Wow hello it's time for another year of IMO review. I even didn't realize it is mid-July instead of early August somehow with all the workload in reality where solving math problem is actually more of a relief for me.
As usual, this is a quick attempt on the questions with my thoughts and insights where I didn't try them under formal time limit. I would also feed them to mainstream AI models as a cross section study of AI capabilities. Let's go!
Q1. Wow a question on lcm and gcd! One of my favourite lemma, as I wrote a
whole entry about it in 2013, is $mn = [m,n](m,n)$ where $[m,n]$ stands for lcm and $(m,n)$ stands for gcd.
We are replacing $m,n$ by $(m,n)$ and $[m,n]/(m,n)$ this time, so not quite the application of that lemma yet the direction is clear -- the way the question is formulated screamed for specific approaches.
When they asked for finite termination, this is asking for a metric that is strictly monotone upon the operation. Is there any metric better than products of all numbers?
Notice that every operation either (1) turns a number into 1 (when $(m,n)=1$) or (2) decreases the product (when $(m,n)>1$). Since the number count and the product are both finite this is done.
They also asked for the final number standing is fixed given the starting parameters. This is to find an invariant. To this end we look at the prime powers of each number -- i.e. we look at $v_p()$ of the 2026 numbers. We are sending $v_p(m), v_p(n)$ to $\min (v_p(m), v_p(n))$ and $\max (v_p(m), v_p(n)) - \min (v_p(m), v_p(n))$. But wait! This is the operation in Euclidean algorithm! And what's invariant? The gcd across all prime exponents.
When there is one number above 1 left, it must take all the prime powers (the prime powers of all the rest are zero -- think about how Euclidean algorithm works a step further to turn the two numbers $(m,n)$ and zero). Since the question does not ask to determine $M$ we don't even need to write it down! (But you can write it down right?)
Q2. Oh coordinate geometry. Not my taste and probably won't be able to brute force it under exam environment. When I checked on AoPS, it seems like this question is relatively hard as Q2 and have deep relation to projective geometry? Interesting.
Q3. While Q2 is surprisingly difficult (average score of 1.6), Q3 is actually relatively easy. In fact, one very rare combinatorics Q3/6 that I solved without trying hard.
It boils down to the essence of equilibrium where it is reached only when you make a move that opponent's move is indifferent. In this case, indifference means even by not cutting the rod. Technically you can't do that given the rules but it is always possible to cut infinitesimally small portions.
First a simple note: when the cut is done, it is clear that player A will always take the 1st, 3rd, 5th... largest portions and B will take the rest. We are therefore minmaxing the total length of the odd ordered pieces.
Consider the case $n=1$ where player A makes a cut then player B makes a cut. Where would player A choose? The answer is to cut the rod into 1/3 and 2/3 and the claim is A will be claiming 2/3 of the rod.
Case 1: If player B cuts on the 1/3 portion then A can simply take the 2/3 portion and done.
Case 2: If player B cuts on the 2/3 portion into $x, 2/3-x$ then either portion will be at least 1/3. Take the larger one so that B will take 1/3, and you will take the rest for 2/3.
The key is to realize that "cutting the 1/3 portion" has the same minmax result compared to "cutting the 2/3 portion". For case 2 above A is guaranteed to have 2/3, no more, no less. It sounds like A would get more than 2/3 in case 1 but not really if B cuts an infinitesimally small piece on the 1/3 portion leaving A with $2/3 + \varepsilon$ in total. It is easy to show that 1/3 is the sweet spot making the two cases indifferent.
Can we generalize this? Absolutely. The first cut would divide the rod into $\alpha < 1-\alpha$ such that if you don't break the $1-\alpha$ portion A will take the whole portion, with $1-\alpha$ being the part A can always get regardless of B's choice.
With the first cut being made, this forces B's first cut to land on the $1-\alpha$ portion -- and the best he can do is to split the portion into 2 equal parts or else A would claim the larger part. How can we guarantee that he won't make multiple cuts on this portion? Again, indifference -- divide the smaller portion in a way such that making multiple cuts on the already halved $1-\alpha$ portion is equivalent to making cuts on smaller portions. A can simply make a $(1-\alpha)/2$ portion out of the $\alpha$ portion which is as much as the halved $(1-\alpha)$ portion, and so on. That gives a geometric series of division, foul proof and hence optimal.
Do you realize the solution already? Once you write down the partition the rest of the proof is easy case by case argument.
One question I really like, probably very much to my taste although students think otherwise as reflected by the average score of 0.58.
