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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Jul 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.
Posting Guidelines
Update on Basic Forum Rules
What belongs on this forum?
How do I write a thorough solution?
How do I get a problem on the contest page?
How do I study for mathcounts?
Mathcounts FAQ and resources
Mathcounts and how to learn

As always, if you have any questions, you can PM me or any of the other Middle School Moderators. Once again, if you see spam, it would help a lot if you filed a report instead of responding :)

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
Areas of triangles AOH, BOH, COH
Arne   72
N 4 minutes ago by mudkip42
Source: APMO 2004, Problem 2
Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Prove that the area of one of the triangles $AOH$, $BOH$ and $COH$ is equal to the sum of the areas of the other two.
72 replies
Arne
Mar 23, 2004
mudkip42
4 minutes ago
Cauchy-Schwarz proof
Nguyenhuyen_AG   1
N 7 minutes ago by Nguyenhuyen_AG
Let $a, \, b, \, c$ be non-negative real numbers. Prove that
\[\frac{8a}{b+2c}+\frac{48b}{c+3a}+\frac{c}{a+4b} \geqslant 4.\]\[\frac{16a}{b+2c}+\frac{85b}{c+3a}+\frac{2c}{a+4b} \geqslant 8.\]hide
1 reply
Nguyenhuyen_AG
21 minutes ago
Nguyenhuyen_AG
7 minutes ago
inequality
SunnyEvan   3
N 11 minutes ago by SunnyEvan
Source: Own
Let $ x \in [\frac{\pi}{2}-1, 1) $, try to prove or disprove that :
$$ \frac{(\sqrt2 cosx -1)^2}{cos2x+tan\frac{\pi}{8}}-\frac{(\sqrt2 sinx -1)^2}{cos2x-tan\frac{\pi}{8}} \geq \frac{1}{2}(\frac{tanx-1}{tanx+1})^2 $$
3 replies
SunnyEvan
Yesterday at 1:24 PM
SunnyEvan
11 minutes ago
Easy Combinatorial Geometry
EthanWYX2009   0
11 minutes ago
Source: 2025 February 谜之竞赛-4
Given integer $n\ge 2$. Define set
\[V_n:=\{(a_1,\cdots ,a_n)\in\mathbb R^n\mid a_1,\cdots ,a_n\ge 0\text{ and }a_1+\cdots +a_n=1\}.\]For $\alpha =(a_1,\cdots ,a_n)$, $\beta =(b_1,\cdots ,b_n)\in V_n$, define $d(\alpha ,\beta):=\sum\limits_{i=1}^n|a_i-b_i|.$

Determine the minimum real number $\lambda$, such that for all $v_1,v_2,\cdots ,v_{n+2}\in V_n$, there exists $1\le i<j\le n+2$ such that $d(x_i,x_j)\le\lambda$.

Created by Cheng Jiang
0 replies
+1 w
EthanWYX2009
11 minutes ago
0 replies
9 Pythagorean Triples
ZMB038   83
N Jul 8, 2025 by OWOW
Please put some of the ones you know, and try not to troll/start flame wars! Thank you :D
83 replies
ZMB038
May 19, 2025
OWOW
Jul 8, 2025
segment of midpoints in a trapezoid
FrancoGiosefAG   1
N Jun 9, 2025 by miles888
The bases of a trapezoid measure $18$ cm and $8$ cm, and the other two sides measure $8$ cm and $6$ cm. Find the length, in centimeter, of the segment that joins the midpoints of the bases.
1 reply
FrancoGiosefAG
Jun 9, 2025
miles888
Jun 9, 2025
The best math formulas?
anticodon   20
N May 10, 2025 by valisaxieamc
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
20 replies
anticodon
May 8, 2025
valisaxieamc
May 10, 2025
2023 AMC 12A Problem 18 Question
Kempu33334   2
N Sep 30, 2024 by Kempu33334
Why does the solution not work if we use pythagorean theorem on $A,O$ and the intersection of $C_1, C_2, C_4$?

