $n$, how many orderly colorings are there?

Proposed by Alec Sun
26 replies
GrantStar
Mar 21, 2024
joshualiu315
Saturday at 6:00 AM
rows are DERANGED and a SOCOURGE to usajmo .
G H J
G H BBookmark kLocked kLocked NReply
Source: USAJMO 2024/4
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GrantStar
812 posts
#1 • 13 Y
Y by sixoneeight, meth4life2020, Danielzh, AtharvNaphade, ex-center, OronSH, alsk, pikapika007, megarnie, KevinYang2.71, qwerty123456asdfgzxcvb, crazyeyemoody907, ihatemath123
Let $n \geq 3$ be an integer. Rowan and Colin play a game on an $n \times n$ grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid and Colin is allowed to permute the columns. A grid coloring is orderly if:
  • no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and
  • no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring.
In terms of $n$, how many orderly colorings are there?

Proposed by Alec Sun
This post has been edited 3 times. Last edited by GrantStar, Mar 21, 2024, 7:37 PM
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GrantStar
812 posts
#2 • 7 Y
Y by thinkcow, bachkieu, OronSH, Sagnik123Biswas, KevinYang2.71, ihatemath123, lelouchvigeo
The answer is $2n!+2$. This achieved by
  • The one configuration that's all red and the one configuration that's all blue
  • The $n!$ configurations that have exactly $1$ blue in each row and column
  • The $n!$ configurations that have exactly $1$ blue in each row and column
It is easy to see that all of these work. Now, we show only these work.

Claim: The amount of red cells in any row and column must be at most $1$ or at least $n-1$.
Proof. The claim is clearly true for $n=3$. If $n>3$, then say there are $2\leq k \leq n-2$ red cells in some column. Rowan can then mix up the $n$ cells in that column in $\binom n k$ ways, each of which must appear as a column in the original coloring. But for $\binom n k\geq \binom n2 >n$ for $n>3$, a contradiction. $\blacksquare$

This means each column has $0,1,n-1,n$ red cells. If a column is fully red or fully blue, then Colin can move this column where ever he wants. No matter how Rowan permutes the rows now, this column is always a solid color. As Colin can move the original column any where, we conclude the whole grid is one color.

Else, without loss of generality assume there is a column with exactly $1$ red square in it. Colin can then move this column anywhere, so like before, we conclude all columns have exactly $1$ red square. We easily eliminate the case of one all red row and one row with $n-1$ red squares as we can move the large row to any other row and get a contradiction. Thus each row and column has exactly $1$ red square in it as desired.
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KevinYang2.71
392 posts
#3 • 9 Y
Y by GrantStar, EpicBird08, Orthogonal., LostDreams, OronSH, megarnie, bjump, deduck, qwerty123456asdfgzxcvb
@above solution so simple

why did i not think of that (i actually wrote this in contest)

Let $G=(V,E)$ be the bipartite graph with bipartition $V=R\sqcup C$ where $R$ and $C$ are the set of rows and columns of the $n\times n$ grid, and $e=rc$ (with $(r,c)\in R\times C$) is an edge if and only if square $(r,c)$ of the grid is red. Assume $G$ is orderly. Then $\mathrm{Aut}(G)$ acts transitively on $R$ and $C$. Thus the degree of each vertex in $R$ is the same and the degree of each vertex in $C$ is the same. Since $|R|=|C|=n$, $G$ is regular, say $d$-regular.

For all $r_1,r_2,r_3\in R$, we must have $|N(r_1)\cap N(r_2)|=|N(r_1)\cap N(r_3)|$ due to the transposition switching $r_2$ and $r_3$. It follows that $|N(r_1)\cap N(r_2)|$ is constant over all $r_1,r_2\in R$. Let this constant be $k_r$. Similarly, let $k_c$ be the constant such that $|N(c_1)\cap N(c_2)|=k_c$ for all $c_1,c_2\in C$.

Fix a vertex $r\in R$. We double count $|E(N(r),R\setminus\{r\})|$. On one hand, each vertex in $N(r)$ has degree $d$ so $|E(N(r),R\setminus\{r\})|=d(d-1)$. On the other hand, each vertex in $R\setminus\{r\}$ has $k_r$ common neighbors with $r$ so $|E(N(r),R\setminus\{r\})|=k_r(n-1)$. Thus $k_r(n-1)=d(d-1)$.

