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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.

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)!

Be sure to mark your calendars for the following upcoming events:
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[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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 A Letter to MSM
Arr0w   23
N Sep 19, 2022 by scannose
Greetings.

I have seen many posts talking about commonly asked questions, such as finding the value of $0^0$, $\frac{1}{0}$,$\frac{0}{0}$, $\frac{\infty}{\infty}$, why $0.999...=1$ or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.
[list]
[*]Firstly, the case of $0^0$. It is usually regarded that $0^0=1$, not because this works numerically but because it is convenient to define it this way. You will see the convenience of defining other undefined things later on in this post.

[*]What about $\frac{\infty}{\infty}$? The issue here is that $\infty$ isn't even rigorously defined in this expression. What exactly do we mean by $\infty$? Unless the example in question is put in context in a formal manner, then we say that $\frac{\infty}{\infty}$ is meaningless.

[*]What about $\frac{1}{0}$? Suppose that $x=\frac{1}{0}$. Then we would have $x\cdot 0=0=1$, absurd. A more rigorous treatment of the idea is that $\lim_{x\to0}\frac{1}{x}$ does not exist in the first place, although you will see why in a calculus course. So the point is that $\frac{1}{0}$ is undefined.

[*]What about if $0.99999...=1$? An article from brilliant has a good explanation. Alternatively, you can just use a geometric series. Notice that
\begin{align*}
\sum_{n=1}^{\infty} \frac{9}{10^n}&=9\sum_{n=1}^{\infty}\frac{1}{10^n}=9\sum_{n=1}^{\infty}\biggr(\frac{1}{10}\biggr)^n=9\biggr(\frac{\frac{1}{10}}{1-\frac{1}{10}}\biggr)=9\biggr(\frac{\frac{1}{10}}{\frac{9}{10}}\biggr)=9\biggr(\frac{1}{9}\biggr)=\boxed{1}
\end{align*}
[*]What about $\frac{0}{0}$? Usually this is considered to be an indeterminate form, but I would also wager that this is also undefined.
[/list]
Hopefully all of these issues and their corollaries are finally put to rest. Cheers.

2nd EDIT (6/14/22): Since I originally posted this, it has since blown up so I will try to add additional information per the request of users in the thread below.

INDETERMINATE VS UNDEFINED

What makes something indeterminate? As you can see above, there are many things that are indeterminate. While definitions might vary slightly, it is the consensus that the following definition holds: A mathematical expression is be said to be indeterminate if it is not definitively or precisely determined. So how does this make, say, something like $0/0$ indeterminate? In analysis (the theory behind calculus and beyond), limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits. However, if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. For example, we could say that $0/0$ is an indeterminate form.

But we need to more specific, this is still ambiguous. An indeterminate form is a mathematical expression involving at most two of $0$, $1$ or $\infty$, obtained by applying the algebraic limit theorem (a theorem in analysis, look this up for details) in the process of attempting to determine a limit, which fails to restrict that limit to one specific value or infinity, and thus does not determine the limit being calculated. This is why it is called indeterminate. Some examples of indeterminate forms are
\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]etc etc. So what makes something undefined? In the broader scope, something being undefined refers to an expression which is not assigned an interpretation or a value. A function is said to be undefined for points outside its domain. For example, the function $f:\mathbb{R}^{+}\cup\{0\}\rightarrow\mathbb{R}$ given by the mapping $x\mapsto \sqrt{x}$ is undefined for $x<0$. On the other hand, $1/0$ is undefined because dividing by $0$ is not defined in arithmetic by definition. In other words, something is undefined when it is not defined in some mathematical context.

WHEN THE WATERS GET MUDDIED

So with this notion of indeterminate and undefined, things get convoluted. First of all, just because something is indeterminate does not mean it is not undefined. For example $0/0$ is considered both indeterminate and undefined (but in the context of a limit then it is considered in indeterminate form). Additionally, this notion of something being undefined also means that we can define it in some way. To rephrase, this means that technically, we can make something that is undefined to something that is defined as long as we define it. I'll show you what I mean.

