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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
jmo cutoff
cakeiswaybetterthancookie   41
N 5 minutes ago by Lhaj3
Source: me
You guys NEED to stop freaking out about JMO cutoffs Its likely going to be in the early-mid 210's assuming an average level aime( 2021 aime?) So if you have a 110+ on the amc 10, you are in contention for USAJMO- its not LIKELY but its possible. My guess is maybe 15ish percent? really, too many people are freaking out about cutoffs- at the end of the day, most of you have next year and the year after that and can always keep improving. You got this!
41 replies
cakeiswaybetterthancookie
Today at 1:10 AM
Lhaj3
5 minutes ago
Counterexample
avn   38
N 5 minutes ago by mhgelgi
Source: 2019 AMC 10B #2 / 12B #2
Consider the statement, "If $n$ is not prime, then $n-2$ is prime." Which of the following values of $n$ is a counterexample to this statement?

$\textbf{(A) } 11 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 27$
38 replies
avn
Feb 14, 2019
mhgelgi
5 minutes ago
Totally AMC
MathNerdRabbit103   0
6 minutes ago
I don’t want a lot for Christmas
There is just one thing I need…
0 replies
MathNerdRabbit103
6 minutes ago
0 replies
2024 Fall Lexington Math Tournament
golue3120   1
N 11 minutes ago by Awesomeness_in_a_bun
The Lexington High School Math Team is proud to announce LMT Fall 2024! LMT is a competition for middle school students interested in math. Students can participate individually, or on teams of 4-6 members.

The competition will take place from 8:30AM-5:00PM on Saturday, December 14, at Lexington High School, 251 Waltham St., Lexington, MA 02421.

The competition will include two individual rounds, a Team Round, and a Guts Round, with a break for lunch and mini-events. A detailed schedule is available at https://lhsmath.org/LMT/Schedule.

There will be a talk for parents and coaches by Dr. Po-Shen Loh, former IMO Coach, Professor at Carnegie Mellon University, and founder of Expii.

There is a $15 fee per participant, paid on the day of the competition. Pizza will be provided for lunch, at no additional cost.

Register for LMT at https://lhsmath.org/LMT/Registration/Home.

More information is available on our website: https://lhsmath.org/LMT/Home. Email lmt.lhsmath@gmail.com with any questions.

IMAGE
1 reply
+1 w
golue3120
an hour ago
Awesomeness_in_a_bun
11 minutes ago
No more topics!
Sakupen circles
EpicBird08   28
N Oct 29, 2024 by JH_K2IMO
Source: 2023 AMC 10A Problem 15
An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed below. What is the least number of circles needed to make the total shaded area at least $2023\pi$?
28 replies
EpicBird08
Nov 9, 2023
JH_K2IMO
Oct 29, 2024
Sakupen circles
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Source: 2023 AMC 10A Problem 15
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EpicBird08
1678 posts
#1 • 1 Y
Y by centslordm
An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed below. What is the least number of circles needed to make the total shaded area at least $2023\pi$?
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ryanbear
1043 posts
#2
Y by
its E because the shaded region equals to $\pi(1+2+...+n) > 2023\pi$ and then use the sum of arithmetic sequence formula to get that $(n)(n+1) > 4046$ then plug in the answer choices
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wuwang2002
1129 posts
#3
Y by
i got E (64)
This post has been edited 1 time. Last edited by wuwang2002, Nov 9, 2023, 5:04 PM
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ostriches88
1455 posts
#4
Y by
can confirm E
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MrThinker
608 posts
#5
Y by
64 really ez
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OlympusHero
17012 posts
#6
Y by
This was E, pretty easy problem to silly though if you miswrite the summation
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gicyuraok2
1059 posts
#7
Y by
It is definitely E bad solution

sakupen circles :skull:
This post has been edited 1 time. Last edited by gicyuraok2, Nov 9, 2023, 7:41 PM
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HonestCat
950 posts
#8
Y by
I got 46 :(
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Alex-131
5152 posts
#9
Y by
gicyuraok2 wrote:
It is definitely E bad solution

I got the correct inequality and though there is no way n = 48 or answer choices works and skipped. I forgot to read all the circles
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peace09
5314 posts
#10 • 12 Y
Y by brainfertilzer, ryanbear, ihatemath123, mahaler, gracemoon124, Zhaom, Significant, EpicBird08, mathmax12, akliu, centslordm, Alex-131
Click to reveal hidden text

Y'all really need to take the WMC.
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think4l
342 posts
#11
Y by
Solution: https://primesets.blogspot.com/2023/11/2023-amc-10a-15.html
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MC413551
2225 posts
#12
Y by
2^2-1^2+4^3-3^2+...
Factorize to 1+2+3+4+...
n(n+1)/2>2023
n=64
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joshualiu315
2431 posts
#14
Y by
peace09 wrote:
Click to reveal hidden text

