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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
Logical guessing game!
Mathdreams   22
N an hour ago by JH_K2IMO
Source: 2021 Fall AMC10B P10
Fourty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?

$\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67$
22 replies
Mathdreams
Nov 17, 2021
JH_K2IMO
an hour ago
Possibility of USAMO?
MathXplorer10   4
N an hour ago by MathXplorer10
Hi guys!


I got a 118.5 on the 12B test this year. I am wondering if it is possible to make USAMO (what do you think the cutoffs would be this year?)

For some background, I got 121.5/127.5 on the 10s last year, and got a 7 on AIME with no extra prep. Is it possible to go from a 7 to a 10 (or whatever I need to get on AIME)?

Thank you!
4 replies
MathXplorer10
4 hours ago
MathXplorer10
an hour ago
10a vs 10b
golden_star_123   111
N an hour ago by happyfish0922
Post the difference between your 10a and 10b score!
111 replies
1 viewing
golden_star_123
Wednesday at 6:24 PM
happyfish0922
an hour ago
What do next?
FuturePanda   2
N an hour ago by Tem8
Hi everyone,

I think I got an 81 and 102 for 12A and 10B, sillying way too much on both. I read all of the AOPS books, and I know most of the theorems for the AMC’s I just don’t know which ones to apply to solve the problems. Additionally, I suck at trig, complex, and logarithms. What should I do to improve?

For example, should I be grinding past AIME’s?
I plan on reading most of the Awesomemath books for L3
2 replies
FuturePanda
3 hours ago
Tem8
an hour ago
No more topics!
Swap this with P10 (mean/median/range problem)
PEKKA   15
N Yesterday at 11:00 PM by asdf334
Source: 2024 AMC 12B Problem 23
A right pyramid has regular octagon $ABCDEFGH$ with side length $1$ as its base and apex $V.$ Segments $\overline{AV}$ and $\overline{DV}$ are perpendicular. What is the square of the height of the pyramid?

$
\textbf{(A) }1 \qquad
\textbf{(B) }\frac{1+\sqrt2}{2} \qquad
\textbf{(C) }\sqrt2 \qquad
\textbf{(D) }\frac32 \qquad
\textbf{(E) }\frac{2+\sqrt2}{3} \qquad
$
15 replies
PEKKA
Wednesday at 5:15 PM
asdf334
Yesterday at 11:00 PM
Swap this with P10 (mean/median/range problem)
G H J
Source: 2024 AMC 12B Problem 23
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PEKKA
1791 posts
#1
Y by
A right pyramid has regular octagon $ABCDEFGH$ with side length $1$ as its base and apex $V.$ Segments $\overline{AV}$ and $\overline{DV}$ are perpendicular. What is the square of the height of the pyramid?

$
\textbf{(A) }1 \qquad
\textbf{(B) }\frac{1+\sqrt2}{2} \qquad
\textbf{(C) }\sqrt2 \qquad
\textbf{(D) }\frac32 \qquad
\textbf{(E) }\frac{2+\sqrt2}{3} \qquad
$
This post has been edited 5 times. Last edited by jlacosta, Wednesday at 6:22 PM
Reason: updated to official wording
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Bluesoul
756 posts
#2
Y by
Denote the center of the octagon is the origin of the 3-D plane.

We could easily express the coordinates of $A$ and $D$, which are $(-\frac{1}{2},\frac{1}{2\tan(22.5^{\circ})},0); (\frac{1}{2\tan(22.5^{\circ})},-\frac{1}{2},0)$

