The AMC contests are much less reliant on obscure theorems than I would have thought at first. For success you probably need to know just a few principles for each topic but then learn a few general strategies, at least one of which should help if you’re stuck. This is a small collection of some of the most common and useful ones that will take you a long way, which I have realized by working through a few good examples which have become some of my favorite math questions of all time.

The examples I use are pretty hard questions, especially the last two, so if you aren’t experienced with math contests, don’t expect to get them even with hints. At the same time, they technically shouldn’t require anything you didn’t learn by high school.

1. Setting up a good system of equations

This one shows up in physics a lot too. Problem number 4 from the 2012 AIME I is a really nice example.

There are plenty of conserved quantities between Butch and Sundance. We basically need to find when they align. If they are at the same place at the same time, then a whole lot of things are the same for both of them: obviously total time, and not only just total distance, but the miles Butch walked/rode are the exact same as the miles Sundance rode/walked. This gives us 3 variables to relate to each other. We have no idea what the total time is, and don’t need to know, so we can eliminate it by making two equations that solve for time in two different ways. Think about time from Butch and Sundance’s different points of view. This gives us a linear equation with two variables, which has infinite solutions. Which one are we looking for?

2. Find the inverse

For the next two, here is the USAMO 2012/1.
We all know about counting the complement for probability. It is sometimes easier to find the probability that something is true by finding when it is false (example, for 1 of 2 independent events occuring). This totally applies to proofs as well. In this problem, we are told to find the acute triangle in every single possible list to prove that all of them have one. This is absurd to do as it is written. Much easier is trying to find a single possible list that does not contain acute triangles, which would then instantly prove that not all of them have one. On the other hand, if it is impossible to find a list without an acute triangle, then all of them have one.

3. Without loss of generality

This is a fancy way of saying that we are modifying the question in ways that will not affect our answer. This can save a lot of headache, especially for these generic lists that are so common in math contests. Let’s use the problem above again. There are definitely ways to modify the list without changing whether or not the triangles formed are acute. For one, we can change the order of items in the list into something more preferable, like ascending order. We could even multiply everything by 1/a₁ so that a₁ would equal 1 without changing any angles.

Can you solve this problem now? Remember those Pythagorean inequalities that can tell you whether it is acute, right, or obtuse, depending only on the side lengths.

4. One to one correspondence (or counting in a smarter way)

Last is the IMO 2011/4. This one was pretty mindblowing to understand and changed how I approach questions like these.

If we are counting in the way that the person is placing the weights, the restriction causes us a whole lot of pain, as there doesn’t seem to be any good way to know how many possible weights you can put on the right pan. If the weight you place is the new heaviest, it has to go on the left pan, and if it isn’t, then it doesn’t matter. Good luck trying to find the average number of times you will place a new heaviest, though.

Is there a better way to model this situation than from the point of view of putting the weights on? The answer is yes. Chronological order will have to come second. Let’s arrange them in a chronological line before placing them on the scale, starting with the heaviest and then going down to the lightest. For each weight, we will decide where it is in the line compared to the previous weights we have put, as well as whether it is on right or left on the scale.

Why would this work better? Because we are putting these weights in the line in descending order, each successive weight will be the new lightest compared to what you have already placed in the line, but the heaviest compared to what you have yet to place. This means that if you put it in the front, it must go on the left because no amount of the smaller weights you haven’t ordered yet will balance it out, and if you don’t put it in the front, it will not tip the scale because there was a bigger previous weight. Same is true for the rest of the weights.

Once we are done with making our line of weights, we have also made a valid configuration to follow through and put on the scales. Every possible line is then every valid way to place the weights normally.