What’s USAMTS?
USAMTS is a free, proof-based math contest for high schoolers, funded by the NSA. It consists of three rounds, each with five problems worth five points each. The contest is fully remote and runs for a month per round — the problems are hard enough that you genuinely need that time. It’s a great way to keep your brain constantly churning on interesting problems and sharpen your proof-writing skills.
Problem 1
Not too much to say here, except I probably could have approached this more elegantly. I ended up doing trial and error with minor casework until I landed on the answer.That said, there’s a useful observation right at the start: the circle with diameter 5 that’s given immediately tells you that three diameter-1 circles fit inside it. That gives you a solid foothold. From there, I did casework on where the diameter-7 circles could go and arranged the rest of the circles around that. There’s almost certainly a cleaner method, but this one took the least mental overhead for me. The correct configuration is shown above.
Problem 2
Full disclosure: I didn’t look at this problem until the night before the deadline. Apparently that’s when my productivity peaks.
To make matters worse, I completely misread the problem. I missed the part where the areas of the black and white regions had to be equal — I thought it was just about the number of regions. So I spent a while completely baffled, thinking “literally everything works, how is this even a problem?” Then I reread it three hours before submission and was devastated. I still gave it a shot though.
To get some spatial intuition, I used a tennis ball and rubber bands — spatial reasoning is not my strong suit. The cleanest approach is to just draw out the n = 1, 2, 3, and 4 cases. n = 1 is easy since a single great circle just divides the sphere in half. n = 2 is easy to disprove — shrink the angle between the two arcs and the areas become unequal.
For n = 4, as shown above, you can see immediately that any even n can be configured to work. I used that as the foundation to prove that odd n also works: start with an even configuration, then add another great circle and observe that the colors on one half invert while two new regions are created.
Problem 3
This one looks intimidating at first glance, and my approach was certainly not the most efficient — but I’ll walk through the better way to think about it.
So we notice that the numbers described are awfully specific, which calls for a generalization! The 1000 seems to specific. I noticed this after playing around with wolfram alpha for a bit, which I suggest you do for these types of problems. Gaining intuition. So we can generalize the $2^{1000}$ to $2^n$ and use induction which is straightforward from there.
A tidbit that is interesting to notice that if you take the expression mod $2^{12}$ let’s say, then everything from $2^{13}$ and above cancels. That is how I centered my proof, and proceeded to assume by contradiction that there is some point $2^k$ at where you can’t have 1 or 2 for that term and disprove that. It was a very long and ugly proof but it got the job done.
Problem 4
Now this problem was a woozy. For this, I used actual cards and played the game with my brother to get a sense of what was happening. After playing for a while, you can see that Grogg and Winnie will just start alternating cards at one point.
Let’s take the n = 4 scenario. So we have the cards 1, 2, 3 and 4. We know that Grogg goes first and let’s just say`that he puts down the 3 first in one pile. So Winnie’s logical move would be to put 4 next in that same pile. We see that if Grogg puts down a 3 or 4, that’s bad. To minimize this his optimal move could be to put down 1 first. Then we would have Winnie put the 4 with the 1. This leaves Grogg to put 3 in the other pile, and Winnie puts 2 with the 1 and 4. This makes a final sum of 5.
In this similar fashion, I made the grand mistake of thinking that the optimal move for Grogg was to put down a 1 first and then Winnie puts down a 50, Grogg puts a 49, Winnie puts a 48 and so on. Although! I didn’t account for the fact that no matter what Winnie will want to make sure that the 50 and 49 are in the same pile. So it’s actually Grogg’s best move to put 50 down first which is something that I totally overlooked.
From when Grogg puts down the 50, then Winnie and Grogg start to “alternate”. So I was almost there but not quite.
Problem 5
This problem was my favorite by far! It was the first one I started working on even before the puzzle.
I liked this mostly because of how straightforward it is and how simple the proofs are. Not to say that they were easy to think of, but they were really clean.
My solution here for 5b is different than the one presented in the AOPS math jam, although I think it’s pretty good. For 5a it’s quite simple to see the polynomial $n^2 + 2n + 37$ after some tinkering around with $p^2 + (p+6)^2 + n^2 + 1$ by expanding it out. Although how to prove it’s irreducible? It’s obvious it is, but that itself is not a proof. A way of suggestion would be that the discriminant is negative or by using Einstein’s Irreducibility Criterion. I did the second one just for fun.
For 5b, I first saw that 27 was a pseudo sixish number first by just trial and error. I used wolfram alpha to make the calculations less painful and to just see straight up. This is what led me to believe it was the only pseudo sixish number because after going up to 100 or something I didn’t reach any others.
Proof Sketch: So I wrote out the divisors of n as $(d_1)(d_2)(d_3)…(d_n)$ and then I wrote out the sum of squares of the divisors as $(d_1)^2 + (d_2)^2 … + (d_k)^2$. I wrote n as a product of some two divisors $i$ and $j$ such $n = (d_i)(d_j)$ and so $2(d_i)(d_j) + 36 = (d_1)^2 + (d_2)^2 … + (d_k)^2$. So we can write $36 = (d_i - d_j)^2 + (S_{rest})$ where $S_{rest}$ is the sum of the rest of the divisors besides $d_i$ and $d_j$, now we’ve reduced the problem to just finding the sum of squares that add up to 36.
There. The problem is simple with some casework and then you’ll find that 27 is the only pseudo sixish number.
Final Thoughts
This round felt noticeably more approachable than I expected, though I’m not complaining. Glancing at the Round 2 problems though — that’s a different story. There’s another game theory problem, and they brought back Grogg as a character. No originality with names, but then again, Grogg was my favorite character in the Beast Academy books, so I’ll let it slide.