https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_18
2 replies
Kempu33334
Sep 30, 2024
Kempu33334
Sep 30, 2024
Pythagorus Theorem Proof
SomeonecoolLovesMaths   39
N Jun 18, 2024 by dbnl
Are there any interesting proof of the Pythagorus Theorem that stand out from the rest? If yes then please share them with me.
39 replies
SomeonecoolLovesMaths
May 20, 2024
dbnl
Jun 18, 2024
Geometry Proofs!
Raymonm   1
N Nov 22, 2023 by AndrewTom
Prove any geometry theorems.
I’ll start.
Pythagorean Theorem
We attach four right triangles so that we make a smaller square on the inside and a larger square on the outside. Let a and b be the legs of triangles and c is the hypotenuse. Therefore, the area of the outer square is equal to $(a+b)(a+b)=a^2+2ab+b^2$.The areas of all the right triangles is $4(ab/2)=2ab$. The area of the inner square is the area of the outer square minus the area of all the triangles, or $a^2+2ab+b^2-2ab=a^2+b^2$. Then $\sqrt{a^2+b^2}=c$. Squaring both sides gives us $a^2+b^2=c^2$.
1 reply
Raymonm
Nov 22, 2023
AndrewTom
Nov 22, 2023
Prove to me why the pytheogrean theorm works
scottmiddle2023   19
N Jul 28, 2023 by mathmax12
Just curious.
19 replies
scottmiddle2023
Jul 12, 2023
mathmax12
Jul 28, 2023
400 post math story
DhruvJha   71
N Jul 8, 2023 by Aopsauser9999
I know im not as orz as most of yall but heres my math story:


pre-k: learned to count occazionaly skipping numbers like 1,2,3,5,4,8,9,12....


kinder- actually learned to count through the torchure of kumon. still not great at math and was a very naughty kid... i remember getting put in a hallway for not writing a story on fish sus.

first- learned the times tables and the basics of multiplication due to annoying kumon which made me cry every day cuz i had a skill issue. known as the class genius.

2nd- still doing kumon aka torture learning things like fractions and long division. I hated long division. known as the class clown and stuff and also for being smart. I was trying to be the "cool kid".....

3rd- Kumon had gotten me into advanced algebra(in my 3rd grade eyes). This was the year i made the bigust jump in mathematica doing things like linear systems and by the end of the year doing Pythagorean theorem. lol i remember while i was learning the pythagorean theorem my kumon teacher completely ruined my knowledge as she was not the brightest. She would tell me that a^2+b^2=c^2. She said she could take the squares common to get a+b=c making me think that 3,4,7 would be a right triangle lol. I was also a maps tets grinder and made my biggest jump of 31 in a single session from being 215 in fall to 246 in winter lol my parents didn believe me till the scores were emailed home.

4th- Mastering all the topics from last and was known as the smartest kid in my class and stuf. maps test score was near 260 by the end of year as i was pro grinder. kumon had gotten me into basic factoring, quadratics. Year was cut short due to covid in febuary or smthn. i never did any of the virtual assignments lol.

5th- Kumon had gotten me into advanced things such as complex numbers... and made me do basically the full algebra 2 curriculum and by the end of year doing hard stuff. I was virtual for this full year and never turned on the camera so the teacher thought i was playing amogus.

6th- quit kumon.. as integration and difrentiation was too complex for my dumb mind to handle... this was one of my down years... i never knew there was a such thing as mathcounts and even when i was eligible i didnt signup because im unorz. still grind maps tests scores were like 280.

7th- joined mathcounts- throughout the year was very dominant and by the end was tied for first with my bestie. yays orz. In chapter competition placed second individually(lost to a 8th grader)... then state competition came which meant disaster. I placed 5th in the state losing to my bestie. I held back tears in the bus back home and was depressed... I found out 3 days later that our scores were tied but they gave him the nod for some reason i dont know why.

