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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 17 minutes ago by lpieleanu
Does anybody know where we can find old PUMaC results/placing cutoffs
1 reply
mathkiddus
2 hours ago
lpieleanu
17 minutes ago
study for AIME to qual for USJMO
Logicus14   0
29 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
29 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!
Cute DJ-esque Geo
peace09   29
N Nov 13, 2024 by Spacepandamath13
Source: 2024 AMC 10B #11 / 12B #7
In the figure below $WXYZ$ is a rectangle with $WX=4$ and $WZ=8$. Point $M$ lies $\overline{XY}$, point $A$ lies on $\overline{YZ}$, and $\angle WMA$ is a right angle. The areas of $\triangle WXM$ and $\triangle WAZ$ are equal. What is the area of $\triangle WMA$?

IMAGE

$
\textbf{(A) }13 \qquad
\textbf{(B) }14 \qquad
\textbf{(C) }15 \qquad
\textbf{(D) }16 \qquad
\textbf{(E) }17 \qquad
$
29 replies
peace09
Nov 13, 2024
Spacepandamath13
Nov 13, 2024
Cute DJ-esque Geo
G H J
Source: 2024 AMC 10B #11 / 12B #7
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peace09
5313 posts
#1 • 4 Y
Y by ostriches88, gracemoon124, clarkculus, golue3120
In the figure below $WXYZ$ is a rectangle with $WX=4$ and $WZ=8$. Point $M$ lies $\overline{XY}$, point $A$ lies on $\overline{YZ}$, and $\angle WMA$ is a right angle. The areas of $\triangle WXM$ and $\triangle WAZ$ are equal. What is the area of $\triangle WMA$?

[asy]
pair X = (0, 0);
pair W = (0, 4);
pair Y = (8, 0);
pair Z = (8, 4);
label("$X$", X, dir(180));
label("$W$", W, dir(180));
label("$Y$", Y, dir(0));
label("$Z$", Z, dir(0));
draw(W--X--Y--Z--cycle);
dot(X);
dot(Y);
dot(W);
dot(Z);
pair M = (2, 0);
pair A = (8, 3);
label("$A$", A, dir(0));
dot(M);
dot(A);
draw(W--M--A--cycle);
markscalefactor = 0.05;
draw(rightanglemark(W, M, A));
label("$M$", M, dir(-90));
[/asy]

$
\textbf{(A) }13 \qquad
\textbf{(B) }14 \qquad
\textbf{(C) }15 \qquad
\textbf{(D) }16 \qquad
\textbf{(E) }17 \qquad
$
This post has been edited 13 times. Last edited by peace09, Nov 13, 2024, 6:56 PM
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yambe2002
1642 posts
#2
Y by
is it 15$  $
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mathwiz_1207
41 posts
#3
Y by
can confirm 15
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saturnrocket
1305 posts
#4
Y by
15 i pythag bashed it but apparently there's sim triangles
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peace09
5313 posts
#5
Y by
The area condition implies that $\tfrac{BE}{DF}=\tfrac{1/AB}{1/AD}=\tfrac{CB}{CD}$, meaning $CEF\sim CBD$ (and $ABE$). Then $BE=\tfrac{AB}{2}=2$, $CE=BC-BE=6$, and $CF=\tfrac{CE}{2}=3$, and so $[AEF]=\tfrac{2\sqrt{5}\cdot3\sqrt{5}}{2}=\boxed{\textbf{(C) }15}$.
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elizhang101412
1060 posts
#6
Y by
i wrote a quadratic, got -4 and -1 as solutions, then just assumed it was 1 and got it right
fun
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pingpongmerrily
2424 posts
#7
Y by
15
this took me 15 minutes bc i couldn't stop messing up the algebra and doing dumb things
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gracemoon124
871 posts
#8
Y by
my favorite problem on the test lol >.<
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fruitmonster97
2239 posts
#9
Y by
also 12A #7(I think?)
This post has been edited 1 time. Last edited by fruitmonster97, Nov 13, 2024, 5:59 PM
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yambe2002
1642 posts
#10
Y by
genuinely took the most time for me on the test even surpassing the dumb sequence problem for some reason lol