The apex is $(0,0,x)$, set the dot product to 0 and we have $x^2=\frac{1}{2 \tan(22.5^{\circ})}=\frac{1+\sqrt{2}}{2}$
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vsamc
3768 posts
#3 • 2 Y
Y by ihatemath123, centslordm
scale up by sqrt(2-2cos 135), then the radius is 1. Then A = <1, 0, 0>, B = <-sqrt2/2, sqrt2/2, 0>, C = <0,0,h1>. Then, <1, 0, -h1> dot <-sqrt2/2, sqrt2/2, -h1> = 0, so h1^2 = sqrt2/2. Now, scale down, it becomes sqrt2/2 * 1/(2-2cos 135) = sqrt2/2 * 1/(2-sqrt2) = sqrt2/2 * (2+sqrt2)/2 = (sqrt2+1)/2 -> B
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Jack_w
59 posts
#4
Y by
Extend $AB$ and $CD$ to meet at $P$. Then $\triangle{APD}$ and $\triangle{AVD}$ are both right isosceles, so $AV = \frac{2 + \sqrt{2}}{2}$. Then find the height
This post has been edited 1 time. Last edited by Jack_w, Wednesday at 5:19 PM
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axusus
783 posts
#5
Y by
PEKKA wrote:
A right pyramid has regular octagon $ABCDEFGH$ with side length $1$ as its base and apex $V.$ Segments $AV$ and $DV$ are perpendicular. What is the square of the height of the pyramid?

$(A) 1$ $(B) \frac{1+\sqrt2}{2}$ $(C) \sqrt2$ $(D) \frac 32 $ $(E) \frac{2+\sqrt2}{3}$

I did
AD = 1 + sqrt2
AV or AD = (1+sqrt2)/sqrt2
so the height of the pyramid was centered at the midpoint is (1+sqrt2)/2
connecting center to midpoint of AD is 1/2 length
(1/2)^2 + x^2 = ((1+sqrt2)/2)^2

and I think I got the answer from there
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shiamk
100 posts
#6
Y by
I got option B
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ihatemath123
3338 posts
#7 • 1 Y
Y by centslordm
just coordinatize ABCDEFGH and use dot product
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evanhliu2009
1031 posts
#8 • 1 Y
Y by akqbstr
This problem was lowkey free for 23
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PEKKA
1791 posts
#9
Y by
My Sol:
1. Find AD which gives AV, DV, distance from V to each vertex. (Done with half angles)
2. Draw triangle AEV and altitude from V
3. Pythagorean theorem
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axusus
783 posts
#10
Y by
evanhliu2009 wrote:
This problem was lowkey free for 23

agreed but I almost panicked because I didn't do it until last few minutes
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mathprodigy2011
57 posts
#11
Y by
BRO, I got this and bubbled the wrong answer
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axusus
783 posts
#12
Y by
mathprodigy2011 wrote:
BRO, I got this and bubbled the wrong answer

i feel you, when you tell others their like "huh what" but you actually just had a brainfart and brain decided to commit whoopsie
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dumplingsun21
17 posts
#13
Y by
this has to be the easiest problem 23 I've ever seen like what
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Shreyasharma
589 posts
#14
Y by
Why so trivial
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brainfertilzer
1802 posts
#15
Y by
didnt read this till now; looks like people were not exaggerating how misplaced this is

Let $A = (r,0,0)$ and $D = (-r/\sqrt{2}, r/\sqrt{2},0)$ and $V = (0, 0, v)$ (here $r$ is obivously the radius of a sidelength-$1$ regular octagon). Then we have
\[ AV\perp DV\implies (A-V)\cdot (D-V) = 0\implies (r,0,-h)\cdot (-r/\sqrt{2}, r/\sqrt{2}, -h) = 0\implies h^2 = \frac{r^2}{\sqrt{2}}.\]We can easily derive $2r^2 - 2r^2(1/\sqrt{2}) = 1$ with the law of cosines, so solving for $r^2$ and plugging it into the above, we get
\[h^2 = \frac{1}{\sqrt{2}}\left(\frac{1}{2(1 - \frac{1}{\sqrt{2}})}\right) = \frac{1}{2(\sqrt{2} - 1)} = \boxed{\frac{\sqrt{2} + 1}{2}}.\]
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asdf334
7547 posts
#16
Y by
lmao just guess !B
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