Now- in the summer of 7th grade grinding mathcounts and aops amc stuff.

71 replies
DhruvJha
Jul 1, 2023
Aopsauser9999
Jul 8, 2023
1K Post Milestone Reached!
shaayonsamanta   28
N Jun 15, 2023 by naenaendr
I reach one thousand posts!

I didn't feel like giving a math story, because I decided I will once I reach one year with an AoPS Account. So, here are ten problems, each representing 100 posts!

P1: Ratios and Algebraic Equations
P2: Complex Numbers
P3: Fractions and Decimals
P4: Percentages and Roots
P5: Angles and Polygons
P6: Simultaneous Equations and Algebraic Inequalities
P7: Sums, Averages, and Ranges
P8: Triangles and the Pythagorean Theorem
P9: Counting
P10: Age Trick
P11: Rationality of 3
P12: Digits of Powers
28 replies
shaayonsamanta
Apr 24, 2023
naenaendr
Jun 15, 2023
Pythagorean Theorem cubed?
juicetin.kim   21
N Jun 14, 2023 by alpha_2
We know that for any right triangle,
$a^2+b^2=c^2$
Now, let a, b, c, and d be positive integers.
Find a possible solution for the following equation.
EDIT: If there are no solutions write no solutions.
$a^3+b^3+c^3=d^3$

I'm just curious about this.
21 replies
juicetin.kim
Jun 14, 2023
alpha_2
Jun 14, 2023
Graph Process Problem
Maximilian113   15
N Jun 13, 2025 by heheman
Source: CMO 2025 P1
The $n$ players of a hockey team gather to select their team captain. Initially, they stand in a circle, and each person votes for the person on their left.

The players will update their votes via a series of rounds. In one round, each player $a$ updates their vote, one at a time, according to the following procedure: At the time of the update, if $a$ is voting for $b,$ and $b$ is voting for $c,$ then $a$ updates their vote to $c.$ (Note that $a, b,$ and $c$ need not be distinct; if $b=c$ then $a$'s vote does not change for this update.) Every player updates their vote exactly once in each round, in an order determined by the players (possibly different across different rounds).

They repeat this updating procedure for $n$ rounds. Prove that at this time, all $n$ players will unanimously vote for the same person.
15 replies
Maximilian113
Mar 7, 2025
heheman
Jun 13, 2025
Graph Process Problem
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Source: CMO 2025 P1
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Maximilian113
587 posts
#1
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The $n$ players of a hockey team gather to select their team captain. Initially, they stand in a circle, and each person votes for the person on their left.

The players will update their votes via a series of rounds. In one round, each player $a$ updates their vote, one at a time, according to the following procedure: At the time of the update, if $a$ is voting for $b,$ and $b$ is voting for $c,$ then $a$ updates their vote to $c.$ (Note that $a, b,$ and $c$ need not be distinct; if $b=c$ then $a$'s vote does not change for this update.) Every player updates their vote exactly once in each round, in an order determined by the players (possibly different across different rounds).

They repeat this updating procedure for $n$ rounds. Prove that at this time, all $n$ players will unanimously vote for the same person.
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Maximilian113
587 posts
#2
Y by
bruh
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awesomeming327.
1780 posts
#3
Y by
Solution Sketch:

The idea is that there is this "center cycle" that will persist until it becomes a self-loop. Each time a member of the center cycle changes vote, the center cycle's size decreases by 1. One can find that this means that each time the procedure is run through, the center cycle decreases by at least half. This will take $\lfloor \log_2(n)\rfloor$

After that, we can consider distances to that self-loop. Each individual path decreases in a way very similar to that cycle we mentioned earlier, and so the worst case scenario is that the path size is $n-1$, whence it will take $\lfloor \log_2(n-1)\rfloor$

Summing these two together, one realizes that it is at most $n$.