solved it at 5 minutes left
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andrewcheng
482 posts
#11 • 1 Y
Y by yambe2002
confirm this was harder than P25 ans is 15
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YaoAOPS
1317 posts
#12
Y by
[asy]
pair X = (0, 0);
pair W = (0, 4);
pair Y = (8, 0);
pair Z = (8, 4);
label("$X$", X, dir(180));
label("$W$", W, dir(180));
label("$Y$", Y, dir(0));
label("$Z$", Z, dir(0));
draw(W--X--Y--Z--cycle);
dot(X);
dot(Y);
dot(W);
dot(Z);
pair M = (2, 0);
pair A = (8, 3);
label("$A$", A, dir(0));
dot(M);
dot(A);
draw(W--M--A--cycle);
markscalefactor = 0.05;
draw(rightanglemark(W, M, A));
label("$M$", M, dir(-90));
[/asy]
This post has been edited 2 times. Last edited by YaoAOPS, Nov 13, 2024, 6:01 PM
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peace09
5313 posts
#13
Y by
@4above: done.
@above: ???
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YaoAOPS
1317 posts
#14
Y by
Don't know what to tell you these were the labels on 12B, problem 7.
Attachments:
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MathRook7817
234 posts
#15
Y by
15 confirmed
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Squidget
372 posts
#16
Y by
I “cheated” and inputted 1 and 2 for DF and BE respectively(because we know BE is double DF) and just bashed I guess? MAA loves similar triangles, I’m not sure why I didn’t realize that.
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peace09
5313 posts
#17
Y by
YaoAOPS wrote:
Don't know what to tell you these were the labels on 12B, problem 7.
I see, I was referring more to the time when $A$ didn't exist on your diagram :blush:
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axusus
783 posts
#18
Y by
Squidget wrote:
I “cheated” and inputted 1 and 2 for DF and BE respectively(because we know BE is double DF) and just bashed I guess? MAA loves similar triangles, I’m not sure why I didn’t realize that.

I actually solved it but a man with a rectangular mustache right next to me did this. he said it isn't cheating, its "wishful thinking"
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andrewcheng
482 posts
#19
Y by
peace09 wrote:
In the figure below $WXYZ$ is a rectangle with $WX=4$ and $WZ=8$. Point $M$ lies $\overline{XY}$, point $A$ lies on $\overline{YZ}$, and $\angle WMA$ is a right angle. The areas of $\triangle WXM$ and $\triangle WAZ$ are equal. What is the area of $\triangle WMA$?

[asy]
pair X = (0, 0);
pair W = (0, 4);
pair Y = (8, 0);
pair Z = (8, 4);
label("$X$", X, dir(180));
label("$W$", W, dir(180));
label("$Y$", Y, dir(0));
label("$Z$", Z, dir(0));
draw(W--X--Y--Z--cycle);
dot(X);
dot(Y);
dot(W);
dot(Z);
pair M = (2, 0);
pair A = (8, 3);
label("$A$", A, dir(0));
dot(M);
dot(A);
draw(W--M--A--cycle);
markscalefactor = 0.05;
draw(rightanglemark(W, M, A));
label("$M$", M, dir(-90));
[/asy]

$
\textbf{(A) }13 \qquad
\textbf{(B) }14 \qquad
\textbf{(C) }15 \qquad
\textbf{(D) }16 \qquad
\textbf{(E) }17 \qquad
$

did a funny quadratic via pythag to get x=2 as the base of the 4 height triangle and then solved to get C
This post has been edited 1 time. Last edited by andrewcheng, Nov 13, 2024, 6:20 PM
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Countmath1
113 posts
#20
Y by
Confirmed 15
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LostDreams
125 posts
#21
Y by
similar triangles and use the fact that the area of the 2 triangles are equal should give you C
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pingpongmerrily
2424 posts
#22
Y by
i had kendrick stuck in my head for this problem and couldn't focus

still managed to get it after a while tho
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LTristan
236 posts
#23
Y by
LMAOOO I LITERALLY WAS TRYING TO RULER APPROACH THIS

realized that diagram was not to scale, so I DREW MY OWN DIAGRAM and somehow got it right (edge of ruler is perfect 90 degree angle)
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djmathman
7928 posts
#24 • 1 Y
Y by peace09
@title: not dj geo! I only had one problem on the B tests, as far as I'm aware.
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Shreyasharma
589 posts
#25
Y by
I couldn't solve the system first time around, making this my last solve on the test.
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eg4334
391 posts
#26
Y by
Notice that $\triangle WXM, \triangle MAY$ are $1-2-\sqrt{5}$ triangles. $WM = 2\sqrt{5}$ and $AM = 3\sqrt{5}$, giving $\boxed{15}$
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pieMax2713
4131 posts
#27
Y by
i initially wasted 5 minutes on this then i just guessed XM=2 and that worked
C
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Mr.Sharkman
386 posts
#28
Y by
Let $XM = x.$ Then, $MY = 8-x.$ Since $\triangle WXM \sim \triangle MYA$ by $AA$ similarity, we get $YA = \frac{8x-x^{2}}{4},$ so $AZ = \frac{16-8x+x^{2}}{4}.$ Now, $AZ \cdot WZ = WX \cdot XM,$ so $x=2.$ Thus, we have that $MA = 3\sqrt{5},$ and $WM = 2\sqrt{5},$ for an answer of $15.$
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exp-ipi-1
965 posts
#29
Y by
I just set AZ as x and XM as 2x and solved from there by similar triangles XWM and MAY which resulted in a quadratic
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Spacepandamath13
201 posts
#30
Y by
I hate this test. got 17 for that one :wallbash_red:
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