It's February and we'd love to help you find the right course plan!

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k a February Highlights and 2025 AoPS Online Class Information
jlacosta   0
Feb 2, 2025
We love to share what you can look forward to this month! The AIME I and AIME II competitions are happening on February 6th and 12th, respectively. Join our Math Jams the day after each competition where we will go over all the problems and the useful strategies to solve them!

2025 AIME I Math Jam: Difficulty Level: 8* (Advanced math)
February 7th (Friday), 4:30pm PT/7:30 pm ET

2025 AIME II Math Jam: Difficulty Level: 8* (Advanced math)
February 13th (Thursday), 4:30pm PT/7:30 pm ET

The F=ma exam will be held on February 12th. Check out our F=ma Problem Series course that begins February 19th if you are interested in participating next year! The course will prepare you to take the F=ma exam, the first test in a series of contests that determines the members of the US team for the International Physics Olympiad. You'll learn the classical mechanics needed for the F=ma exam as well as how to solve problems taken from past exams, strategies to succeed, and you’ll take a practice F=ma test of brand-new problems.

Mark your calendars for all our upcoming events:
[list][*]Feb 7, 4:30 pm PT/7:30pm ET, 2025 AIME I Math Jam
[*]Feb 12, 4pm PT/7pm ET, Mastering Language Arts Through Problem-Solving: The AoPS Method
[*]Feb 13, 4:30 pm PT/7:30pm ET, 2025 AIME II Math Jam
[*]Feb 20, 4pm PT/7pm ET, The Virtual Campus Spring Experience[/list]
AoPS Spring classes are open for enrollment. Get a jump on 2025 and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile! Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!

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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Special AIME Problem Seminar B
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F=ma Problem Series
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0 replies
jlacosta
Feb 2, 2025
0 replies
P6 Geo Finale
math_comb01   7
N 2 minutes ago by GuvercinciHoca
Source: XOOK 2025/6
Let $ABC$ be a triangle with incenter $I$ and excenters $I_A$, $I_B$, $I_C$ opposite to $A,B,C$ respectively. Suppose $BC$ meets the circumcircle of $I_AI_BI_C$ at points $D$ and $E$. $X$ and $Y$ lie on the incircle of $\triangle ABC$ so that $DX$ and $EY$ are tangents to the incircle (different from $BC$). Prove that the circumcircles of $\triangle AXY$ and $\triangle ABC$ are tangent.

Proposed by Anmol Tiwari
7 replies
math_comb01
Feb 10, 2025
GuvercinciHoca
2 minutes ago
A functional equation
super1978   1
N 13 minutes ago by pco
Source: Somewhere
Find all functions $f: \mathbb R \to \mathbb R$ such that:$$ f(f(y-x)-xf(y))+f(x)=y(1-f(x)) $$for all $x,y \in \mathbb R$
1 reply
super1978
an hour ago
pco
13 minutes ago
Sequences Handout
M11100111001Y1R   4
N 17 minutes ago by MR.1
Source: Own
Hi everyone, I wrote this handout about sequences in NT.
Hope you enjoy!
4 replies
+2 w
M11100111001Y1R
Oct 19, 2022
MR.1
17 minutes ago
[Handout] 50 non-traditional functional equations
gghx   2
N 44 minutes ago by GreekIdiot
Sup guys,

I'm retired. I love FEs. So here's 50 of them. Yea...

Functional equations have been one of the least enjoyed topics of math olympiads in recent times, mostly because so many techniques have been developed to just bulldoze through them. These chosen problems do not fall in that category - they require some combi-flavoured creativity to solve (to varying degrees).

For this reason, this handout is aimed at more advanced problem solvers who are bored of traditional FEs and are up for a little challenge!

In some sense, this is dedicated to the "covid FE community" on AoPS who got me addicted to FEs, people like EmilXM, hyay, IndoMathXdZ, Functional_equation, GorgonMathDota, BlazingMuddy, dangerousliri, Mr.C, TLP.39, among many others: thanks guys :). Lastly, thank you to rama1728 for suggestions and proofreading.

Anyways...
2 replies
gghx
Sep 23, 2023
GreekIdiot
44 minutes ago
No more topics!
Clutched problem
mathscrazy   8
N Jan 19, 2025 by atrax.i
Source: STEMS 2025 Category A1
Alice and Bob play a game. Initially, they write the pair $(1012,1012)$ on the board. They alternate their turns with Alice going first. In each turn the player can turn the pair $(a,b)$ to either $(a-2, b+1), (a+1, b-2)$ or $(a-1, b)$ as long as the resulting pair has only nonnegative values. The game terminates, when there is no legal move possible. Alice wins if the game terminates at $(0,0)$ and Bob wins if the game terminates at $(0,1)$. Determine who has the winning strategy?

Proposed by Shashank Ingalagavi and Krutarth Shah
8 replies
mathscrazy
Dec 29, 2024
atrax.i
Jan 19, 2025
Clutched problem
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G H BBookmark kLocked kLocked NReply
Source: STEMS 2025 Category A1
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mathscrazy
113 posts
#1
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Alice and Bob play a game. Initially, they write the pair $(1012,1012)$ on the board. They alternate their turns with Alice going first. In each turn the player can turn the pair $(a,b)$ to either $(a-2, b+1), (a+1, b-2)$ or $(a-1, b)$ as long as the resulting pair has only nonnegative values. The game terminates, when there is no legal move possible. Alice wins if the game terminates at $(0,0)$ and Bob wins if the game terminates at $(0,1)$. Determine who has the winning strategy?