One example of making something undefined into something defined is the extended real number line, which we define as
\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]So instead of treating infinity as an idea, we define infinity (positively and negatively, mind you) as actual numbers in the reals. The advantage of doing this is for two reasons. The first is because we can turn this thing into a totally ordered set. Specifically, we can let $-\infty\le a\le \infty$ for each $a\in\overline{\mathbb{R}}$ which means that via this order topology each subset has an infimum and supremum and $\overline{\mathbb{R}}$ is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In $\overline{\mathbb{R}}$ it is perfectly OK to say that,
\begin{align*}
a + \infty = \infty + a & = \infty, & a & \neq -\infty \\
a - \infty = -\infty + a & = -\infty, & a & \neq \infty \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\
\frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\
\frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\
\frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0).
\end{align*}So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,
\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]So while we define certain things, we also left others undefined/indeterminate in the process! However, in the context of measure theory it is common to define $\infty \times 0=0$ as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by $0$ is undefined still! Is there a place where it isn't? Kind of. To do this, we can extend the complex numbers! More formally, we can define this extension as
\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, $\tilde{\infty}$ means complex infinity, since we are in the complex plane now. Here's the catch: division by $0$ is allowed here! In fact, we have
\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]where $\tilde{\infty}/\tilde{\infty}$ and $0/0$ are left undefined. We also have
\begin{align*}
z+\tilde{\infty}=\tilde{\infty}, \forall z\ne -\infty\\
z\times \tilde{\infty}=\tilde{\infty}, \forall z\ne 0
\end{align*}Furthermore, we actually have some nice properties with multiplication that we didn't have before. In $\mathbb{C}^*$ it holds that
\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]but $\tilde{\infty}-\tilde{\infty}$ and $0\times \tilde{\infty}$ are left as undefined (unless there is an explicit need to change that somehow). One could define the projectively extended reals as we did with $\mathbb{C}^*$, by defining them as
\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]They behave in a similar way to the Riemann Sphere, with division by $0$ also being allowed with the same indeterminate forms (in addition to some other ones).
23 replies
Arr0w
Feb 11, 2022
scannose
Sep 19, 2022
k i Marathon Threads
LauraZed   0
Jul 2, 2019
Due to excessive spam and inappropriate posts, we have locked the Prealgebra and Beginning Algebra threads.

We will either unlock these threads once we've cleaned them up or start new ones, but for now, do not start new marathon threads for these subjects. Any new marathon threads started while this announcement is up will be immediately deleted.
0 replies
LauraZed
Jul 2, 2019
0 replies
k i Basic Forum Rules and Info (Read before posting)
jellymoop   368
N May 16, 2018 by harry1234
f (Reminder: Do not post Alcumus or class homework questions on this forum. Instructions below.) f
Welcome to the Middle School Math Forum! Please take a moment to familiarize yourself with the rules.

Overview:
[list]
[*] When you're posting a new topic with a math problem, give the topic a detailed title that includes the subject of the problem (not just "easy problem" or "nice problem")
[*] Stay on topic and be courteous.
[*] Hide solutions!
[*] If you see an inappropriate post in this forum, simply report the post and a moderator will deal with it. Don't make your own post telling people they're not following the rules - that usually just makes the issue worse.
[*] When you post a question that you need help solving, post what you've attempted so far and not just the question. We are here to learn from each other, not to do your homework. :P
[*] Avoid making posts just to thank someone - you can use the upvote function instead
[*] Don't make a new reply just to repeat yourself or comment on the quality of others' posts; instead, post when you have a new insight or question. You can also edit your post if it's the most recent and you want to add more information.
[*] Avoid bumping old posts.
[*] Use GameBot to post alcumus questions.
[*] If you need general MATHCOUNTS/math competition advice, check out the threads below.
[*] Don't post other users' real names.
[*] Advertisements are not allowed. You can advertise your forum on your profile with a link, on your blog, and on user-created forums that permit forum advertisements.
[/list]

Here are links to more detailed versions of the rules. These are from the older forums, so you can overlook "Classroom math/Competition math only" instructions.
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What belongs on this forum?
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Mathcounts and how to learn

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Marathons!
Relays might be a better way to describe it, but these threads definitely go the distance! One person starts off by posting a problem, and the next person comes up with a solution and a new problem for another user to solve. Here's some of the frequently active marathons running in this forum:
[list][*]Algebra
[*]Prealgebra
[*]Proofs
[*]Factoring
[*]Geometry
[*]Counting & Probability
[*]Number Theory[/list]
Some of these haven't received attention in a while, but these are the main ones for their respective subjects. Rather than starting a new marathon, please give the existing ones a shot first.

