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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 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
How to be wide awake during math competitions.
orangebear   13
N 2 minutes ago by studymoremath
As three years of taking math competitions, I have noticed that some of the competitions that I did really bad in " I mean really bad in" where the cases where I was sleepy, and not fully awake "examples like the AMC 10 B 2024, AMC 10 A 2023, AMC 8 2023, AMC 8 2024, Mathcounts chapter 2024, etc." and the ones where I was awake and, alert like "AMC 8 2022, National Math-con 2024 "not mathcounts", Mathkangaroo 2021, Mathcounts state, etc" I did medium - too surprising myself. Do any of you people have any tips on how to be alert during math competitions?
13 replies
orangebear
Today at 1:51 AM
studymoremath
2 minutes ago
9 How do AMC-10 scores translate to MATHCOUNTS States
pingpongmerrily   49
N 4 minutes ago by bwu_2022
How do AMC scores translate to MathCounts States, given that the problems are pretty similar?

Use your own individual experience/guesses to determine what you think..

NOTE: THIS IS FOR STATES!!!
49 replies
pingpongmerrily
Nov 14, 2024
bwu_2022
4 minutes ago
AMC Music Video - Orz to the Legends (AMC 10/12 Version)
megahertz13   22
N 18 minutes ago by ijco
Orz to the Legends (Music, AMC version - MegaMath Channel original)

Get pumped up for the AMCs!

[youtube]https://www.youtube.com/watch?v=KmiNI00uo-s&ab_channel=MegaMathChannel[/youtube]

Join our discord! https://discord.gg/hh7vntTb2E

Math Problem (AMC 10/12): Let $M$ be the smallest positive integer satisfying the property that $M^6$ is a multiple of both $2024^2$ and $2025^3$. How many positive divisors does $M$ have?

James has a geometric series with first term $1$ and common ratio $0<r<1$. Andy multiplies the first term by $\frac{7}{5}$, and squares the common ratio. If the sum of the first series equals the sum of the second series, the common sum can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
22 replies
megahertz13
Oct 13, 2024
ijco
18 minutes ago
9 AIME Qualification AMC 12A
studymoremath   1
N 19 minutes ago by studymoremath
Now with the AMC's over, I think everyone has more of an idea of cutoff ranges and difficulty levels this year. Are my chances high enough that I should start studying for AIME?
1 reply
studymoremath
Yesterday at 6:43 PM
studymoremath
19 minutes ago
No more topics!
Problem 21
evt917   16
N Nov 13, 2024 by brainfertilzer
Source: AMC 12B 2024 Problem 21
The measures of the smallest angles of three different right triangles sum to $90^\circ$. All three triangles have side lengths that are primitive Pythagorean triples. Two of them are $3-4-5$ and $5-12-13$. What is the perimeter of the third triangle?

$
\textbf{(A) }40 \qquad
\textbf{(B) }126 \qquad
\textbf{(C) }154 \qquad
\textbf{(D) }176 \qquad
\textbf{(E) }208 \qquad
$
16 replies
1 viewing
evt917
Nov 13, 2024
brainfertilzer
Nov 13, 2024
Problem 21
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G H BBookmark kLocked kLocked NReply
Source: AMC 12B 2024 Problem 21
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evt917
1514 posts
#1 • 1 Y
Y by clarkculus
The measures of the smallest angles of three different right triangles sum to $90^\circ$. All three triangles have side lengths that are primitive Pythagorean triples. Two of them are $3-4-5$ and $5-12-13$. What is the perimeter of the third triangle?

$
\textbf{(A) }40 \qquad
\textbf{(B) }126 \qquad
\textbf{(C) }154 \qquad
\textbf{(D) }176 \qquad
\textbf{(E) }208 \qquad
$
Z K Y
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PEKKA
1791 posts
#2
Y by
154, its a 33-56-65 triangle.
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HonestCat
947 posts
#3
Y by
PEKKA wrote:
154, its a 33-56-65 triangle.