Similarly, $k_c(n-1)=d(d-1)$ so $k_r=k_c=:k$. Fix $r_1,r_2\in R$. Let
\begin{align*}
C_1&:=N(r_1)\setminus N(r_2)\\
C_2&:=N(r_2)\setminus N(r_1).
\end{align*}In order to recover $G$, Colin's permutation $\sigma$ must map $C_1$ to $C_2$ and $C_2$ to $C_1$.

If $k\ge d-1$ then $k=d-1$ or $k=d$ so $d\in\{0,1,n-1,n\}$. If $d=0$ or $d=n$ then $G$ is $\overline{K_{2n}}$ or $K_{n,n}$. Clearly these are orderly. If $d=1$ then $G$ is a matching of size $n$, which is orderly. There are $n!$ such matchings. If $d=n-1$ then $G$ is $K_{n,n}$ with the edges of a matching of size $n$ removed, which is orderly. There are $n!$ such graphs.

So we may assume $k<d-1$. It follows that $C_1$ and $C_2$ are nonempty. Pick $v\in C_1$ and let $u:=\sigma(v)$. Since $R\setminus\{r_1,r_2\}$ is untouched, $N(v)\cap(R\setminus\{r_1,r_2\})=N(u)\cap(R\setminus\{r_1,r_2\})$. Thus $|N(v)\cap N(u)|=d-1>k$, contradicting the definition of $k$. Thus there are no orderly graphs in this case.

Combining the above, there are $\boxed{2+2n!}$ orderly graphs. $\square$
This post has been edited 3 times. Last edited by KevinYang2.71, Mar 22, 2024, 7:19 AM
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pi_is_3.14
1437 posts
#4
Y by
is this right? way harder than the 2 above sol but I think should still work
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CatCatHead
27 posts
#5
Y by
I also get 2(n!+1) for n not equal 4, but I think when n=4, there will be more kinds, so I get 74 for n=4, different from others.
Like brbb/rbbb/rrrb/rrbr for n=4
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MY-2
380 posts
#6
Y by
CatCatHead wrote:
I also get 2(n!+1) for n not equal 4, but I think when n=4, there will be more kinds, so I get 74 for n=4, different from others.
Like brbb/rbbb/rrrb/rrbr for n=4

If Rowan swaps the first and fourth rows this breaks I believe.
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CatCatHead
27 posts
#7
Y by
MY-2 wrote:
CatCatHead wrote:
I also get 2(n!+1) for n not equal 4, but I think when n=4, there will be more kinds, so I get 74 for n=4, different from others.
Like brbb/rbbb/rrrb/rrbr for n=4

If Rowan swaps the first and fourth rows this breaks I believe.

yeah, i mistake at this condition, so the answer is still 2(n!+1)
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beibeizhu
98 posts
#8
Y by
GrantStar wrote:
It is easy to see that all of these work.

if they take that, i dearly hope they'll take my "hurr durr bijection hurr durr use strings to tie exactly one row to exactly one column hurr durr it's obvious" - i had no idea how to rigorously prove it for the case where there's 1 red or 1 blue because it's so obvious
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A_MatheMagician
2251 posts
#9 • 1 Y
Y by fantasticcrow
beibeizhu wrote:
GrantStar wrote:
It is easy to see that all of these work.

