The USAMO

I took the USAMO on Tuesday and Wednesday. Here is my detailed analysis of each problem. (Note: there are some spoilers here.)

Pro(j)blem 1: This pro(j)blem was silly - far easier, in my opinion, than last year's #1. Although I hear many complaints about #1 last year, I think this problem was a worse choice than that one.

The complaint I hear most about last year's problem #1 is that it was trivialized by certain techniques. In comparison to 2010 #1, I have to say that this is very not true. Even after thinking of this technique, it is still necessary to notice some things.

On the other hand, this problem was a straight angle chase. (Anyone who says it requires Simson lines is wrong: just look here.) I am strongly opposed to this, because while angle chasing is a useful skill to have, this problem encourages trying to solve geometry problems using just angle chasing.

(It could have been made less silly if it were not specified that $$AB$$ was the diameter, and if $$O$$ were redefined as the center of the circle. I am told that this interpretation is still a true statement, but I haven't worked through it yet.)

Problem 2: This was an interesting and nice problem. I thought it was pretty easy for a #2, though. I solved it pretty quickly after some playing around with small cases.

What disappointed me about this problem was the existence of an easy solution using induction directly on the problem statement. (I don't really appreciate induction, because I feel that it kills the beauty of a lot of problems. But the solution I actually submitted for this problem used some induction, so maybe I should shut up.)

Also, note that while the problem may have been easy for a #2, it was by no means the easiest recent USAMO #2. Although it may seem "trivial" due to many people solving it, note that many of their solutions are incorrect.

Problem 3: This was disappointing for me. Even though I had four hours to solve this problem, I got completely "trivialized" by it.

Well, maybe not. I, apparently, got pretty close to the construction for the extremely guessable answer $$4019 \cdot 4015 \cdot 4011 \cdots 3$$ (enough to get 1 point, for sure, and 2 if I'm really lucky) but failed to follow through with it.

I wasn't really expecting to solve #3/#6 anyway, so maybe it's somewhat okay.

Pro(j)blem 4: Let's just say this: If a pro(j)blem can be bashed by the angle bisector theorem and repeated applications of the Pythagorean theorem, that's already a pretty good reason to not like it.

I found a nice trig solution (no, not the one where you use law of cosines on BIC)

Problem 5: My solution was just a sequence of five ridiculous partialfractionizings, cancellations and reductions that worked EVERY TIME when $$p \equiv 2 \pmod 3$$. Then a sequence of five ridiculous cancellations and reductions that worked EVERY TIME when $$p \equiv 1 \pmod 3$$. Then a simple check of the case $$p \equiv 0 \pmod 3$$. It ended up being 7 pages.

I will probably lose a point for not specifying that I was considering the homomorphism $$ \varphi: S \mapsto F_p$$ defined by $$\varphi(a/b) = ab^{-1}$$, where $S$ is the set of rationals with the denominator not divisible by $$p$$. Well, I guess it's my fault for using fields in the first place without having a strong background in field theory.

Problem 6: I didn't really try this problem at all. I was too busy pwning #5.

My predictions for the cutoffs are:

Red: 15
Blue: 22 (or maybe 23, but I doubt it)
Black: 30 (it could be 29, but I am not really sure)

(One may notice that these predictions are significantly lower than some of the ones posted on the AoPS forum. Please note, however, that many of the predictions posted on the AoPS forum are complete nonsense. Read my post here for more information.)