You can also view marathons via the Marathon tag.

Think this list is incomplete or needs changes? Let the mods know and we'll take a look.
368 replies
jellymoop
May 8, 2015
harry1234
May 16, 2018
Inequality
lgx57   0
4 minutes ago
Source: Own
$a,b>0$,$a^4+a^2b^2+b^4=k$.Find the min of $4a^2-ab+4b^2$.

$a,b>0$,$a^4-a^2b^2+b^4=k$.Find the min of $4a^2-ab+4b^2$.
0 replies
lgx57
4 minutes ago
0 replies
Inequality
Sadigly   2
N 12 minutes ago by sqing
Source: Azerbaijan Junior MO 2025 P5
For positive real numbers $x;y;z$ satisfying $0<x,y,z<2$, find the biggest value the following equation could acquire:


$$(2x-yz)(2y-zx)(2z-xy)$$
2 replies
+2 w
Sadigly
an hour ago
sqing
12 minutes ago
Inspired by lgx57
sqing   3
N 18 minutes ago by sqing
Source: Own
Let $ a,b>0, a^4+ab+b^4=10  $. Prove that
$$ \sqrt{10}\leq a^2+ab+b^2 \leq 6$$$$ 2\leq a^2-ab+b^2 \leq  \sqrt{10}$$$$  4\sqrt{10}\leq 4a^2+ab+4b^2 \leq18$$$$  12<4a^2-ab+4b^2 \leq14$$
3 replies
sqing
Yesterday at 2:19 PM
sqing
18 minutes ago
Interesting functional equation with geometry
User21837561   1
N 20 minutes ago by User21837561
Source: BMOSL 2025 G7
For an acute triangle $ABC$, let $O$ be the circumcentre, $H$ be the orthocentre, and $G$ be the centroid.
Let $f:\pi\rightarrow\mathbb R$ satisfy the following condition:
$f(A)+f(B)+f(C)=f(O)+f(G)+f(H)$
Prove that $f$ is constant.
1 reply
User21837561
34 minutes ago
User21837561
20 minutes ago
The best math formulas?
anticodon   14
N 6 hours ago by Soupboy0
my math teacher recently offhandedly mentioned in class that "the law of sines is probably in the top 10 of math formulas". This inspired me to make a top 10 list to see if he's right (imo he actually is...)

so I decided, it would be interesting to hear others' opinions on the top 10 and we can compile an overall list.

Attached=my list (sorry if you can't read my handwriting, I was too lazy to do latex, and my normal pencil handwriting looks better)

the formulas
14 replies
anticodon
Yesterday at 11:00 PM
Soupboy0
6 hours ago
sleep tips
Soupboy0   15
N Today at 2:18 AM by giratina3
can someone help me learn how to fall asleep faster bc I'm nervous/excited bc nats is upcoming
15 replies
Soupboy0
Yesterday at 4:20 PM
giratina3
Today at 2:18 AM
9 AMC 10 Prep
bluedino24   32
N Today at 2:17 AM by bluedino24
I'm in 7th grade and thought it would be good to start preparing for the AMC 10. I'm not extremely good at math though.