if they take that, i dearly hope they'll take my "hurr durr bijection hurr durr use strings to tie exactly one row to exactly one column hurr durr it's obvious" - i had no idea how to rigorously prove it for the case where there's 1 red or 1 blue because it's so obvious
here's the proof i put in my solution:
For the all blue and all red coloring, the grid cannot change from its original coloring. WLOG, assume there is $1$ red square in each row/column (the proof for $1$ blue square is similar). No matter how Rowan permutes the rows, there will still always be $1$ red square in each row/column. Colin can take the column with the red square in the same row as the red square in the first row of the original coloring and move that column to the position of the first column. He then takes the column with the red square in the row that the red square in the second column of the original coloring and moves that column to the position of the second column, and so on. Thus, Colin can always restore the original coloring.
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Danielzh
476 posts
#10
Y by
Will there be any deductions for just writing $1+n!+n!+1=2n!+2$ in terms of calculation for the number of coloring? No explanation for how there is only $1$ way or $n!$ ways (e.g. $n$ ways for the first column, $n-1$ ways for the second column, ..., $1$ way for the $n$th column).
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bachkieu
113 posts
#11 • 1 Y
Y by Danielzh
@above probably not if you clearly state what you're counting. fun question tho
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meduh6849
354 posts
#12
Y by
will it be 5 or 2 points if no proof that the $n!$ cases work is provided
somehow forgot that :wallbash_red: anyways basically gave up on JMO for this year
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vsamc
3783 posts
#13
Y by
meduh6849 wrote:
will it be 5 or 2 points if no proof that the $n!$ cases work is provided
somehow forgot that :wallbash_red: anyways basically gave up on JMO for this year

I think that's a 5 or maybe a 6 if the graders are generous, that's not really the main idea of the problem and it's kind of easy to see that they all work, but it is a significant detail that shouldn't be glossed over (and it's not like completely trivial that all the n! configs work, although it is very easy to verify)
This post has been edited 1 time. Last edited by vsamc, Mar 21, 2024, 12:25 PM
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peace09
5409 posts
#14 • 1 Y
Y by OronSH
:noo:

The answer is $2(n!+1)$, attained by and only by grids which are either uniform or are a permutation of a ``Sudoku." Clearly these work.* Necessity hinges on the following:

Claim. No row or column can have both $\ge2$ $R$s and $\ge2$ $B$s.
Proof. Suppose otherwise, and WLOG arrange them in a row in the order $R_1B_1R_2B_2$. We build up the grid as follows:
  1. First, swap the columns with $R_1$ and $B_1$. To revert, there should exist a second row as shown, where the dashed sections are the same in both rows:
    [asy]
label("$R_1$",(-3,1));
label("$B_1$",(-1,1));
label("$R_2$",(1,1));
label("$B_2$",(3,1));
label("$B$",(-3,-1));
label("$R$",(-1,-1));
label("$R$",(1,-1));
label("$B$",(3,-1));
draw((-4,1)--(-3.3,1),dashed);
draw((-2.5,1)--(-1.5,1),dashed);
draw((-0.5,1)--(0.5,1),dashed);
draw((1.5,1)--(2.5,1),dashed);
draw((3.5,1)--(4,1),dashed);
draw((-4,-1)--(-3.3,-1),dashed);
draw((-2.5,-1)--(-1.5,-1),dashed);
draw((-0.5,-1)--(0.5,-1),dashed);
draw((1.5,-1)--(2.5,-1),dashed);
draw((3.5,-1)--(4,-1),dashed);
[/asy]
  2. Next, swap these rows. To revert, it is necessary and sufficient to swap the columns with $R_1$ and $B_1$, so that the solid sections are the same in both columns:
    [asy]
label("$R_1$",(-3,1));
label("$B_1$",(-1,1));
label("$R_2$",(1,1));
label("$B_2$",(3,1));
label("$B$",(-3,-1));
label("$R$",(-1,-1));
label("$R$",(1,-1));
label("$B$",(3,-1));
draw((-4,1)--(-3.3,1),dashed);
draw((-2.5,1)--(-1.5,1),dashed);
draw((-0.5,1)--(0.5,1),dashed);
draw((1.5,1)--(2.5,1),dashed);
draw((3.5,1)--(4,1),dashed);
draw((-4,-1)--(-3.3,-1),dashed);
draw((-2.5,-1)--(-1.5,-1),dashed);
draw((-0.5,-1)--(0.5,-1),dashed);
draw((1.5,-1)--(2.5,-1),dashed);
draw((3.5,-1)--(4,-1),dashed);
draw((-3,-4)--(-3,-1.5));
draw((-3,-0.5)--(-3,0.5));
draw((-3,1.5)--(-3,2));
draw((-1,-4)--(-1,-1.5));
draw((-1,-0.5)--(-1,0.5));
draw((-1,1.5)--(-1,2));
[/asy]
  3. Finally, swap the columns with $R_2$ and $B_2$. To revert, there should exist a third row* as shown, where the dashed sections are the same in both rows (in order to match the first row):
    [asy]
label("$R_1$",(-3,1));
label("$B_1$",(-1,1));
label("$R_2$",(1,1));
label("$B_2$",(3,1));
label("$B$",(-3,-1));
label("$R$",(-1,-1));
label("$R$",(1,-1));
label("$B$",(3,-1));
label("$R$",(-3,-3));
label("$B$",(-1,-3));
label("$B$",(1,-3));
label("$R$",(3,-3));
draw((-4,1)--(-3.3,1),dashed);
draw((-2.5,1)--(-1.5,1),dashed);
draw((-0.5,1)--(0.5,1),dashed);
draw((1.5,1)--(2.5,1),dashed);
draw((3.5,1)--(4,1),dashed);
draw((-4,-1)--(-3.3,-1),dashed);
draw((-2.5,-1)--(-1.5,-1),dashed);
draw((-0.5,-1)--(0.5,-1),dashed);
draw((1.5,-1)--(2.5,-1),dashed);
draw((3.5,-1)--(4,-1),dashed);
draw((-3,-4)--(-3,-1.5));
draw((-3,-0.5)--(-3,0.5));
draw((-3,1.5)--(-3,2));
draw((-1,-4)--(-1,-1.5));
draw((-1,-0.5)--(-1,0.5));
draw((-1,1.5)--(-1,2));
draw((-4,-3)--(-3.3,-3),dashed);
draw((-2.5,-3)--(-1.5,-3),dashed);
draw((-0.5,-3)--(0.5,-3),dashed);
draw((1.5,-3)--(2.5,-3),dashed);
draw((3.5,-3)--(4,-3),dashed);
[/asy]
    The $R$ and $B$ in corresponding sections of the solid sections give us a contradiction. $\square$