Yep, I sillied but I can confirm
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ihatemath123
3339 posts
#4
Y by
why was this placed as #21
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Toinfinity
588 posts
#5
Y by
Complex bashing (4+3i)(12+5i)(a+bi) has to be imaginary, (4+3i)(12+5i)=(33+56i), so a+bi=56+33i
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balllightning37
351 posts
#6
Y by
Why not just do tan(arctan(3/4)+arctan(5/12))
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vsamc
3769 posts
#7 • 1 Y
Y by centslordm
tan(90-tan^(-1)(3/4)-tan^(-1)(5/12))
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centslordm
4449 posts
#8 • 4 Y
Y by clarkculus, mygoodfriendusesaops, idksomething, Riemann123
Computing $\arctan \tfrac 34 + \arctan \tfrac 5{12} = \arctan \left( \frac{\frac 34 + \frac 5{12}}{1 - \frac 34 \cdot \frac 5{12}}\right) = \arctan \tfrac {56}{33},$ we see that the triple is $33{-}56{-}65,$ so $\boxed{154}.$
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mathical8
295 posts
#9
Y by
I did cos(x)=sin(arcsin(3/5)+arcsin(5/13)). From there I got 33, 56, 65, so 154.
This post has been edited 1 time. Last edited by mathical8, Nov 13, 2024, 5:37 PM
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mathprodigy2011
58 posts
#10
Y by
this one was so free imo
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Countmath1
113 posts
#11
Y by
Complex number mult. Need $\arctan \frac{3}{4} + \arctan \frac{5}{12} + \arctan\frac{p}{q} = \frac{\pi}{2},$ set slope to be undefined, etc.
This post has been edited 1 time. Last edited by Countmath1, Nov 13, 2024, 6:20 PM
Reason: latex
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Shreyasharma
589 posts
#12
Y by
Favorite problem on test.

Complex gives,
\begin{align*}
(4 + 3i)(12 + 5i)(m + ni) \in i\mathbb{R}
\end{align*}and it just works.
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Countmath1
113 posts
#13
Y by
This is like that one AIME problem with the arctans, just remembered
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Shreyasharma
589 posts
#14
Y by
Yea, funny story: In OTIS somebody has posted a problem which involved the following identity the day before the test,
\begin{align*}
\tan x + \tan y + \tan z = \tan x \cdot \tan y \cdot \tan z
\end{align*}if $90 \mid x + y + z$ or something similar. In an offhand remark somebody mentioned that a proof of this is to take a complex number interpretation which is why I solved this problem.

tldr; browsing OTIS instead of doing actual test prep is good.
This post has been edited 1 time. Last edited by Shreyasharma, Nov 13, 2024, 6:44 PM
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megarnie
5307 posts
#15
Y by
maybe i should've done 12b

Cosine of the smallest angle of the first one is $ \frac 45$ and $\frac{12}{13}$ for the second one. Their sines are $\frac 35$ and $\frac{5}{13}$, respectively. Therefore, the cosine of the sum of the angles is $\frac{48 - 15}{65} = \frac{33}{65}$, so the sine of the smallest angle of the other triangle is $\frac{33}{65}$. Note that $33, 56, 65$ is the primitive Pythagorean triple satisfying these requirements, so it's just $33 + 56 + 65= \boxed{\textbf{(C)}\ 154}$.
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pieMax2713
4131 posts
#16
Y by
i literally thought of doing tan sum and i got $\frac{56}{33}=\tan(90-\theta)$ but i was too stupid to realize that tan 90-θ is literally just from the other perspective
This post has been edited 1 time. Last edited by pieMax2713, Nov 13, 2024, 7:15 PM
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brainfertilzer
1802 posts
#17
Y by
the angle is $\tan \theta = \cot (\alpha + \beta)$ where $\alpha$, $\beta$ are the small angles of the 345 and 51213 triangles. just use tangent addition to compute that and then you can solve it easily
This post has been edited 2 times. Last edited by brainfertilzer, Nov 14, 2024, 12:35 AM
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