Not sure when but they finally decided to announce the result altogether with the full problem set. As a tradition allow me to first comment on paper C which I did not get my hands on without looking at the solution, then we will wrap up with my comments to the solution.
C1. This is actually a good Q1. Algebraically less trivial but easy once you understand nature of the question.
One direction is simply by rotational symmetry (how long since you last hear this term in college+ maths?) plus origin. The other direction? Let's cheat. All lattice points have a rational x-y ratio. Now if we rotate the square so that the slope of the sides are irrational, it is going to takes the lattice points one at a time each quadrant.
C2. uhhh...factorization?
If this is an IMO question I am 100% sure those 'bulls' would have factorize it without any problem. $f(f(r))-r$ is a polynomial of degree 9 with a very simple factor $(r-1)$. If, assuming in good will, that there are other factors in $\mathbb{R}[r]$, it should be in $\mathbb{Z}[r]$ as well.
The leading coefficient is 8 and the constant is 25, giving us pretty restricted scope of searching. Linear coefficients didn't work, but searching the quadratic quickly gives $(2r^2-4r+1)(2r^2-4r+5)$, and the remaining polynomial clearly has no real root.
Admittedly this is painful. So what? It really boils down to instinct -- if you draw the graph of $f(r)$ you can see it looks pretty *rotational* symmetric around $r = 1$, and substitution would verify that guess. With that in mind, we perform a substitution of $r = s + 1$ and finds the much, much simpler formula of
$f(f(r)) - r = \frac{1}{8}s(2s^2-1)(2s^2+3)(2s^4+s^2+2)$.
Notice the complete lack of odd degrees - this is a nicely odd function!
C3. A classic question? Not that I have seen it before, but certainly feels like so.
If you know Sierpinski's triangle fractal, everything is about reciting that back onto your answer sheet because you can compute all $a_n$'s explicitly for the sum. I've only got 2 side comments:
First, do you remember the convergence in power series test at the boundary case? This is quite important in this question because you have to handle that as well. Many year 1 students don't do that, oh my.
Second, when did I first came across to the Sierpinski's triangle? Not in competitive training nor in fractal or chaos theory, but in a general education math course I had to take to fulfill graduation requirement, and I still feel guilty about taking the course and claiming the first in class.
C4. Oh GENERATING FUNCTION HELLO AGAIN.
I made a comment last year that they should stop making generating function questions as it has become a tradition and is overly predictable. On top of that once you know you use generating function there is only one way to solve the problem with no surprise at all.
Take this question for example. Would you be surprised that the ratio converges? Even without using generating functions I would not doubt about that the slightest bit. This is just not fun as a Q4.
The updated list of generating functions are now 18A4, 22C3, 23B2 and 24C4. Two damn question 4 among them. Just wow.
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Wow when I look at the problem proposers I saw a few familiar names, more than one of them private friend of mine. But I am not going to point them out because you know, some Simon Marais questions are really bad, right?
A1. Overly trivial to comment.
A2. As pointed out, this is a simplified version of a question from a high school MO (junior level even). While the full version requires an in-depth bashing around base 3 numbers, the small parameter here does not require that. Students could simply count case by case without difficulty, and both provided solutions here did exactly that.
A3. Nice analysis assignment as I said, but thank you for laying out all the details...like an assignment handed in.
A4. Hmm. Round of applaud to this solution because it avoids my criticism against PNT related skill checking. Although the solution still uses the 'well-known' fact of $\prod (1-p^{-1})$ converging to zero which I am not sure given the level of this tournament.
The nature of this question is that as long as the subset of integers (prime numbers here) is not sparse enough the probability will be 1. But the next question is how tight is the bound? Clearly the product $\prod (1-a_n^{-1})$ could converge to zero for some increasing integer sequence $(a_n)$. Cooperating that we might have a proper B4 level open question...
B1. Shame on me who got it wrong, and I was wrong because I didn't know I have to tap one more time at the end for the correct answer! Yes you need 4 taps to know the two boxes that contain coins, then one more tap to kill the game.
B2. Yup standard.
B3. This is the kind of abstract problem we would expect in Putnam, and nice to see that here. I wish we have more of that in the future. It's also nice to have some further comments from the proposer since it really links to higher maths...
B4. Error correcting codes! The world of coding using linear algebra and finite field has been mesmerizing, despite that everyone's onto turbo codes now days...
C1. Yes yes yes. I am glad the 'cheat' is indeed the way to go. Is it possible to construct a square that contains exactly $4n+1$ lattice points for each $n$? Possibly yes but more hassle involved. Do you remember counting lattice points from the sum $\sum [px/q]$ from the proof of quadratic reciprocity? You can do the same here although it's again faster to reside back to the slope argument: pick $(p,q)$ primes both larger than $n$, then the edge would not coincide multiple lattice points at a time (in a single quadrant) before we obtain a square containing exactly $4n+1$ lattice points. An irrational slope is just like the ratio of two infinitely large primes that does the job for arbitrary $n$.
C2. I am also aware that $f$ is increasing hence the fix point approach, but that didn't come to my mind when I typed the my impressions on C1-C4 in 30 minutes.
I know it doesn't make sense to factorize a degree 9 polynomial, even with intuition, but I believe my second approach is sensible enough, especially when you realize that $f(f(r))-r = rg(r^2)$ for some $g \in \mathbb{Z}[r]$.
Nothing much to talk about C3. And for C4 the average score is higher than A4 and B4, and my bold guess is that a group of schools knows precisely they will get generating functions, trained well and get rewarded.
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And that's it!
It's been 3 months since the tournament and 2 months since I posted my first impression on paper A and B, making it very difficult for me to give a general comment.
The scope has not been changing much. We still get lots of analysis, some linear algebra (this time well-hidden like in B4), probability and game theory. I miss the calculus/numerical questions sometimes. And when are we getting true combinatorics or graph theory questions? How about geometry?
Difficulty has always been a complaint but I like the questions this year actually, B3 and C3 in particular. I really wish they put more questions like this because more often than not this is the question that separates the elites from the 'middle class', which is essential in tournaments like this.
Alas, we will see again in 9 months of time when Simon Marais 2025 arrives!
