The time is now - Spring classes are filling up!

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k a AMC 10/12 A&B Coming up Soon!
jlacosta   0
Nov 1, 2024
There is still time to train for the November 6th and November 12th AMC 10A/12A and AMC 10B/12B, respectively! Enroll in our weekend seminars to be held on November 2nd and 3rd (listed below) and you will learn problem strategies, test taking techniques, and be able to take a full practice test! Note that the “B” seminars will have different material from the “A” seminars which were held in October.

[list][*]Special AMC 10 Problem Seminar B
[*]Special AMC 12 Problem Seminar B[/list]
For those who want to take a free practice test before the AMC 10/12 competitions, you can simulate a real competition experience by following this link. As you assess your performance on these exams, be sure to gather data!

[list][*]Which problems did you get right?
[list][*]Was the topic a strength (e.g. number theory, geometry, counting/probability, algebra)?
[*]How did you prepare?
[*]What was your confidence level?[/list]
[*]Which problems did you get wrong?
[list][list][*]Did you make an arithmetic error?
[*]Did you misread the problem?
[*]Did you have the foundational knowledge for the problem?
[*]Which topics require more fluency through practice (e.g. number theory, geometry, counting/probability, algebra)?
[*]Did you run out of time?[/list][/list]
Once you have analyzed the results with the above questions, you will have a plan of attack for future contests! BEST OF LUCK to all competitors at this year’s AMC 10 and AMC 12!

Did you know that the day after both the AMC 10A/12A and AMC 10B/12B you can join a free math jam where our AoPS team will go over the most interesting problems? Find the schedule below under “Mark your calendars”.

Mark your calendars for these upcoming free math jams!
[list][*]November 20th: Amherst College Info Session, 7:30 pm ET: Matt McGann, Dean of Admission and Financial Aid at Amherst College, and Nathan Pflueger, math professor at Amherst College, will host an info session exploring both Amherst College specifically and liberal arts colleges generally. Topics include opportunities in math, the admission process, and financial aid for both US and international students.
[*]November 7th: 2024 AMC 10/12 A Discussion, Thursday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 A, administered November 6. We will discuss some of the most interesting problems from each test!
[*]November 13th: 2024 AMC 10/12 B Discussion, Wednesday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 B, administered November 12. We will discuss some of the most interesting problems from each test![/list]
AoPS Spring classes are open for enrollment. Get a jump on the New Year 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!

Whether you've taken our classes at AoPS Online or AoPS Academies or reached the top echelons of our competition training with our Worldwide Online Olympiad Training (WOOT) program, you can now add your AoPS experience to the education section 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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0 replies
jlacosta
Nov 1, 2024
0 replies
A colouring game on a grid
Tintarn   0
a minute ago
Source: Baltic Way 2024, Problem 8
Let $a$, $b$, $n$ be positive integers such that $a + b \leq n^2$. Alice and Bob play a game on an (initially uncoloured) $n\times n$ grid as follows:
- First, Alice paints $a$ cells green.
- Then, Bob paints $b$ other (i.e.uncoloured) cells blue.
Alice wins if she can find a path of non-blue cells starting with the bottom left cell and ending with the top right cell (where a path is a sequence of cells such that any two consecutive ones have a common side), otherwise Bob wins. Determine, in terms of $a$, $b$ and $n$, who has a winning strategy.
0 replies
Tintarn
a minute ago
0 replies
Tiling squares in 2024 is harder than in 2025, right?
Tintarn   0
3 minutes ago
Source: Baltic Way 2024, Problem 7
A $45 \times 45$ grid has had the central unit square removed. For which positive integers $n$ is it possible to cut the remaining area into $1 \times n$ and $n\times 1$ rectangles?
0 replies
Tintarn
3 minutes ago
0 replies
A magical labyrinth with a lot of caves
Tintarn   0
4 minutes ago
Source: Baltic Way 2024, Problem 6
A labyrinth is a system of $2024$ caves and $2023$ non-intersecting (bidirectional) corridors, each of which connects exactly two caves, where each pair of caves is connected through some sequence of corridors. Initially, Erik is standing in a corridor connecting some two caves. In a move, he can walk through one of the caves to another corridor that connects that cave to a third cave. However, when doing so, the corridor he was just in will magically disappear and get replaced by a new one connecting the end of his new corridor to the beginning of his old one (i.e., if Erik was in a corridor connecting caves $a$ and $b$ and he walked through cave $b$ into a corridor that connects caves $b$ and $c$, then the corridor between caves $a$ and $b$ will disappear and a new corridor between caves $a$ and $c$ will appear).

