It’s July again: the month where the brightest mathematical minds all over the world celebrates their talent on the International Mathematical Olympiad again. As a retired contestant, I could hardly break the habit to read the problems, take a deep breath, appreciate the ingenuity of the problem author and the magnificent result embedded in the problems, before finally tackling them by working my mind hard. Yet the depression from the 2017 IMO disillusioned me from the vanity of wanting to sit the paper live and hoping (against hope as of the 2017 case) to vanquish them as quickly as possible.
On Monday and Tuesday mornings, I read the paper and decided to casually squeeze some free time in the week to do it (I managed to get 5 problems under the time-relaxed condition, perhaps 4 under the typical competition pressure), so here’s it:
Problem 1 is Geometry, the area that I took pride of because it’s the most attractive for practice. Yet the surprise came in as I couldn’t find a way to get around it except to resort to my forte: trigonometric computation. The only happiness in disguise was the 3 likes on the Art of Problem Solving forum due to the unique approach 😉
The first impression on Problem 2 looked daunting, perhaps due to my diffidence on sequence-like Algebra problems (especially when there were just too many constraint to follow). Yet the problem couldn’t have been made more vulnerable when the first step I needed to do was to play with the signs of the sequences. Two consecutive positive terms, one term zero: these are quickly eliminated. The only tricky case to consider was one negative term in between two positive terms. The +1 makes the terms following them pretty uncertain, but thanks to the cyclic structure of the sequence we can do some ‘backtracking’: it turns out that the sequence must have alternating positive and negative terms throughout, and the negative terms will only go increasingly negative down the line – hence the contradiction B) The only remaining possibility is – – + – – + repeat, which means 3 divides n. What about n = 3? It took a while, but I found that -1, -1, 2 works.
Problem 3 was the only problem I haven’t attempted — my fear for combinatorics stayed me away…but…isn’t this the exact problem I got on Google Games (use a computer to solve for a triangle with 5 rows). It turned out that this problem has been discussed in a few contexts before, and it’s therefore interesting to think how did the problem end up on the IMO…
The first impression on Problem 4 was…just…difficult (come on are you killing students by disguising this as something ‘easy’?) The first thing I understood: no two red stones are associated by a knight’s move on the chessboard, which are of grids of different colours. So putting it on all white squares on the chessboard should be good: there are 200 of them and Amy can occupy at least half (100) of them. But how does Ben prevent her from progressing any further? “Knight-like bijection, maybe”, I conjectured, but, easier said than done. I swiftly drew myself a 4 x 4 chessboard and see if I could get much from it, only to end with uncertainty. The required mapping only crystallized only after more than 5 times drawing and erasing the 4 x 4 chessboard, shown as below:
TL;DR whenever Alice moves, Ben put a blue stone on a spot of the same colour as Alice’s previous move such that the other two spots of the same colour are no longer accessible by Alice.
Problem 5 won my heart instantly: how on earth can one discover the identity behind this simple problem formulation that would render everyone hands-down? The first (and classical) approach was simply to expand the sequence by 1, and work from there. Find feasible a_1 based on a_n and a_(n+1)? Sure but this wouldn’t get us far since some funny combinations can work (e.g. a_1=18, a_n=77, a_(n+1)=99). Solving for a_(n+1) based on a_1 and a_n? Oh well…no thanks since there is quadratic equation involved. In view of the feasibility of a_(n+1) := a_1 or a_n there needs to be a bit of change. How about fixing a_1? Things now became interesting: the denominator of a_n/a_(n+1) in reduced form cannot exceed a1 (!) That’s a big step, and what about the extreme where a_(n+1) = a_1 * a_n? This seems fishy since we would need (a_n)-1 to be divisible by a_1. Here comes the conjecture: the gcd(a_(n+1), a_1) must be divisible by gcd(a_n, a_1), and it’s true – so at one point this gcd must remain constant as n increases. Things got easy henceforth, when we must have a_n divisible by a_(n+1) so the sequence must be non increasing from that point, and hence eventually constant.
Problem 6 was another reason to be happy of, since there are finally two geometry problems on this IMO (after its rarity for two years). The angle condition gives rise of two circles on which the point X must lie on, so it’s the intersection of the two circles. A couple more angle conditions arose, but they seemed to lead me deeper into the black hole without me noticing any way out. Anxious. Lost. I couldn’t help but to whine about the inability of my trigonometry magic to defeat it. Then I realized…wouldn’t it be great to investigate any alchemy arising from the length condition? Apollonius circle seemed to be the panacea to problems of this type, but the key never came to me until more than two hours have elapsed where I noticed the ultimate clue arising after I took the second intersection (named Y) of the circles ABX and CDX: Y lies on BD with angles related linearly to those formed with X! Now that Y, A, C and the centre of the Apollonius circle of ABD are on the same circle, more equal angles emerged and that comes to the end of my three-hour battle with it. 😀
The abnormally high medal threshold this time round took me into disbelief, especially when I was certain of the difficulty of problem 4, convinced by the non-triviality of problem 6 and unsure of that of problems 2 and 5 (which turned out to be quite ‘easy’ as medium problems). On a positive side, the return of a double geometry assured me that my previous paranoia (and despair) that “geometry has been officially de-emphasized” was false. Perhaps on a more unbiased viewpoint (even as a fan of Euclidean geometry), I hope that problems on the IMO shortlist are not discriminated for (or against) just because of their topics (beyond the requirement of distinct topics must appear in the sets (1, 2, 4, 5), (1, 2, 3), (4, 5, 6), (3, 6) ). Beauty of a problem comes first, shouldn’t it?
Finally, here comes the second round of work!