Friday 21 July 2023

IMO 2023 quick thoughts

Ok the yearly event to review, IMO 2023. This time it was in Japan, hosted exactly at where comiket is held. This is kind of strange huh?

Q1. Many different approach towards the right idea. My first thought is that prime powers certainly works. Then the next simplest type of numbers to check is of the form $pq$ which...immediately fails with very simple numbers like 6 and 15. When going for 30, the product of 3 distinct primes, you quickly realizes where the problem is, then question solved.

I'm quite shocked by how easy this is due to how approachable the idea is. Here is the approach taken by most as on AoPS: observe that $d_{n-2}|d_n$ to force a chain of divisibility. It is also easy because how easy is it to check the examples and counterexamples, starting from small integers. The question gets an average score of almost 6 (5.85) shows how easy this is.

Q2. Okay geometry...so I am not going to act like smart ass and solve. But frankly this is possibly one of the few geometry questions that I can actually solve at this level. There are too many parallel and perpendicular lines, trivializing angle and similar angle tracing. The extra triangle to construct is also pretty straightforward.

Q3. Very nice looking question. FToA strikes when it comes to solution of polynomials (just like the Vandermonde matrix), so you get ideas of recurring sequence (this is already non-trivial). Solution can proceed from here, details not discussed. After all this is quite a respectable Q3.

Q4. Yes finally, inequality is back! Also not in the traditional way with difficulty stacked because you guys can do AM-GM pretty easily with 3 terms but not with 2022+n terms.

The idea is obvious here. $a_n$ is strictly increasing and also increasing by at least 1 each time. But the number in question is 3034, indicating an increment of 1.5 per term -- or 3 every two terms! It is easy to calculate how much the increment needs to be to achieve an increment of 1, and this is where the pairwise different condition kicks in (which is pretty odd before this step).

Of course, those inequality gurus already spot the obvious pattern of Cauchy-Schwatz: square roots and reciprocals. This is solvable directly using C-S, or AM-GM, or a combination of two. In overall a suitable and nice looking Q4.

Q5. Combinatorics question and I skipped, but I saw the word ninja. I saw what you did, Japan ;)

Q6. The moment I saw the condition about 480 degrees and scalene triangles (and it took me some time to realize what's a scalene triangle) I knew immediately this a devilish question. Not only the angle sum condition is absurd, the graph itself is absurd too: $AA_1A_2$ are almost colinear, making the corresponding circumcircle (as well as the other two) incredibly huge.

The thing I notice is that why two? What happens if we don't have a scalene triangle -- well then the almost colinear points becomes colinear and the problem collapses. From here the radical axis would come to help. Of course, proving two is another difficult thing: you prove one, then prove another, but you also need to prove that the two are distinct. Proving one seems to be the harder part though.

I am of course not qualified to talk about this problem, but quite funny that we find many possible solutions: pure geometry, inversion, coordinate geometry, bary bash and so on. The fact that the problem allows so many solution yet not many solved right (second hardest pure geometry Q6 since 2000), is the beauty of it.

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The only thing I can say is that geometry is not my cup of tea, but I can feel the beauty of the two questions assigned this year. Yet, I always love to see a Q6 in number theory. The most elegant but lethal problems.

I am less focused on competitive maths this year, but Simon Marais is again something I will eagerly wait for.