Proposed by Shashank Ingalagavi and Krutarth Shah
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MathematicalArceus
34 posts
#2
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Pray this is not fakesolve
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HoRI_DA_GRe8
584 posts
#3
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MathematicalArceus wrote:
Pray this is not fakesolve

But bob can change the parity right?
Since bob has to win in $(0,1)$ and you can't guarantee that $b$ doesn't change parity in bobs move , how can you say $(0,1)$ cannot be reached.

However here's the fix,note that the sum decreases by $1$ in each move.So when the sum becomes $3$ the last move is played by Alice.Since she doesn't change parity of $b$ it's even , so the pair will be $(1,2)$ .From here Alice is prolly unstoppable.
This post has been edited 1 time. Last edited by HoRI_DA_GRe8, Dec 30, 2024, 6:03 AM
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MathematicalArceus
34 posts
#5
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HoRI_DA_GRe8 wrote:
MathematicalArceus wrote:
Pray this is not fakesolve

But bob can change the parity right?
Since bob has to win in $(0,1)$ and you can't guarantee that $b$ doesn't change parity in bobs move , how can you say $(0,1)$ cannot be reached.

However here's the fix,note that the sum decreases by $1$ in each move.So when the sum becomes $3$ the last move is played by Alice.Since she doesn't change parity of $b$ it's even , so the pair will be $(1,2)$ .From here Alice is prolly unstoppable.

oh, I was claiming that if Bob plays $(a-2,b+1)$, the only move that changes parity, Alice can always play that too, but I see the mistake. I have no fix for when $(a-1,b)$ is being played except for the fact that Alice can play $(a+1,b-2)$, but I didn't specify it. I hope I will get some marks.
This post has been edited 1 time. Last edited by MathematicalArceus, Dec 30, 2024, 8:08 AM
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SimplisticFormulas
74 posts
#6
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huh
I thought Alice has a winning strat cuz she can keep doing $(a+1,b-2)$ and no matter what bob does the no. on the right is $\le$ no. on the left, so the no. on the right reaches $0$ first. So the game can end only in $(0,0)$.
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HoRI_DA_GRe8
584 posts
#8
Y by
SimplisticFormulas wrote:
huh
I thought Alice has a winning strat cuz she can keep doing $(a+1,b-2)$ and no matter what bob does the no. on the right is $\le$ no. on the left, so the no. on the right reaches $0$ first. So the game can end only in $(0,0)$.

Then bob will keep doing $(a-2,b+1)$ to equalize.When the game reaches $(1,1)$ Alice cannot get to $(0,0)$ from here :|
This post has been edited 1 time. Last edited by HoRI_DA_GRe8, Dec 30, 2024, 9:10 AM
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Int_guy
1 post
#9
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Answer: Alice has the winning strategy
It is obvious that if (a,b) is the no. and (a+b) is even it is Alice's move otherwise it's Bob's.

Claim 1: If (a,b)=(x,0) Alice wins. x not equal to 3 or 4 or 5.
Proof: It suffices to prove Alice wins at (2k-1,0) as Alice can turn (2k,0) to (2k-1,0).
We prove by induction assume true for x less than 2k+1 for some k>3 and x>6
Bob can't turn (2k+1,0) to (2k,0) so he turns (2k+1,0) to (2k-1,1) alice always decrease a by 1 in (a,b) so it turns to (2k-2,1) bob can't increase b by 1 as then b=2 and Alice turns b to 0, so he also always decrease a by 1 so after many iterations it turns to (7,1) where alice can win.

Claim 2: If (a,b)=(x,1) alice wins x not equal to 1 or 2 or 6
Proof: As in above Bob must only decrease a by 1 so (7,1) must be reached.

Claim 3:If (a,b)=(c,c) Alice wins c greater than 7.
Proof: We use induction.
Alice turns (c,c) to (c+1,c-2) Bob can't turn it to (c-1,c-1) so he turns it to (c,c-2) now at every point Alice decreases b by 2 and increase a by 1 regardless of what Bob does. It can be seen that due to the first move and alice's strategy a-b>=2. So the only problem areas occur when final (a,b)= (6,1); (3,0); (4,0) ;(5,0) by backtracking it can be seen that all the problems are fixed when Alice converts (5,3) to (4,3) instead of (6,1). Hence, concluding the proof.

(there are a lot of base cases for induction and backtracking cases which are omitted as they are very messy so you can do them yourself or take my word for it!)

It is my first solution post on Aops kindly tell me if the solution is correct or if the proof writing can be improved.
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SimplisticFormulas
74 posts
#11
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HoRI_DA_GRe8 wrote:
SimplisticFormulas wrote:
huh
I thought Alice has a winning strat cuz she can keep doing $(a+1,b-2)$ and no matter what bob does the no. on the right is $\le$ no. on the left, so the no. on the right reaches $0$ first. So the game can end only in $(0,0)$.

Then bob will keep doing $(a-2,b+1)$ to equalize.When the game reaches $(1,1)$ Alice cannot get to $(0,0)$ from here :|

oof :(
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atrax.i
1 post
#13
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HoRI_DA_GRe8 wrote:
MathematicalArceus wrote:
Pray this is not fakesolve

But bob can change the parity right?
Since bob has to win in $(0,1)$ and you can't guarantee that $b$ doesn't change parity in bobs move , how can you say $(0,1)$ cannot be reached.

However here's the fix,note that the sum decreases by $1$ in each move.So when the sum becomes $3$ the last move is played by Alice.Since she doesn't change parity of $b$ it's even , so the pair will be $(1,2)$ .From here Alice is prolly unstoppable.

You discussed the (1, 2) case but not the (3, 0) case. Bob will turn (3, 0) to (1, 1) forcing a victory. (3, 0) is a forced loss for Alice.
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