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AMC 12 How to Study
krishiam 1
N
by Alex-131
Hi, I am in my junior year and want to study for the upcoming AMC 12, but I'm not sure. I have done AOPS Geometry and Intermediate Algebra, but I forgot everything, and I have just a strong foundation in high school math and a basic understanding of probability, and know just the basic stuff needed. For reference, I scored like somewhere in the 70s AMC 10 last year. How should I prepare? Is grinding practice problems enough to qualify for AIME?
1 reply
funny log fraction
OronSH 13
N
by SomeonecoolLovesMaths
Source: 2024 AMC 12B #8
What value of 
satisfies ![\[\frac{\log_2x\cdot\log_3x}{\log_2x+\log_3x}=2?\]](//latex.artofproblemsolving.com/f/f/e/ffe17a6345042ba69850f77f5df1309773bfb1be.png)
satisfies ![\[\frac{\log_2x\cdot\log_3x}{\log_2x+\log_3x}=2?\]](http://latex.artofproblemsolving.com/f/f/e/ffe17a6345042ba69850f77f5df1309773bfb1be.png)
13 replies
Complex roots solution clarification
yingkai_0_ 2
N
by yingkai_0_
Source: https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_17
on solution 1, how do we get to the conclusion that if we have another root w, then we must have 
and 
, respectively? can't u have 2 other roots like 2 and 3, for example where they are not complex or irrational?
please help :(
and 
, respectively? can't u have 2 other roots like 2 and 3, for example where they are not complex or irrational?please help :(
2 replies
Harmonic Mean
Happytycho 4
N
by elizhang101412
Source: Problem #2 2016 AMC 12B
The harmonic mean of two numbers can be calculated as twice their product divided by their sum. The harmonic mean of 
and 
is closest to which integer?
and 
is closest to which integer?
4 replies
Centroids form Equilateral Triangle
Generic_Username 22
N
by Tetra_scheme
Source: 2019 AMC 12B #25
Let 
be a convex quadrilateral with 
and 
Suppose that the centroids of 
and 
form the vertices of an equilateral triangle. What is the maximum possible value of the area of ?
be a convex quadrilateral with 
and 
Suppose that the centroids of 
and 
form the vertices of an equilateral triangle. What is the maximum possible value of the area of ?
22 replies
Real Analysis on AMC
Generic_Username 50
N
by ESAOPS
Source: 2019 AMC 12B #22
Define a sequence recursively by 
and
![\[x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}\]](//latex.artofproblemsolving.com/9/3/4/934a32343352988b8d8328a7ef93c7553a1ce642.png)
for all nonnegative integers 
Let 
be the least positive integer such that
![\[x_m\leq 4+\frac{1}{2^{20}}.\]](//latex.artofproblemsolving.com/e/8/8/e889e3b4b1d11c0fd3ce0eda55858624351b69fd.png)
In which of the following intervals does lie?
![$\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty]$](//latex.artofproblemsolving.com/6/c/2/6c2f679a426a4923a52cc6f0de03e28c75165307.png)
and![\[x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}\]](http://latex.artofproblemsolving.com/9/3/4/934a32343352988b8d8328a7ef93c7553a1ce642.png)
for all nonnegative integers 
Let 
be the least positive integer such that![\[x_m\leq 4+\frac{1}{2^{20}}.\]](http://latex.artofproblemsolving.com/e/8/8/e889e3b4b1d11c0fd3ce0eda55858624351b69fd.png)
In which of the following intervals does lie?![$\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty]$](http://latex.artofproblemsolving.com/6/c/2/6c2f679a426a4923a52cc6f0de03e28c75165307.png)
50 replies
Is my approach right?
cowcheese 1
N
by YauYauFilter
Source: AMC12B 2019 p11
How many unordered pairs of edges of a given cube determine a plane?