What are some important topics I should study? Please comment below. Thanks! :D
32 replies
bluedino24
May 2, 2025
bluedino24
Today at 2:17 AM
9 zeroes!.
ericheathclifffry   8
N Today at 2:15 AM by giratina3
i personally have no idea
8 replies
ericheathclifffry
May 5, 2025
giratina3
Today at 2:15 AM
Facts About 2025!
Existing_Human1   262
N Today at 2:14 AM by giratina3
Hello AOPS,

As we enter the New Year, the most exciting part is figuring out the mathematical connections to the number we have now temporally entered

Here are some facts about 2025:
$$2025 = 45^2 = (20+25)(20+25)$$$$2025 = 1^3 + 2^3 +3^3 + 4^3 +5^3 +6^3 + 7^3 +8^3 +9^3 = (1+2+3+4+5+6+7+8+9)^2 = {10 \choose 2}^2$$
If anyone has any more facts about 2025, enlighted the world with a new appreciation for the year


(I got some of the facts from this video)
262 replies
Existing_Human1
Jan 1, 2025
giratina3
Today at 2:14 AM
9 What competitions do you do
VivaanKam   23
N Today at 1:24 AM by K124659

I know I missed a lot of other competitions so if you didi one of the just choose "Other".
23 replies
VivaanKam
Apr 30, 2025
K124659
Today at 1:24 AM
MAP Goals
Antoinette14   3
N Today at 1:13 AM by Schintalpati
What's yall's MAP goals for this spring?
Mine's a 300 (trying to beat my brother's record) but since I'm at a 285 rn, 290+ is more reasonable.
3 replies
Antoinette14
Yesterday at 11:59 PM
Schintalpati
Today at 1:13 AM
9 Have you participated in the MATHCOUNTS competition?
aadimathgenius9   53
N Today at 12:19 AM by Math-lover1
Have you participated in the MATHCOUNTS competition before?
53 replies
aadimathgenius9
Jan 1, 2025
Math-lover1
Today at 12:19 AM
9 MathandAI4Girls!!!
Inaaya   13
N Yesterday at 10:28 PM by fossasor
How many problems did y'all solve this year?
I clowned and started the pset the week before :oops:
Though I think if i used the time wisely, I could have at least solved 11 of them
ended up with 9 :wallbash_red:
13 replies
Inaaya
Wednesday at 7:25 PM
fossasor
Yesterday at 10:28 PM
9 What is the best way to learn math???
lovematch13   89
N Yesterday at 9:48 PM by Capybara7017
On the contrary, I'm also gonna try to send this to school admins. PLEASE DO NOT TROLL!!!!
89 replies
lovematch13
May 22, 2023
Capybara7017
Yesterday at 9:48 PM
Peter and Bob
mathuz   16
N Oct 14, 2023 by chakrabortyahan
Source: All Russian 2014 Grade 11 Day 1 P2
Peter and Bob play a game on a $n\times n$ chessboard. At the beginning, all squares are white apart from one black corner square containing a rook. Players take turns to move the rook to a white square and recolour the square black. The player who can not move loses. Peter goes first. Who has a winning strategy?
16 replies
mathuz
Apr 30, 2014
chakrabortyahan
Oct 14, 2023
Peter and Bob
G H J
G H BBookmark kLocked kLocked NReply
Source: All Russian 2014 Grade 11 Day 1 P2
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mathuz
1524 posts
#1 • 2 Y
Y by Adventure10, Mango247
Peter and Bob play a game on a $n\times n$ chessboard. At the beginning, all squares are white apart from one black corner square containing a rook. Players take turns to move the rook to a white square and recolour the square black. The player who can not move loses. Peter goes first. Who has a winning strategy?
This post has been edited 1 time. Last edited by mathuz, May 3, 2014, 1:57 PM
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chaotic_iak
2932 posts
#2 • 3 Y
Y by mathuz, Adventure10, Mango247
Where are the rooks? Do Peter and Bob control one rook each or one rook together or some other number of rooks?
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sayakalah
5 posts
#3 • 3 Y
Y by mathuz, Adventure10, and 1 other user
"the board is white(all cells), only for the CORNER cells- black and they have ROOKS.", I think Peter and Bob are allowed to move one of the rooks in each turn, and there is no rooks on the white cells, but we need to confirm this to the author.
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mathuz
1524 posts
#4 • 2 Y
Y by Adventure10, Mango247
we have $1$-corner cell and $1$ rooks on the corner cell.
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mathuz
1524 posts
#5 • 2 Y
Y by Adventure10, Mango247
chaotic_iak wrote:
Where are the rooks? Do Peter and Bob control one rook each or one rook together or some other number of rooks?
You are right! Sorry I have mistake! Edited. :)

On the board there is one rook on corner cell.
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sabbasi
248 posts
#6 • 1 Y
Y by Adventure10
Someone please check this solution:

Peter can always win by always moving the maximum amount that he can vertically.