My comments on the problem:

It is way, way, way too hard for a P1. This should've been put as P4 instead, maybe.
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hectorleo123
355 posts
#4
Y by
I would say this is at least a P3 from the CMO.
$\textbf{Solution:}$

Let round \( 0 \) be the initial state (before any updates occur). Define \( G_i \) as the directed graph where the \( n \) players are the vertices, and the directed edges represent the votes after round \( i \), for all \( i \in \{0,1,\dots,n\} \).

$\textbf{Lemma:}$ \( G_i \) contains a unique cycle, and this cycle has length at most \( n - i \). Furthermore, any vertex outside the cycle is part of a directed path of length at most \( n - i \).

$\textbf{Proof:}$ We proceed by induction on \( i \).

$\textbf{Base case:}$ \( i = 0 \) (trivial).

$\textbf{Inductive Hypothesis:}$ Suppose the lemma holds for all \( i < k \leq n \).

$\textbf{Inductive Step:}$ Let \( v_1, v_2, \dots, v_t \) be the vertices of the cycle in \( G_{k-1} \) such that \( v_i \to v_{i+1} \) and \( v_{t+1} = v_1 \), where \( v_1 \) is the first vertex to update its vote in round \( k \).

Observe that we now have \( v_1 \to v_3 \), meaning that \( v_2 \) is no longer part of the cycle in \( G_k \). Notice that the only way a vertex \( v_j \) can leave the cycle and become part of a directed path (i.e., no longer in any cycle) is if there exists a vertex \( w \) in the cycle such that \( w \) votes for \( v_j \) and \( v_j \) previously belonged to the cycle...(\(\alpha\)).

Now, we verify that any directed path outside the cycle has length at most \( n - k \). By the inductive hypothesis, all such paths in \( G_{k-1} \) have length at most \( n - k + 1 \). In each round, the length of every path decreases by at least \( 2 \) from its outermost vertex towards the cycle, while it increases by at most \( 1 \) at the end touching the cycle (by (\(\alpha\))). Therefore, the length of any path is at most \( n - k \).

This completes the induction. Applying the lemma to \( G_n \), we conclude that all \( n \) players will eventually vote for the same person. $_\blacksquare$
This post has been edited 1 time. Last edited by hectorleo123, Mar 8, 2025, 12:38 AM
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genius_007
275 posts
#5
Y by
I don't think the solution above works fully, from your assumption that the first person to go lies in the cycle. We show how we can make a path grow by more than $2$ every round. Say we have a cycle $C$ and path $P$ coming off of $C$, where $|C| >> |P|$. Let the vertices in $C$ be $v_1,v_2,\dots,v_k$, where $v_i$ points to $v_{i+1}$. WLOG, let the first vertex $w$ of $P$ go to $v_1$. Update $w$ first, so it now points to $v_2$. Update $v_1$, so $v_1$ points to $v_3$. Update $v_2$, so $v_2$ points to $v_4$. Update $v_3$, so $v_3$ points to $v_5$. We may continue this, where $v_i$ points to $v_{i+2}$. In particular, we get the new path $w,v_2,v_4,\dots,v_{2r}$, where $r = \lfloor\dfrac{k}{2}\rfloor$. We've already updated all of those $r+1$ vertices in $P$, so $P$ has new length at least $r$. For large $C$, we are able to increase the length of $P$, which shows your solution is wrong.
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hectorleo123
355 posts
#6
Y by
Sorry, my mistake, I meant that after the movement of v_1, v_2 left the cycle, then v_1 could disappear, fixed
This post has been edited 1 time. Last edited by hectorleo123, Mar 12, 2025, 9:55 PM
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cursed_tangent1434
737 posts
#7
Y by
Am I fakesolving or is the problem much easier than the above solutions indicate?