a. 12. b. 28 c. 36 d. 42. e. 66
I ended up with the right answer. However, no solutions resembled the way I did it, which is why I'm skeptical. After a few failed attempts with other ways of solving (didn't get an answer in the answer options), I decided to use complementary counting, and counted invalid pairs (pairs that don't satisfy the condition given by the problem). To do that I drew out the cube and found for each side, there are 4 sides that will make an invalid pair with it. Because there are 12 sides to a cube, I did 12*4. But, each distinct pair in that 12*4 is counted twice. So, dividing 12*4 by 2 will give 24 invalid pairs. Now, for total pairs, it is 12 choose 2, which is 66. 66-24 = 42. Hence, the answer is 42.
So, is there anything wrong with the way I did it? Is there a specific reason this wasn't a posted solution?
Thanks!
a. 12. b. 28 c. 36 d. 42. e. 66
I ended up with the right answer. However, no solutions resembled the way I did it, which is why I'm skeptical. After a few failed attempts with other ways of solving (didn't get an answer in the answer options), I decided to use complementary counting, and counted invalid pairs (pairs that don't satisfy the condition given by the problem). To do that I drew out the cube and found for each side, there are 4 sides that will make an invalid pair with it. Because there are 12 sides to a cube, I did 12*4. But, each distinct pair in that 12*4 is counted twice. So, dividing 12*4 by 2 will give 24 invalid pairs. Now, for total pairs, it is 12 choose 2, which is 66. 66-24 = 42. Hence, the answer is 42.
So, is there anything wrong with the way I did it? Is there a specific reason this wasn't a posted solution?
Thanks!
1 reply
9 What motivates you
AndrewZhong2012 70
N
by pingpongmerrily
What got you guys into math? I'm asking because I got ~71 on the AMC 12B and 94.5 on 10A last year. This year, my dad expects me to get a 130 on 12B and 10 on AIME, but I have sort of lost motivation, and I know these goals will be impossible to achieve without said motivation.
70 replies
AMC 12 Question
sadas123 12
N
by jb2015007
Hello! I am a 6th grader this year about to become 7th grade next year. I was wondering if I should take the AMC 12 next year because I think I am ready for it, I was thinking to do AMC 10 A and AMC 12 B, do you think it is a good idea? Here are the courses I finished and now I am working on:
Finished:
1. Intro Algebra
2. Intro Number Theory
3. Intro Counting and Probability
4. Volume 1
Working on:
1. Intermdiate Counting and Probability
2. Three Year Mathcounts Marathon
Upcoming:
1. Intro Geomtery (Next Month)
2. Intro to Alg (May)
3. Pre-calc (Summer)
4. Volume 2???
Stats for AMC 12 (Mocked):
1. AMC 12 A 2024: 100.5
2. AMC 12 B 2024: 105
3. AMC 12 A 2023: 96
The reason why I sometimes I get 100+ is because sometimes I know how to do the first step of the problem but the last step I have to kind of infrence but still i know how to do the problem.
Finished:
1. Intro Algebra
2. Intro Number Theory
3. Intro Counting and Probability
4. Volume 1
Working on:
1. Intermdiate Counting and Probability
2. Three Year Mathcounts Marathon
Upcoming:
1. Intro Geomtery (Next Month)
2. Intro to Alg (May)
3. Pre-calc (Summer)
4. Volume 2???
Stats for AMC 12 (Mocked):
1. AMC 12 A 2024: 100.5
2. AMC 12 B 2024: 105
3. AMC 12 A 2023: 96
The reason why I sometimes I get 100+ is because sometimes I know how to do the first step of the problem but the last step I have to kind of infrence but still i know how to do the problem.
12 replies
Accounting for an overcount
WhaleVomit 61
N
by SomeonecoolLovesMaths
Source: 2017 AMC 10B #17, AMC 12B #11
Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, 3, 23578, and 987620 are monotonous, but 88, 7434, and 23557 are not. How many monotonous positive integers are there?
61 replies
