AMC tips

These are just some things that I've seemed to notice about the AMC in general:

The first 12 problems - These are really just to separate who knows basic math and who doesn't. I would say that the first 5 or so are like SAT math problems. The rest are a bit harder, but really just simple extensions of basic math. If you have any trouble with these, then either you don't know your basic math, or you're not looking at it the right way. If you're confident in your math skills and you're having trouble, then look at it again, you're probably making it harder than it is.

Stupid mistakes - Everybody makes them, and your goal should be to cut down on them as much as possible. A way to reduce a lot of them is to keep your work organized. This makes it easier to keep track of your work. If you get lost in it, then you'll probably pick out a wrong number somewhere along the way (remember, it's non-calculator) and screw up the answer. Also read the question slowly and carefully. A good way to make sure you completely understand it is to underline / box key words. Example:

On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?



Ok, you've read the problem quickly, and start going through the math. If 1 is subtracted from each number with equal chance, then there's 1/2 chance of being 4 odds and 2 evens, and 1/2 chance of being 2 evens and 4 odds. so the probability is:



Ok, that's one of the choices, so mark C and go on...or maybe you should reread the problem quickly, to make sure you read it correctly. Oh, not every number has the same probability of being chosen, just each dot. That changes things a lot. Reread the question after you fill in your answer to make sure you gave them what they wanted.

Obviously it's time consuming to label and explain to yourself every single piece of work, considering you're being timed, so it's reasonable that you don't. However, this means you're more likely to make mistakes. I would think that the best solution is to remain balanced in your work organization and your speediness; that is, don't label every single thing, but perhaps draw arrows to show where work is coming from, and box important numbers in your work.

"Trick" questions - I don't think the AMC really has trick questions persay, but just questions that look difficult, but aren't. For example:

How many ordered triples of integers , with , , and , satisfy both and ?



Ugh, have you ever seen anything like that before? I haven't. Skip it...
...or look at it again. Obviously the answer is one of the multiple choices, so if we do some casework, then worst case, we have to find 4 solutions (this is also an example in which the multiple choice answers give you a hint. Obviously they aren't very useful for probability problems, but for this problem they were incredibly useful). We're given restrictions on each, so it's not too bad:

c = 0, then b = 1 (properties of logs here), and to satisfy the 2nd condition, a = 2004. Thats one solution.

c=1, then a=b, and to satisfy the 2nd condition, a=b=1002. That's two solutions.

c=2, Ugh, is large and ugly. A is at least 2, and b = , making b a ridiculously large number. Obviously no more solutions can satisfy the 2nd condition, so there are only two solutions.

This is a clear example where a little casework goes a long way.

WLOG - If you aren't familiar with that, it stands for "without a loss of generality". It means that by solving a specific case of a problem, you solve all cases of that problem. I don't really have a good example, but a problem where it would be useful is where they don't actually give you values for a problem, but want something like a ratio. In this case, you'd plug in something simple for one of the values, and then solve it. Obviously solving it this way is far from rigorous, but they don't care how complete your solution is, just that you marked the right letter on your answer sheet.

KISS - Another acronym that everybody has seen - "keep it simple, stupid". How does that apply to the AMC? Except for maybe the last couple of problems, the problems are designed to test not only your math skills, but your ability to recognize which ones to apply to a given problem. I say except for the last couple because those ones are a lot harder than the rest. Again, if you're having trouble with one of the later problems (13 - 22), and you're confident in your math skills to solve problems of that difficulty, then look at it again, and make sure you're not overlooking something. The problems can be solved without guess and check, too, but don't mix that up with casework.

Time-management - 25 questions, 75 minutes...so 3 minutes per question? I think it's absurd for even some of the smarter kids to solve the last couple of questions in 3 minutes each. The best way to manage your time is to breeze through the easier ones to get more time for the harder ones. If each of the first 12 takes you 1 minute each, then that leaves around 63 minutes left, or around 5 minutes for each of the other one. Of course, these are rough estimates. But if any of the first 12 are taking you more than a minute, box it and come back to it after looking at some of the other problems. Whatever you do, though, put priority on it. Those first 12's should be gimmes towards your score.

#24, #25 - These are the really hard ones on the test. Honestly, unless you KNOW you're capable of a perfect score, I wouldn't even bother with these, and that's that. Focus on the other ones. Remember, that question asking you to maximize the value of tangent of half of an angle on a triangle based on one of the ambiguous side lengths is worth just as much as that much more appealing one asking you how many bananas Jeff can buy if he has x amount of quarters. If you're sure that you passed the 100 point mark, then you might as well guess on them. Otherwise, I'd leave them blank, you may need that extra 3 points to push you over.