Firstly we can show that every time Bob moves he does so horizontally and the black squares that he draws all either share an edge with the edge of the board or with black squares he previously drew. To prove this, note that it is true for the first move since Peter goes to the end of the board. Now consider some point in the game after Peter has made a move. If Peter reached the end of the board, Bob must move along the end. Otherwise Peter must have reached a black square along a row drawn by Bob. The drawing of this row had a vertical column drawn by Peter before and after it, so all of Bob's squares will be adjacent to the horizontal row (since he is bounded by the vertical columns).

As a result of this, Bob's moves will always stack along the top and bottom of the board, so in other words, no square drawn by Bob can be untouched by the board or another square drawn by him both above it and below it. So this means that when Peter moves, he will always completely fill his column.

The only other thing we have to show is that Peter always will have a column to fill. This is true because if Peter reaches a column he hasn't touched yet, at most n - 1 moves must have passed (since otherwise he moved through every column) but that means Bob has only drawn n-1 rows, so there's no way that this new column could be filled after Bob's move.

Peter always has a move so Bob must be the loser.

Solution 2 : I also just realized that Peter always cuts off a new uncolored section with more rows and columns, which will eventually lead to his win since Bob can only win after Peter's move if Peter cuts off a 1xn rectangle. So I think this works too
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johnkwon0328
188 posts
#7 • 1 Y
Y by Adventure10
sabbasi wrote:
Someone please check this solution:

Peter can always win by always moving the maximum amount that he can vertically.

I think your solution is incorrect. Bob can also be a loser considering n=1.

My solution:

n is odd -> Bob wins, n is even -> Peter wins

case 1: n is even

Peter can always win by moving rook from (i, j) to (n+1-i, j) every time.

case 2: n is odd

WLOG we can assume that at the beginning, rook is on (n,1), and Peter moved it to (n,2).
Bob can win by using the following method. Assume that Peter placed the rook on (i, j).
If i<n, Bob can place rook on (n-i, j).
If i=n, Bob can place rook on nth column, because on Bob's turn, number of white cells in nth column is odd.
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sabbasi
248 posts
#8 • 2 Y
Y by Adventure10, Mango247
n=1 is an exception but in every other case that I tried Peter wins using this strategy.
Also you should provide some proof for your strategies since it's not obvious why/if they work.

Here I'll prove that Peter can win in a 3x3
label the cells from (1,1) to (3,3). Peter starts by going from (1,1) to (1,3).
Case 1: Bob goes to (2,3)
Then Peter moves to (2,1), Bob is forced to move to (3,1) then Peter moves to (3,3) and wins
Case 2: Bob goes to (3,3)
Then peter moves to (3,1) Bob is forced to go to (2,1) then Peter moves to (2,3) and wins.
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BSJL
641 posts
#9 • 2 Y
Y by Adventure10, Mango247
Do you consider the case that Bob goes to (1,2)? :wink:
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sabbasi
248 posts
#10 • 2 Y
Y by Adventure10, Mango247
@BSJL

I do not understand your concern. I suspect there is confusion in understanding the problem. Both players move the rook, and every square the rook moves over turns black, and black squares can't be moved on at all times. At least that is what the version here would suggest: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=587460 but that version was the only way I could think of playing anyways. Since if paths are not drawn, it is only dependent on parity, and if paths are self intersecting the game doesn't need to end. So clearly (2,3) and (3,3) are the only possible moves by Bob.
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johnkwon0328
188 posts
#11 • 1 Y
Y by Adventure10
sabbasi wrote:
At least that is what the version here would suggest: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=587460 but that version was the only way I could think of playing anyways.