To finish, clearly there number of $R$'s in each of the rows and columns is a constant $c$, and the above claim implies that $c\in\{0,1,n-1,n\}$, which produces those and only those mentioned at the beginning. $\blacksquare$
This post has been edited 1 time. Last edited by peace09, Mar 21, 2024, 2:29 PM
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KevinYang2.71
392 posts
#15 • 2 Y
Y by megarnie, deduck
vsamc wrote:
meduh6849 wrote:
will it be 5 or 2 points if no proof that the $n!$ cases work is provided
somehow forgot that :wallbash_red: anyways basically gave up on JMO for this year

I think that's a 5 or maybe a 6 if the graders are generous, that's not really the main idea of the problem and it's kind of easy to see that they all work, but it is a significant detail that shouldn't be glossed over (and it's not like completely trivial that all the n! configs work, although it is very easy to verify)

If I used the graph theoretic solution, it is obvious matchings are orderly. Will i get a point deducted? (my sol is above)
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asdf334
7578 posts
#17
Y by
me when i should have taken jmo

1) Notice that any column has at most $n$ permutations; otherwise, Rowan permutes column $c$ to $c'$ where $c'$ was not a column-coloring in the original figure.
2) Do the same for rows, do some basic logic since now your search space is hugely decreased, and finish.
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EpicBird08
1730 posts
#18 • 1 Y
Y by a_smart_alecks
Video solution
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a_smart_alecks
51 posts
#19
Y by
woah, i could've saved a lot of time with the epic inspirational $n\choose k$ bound instead of stumbling over my words with a worse and more verbose version of peace's solution... but at least i think my sol was correct and thats what matters!!!!! and even an hour of extra time probably wouldn't get me a single point on j6

btw as alec(ks) sun, i can confirm that i intentionally made the wording unclear to troll half of the jmo contestants
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bronzetruck2016
87 posts
#20
Y by
spent a page proving the n choose k bound because I and a bit silly and the graders probably wouldn't like it if I just said that it's obvious
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v_Enhance
6857 posts
#21
Y by
Solution from Twitch Solves ISL:

The answer is $2n!+2$. In fact, we can describe all the orderly colorings as follows:
  • The all-blue coloring.
  • The all-red coloring.
  • Each of the $n!$ colorings where every row/column has exactly one red cell.
  • Each of the $n!$ colorings where every row/column has exactly one blue cell.
These obviously work; we turn our attention to proving these are the only ones.
For the other direction, fix a orderly coloring $\mathcal A$.
Consider any particular column $C$ in $\mathcal A$ and let $m$ denote the number of red cells that $C$ has. Any row permutation (say $\sigma$) that Rowan chooses will transform $C$ into some column $\sigma(C)$, and our assumption requires $\sigma(C)$ has to appear somewhere in the original assignment $\mathcal A$. An example for $n = 7$, $m = 2$, and a random $\sigma$ is shown below.

