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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
AMC Music Video - Orz to the Legends (AMC 10/12 Version)
megahertz13   23
N a minute ago by vincentwant
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 $m+n$.
23 replies
megahertz13
Oct 13, 2024
vincentwant
a minute ago
How to be wide awake during math competitions.
orangebear   13
N 5 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
5 minutes ago
9 How do AMC-10 scores translate to MATHCOUNTS States
pingpongmerrily   49
N 7 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
7 minutes ago
9 AIME Qualification AMC 12A
studymoremath   1
N 22 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
22 minutes ago
No more topics!
complex quadrilateral
StressedPineapple   16
N Today at 1:23 AM by Arrowhead575
Source: 2024 AMC 12B #12
Suppose $z$ is a complex number with positive imaginary part, with real part greater than $1$, and with $|z| = 2$. In the complex plane, the four points $0$, $z$, $z^{2}$, and $z^{3}$ are the vertices of a quadrilateral with area $15$. What is the imaginary part of $z$?

$\textbf{(A)}~\displaystyle\frac{3}{4}\qquad\textbf{(B)}~1\qquad\textbf{(C)}~\displaystyle\frac{4}{3}\qquad\textbf{(D)}~\displaystyle\frac{3}{2}\qquad\textbf{(E)}~\displaystyle\frac{5}{3}$
16 replies
StressedPineapple
Nov 13, 2024
Arrowhead575
Today at 1:23 AM
complex quadrilateral
G H J
Source: 2024 AMC 12B #12
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StressedPineapple
4 posts
#1
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Suppose $z$ is a complex number with positive imaginary part, with real part greater than $1$, and with $|z| = 2$. In the complex plane, the four points $0$, $z$, $z^{2}$, and $z^{3}$ are the vertices of a quadrilateral with area $15$. What is the imaginary part of $z$?

$\textbf{(A)}~\displaystyle\frac{3}{4}\qquad\textbf{(B)}~1\qquad\textbf{(C)}~\displaystyle\frac{4}{3}\qquad\textbf{(D)}~\displaystyle\frac{3}{2}\qquad\textbf{(E)}~\displaystyle\frac{5}{3}$
This post has been edited 1 time. Last edited by jlacosta, Nov 13, 2024, 6:08 PM
Reason: adjusted to official wording
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HonestCat
947 posts
#2
Y by
StressedPineapple wrote:
Let $z$ be a complex number with real part greater than $1$ and $|z| = 2$. In the complex plane, the four points $0$, $z$, $z^{2}$, and $z^{3}$ are the vertices of a quadrilateral with area $15$. What is the imaginary part of $z$?

$\textbf{(A)}~\displaystyle\frac{3}{4}\qquad\textbf{(B)}~1\qquad\textbf{(C)}~\displaystyle\frac{4}{3}\qquad\textbf{(D)}~\displaystyle\frac{3}{2}\qquad\textbf{(E)}~\displaystyle\frac{5}{3}$

3/2
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Bluesoul
756 posts
#3 • 1 Y
Y by Sedro
$\frac{1}{2}(8\sin(\theta)+32\sin(\theta))=15, \sin(\theta)=\frac{3}{4}$, the imaginary part is $2\sin(\theta)=\frac{3}{2}$
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samrocksnature
8781 posts
#4 • 1 Y
Y by aidensharp
NOOOOOOOOOO
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Michw08
48 posts
#5
Y by
I couldn't get this during the test and then headsolved it while I was going home :oops_sign:
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OronSH
1624 posts
#6 • 2 Y
Y by David_He, ivyshine13
best geometry problem on the test ngl
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plang2008
271 posts
#7
Y by
i skipped this at first becuase i thought imaginary part 3/2 implied real part < 1 oops
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mathprodigy2011
58 posts
#8
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I skipped at first but question is trivial once you draw a coordinate plane
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Shreyasharma
589 posts
#9
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This was super creative, probably my favorite problem on the test.
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Sedro
5739 posts
#10 • 1 Y
Y by sami1618
I loved this problem. Let $z = 2e^{i\theta}$. Then, the area of the quadrilateral is $20\sin\theta$ and thus $\sin \theta = \tfrac{3}{4}$. We have $\text{Im}(z) = 2\sin\theta = \tfrac{3}{2}$ for answer choice D.
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LostDreams
125 posts
#11 • 1 Y
Y by Jack_w
This was one of the problems of AMC
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john0512
4144 posts
#12 • 4 Y
Y by pog, Jack_w, ivyshine13, centslordm
By spiral similarity, $0,z^2,z^3$ has four times the area of $0,z,z^2$, so $0,z,z^2$ has area $3$. Now $\frac{1}{2}(2)(4)\sin\theta=3$ so $\sin\theta=\frac{3}{4}$, so the answer is $2\cdot \frac{3}{4}=\frac{3}{2}$.

So clean and conceptual. why can't more amc problems be like this
this is why for computational i main college comps and arml now
This post has been edited 1 time. Last edited by john0512, Nov 13, 2024, 7:24 PM
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apotosaurus
75 posts
#13
Y by
I did not cook...
Due to the given conditions, we can express this as the sum of the areas of triangles $0,z,z^2$ and $0,z^2,z^3$. By Complex Shoelace, this is\[\frac i4 \left(\begin{vmatrix} z & \frac 4z & 1 \\ z^2 & \frac{16}{z^2} & 1 \\ 0 & 0 & 1 \end{vmatrix} + \begin{vmatrix} z^2 & \frac {16}{z^2} & 1 \\ z^3 & \frac{64}{z^3} & 1 \\ 0 & 0 & 1 \end{vmatrix}\right) =\frac i4 \left( \frac{16}{z}-4z+\frac{64}{z}-16z\right) = \frac i4 \left(20\bar z - 20z\right) = 10 \mathrm{Im}(z),\]so the answer is $\boxed{\frac 32}$.
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clarkculus
126 posts
#14 • 1 Y
Y by centslordm
NOOOOO I FORGOT TO MULTIPLY BY TWO
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EaZ_Shadow
280 posts
#15
Y by
Can someone explain to me the sol step by step
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EaZ_Shadow
280 posts
#16
Y by
Nvm nvm nvm I understand now
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Arrowhead575
2270 posts
#17
Y by
This was predicted by 2024 Or(z)IME (which also predicted 3 problems on last year's AIME).
2024 Or(z)IME #4 wrote wrote:
When plotted in the complex plane, the distance from complex number $1$ to $z$ is $2$ and the distance from $z$ to $z^2$ is $3$. Then, the distance from $1$ to $z^2$ can be expressed as $\frac{m}{n}$, for relatively prime integers $m$ and $n$. Find $m+n$.

Similarly, draw the quadrilateral in the complex plan. The distance between the origin and $z$ is 2, the distance between the origin and $z^2$ is 4, the distance between the origin and $z^3$ is 8. Further the angle of rotation is the same each time. Therefore, the area of the quadrilateral is $\frac{1}{2}\sin(\theta) (8*4+4*2) = 20\sin(\theta)$ . Then this is equal to 15, giving $\sin(\theta) = \frac{3}{4}$, and therefore the answer is $2*\frac{3}{4} = \frac{3}{2}$.
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