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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
Past PUMaC results
mathkiddus   1
N 16 minutes ago by lpieleanu
Does anybody know where we can find old PUMaC results/placing cutoffs
1 reply
mathkiddus
an hour ago
lpieleanu
16 minutes ago
study for AIME to qual for USJMO
Logicus14   0
28 minutes ago
hi

so this is my first year doing amc10 (9th grade) and I got a decent enough score (121.5) on amc10b and after some research I found out that I need to get AT LEAST 10 problems right to qualify for USJMO

did anyone get 10+ problems right? If you did how did you study? any general tips?


ty
0 replies
Logicus14
28 minutes ago
0 replies
Problem 2
evt917   47
N an hour ago by MC_ADe
Source: 2024 AMC 12B #2 / AMC 10B #2
What is $10! - 7! \cdot 6!$?

$
\textbf{(A) }-120 \qquad
\textbf{(B) }0 \qquad
\textbf{(C) }120 \qquad
\textbf{(D) }600 \qquad
\textbf{(E) }720 \qquad
$
47 replies
evt917
Nov 13, 2024
MC_ADe
an hour ago
AMC Music Video - Orz to the Legends (AMC 10/12 Version)
megahertz13   21
N an hour ago by vsarg
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 $m+n$.
21 replies
megahertz13
Oct 13, 2024
vsarg
an hour ago
No more topics!
nah i'd coordbash
c_double_sharp   31
N Yesterday at 11:29 PM by hduran23
Source: 2024 AMC 10B Problem 10
Quadrilateral $ABCD$ is a parallelogram, and $E$ is the midpoint of the side $\overline{AD}$. Let $F$ be the intersection of lines $EB$ and $AC$. What is the ratio of the area of quadrilateral $CDEF$ to the area of triangle $CFB$?

$\textbf{(A) } 5 : 4 \qquad \textbf{(B) } 4 : 3 \qquad \textbf{(C) } 3 : 2 \qquad \textbf{(D) } 5 : 3 \qquad \textbf{(E) } 2 : 1$
31 replies
c_double_sharp
Nov 13, 2024
hduran23
Yesterday at 11:29 PM
nah i'd coordbash
G H J
Source: 2024 AMC 10B Problem 10
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c_double_sharp
238 posts
#1
Y by
Quadrilateral $ABCD$ is a parallelogram, and $E$ is the midpoint of the side $\overline{AD}$. Let $F$ be the intersection of lines $EB$ and $AC$. What is the ratio of the area of quadrilateral $CDEF$ to the area of triangle $CFB$?

$\textbf{(A) } 5 : 4 \qquad \textbf{(B) } 4 : 3 \qquad \textbf{(C) } 3 : 2 \qquad \textbf{(D) } 5 : 3 \qquad \textbf{(E) } 2 : 1$
Z K Y
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ostriches88
1441 posts
#2 • 1 Y
Y by hduran23
A by sketchy similar triangles

or basic shoelace
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gladIasked
592 posts
#3
Y by
yeah I can confirm a.

cheese by assuming square
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FuturePanda
106 posts
#4
Y by
cheese by assuming rectangle lol
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Aaronjudgeisgoat
620 posts
#5
Y by
Assume its a square, then similar triangles via parallel lines gives 5:4
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nsking_1209
82 posts
#6
Y by
FuturePanda wrote:
cheese by assuming rectangle lol

Even better. Assume square
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williamxiao
2457 posts
#7
Y by
AFE is similar to CFE with ratio 1:2, finding the areas of all the triangles gives 5/12 : 1/3 is 5:4, A

You don’t even need to assume square because literally nothing changes
This post has been edited 1 time. Last edited by williamxiao, Nov 13, 2024, 5:18 PM
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gracemoon124
871 posts
#8 • 1 Y
Y by hduran23
coordbash by assuming square -> $\textbf{(A)}$
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EthanSpoon
635 posts
#9
Y by
A by area bash(I almost chose E but then I thought it was sus)
This post has been edited 1 time. Last edited by EthanSpoon, Nov 13, 2024, 5:19 PM
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paganiniana
201 posts
#10
Y by
ruler ;) to get A
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alexanderhamilton124
242 posts
#11
Y by
Assume square and win to get $(A)$
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elizhang101412
1060 posts
#12
Y by
i assumed square but still wasted time, at least I got it right
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EpicBird08
1679 posts
#13 • 1 Y
Y by Tem8
all coordbashers are banned
Let $A = [AEF],$ then by similar triangles we get $CF = 2AF.$ Then $[ADC] = \frac{2+1}{1} \cdot \frac{3+1}{1} \cdot [AEF] = 6A,$ so $[CDEF] = 5A.$ We also get $[BFC] = \frac{2}{1} \cdot \frac{2}{1} \cdot [AEF] = 4A,$ and our answer is $\textbf{(A)    } 5 \colon 4$.
This post has been edited 1 time. Last edited by EpicBird08, Nov 13, 2024, 5:38 PM
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Mintylemon66
31 posts
#14
Y by
Assign $A,B,C,D$ mass $1$. Then $E$ has mass $2$, so $EF:FB=1:2$. Connect $CE$. Let $[AEF]=s$, then $[ABCD]=12, [CFB]=4s,[CDEF]=[CEF]+[CDE]=2s+3s=5s$, so the answer is $\boxed{\textbf{(A) }5:4}$.
This post has been edited 2 times. Last edited by Mintylemon66, Nov 13, 2024, 6:05 PM
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pingpongmerrily
2424 posts
#15
Y by
yup just let ABCD=4x4 square and then bash the rest :skull:
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Tem8
154 posts
#16
Y by
No cord bash, just similar triangles.
[asy]
import olympiad;