We prove the claim via induction. In the case where $n=2$ the result is trivial. Now, assume the result holds for some positive integer $k \ge 2$. When there are $k+1$ players, consider the very first update. Say this was an update for player $i$. Player $i$ votes for $i+1$ who votes for $i+2 \pmod{k+1}$ so $i$ will now vote for $i+2$. Note that then, there is no player voting for player $i+1$. Since each move replaces the player each player is voting for by some other player who is being voted for by some player, player $i+1$ will never receive a vote from any player ever again.

Now, player $i+1$ can be 'ignored'. He does not have any effect whatsoever on the players everyone else is voting for. However, the player he votes for changes over time (depending on the choices of the other players). Thus, each round can be considered as the conjunction of a round among $k$ players and an extra move for player $i+1$ (which may happen at any time during the round, in the first round it occurred as the first move). After $k$ rounds, players $1$ to $i$ and $i+2$ to $k+1$ will unanimously vote for the same person (who is within this set). Now, if player $i+1$ also votes for this player we are already done. Else, in the last round player $i+1$ adjusts his vote to match someone within the other set of $k$ players who all vote for the same person, which completes the induction.

Thus, repeating the updating procedure for $n$ rounds, always the $n$ players will end up voting for the same person as desired.
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genius_007
275 posts
#8
Y by
This is a fakesolve. After $i$ changes their vote to $i+1$, we cannot directly use the I.H. on $1,2,\dots,i,i+2,\dots,k+1$, as $i$ cannot have their vote updated (they already updated their vote).
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lw202277
3 posts
#9 • 2 Y
Y by swynca, MS_asdfgzxcvb
Treat the players and their votes as a directed graph $G$.

Claim: There is exactly 1 cycle $C$ in $G$.

This is initially true. Suppose $a \rightarrow b \rightarrow c$ becomes $a \rightarrow c$ and $b \rightarrow c$. The number of new cycles created is the number of directed paths from $c$ to $a$, and the number of cycles removed is the number of directed paths from $b$ to $a$. Since $b \rightarrow c$, this is equal to the number of directed paths from $c$ to $a$. Therefore the number of cycles stays constant.

Claim: The cycle's size decreases by at least 1 each round. (In hindsight, this was a very weak bound and can be replaced with $O(\log_2 n)$)

Just note that whenever a player $a \in C$ changes their vote from $b$ to $c$, $b$ is removed from the cycle unless $b=c$. If $b=c$ then we must have $a=b=c$ or else there would be no directed path from $b=c$ to $a$. This occurs iff the cycle has size 1.

Claim: Any branch (a directed path pointing into the cycle) has length at most 2, at the end of each round.

Note that any branch with length $\geq 2$ decreases its length when a player not voting for an element of $C$ switches their vote. It suffices to show that no branch can increase its length by more than 1 in any round.

Suppose $a \rightarrow b \rightarrow c$ where $a,b,c \in C$, and let $v_1 \rightarrow \dots \rightarrow v_k \rightarrow b$ be a branch. The only way for the branch to increase its length is for $b$ to be ejected from the cycle, which only occurs when $a$ switches their vote from $b$ to $c$. After this, $c$ is now the endpoint of the branch. Note that $c$ cannot be ejected from the cycle since this would require $a$ to switch their vote from $c$ to another player, which is impossible since $a$ already switched their vote. Thus the branch's length increases by at most 1 each round, which is nullified by it decreasing its length by at least 1 each round.

After round $n-1$ the cycle will have size 1, which is a player voting for themselves. All branches will have length at most 2, and each of these branches will reduce their length to 1 on round $n$. This implies that everyone will be voting for the same player. QED
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ray66
83 posts
#10
Y by
Treat the system as a group where elements consist of players and their respective votes. The group operation turns $a \rightarrow b \rightarrow c$ into $a \rightarrow c$ and $b \rightarrow c$.