Since Erik likes designing labyrinths and has a specific layout in mind for his next one, he is wondering whether he can transform the labyrinth into that layout using these moves. Prove that this is in fact possible, regardless of the original layout and his starting position there.
0 replies
Tintarn
4 minutes ago
0 replies
What if you take a bit more than the average?
Tintarn   0
6 minutes ago
Source: Baltic Way 2024, Problem 5
Find all positive real numbers $\lambda$ such that every sequence $a_1, a_2, \ldots$ of positive real numbers satisfying
\[
a_{n+1}=\lambda\cdot\frac{a_1+a_2+\ldots+a_n}{n}
\]for all $n\geq 2024^{2024}$ is bounded.

Remark: A sequence $a_1,a_2,\ldots$ of positive real numbers is \emph{bounded} if there exists a real number $M$ such that $a_i<M$ for all $i=1,2,\ldots$
0 replies
Tintarn
6 minutes ago
0 replies
No more topics!
Find all triangles
Rushil   6
N Nov 13, 2024 by Mr.Sharkman
Source: IMO 1968 A1
Find all triangles whose side lengths are consecutive integers, and one of whose angles is twice another.
6 replies
Rushil
Nov 4, 2005
Mr.Sharkman
Nov 13, 2024
Find all triangles
G H J
Source: IMO 1968 A1
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Rushil
1592 posts
#1 • 2 Y
Y by Adventure10, Mango247
Find all triangles whose side lengths are consecutive integers, and one of whose angles is twice another.
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campos
411 posts
#2 • 2 Y
Y by Adventure10, Mango247
Let $ABC$ be a triangle with such conditions, such that $A=2B$. It's very well-known (or it can be deduced from the cosine law and the double-angle formula, or i think it's somewhere in the forum) that $a^2-b^2=bc$
Then, we just have to check the cases $(a,b,c)=(a,a-1,a-2), (a,b,c)=(a,a-2,a-1)$, and $(a,b,c)=(a,a-1,a+1)$.

$(a,b,c)=(a,a-1,a-2)$ gives $a^2-5a+3=0$, but it has no integer roots.
$(a,b,c)=(a,a-1,a+1)$ gives $a^2-2a=0 \Rightarrow a=2 \Rightarrow a+b=3=c \Rightarrow\Leftarrow$
finally, $(a,b,c)=(a,a-2,a-1)$ gives $a^2-7a+6=0$ and since $a=1 \Rightarrow c=0$, then we must have that $a=6$.
then the triangle that satisfies such conditions is $(a,b,c)=(6,4,5)$ with $A=2B$
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sayantanchakraborty
505 posts
#3 • 2 Y
Y by Adventure10, Mango247
If <A=2<B, some simple manipulating with the sine rule gives \[ /a^2=b(b+c) /\]. Assuming the lengths of the sides to be \[/ (a-1,a,a+1)/\] and considering all possible cases we get the solutions in (a,b,c) as
\[/(1,2,3) and (4,5,6)/\] and obviously all its permutations.
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sayantanchakraborty
505 posts
#4 • 2 Y
Y by Adventure10, Mango247
Campos missed the solution \[(1,2,3)\].
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sunken rock
4339 posts
#5 • 2 Y
Y by Heisenberg09, Adventure10
sayantanchakraborty wrote:
Campos missed the solution \[(1,2,3)\].

That´s NOT a triangle, but a degenerated one!

Best regards,
sunken rock
Z K Y
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suli
1498 posts
#6 • 2 Y
Y by Adventure10, Mango247
Well technically a degenerate triangle is a triangle. The problem should be worded more clearly, or both solutions should be accepted as correct.
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Mr.Sharkman
386 posts
#8
Y by
Bruh AMC 2024 #22
Z K Y
N Quick Reply
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