size(7cm);
pair A=(0,0),
D=(5,0),
C=(7,3),
B=(2,3),
E=(2.5,0),
F=intersectionpoint(A--C, B--E);
draw(A--B--C--D--cycle);
draw(A--C);
draw(B--E);

dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(F);
label("$A$", A, (-.6,-.6));
label("$B$", B, (-.6, .6));
label("$C$", C, (.6,.6));
label("$D$", D, (.6,-.6));
label("$E$", E, (0,-.6));
label("$F$", F, (.7,1.6));
[/asy]

Let $x=BC$ and $h$ be the height from $C$ to $AD$.
By similar triangles $\triangle AEF \sim \triangle CFB$, it follows that $\frac{CF}{FA}=\frac{d(F,BC)}{d(F,AE)}=\frac{BC}{AE}=2$, so $d(F,AE)=\frac{h}{3}$, $d(F,BC)=\frac{2h}{3}$ and $CF:FA=2$.

Now, let's find the areas of $CDEF$ and $CFB$.
\begin{align*}
[CDEF] &= [CDA]-[AEF] = \frac{xh}{2} - \frac{xh}{12} = \frac{5xh}{12} \\
[CFB] &= \frac{1}{2}\cdot\frac{2xh}{3} = \frac{xh}{3}.
\end{align*}Hence, $[CDEF]:[CFB]=\frac{5xh}{12}\cdot\frac{3}{xh}=\boxed{\textbf{(A) }\frac{5}{4}}$
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mathboy282
2966 posts
#17
Y by
5:4

[afe]/[bfc]=1/4. Let afe = x, cfed = y. hence bfc = 4x. now note: [bfc]+[cfed]=3/4 * total area, let this total area = A. so 4x+y=3/4 A.

Also note: [afe] + [cfed]=1/2 A. x+y=1/2 A. Hence x = 1/12 A, y=5/12 A, you want y:4x which is 5:4
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MathRook7817
234 posts
#18
Y by
just us mass point theorem and then similar triangles
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golden_star_123
130 posts
#19
Y by
Assumed square and then coordbashed
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pen_pineapple_apple_pen
8 posts
#20
Y by
I love cheese
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KevinChen_Yay
130 posts
#21
Y by
relatable title
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andrewcheng
482 posts
#22
Y by
cheese assuming square then cordbash
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BS2012
536 posts
#23
Y by
coordbash is not even fast just split the quad and do similar triangles
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pingpongmerrily
2424 posts
#24
Y by
andrewcheng wrote:
cheese assuming square then cordbash

i assumed square, cordbash is not good tho.
i agree with @bove, similar triangles is an easy solve if you're not terrible at geo like me.

i still got it
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giratina3
243 posts
#25
Y by
I tried coord bashing, but after having to rationalize so many denominators, I tried solving it intuitively and somehow got the answer
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Spacepandamath13
201 posts
#26
Y by
I either got 5:4 or 5:3 :furious:
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stjwyl
643 posts
#27
Y by
EthanSpoon wrote:
A by area bash(I almost chose E but then I thought it was sus)

I got E :sob:
At least I cooked on the B more than the A bro
Only 3 wrong :)
Just barely made AIME
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Kempu33334
491 posts
#28
Y by
MathRook7817 wrote:
just us mass point theorem and then similar triangles

What is mass point theorem?
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andrewcheng
482 posts
#29
Y by
pingpongmerrily wrote:
andrewcheng wrote:
cheese assuming square then cordbash

i assumed square, cordbash is not good tho.
i agree with @bove, similar triangles is an easy solve if you're not terrible at geo like me.

i still got it

this was late into the test where I basically gave up on anything slightly intuitive so cordbash was quite appealing esp with the easy lines due to square
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RabtejKalra
12 posts
#30
Y by
Nooooooooooooooooooooo I said E.
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Spacepandamath13
201 posts
#31
Y by
Yep. I'm was very underprepared if I don't even know what bashing is.
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hduran23
4 posts
#32
Y by
gracemoon124 wrote:
coordbash by assuming square -> $\textbf{(A)}$

Same
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N Quick Reply
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