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Does anyone have tips for the AMC (I really want to make the AIME)? I found this site last week. By the way, I have currently done the books in the intro series for Artofproblemsolving, but I find myself overthinking the way to solve questions. For example, there was a question that asked me to find the area of a wrapped region around a cylinder. It took me lots of time since I was thinking about the angle that the strap was wrapped around and things like that. But, I soon realized that I could 'unwrap' the cylinder solve the question. I currently am trying to solve past AMC questions, use alcumus, volume 1, and look at questions on this site. Also, I am currently in 10th grade.

 

I also am trying to do the ASHME questions from 1992: https: //artofproblemsolving.com/wiki/index.php/1992_AHSME_Problems/Problem_25

I am also trying to not cheat on alcumus like others have (based on the questions that people here have asked). Please don't give me answers, but help me understand certain ideas on these tests.

 Jan 4, 2020
 #1
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Hello!

I have qualified for USJMO twice, and USAMO once. Currently I am in 12th grade. I would say that AoPS certainly helps, but I never really used Alcumus, as the questions there are quite easy. Currently, there is around 1 more month left 'till the amc10a, and if you want to make AIME, you should start spamming tests as of now. I would say 5 amc10 tests MINIMUM a day should prepare you fine. However, you should make sure you understand the concepts to solving the problems. Furthermore, I would say you should get some AoPS books, the introduction to alegebra being the most important. If you want to make USJMO and above, I would say you'd need the precalculus book, which has much information in it.

Now, often times there is time pressure, along with the difficulty of the problems. I would suggest "ftw" on AoPS to help increase your speed in solving and skimming the problems. Note that amc10 questions 1-10 should take each maximum of 2 mins per question. Questions 10-20 take some time, and usually for me, require around 4-5 min to figure out 17-20. I suggest taking some of the newest tests, to get an idea of the difficulty of the current tests. If you can score a 120 and above consistently on amc10 tests, i would say you are fine to make AIME.

Some important concepts for the AMC tests include but are not limited to:

AM-GM

Cauchy-Schwartz Inequality

Triangluar Inequality

Pythagorean Theorem

Shoelace formula

Menelaus Theorem

Stewarts Theorem

Angle-Bisector Theorem

Basic trig (Sum of sins, cosines, law of sines, etc)

Veita's formula (extended)

Geometric and Arithemetic series

Difference and Sum of squares

Binomial Theorem

Telescoping

Minimum/Maximum value

Basic Log

Abs Value

Floor and Ceiling

Recursion

Advanced Factorial Concepts

Combination & Purmutation

Hockey Stick Identity

Mass point geometry

Stars and Bars

Descartes Circle theorem

Roots of Unity

That is all of I can think of now. These concepts are extremely useful in doing well in the AMC tests. Good luck! :_)

 Jan 4, 2020
edited by USAMO2019  Jan 4, 2020
 #2
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I have done the whole intro series, but when I do questions on the AMC, I get stuck around question 18. I find the difficult questions hard to navigate. Anyways, on Alcumus, I usually get stuck on questions that are level 25, and waste lots of time to get the answer wrong. Can you please explain (but not solve) how to relate the angles on ashme question 20 (1992).

Mathc1  Jan 4, 2020
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Sol for ASHME Q20 1992:

Since it is an n-pointed star, we have a 2n-polygon. As we know, the sum of angles in a polygon formula is (n-2)(180).

Thus, the sum of angles in our n-pointed star is 180(2n-2). If the angle at each vertex A(n) is some number A(i) (Note that the () represent subscripts) and the angle at each vertex B(n) equals B(i), we have the following equation:

n*A+n*B=180(2n-2). Note that the acute angle at B(i) is 360-B, so therefore A(i) is simply 360-B-10=350-B Hence, A+B=350. Therefore we can set up the following equation:

350*n=180(2n-2)

After we solve for n, we get that n=36!

 Jan 4, 2020
 #4
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How is nA+nB equal to 180(2n-2)? Angles b(1),...,b(n) look like exterior angles.

Mathc1  Jan 4, 2020
 #5
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n*A+n*B suffices because angle B is represented as the interior angle in each vertex

 Jan 4, 2020
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As I am thinking about it, I notice that for any b(i), the angle for b is 180 minus the reflex angle. So, nA+180n-n(times the reflex angles) is some quantity. But, I am not sure how to procced.

Mathc1  Jan 4, 2020

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