$\textbf{Claim: }$ There exists only one cycle in each element

$\textbf{Claim: }$ Each round of operations decreases the length of the cycle by at least 1

$\textbf{Claim: }$ If the length of the cycle is $0$ we finish

$\textbf{Proof: }$ From the starting element (a circular graph) there is only 1 cycle. Each node in a 1-cycle graph can be characterized into nodes that comprise the cycle and nodes that feed into the cycle. The starting element only has nodes that comprise the cycle, and if $a$ and $b$ are both cycle nodes ($a$ and $b$ exist because the cycle length is at least 1), the operation $a \rightarrow b \rightarrow c$ into $a \rightarrow c$ and $b \rightarrow c$ turns $b$ into a feeding node while $a$ and $c$ remain cycle nodes.

$\textbf{Claim: }$ Feeding nodes always point into the cycle

$\textbf{Claim: }$ Each round of operations decrease the length of a feeding chain by at least 1

Because all processes must terminate and there exists exactly 1 cycle in each graph, everyone will eventually vote for the same player.
This post has been edited 1 time. Last edited by ray66, Mar 13, 2025, 3:13 PM
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Ru83n05
172 posts
#11
Y by
Here is how to achieve an $O(n)$ bound (which solutions above improve to $\lfloor\log_2 n\rfloor + 1$).

Call a player a leaf if no other player is voting for them. Once a leaf, they will always be a leaf. We claim that after $k\leq n-1$ rounds, there will be at least $k$ leaves.

We prove this by induction. In the first round, the first time a player updates their vote, the person they used to vote for becomes a leaf. Hence this is true when $k=1$.

Now, suppose we know the statement is true for $k$. If there exists a player that is only voted by leafs, that player will become a leaf at the end of the round. Otherwise, all non-leaf players are being voted by at least one non-leaf, and therefore exactly one non-leaf. In that case, the first time a non-leaf (in a nontrivial cycle) updates their vote, we get a vertex that is only voted by leafs. Hence by the end of the round we always get an extra leaf.

In particular, after $n-1$ rounds, all players will be voting for the same person.
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heheman
1090 posts
#13
Y by
how are u guys so genius bro teach me how i cant solve the first problem im gonna rage
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genius_007
275 posts
#14 • 1 Y
Y by heheman
tbf this was arguably the hardest question on CMO this year, a lot of people were not able to solve
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heheman
1090 posts
#15
Y by
guys, i think i got it!!!!

First draw graph with vertices representing players, and if $a$ votes for $b$ then there's arrow directed from $a$ to $b$. call the graph "connected" if the graph is connected when viewing the directed edges as undirected.

Observation 1: graph is always connected
Proof: note that every operation preserves it

Thus there is only ever 1 cycle in the whole graph, since if there is $2$ the only way it can exist if they are disjoint. not true by our observation.

Now note that the cycle decreases by at least $1$ for every turn. Note that during any turn it will always preserve the cycle structure. and at the end there are 2 arrows pointing to the same vertex by pigeonhole so one of them must have no arrow pointing to so it loses at least 1. note

anyways, let $k$ be the largest distance from a point traveling from outside the cycle needs to enter the cycle. Note that $k$ goes to $k-1+[$# of pts removed from the cycle}$]$ after every turn (or simply $k+[$# of pts removed from the cycle$]$ if $k$ was $0$ which is only at first turn, or $1$), at the worst case.

But every time some x = [# pts is removed from the cycle] is removed, it buys us $x-1$ turns of extra time in the future so we can subtract all of it from $k$

class ended so i am need to be brief
This post has been edited 1 time. Last edited by heheman, Jun 10, 2025, 10:47 PM
Reason: indexing error(was $0$ which is only at first turn -> was $0$ which is only at first turn, or $1$.)
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awesomeming327.
1780 posts
#16 • 1 Y
Y by heheman
I think most people who tried to do arguments of the form "cycle decreases by at least 1 each turn" got a 2/7

lol
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heheman
1090 posts
#17
Y by
awesomeming327. wrote:
I think most people who tried to do arguments of the form "cycle decreases by at least 1 each turn" got a 2/7

lol

o sadge

do u know wut errors they might've made, idk if i fakesolve
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