Q4. Another game?
At first I was thinking about countability argument like is there a way to iterate all rational angle, but if they asks for the largest possible set of angle why not just $\mathbb{Q}[\pi]$? Why not $\mathbb{Q}[\pi, \sqrt{2}]$? ...then you know this isn't the way to go.
It reduces to something very simple: it only works it integral divisions of 180 degrees where you can divide the angle $k \theta$ into lower multiples of $\theta$ and there is no stopping it. If the initial triangle contain no such angle you can create that by dividing the 180 degree angle. On the other hand it is easy to prove that if a triangle contains no integral multiple of $\theta$ then there is no move that grants you a win within a single move, then induction does the job. Do you realize what is needed for the inductive step? External angle of triangle! This is such cute geometry trick when you don't have to summon huge block of geometry argument like Q2.
Q5. Functional inequality! Always nice to have them. The more complicated it looks, the more important it is to find the right substitution to draw useful information out of it. The given relation involves variable with and without passing into the function $f$, so the natural approach is to substitute $(x,y)\mapsto (f(x),x)$ so find the relation between $x$ and $f(x)$: $f(f(x)) = 2f(x) - x$.
A linear relation!
That screams a simple linear function being the sole answer, although there is a lot to do before we can reach that conclusion. The fact that recurred linear function is also linear with the sequence of $f^n$ being arithmetic themselves is also covered in
another entry, this time dating back to 2011 (probably not worthy for a read but yeah).
By induction we have $f^n(x) = nf(x) - (n-1)x$, showing that $f(x)\geq x$ for all $x$ by taking $x\to \infty$. The rest is probably the hardest -- to argue $f(x)-x = g(x)$ is constant by equating $g(x)$ and $g(y)$ using the given inequality involving quadratic mean. Perhaps I can do it given the time, but I didn't spend too much time thinking into it.
Q6. A NT Q6 is always the ultimate thriller like 1988 Q6, and this one is of no difference. What an elegant formulation.
At first, I even struggle to come up with a single example where the sequence isn't simple arithmetic. I was trying larger numbers with no luck, before I finally ran into $a_1 = 15$. Then I ran into another problem: isn't it strictly arithmetic after a while, and if that was true the demanded statement is false right?!?
Well no. The sequence starting from 15 looks like 15, 18, 20, 24, 30, 36, 40, 42, 45, 48,... with an increment of 30 and period 8.
Think it this way. $a_2-a_1$ must be the smallest prime factor of $a_1$ denoted $p$. The sequence of $(a_1 + (n-1)p)$ clearly satisfies the gcd requirement, and it only fails only when an increment below $p$ somehow satisfies the gcd requirement as well. For example when $a_1 =15$ you know $a_2 = a_1+3$, but somehow an increment of 2 worked for $a_3$ breaking the pattern. It is clear that it's the small prime that matters.
That is, we claim that only prime factors at most $a_1$ would matter. Something is wrong if you have an increment that is bigger the products of these smaller primes. With that you can argue periodicity by finiteness (of course we need to be careful to show that $a_1$ is mod-repeated in the sequence).
The details aren't precisely straightforward -- of course that should be the case considering how this is Q6 not Q3 or Q5. I don't think I can write a complete proof under exam environment either. The average score of 0.38 defends the family of NT Q6 on the harder end among boss questions, although I feel like 1 or 2 points based on nature of increments shouldn't be hard.
*
How do I feel about the questions this year? It feels strangely familiar.
If we ignore the geometry question, every single question points to something familiar on my side directly applicable towards the intended solution. Writing the full proof is one thing, but being able to analyze the question without wasting time on wrong directions would be a huge bless after all. Although there's also a bad side for being familiar with the questions -- I can't really evaluate the difficulty of these questions accurately. With that being said, average score shows that the difficulty this year is surely not out of the blue.
As for AI models, it is not surprising that GPT 5.6 Sol has managed to solve it all. When someone like me managed to get all directions correct easily it is madness to expect difficulty for leading models to do the same.
...leading thinking models only, of course.
It is still far away for mid level LLMs to perform up to that level. GPT Terra had problem from Q3, Grok can't even solve Q1 properly. Our poor follow Gemini Pro 3.1 managed to solve Q1-4 properly, completely messed up Q5 but solved Q6 with minor mistakes.
It has been a spectacular year of AI progress, so much to the point it becomes fearsome. But for now, there is no better timing to start embrace the use of AI to relief us from the calculation. It has been shown that once we get the direction right, AI can do the rest for us.
Let's meet again in 2027.
