IMO 2017

Let’s aim for a 35: this is what I should have gotten in IMO 2014 if I didn’t screw up my geometry problem 3 because I solved it eventually in a rather simple way. Forget it, 35 is only ‘reserved’ for the top 3 in the contest, as it turns out.

As usual, I ‘took’ the IMO 2017 paper. As usual, I never did well on the IMO when I try it at home instead of at the contest (IMO 2016 was still considered okay since I could have gotten a gold if not because of disbelieving my initial construction on problem 5). But the depression from this IMO is so strong that I doubt if I myself can even take it.

As it turns out, I was not alone. Problem 1 has the highest mean score in the recent contests, but the complete opposite also happened on Problem 3. The extreme difficulty on Problems 2 and 5 simply made the IMO unbelievably crazy and pull the gold medal boundary down to an unprecedentedly low of 25.

Day 1 : “okay, I am ready. On the goal to get 21 on it?” I opened the AoPS forum, only to be disappointed with the dearth of geometry on the paper.

My face when read the IMO Day 1 paper

I went on, nevertheless, and instantly got Problem 1 in two minutes after trying out small cases (like 3k, 3k+3, … that will give you the sequence of 3,6,9,3,6,9,… eventually, and 3k+1, 3k+4,… that will eventually make 3m+2 into the sequence and the numbers will not repeat themselves again!) Oh yeah, life is good.

But Problem 2 (for the non-constant case) wasn’t a safe sailing, unfortunately (I only get the case of f:integer to integer, which should merit a few marks I guess?) The first thing to do is to substitute 0 into both x and y (which served me right and I got f(c^2)=0 if c=f(0) ), and a few more experimenting gave me f(kc^2)=(1-k)c. “If only the question asks for integers to integers instead of reals to reals”, I whined silently to myself. I kept playing around with the substitution, got a whirl of bizarre terms like f(c^k)=0 for every k a power of 2, only to notice that they wouldn’t help at all (unless we are told that c is an integer). It took me long to get f(1)=0, after which I didn’t feel like proceeding any further.

It’s not until I saw (from a solution) the crucial step of proving that f(b)=0->b=1 by the slick substitution of (b, b/(b-1)) which yields f(f(b)f(b/(b-1))=0 (imagine if f(b) is 0 and b is not 1, then f(0)=0->f is a constant!) There were a few more steps to go, but looks like this step is bringing us there. Whatever, I am too tired/frustrated to try it.

Meme extracted randomly πŸ˜›

Problem 3? A beautiful problem I should concede, but the fact that only 7 contestants attained nonzero score (with a lowest-ever mean score of 0.042/7) dissuaded me from trying it. πŸ˜›

Day 2: looks more bearable, with geometry as Problem 4. For some reason it took me quite long (35 minutes I guess?) until I saw the fact that AT and RK are parallel (via some angle chasing), and the solution can be completed using some properties of triangle similarities.

Now the next step (according to my plan) wasn’t to try Problem 5 (some algorithmic combinatorics) but Problem 6 instead (number theory, which matches more closely in my favour). In fact, that was the problem that kept me awake at 4am when all the computations happened in my brain. Here I went:
-The determinant (name it K) of the square matrix with x^(n-1), x^(n-2)y, …, y^(n-1) on each column (for n different pairs of (x,y) ) is the product of (x[i]y[j]-x[j]y[i]), running over all possible i and j with i<j (or the other way round?) This means it cannot be zero so the system of equation has a unique rational solution (Yay, no?)
-Now let’s try to expand n to make it more reasonable. Honestly speaking, I thought the struggle would end really really soon when I (mistakenly) conjectured that all we need is to find an n big enough (say phi(K) ) and a and b such that ax+by is relatively prime to K for all pairs of such (x,y). Therefore the system of equation (i.e. (ax+by)^n) has remainder 1 when divided by K, and we can play around with the coefficients to make the values actually 1. Oh well, this is wrong. We need it to have remainder 1 when divided by Kx^(n-g) where g is the number of such pairs. (Neh…that sounds impossible)

Finally, with much chagrin I should say that I haven’t found an algorithm to Problem 5, but a few pitfalls instead. (Oh well this isn’t exactly the type of IMO paper I want to see…). Thanks to the generous partials given on Problem 2 can I be sure to say that I could medal with 16 points (or maybe silver at 19, if I am lucky enough to get garbage points in Problem 6 as well). RIP gold.

Congrats to my Team Malaysia which improved it ranking from 56 to 37, in any case. πŸ™‚

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