[asy]
defaultpen(fontsize(14pt)); pen bo = black + 1.4; filldraw(shift(0,0)*unitsquare, lightcyan, bo); filldraw(shift(0,1)*unitsquare, lightred, bo); filldraw(shift(0,2)*unitsquare, lightcyan, bo); filldraw(shift(0,3)*unitsquare, lightcyan, bo); filldraw(shift(0,4)*unitsquare, lightred, bo); filldraw(shift(0,5)*unitsquare, lightcyan, bo); filldraw(shift(0,6)*unitsquare, lightcyan, bo); label("$C$", (0.5,0), dir(-90));
pen gr = deepgreen; real h = 8; draw((1,1.5)--(h,4.5), gr, EndArrow, Margins); draw((1,4.5)--(h,3.5), gr, EndArrow, Margins); draw((1,2.5)--(h,1.5), gr, EndArrow, Margins); draw((1,0.5)--(h,0.5), gr, EndArrow, Margins); draw((1,3.5)--(h,6.5), gr, EndArrow, Margins); draw((1,6.5)--(h,5.5), gr, EndArrow, Margins); draw((1,5.5)--(h,2.5), gr, EndArrow, Margins);
filldraw(shift(h,0)*unitsquare, lightcyan, bo); filldraw(shift(h,1)*unitsquare, lightcyan, bo); filldraw(shift(h,2)*unitsquare, lightcyan, bo); filldraw(shift(h,3)*unitsquare, lightred, bo); filldraw(shift(h,4)*unitsquare, lightred, bo); filldraw(shift(h,5)*unitsquare, lightcyan, bo); filldraw(shift(h,6)*unitsquare, lightcyan, bo); label("$\sigma(C)$", (h+0.5,0), dir(-90)); label("$\sigma$", ((1+h)/2, 0), dir(-90), gr);
[/asy]
On the other hand, the number of possible patterns of $\sigma(C)$ is easily seen to be exactly $\binom{n}{m}$ --- and they must all appear. In particular, if $m \notin \{0, 1, n-1, n\}$, then we immediately get a contradiction because $\mathcal A$ would need too many columns (there are only $n$ columns in $\mathcal A$, which is fewer than $\binom{n}{m}$). Moreover, if either $m=1$ or $m=n-1$, these columns are all the columns of $\mathcal A$; these lead to the $2n!$ main cases we found before.
The only remaining case is when $m \in \{0,n\}$ for every column, i.e.\ every column is monochromatic. It's easy to see in that case the columns must all be the same color.
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Shreyasharma
666 posts
#22
Y by
Claim: All rows have the same number of red tiles.
Proof. Obvious, as otherwise if Rowan permutes rows $i$ and $j$ with different numbers of red tiles, Colin can never return to the original configuration, because no matter how he moves row $j$ will always have a different number of tiles than it's starting configuration. $\square$

The same bound works for columns, so each row and column has the same number of elements from an easy double count. Now note that after any given move that Rowan makes, we require the columns to simply be a permutation of the original grid coloring. This occurs if and only if upon swapping any two rows,
  • The two columns ``swap" upon such a switch
  • The two rows are identical.
If the first scenario occurs then each row can have at most $1$ colored tile, or at least $n - 1$ colored tiles, and this expands to the columns. In the second, this forces all rows to be identical, and all column's to be identical, which forces the entire grid to be colored. This is exactly the claimed solution set which can easily be observed to work by thinking of it as moving individual blocks. We are done.
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awesomeguy856
7255 posts
#23
Y by
My proof that the $k=1$ case works:

There are $n!$ ways to permute either the rows or columns, each permutation being distinct
There are $n!$ ways to color the grid such that each row and each column has exactly one red square

Thus the original coloring is always achievable from any new coloring
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cakeiswaybetterthancookie
190 posts
#24
Y by
I think this works???.... sombody please let me know.