Y'all really need to take the WMC.

lmao real. i managed to silly that wmc problem sadly

The formula is simply $1+2+\dots+n$ for even $n$. Thus, $n=\boxed{\textbf{(E) } 64}$.
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Shreyasharma
590 posts
#15
Y by
This was eh.
Note that we basically have $\pi \cdot \left( (2^2 - 1^2) + (4^2 - 3^2) + (6^2 - 5^2) + \dots \right)$. The terms in the parentheses form a arithmetic series of the form $\{3, 7, 11, \dots\}$. Thus we have the total area after $2k$ circles is,
\begin{align*}
\pi \cdot \frac{(4k+2) \cdot k}{2} = \pi \cdot (2k+1) \cdot k
\end{align*}Now noting that $k \approx 32$ is when the sum above exceeds $2023 \pi$, we find $2k=64$.
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ryanbear
1043 posts
#17
Y by
https://www.quora.com/How-do-you-find-the-sum-of-series-100%C2%B2-99%C2%B2-98%C2%B2-97%C2%B2-96%C2%B2-95%C2%B2-2%C2%B2-1%C2%B2
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HighWater
379 posts
#18
Y by
I got (E) 64

Also why Sakupen Circles
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I-_-I
2047 posts
#19 • 6 Y
Y by ihatemath123, centslordm, pog, akliu, mathmax12, rayford
5 seconds
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MathKid16
70 posts
#20 • 1 Y
Y by thegamer1049
no way gometry dash reference??? AMC10 was sponsoredby robtop??
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xHypotenuse
571 posts
#21
Y by
Is that a geometry dash reference????????


oops sniped
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gracemoon124
871 posts
#22 • 1 Y
Y by peace09
edit problem statement to be the below, thanks


An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed below. What is the least number of circles needed to make the total shaded area at least $2023\pi$?
[asy]
size(6cm);
pen greywhat;
greywhat = RGB(105,105,105);
filldraw(circle((8, 0), 8), greywhat);
filldraw(circle((7, 0), 7), white);
filldraw(circle((6, 0), 6), greywhat);
filldraw(circle((5, 0), 5), white);
filldraw(circle((4, 0), 4), greywhat);
filldraw(circle((3, 0), 3), white);
filldraw(circle((2, 0), 2), greywhat);
filldraw(circle((1, 0), 1), white);
[/asy]
$\textbf{(A) }46\qquad\textbf{(B) }48\qquad\textbf{(C) }56\qquad\textbf{(D) }60\qquad\textbf{(E) }64$
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akliu
1705 posts
#23
Y by
Somehow managed to not silly the WMC question, but sillied and got computationally blocked on this, tilting my entire testing rhythm.

There's always going to be that one question on the test though unless you're @epicbird08 :o
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StevePlayer
7 posts
#24
Y by
E, i swear i saw this problem a few years ago
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FireK47
140 posts
#25
Y by
How I did it
A lot of text but in my mind it went quickly
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programmeruser
2455 posts
#26
Y by
dᅟamn i thought this was some weird circle geo, should've taken more than a 5 second read on this
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ryanbear
1043 posts
#27
Y by
E, i swear i saw this problem a few years ago
2003 AIME I P2
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vincentwant
1158 posts
#28
Y by
I went from P15 to P1 on 10A so I started with this one lol

nice title
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paixiao
1704 posts
#29
Y by
I-_-I wrote:
5 seconds

May I ask how I got that?
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bachisnotcool7
391 posts
#30
Y by
Work through a few small cases and see that the area increases at a rate increasing by 4, so you can create an equation.
$\frac{n^2+n}{2}=A$, where $A$ is area in $\pi$ and $n$ is the number of circles. We can clearly see $n=63$ is just low and $n=64$ is at least $2023\pi$ and $64$ is even so it works $(E)$
This post has been edited 1 time. Last edited by bachisnotcool7, Dec 30, 2023, 12:36 PM
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JH_K2IMO
60 posts
#32
Y by
The shaded areas can be described as follows: starting from the rightmost shaded region, the area for the first region is calculated as 4π−1π=3π, for the second region as 16π−9π=7π, and for the third region as 36π−25π=11π. Observing this pattern, it can be deduced that the area of the n-th shaded region is given by the formula 4n−1.

Thus, the total shaded area can be expressed as:(4×1−1)+(4×2−1)+(4×3−1)+…+(4×n−1)=4{n(n+1)/2}-n
To determine the minimum n such that the total shaded area is at least 2023π, the following inequality must hold: 2n^2+n≥2023.
Solving this quadratic inequality will yield the smallest integer value for n. After performing the necessary calculations, it is found that the minimum
n satisfying this condition is 32. Therefore, the total number of circles is given by 2n, resulting in 32×2=64.

Thus, the answer is 64.
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