Thanks for the link. I had some kind of confusion understanding this problem. I didn't realized that the path of the rook was colored.(the problem wasn't clear in the version of this post, actually)

But I cannot understand why it depends on parity, if the path is not drawn. I think that this game can come to the end before every cell is colored...
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sabbasi
248 posts
#12 • 1 Y
Y by Adventure10
Sorry that was a comment I made because I thought I had a strategy. But yeah this problem statement isn't clear.
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Niosha
18 posts
#13 • 4 Y
Y by SockFoot, newovertimee, Alireza_Amiri, Adventure10
I'm agree with Johnkwon. I had another strategy:
If n is odd: Bob wins. it's enough to divide our chessboard into $1\times 2$ rectangles. And then when Peter moves Bob can move to the second square.
If n is even: Peter wins . Like first situation just in the first move Peter goes to the (1,n) and then we should divide remain squares into $1\times 2$ rectangles. So Peter wins :whistling:
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mathtastic
3258 posts
#14 • 1 Y
Y by Adventure10
In fact, they shouldn't divide the board into dominoes because that makes their strategies obvious to the other player. The players can just pair the squares randomly :).
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PCChess
548 posts
#15 • 1 Y
Y by Mango247
I claim that Peter wins if $n$ is even and Bob wins when $n$ is odd. For simplicity, let the squares be denoted $(i,j)$, where a square is located in the $i$th column from the left and $j$th row from the bottom. WLOG, let the rook start at $(n,n)$ and let Peter's first move be along the top row.

First we will prove that Peter wins when $n$ is even. Consider the vertical line $m$ between the $n/2$ and $n/2+1$ column. For whatever Bob's first move is, Peter will just move the rook to the square that is the reflection of Bob's last move across $m$. Peter's move is always legal, since each square has exactly one reflection and each square is in the same row as its reflection across $m$. Since Peter can always play the mirrored square of Bob, Peter will win.

Now consider when $n$ is odd. After Peter's first move, let Bob move to another square in the top row. Consider the horizontal line $\ell$ that is between the $(n-1)/2$ and $(n+1)/2$ row. If Peter moves to a square in a top row, Bob will move to a square in the top row. If Peter moves to a square not in the top row, Bob will move to the reflection of Peter's square over line $\ell$. Note that there are an even number of squares in the top row if the starting square $(n,n)$ is not considered, so if Peter moves to a square in the top row Bob will always be able to move to a square in the top row. Moreover, for the squares not in the top row, each square corresponds to exactly one another square that is its reflection over $\ell$, and these two squares are always in the same column. Hence, if Peter moves to a square not in the top row, Bob will always be able to move to its reflection across $\ell$. Since Bob always has a response to whatever square Peter moves to, Bob wins for odd $n$.
This post has been edited 1 time. Last edited by PCChess, Mar 30, 2021, 4:14 AM
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IAmTheHazard
5001 posts
#16 • 7 Y
Y by centslordm, Fatemeh06, EpicNumberTheory, KK_1729, quantam13, Eka01, guruguha9
El clásico

Peter wins iff $n$ is even. If $n$ is even, his strategy is to divide the board into $1 \times 2$ rectangles; whenever a rook enters a rectangle (including on the first move) he moves it to the other cell. If $n$ is odd, Bob's strategy is to divide the board except for the starting cell into $1 \times 2$ rectangles and employ the same strategy. $\blacksquare$
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chakrabortyahan
381 posts
#17 • 1 Y
Y by iamahana008
I saw this problem 3 times last week USSR 78, Bmo 1 2023 and now this
Peter has a winning strategy if $n$ is even and Bob has one if $n$ is odd. We break the whole board into $2 \cdot 1 $ dominoes and note that if $n$ is even then $\text{Peter}$ has always a move if $\text{Bob}$ has a move because the square where $\text{Bob}$ moves is part of some domino and by his move $\text{Bob}$ bridges two dominoes always. So, B will be exhausted of his moves.
Similarly, for $n$ odd $\text{Bob}$ has a similar winning strategy .
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