Answer: 2n!+2. Here is how we found the answer:
An all-blue coloring
An all-red coloring
n! ways for there to be exactly 1 blue in each row/column
n! ways for there to be exactly 1 blue in each row/column

(I think) this is pretty easy to see and will not get explained further in the proof

Now, lets prove these are the only cases that are possible.

Claim : There is at most 1 red chip per row or at least n-1 red chips

Supose that the claim is false and that there are other possbilities. Let's define k to the number of red chips in a row. k will be between 2 and n-2 chips(since we are assuming the claim is false and that indeed there are other possibles). There are n choose k ways to permute the chips in a given row. We know that n choose k is greater than or equal to n choose 2 which is greater than n for values of n greater than or equal to 3. So, there are not enough room to fit all the chips. Therefore, we are complete with the proof.
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blueprimes
300 posts
#25
Y by
We claim the answer is $2n! + 2$, which counts the number of configurations where:
  • All cells are red/blue, which counts $2$ configurations.
  • Exactly one cell in each row/column are red/blue, which counts $2n!$ configurations.
It can be verified that these work.

Now we show these are the only ones. Since every row and column have an equal number of red cells, denote it by $k$. For $n = 3$, note that any value of $k$ is already classified in the configurations listed earlier, so we assume $n \ge 4$. For the sake of contradiction assume an orderly coloring satisfies $2 \le k \le n - 2$. Note that for a particular column, there are $\binom{n}{k}$ ways to change it, for which it must be a column in the original coloring. This implies the bound
$$\binom{n}{k} \le n \implies n - 1 \le 2 \implies n \le 3$$which is a contradiction. Then we must have $k = 0, 1, n - 1, n$. These imply the constructions we described, so we are done.
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somebodyyouusedtoknow
255 posts
#26
Y by
o1 could actually get the right number for this problem.
Runtime: 1m 2s

o1's response
Its argument for no other solutions is not acceptable, though.

o1's thinking process

Just curious what score would this solution obtain.
This post has been edited 2 times. Last edited by somebodyyouusedtoknow, Dec 5, 2024, 9:24 PM
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eg4334
598 posts
#27
Y by
The answer is $2n!+2$, which counts the two configurations of the same color, those with exactly one blue cell in each row/column, and those with exactly one red cell in each row/column. The latter two each have $n!$ clearly. The $n=3$ case can easily be manualyl checked so for the rest of this solution assume $n \geq 4$.
Claim: Each row/column cannot have two or more of both colors.
Proof: Assume there was. Then its clear that when looking at some particular row there are at least $\binom{n}{2}$ conigurations that must appear in our original grid. But this exceeds $n$ always, which finishes.

These easily imply our originally mentioned constructions.
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joshualiu315
2513 posts
#28
Y by
for no reason, i struggled to write a proper solution for this problem both in contest and right now :oops:


The answer is $\boxed{2 \cdot n!+2}$ achieved in the following configurations:
  • All squares are red or blue.
  • Each row and each column has exactly one red square.
  • Each row and each column has exactly one blue square.

All four of these colorings can easily be check to be orderly.

Let $R = (r_1, r_2, \dots, r_n)$ denote the number of red squares in each of the rows. It is easy to see that Colin cannot permute any of the values in $R$. Hence, if any $r_i \neq r_j$, Rowan could simply switch those two rows, and Colin would not be able to return the grid back to its original state. Hence, in any orderly coloring, the number of red squares in each row is equal, and an analogous argument shows that the number of blue squares in each column is equal as well. Suppose that there are exactly $k$ red squares in each row and column.

Consider an arbitrary column. Rowan can shuffle the rows of that column in $\tbinom{n}{k}$ ways. In order for Colin to be able to return the grid to its original coloring, this new permuation of the column must be equivalent to another column. Thus, Rowan can shuffle the rows in at most $n$ ways, which gives us $\tbinom{n}{k} \le n$. It is clear that this is satisfied if and only if $k \le 1$ or $k \ge n-1$, which provide all of the above constructions.
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