Preparation - I put this last because this applies to me last, I pretty much know the basics that I need to do well on the AMC. But if you're curious, the common topics for the problems seem to be:

Arithmetic, Quadratics, Absolute Values
Algebraic Techniques
Number Theory
Polynomials, Logarithms, and other Functions
Binomial Expansion, Inequalities, Optimization, Systems of Equations
Counting and Probability
Statistics, Sequences, and Series
Complex Numbers, Trigonometry, Law of Cosines
Right Triangles, Quadrilaterals, Pick's Theorem
Circles and Spheres
Triangles, Polygons, Polyhedra
Coordinates, Graphs, and Geometric Inequalities

I actually copy pasta'd that from an amc prep class syllabus. But those are the general things you should know. They threw in some specifics in there because they're common (I dunno about Pick's theorem though, I don't think I've EVER seen that on any amc problems, maybe one AIME problem). Geometric inequalities are things like the maximum side length of a triangle given this this and this, and other similar problems. Polyhedra are those ugly 3-dimensional problems they always ask. Number theory is properties of integers. I'm sure you know what the rest are. But if you're interested in preparing for the amc, those are the things you should be familiar with.

If I think of anything else, then I'll add it. Otherwise, that's all that comes to mind.

AIME problem - Equations

The solutions to the system of equations:






are and . Find .

Start with some substitution:

Let
Let

Our system of equations now becomes:




Manipulate the first equation to get and then substitute into the second:









Quadratic equation:



We can now solve for b:







Now we can solve for x and y.





y is solved for similarly, I won't bother showing all the steps:



Now we have to find what the question asked for, which is:





Realize that



From our previous statement, we can see this is equal to .

Therefore:



Solved.

Number Theory Olympiad Problem

Heres a beasty number theory problem. Actually, I'm sure that to seasoned number...uh, theorists, that it's pretty easy.

Prove that every integer has a multiple which consists of only the digits 1 and 0

Consider the first n+1 integers which consist of only 1's:

. Where the (n+1)th integer has n+1 1's.

Now consider taking a number modulo n. The only possible values (after simplification) are , for a total of n possible "remainders".

Now consider taking each of our strings of 1's modulo n. There are n+1 of these strings, but only
n possible remainders. Therefore, two of them must have the same remainder when divided by n.

Let these two strings of 1's be:



And let their remainder, when divided by n, be r.

First, it is clear that will only consist of 1's and 0's. Specifically, the first digits will be 1's, and the last digits will be 0's.

Also consider taking this difference modulo n. We know they have the same remainder, r, when divided by n.



Since it has a remainder of 0 when divided by n, it must be a multiple of n. Problem solved.

Trigonometry and complex numbers (they're related???)

Hopefully you know that they're related. Anyway, some random stuff. The AIME class is pretty good. This week focused on solving difficult equations. The class is mostly strategies for solving problems. I have to admit that it helped, most of the practice problems we were given afterwards I probably wouldn't have been able to do before the class.

School has a couple weeks left. Really not much else to say, nothing to rant about. So I guess I'll go right to the problem.





Ooomph. I mentioned complex numbers in the title, and I'm going to incorporate them. When you first look at this, complex numbers are most definitely NOT what come to mind immediately. After all, theres not a single square-root-of-negative-one in there. however, the cos and the n inside of the function should eventually tip you off. I'll show you why first.

First off, rewrite that with . It'll make sense soon.

Now consider the cis function. You should be familiar with that, its the polar representation of complex numbers, and comes up a lot when dealing with them. If you aren't familiar with it, then you probably won't have much luck when looking at complex number problems (With some exceptions, of course. Then again, if you aren't, then I doubt you'll be able to do the harder problems which don't use it).



If you're familiar with powers of complex numbers, then you also know that:



Now you should see that the real element of this is very similar to our expression, and if r = 1/2, then it IS our expression. This means that we can rewrite our summation as:






The summation is the definition of an infinite geometric series, so we can rewrite it as:



We know how to express cis, so we just need to find sin of theta to complete it, using the trigonometric identity:













Now we can find our infinite series:























We only want the real element of this, so you should see that there's no point in calculating the unreal part now.



And theres the answer.

I'd like to take a moment to talk about the problem solving aspect of this. I'll admit that I had a hint, it was in the "Complex Numbers" section of my book, so I knew I would be using those to solve it. But I could see somebody looking at this and trying to use trig identities to figure it out. Obviously you can't do this up till infinity.

The main part of solving this consisted of recognizing that it was the real element of a series of complex powers. Once that was figures out, then the rest was simple. However, that first step is usually the hardest part of solving a problem.

We can learn how to solve generic types of problems, and how to apply these skills to slightly different problems. However, solving problems which we've never seen before is usually a lot harder. This was something we discussed in our first AIME class:

"It is one thing to understand a few methods for solving equations -- even odd equations. It is another thing to creatively manipulate equations into forms that we can apply these methods to. When examining a difficult or unusual equation we should think about what kind of equation we have and what similar kinds of equations we know we can solve."

Applying this to the previous problem, recognizing that the expression was similar to powers of cis was key.