# IMO 2019 Aftermath

Here is my commentary for the 2019 International Math Olympiad, consisting of pictures and some political statements about the problem.

## Summary

This year’s USA delegation consisted of leader Po-Shen Loh and deputy leader Yang Liu. The USA scored 227 points, tying for first place with China. For context, that is missing a total of four problems across all students, which is actually kind of insane. All six students got gold medals, and two have perfect scores.

1. Vincent Huang 7 7 3 7 7 7
2. Luke Robitaille 7 6 2 7 7 6
3. Colin Shanmo Tang 7 7 7 7 7 7
4. Edward Wan 7 6 0 7 7 7
5. Brandon Wang 7 7 7 7 7 1
6. Daniel Zhu 7 7 7 7 7 7

Korea was 3rd place with 226 points, just one point shy of first, but way ahead of the 4th place score (with 187 points). (I would actually have been happier if Korea had tied with USA/China too; a three-way tie would have been a great story to tell.)

## Contest analysis

You can find problems and my solutions on my website already, and this year’s organizers were kind of enough to post already the official solutions from the Problem Selection Committee. So what follows are merely my opinions on the problems, and my thoughts on them.

First, comments on the individual problems. (Warning: spoilers follow.)

1. This is a standard functional equation, which is quite routine for students with experience. In general, I don’t really like to put functional equations as opening problems to exams like the IMO or USAJMO, since students who have not seen a functional equation often have a difficult time understanding the problem statement.
2. This is the first medium geometry problem that the IMO has featured since 2012 (the year before the so-called “Geoff rule” arose). I think it’s genuinely quite tricky to do using only vanilla synthetic methods, like the first official solution. In particular, the angle chasing solution was a big surprise to me because my approach (and many other approaches) start by eliminating the points ${P_1}$ and ${Q_1}$ from the picture, while the first official solution relies on them deeply. (For example one mightt add ${X = PB_1 \cap AB}$ and ${Y = QA_1 \cap AB}$ and noting ${P_1CXA}$ and ${Q_1CYB}$ are cyclic so it is equivalent to prove ${T = PX \cap QY}$ lies on the radical axis of ${\triangle CXA}$ and ${\triangle CYB}$). That said, I found that the problem succumbs to barycentric coordinates and I will be adding it as a great example to my bary handout. The USA students seem to have preferred to use moving points, or Menelaus theorem (which in this case was just clumsier bary).
3. I actually felt the main difficulty of the problem was dealing with the artificial condition. Basically, the problem is about performing an operation while trying to not disconnect the graph. However, this “connectedness” condition, together with a few other necessary weak hypotheses (namely: not a clique, and has at least one odd-degree vertex) are lumped together in a misleading way, by specifying 1010 vertices of degree 1009 and 1009 vertices of degree 1010. This misleads contestants into, say, splitting the graph into the even and odd vertices, while hiding the true nature of the problem. I do think the idea behind the problem is quite cute though, despite being disappointed at how it was phrased. Yang Liu suggested to me this might have been better off at the IOI, where one might ask a contestant to output a sequence of moves reducing it to a tree (or assert none exist).
4. I liked the problem (and I found the connection to group theory amusing), though I think it is pretty technical for an IMO 1/4. Definitely on the hard side for inexperienced countries.
5. This problem was forwarded to the USAMO chair and then IMO at my suggestion, so I was very happy to see it on the exam. I think it’s a natural problem statement that turns out to have an unexpecetdly nice solution. (And there is actually a natural interpretation of the statement via a Turing machine.) However, I thought it was quite easy for the P5 position (even easier than IMO 2008/5, say).
6. A geometry problem from one of my past students Anant Mudgal. Yang and I both solved it very quickly with complex numbers, so it was a big surprise to us that none of the USA students did. I think this problem is difficult but not “killer”; easier than last year’s IMO 2018/6 but harder than IMO 2015/3.

For gold-level contestants, I think this was the easiest exam to sweep in quite a few years, and I confess during the IMO to wondering if we had a small chance at getting a full 252 (until I found out that the marking scheme deducted a point on P2). Problem 2 is tricky but bary-able, and Problem 5 is quite easy. Furthermore neither Problem 3 or Problem 6 are “killers” (the type of problem that gets fewer than 20 solves, say). So a very strong contestant would really have 3 hours each day to work on a Problem 3 or Problem 6 which is not too nightmarish. I was actually worried for a while that the gold cutoff might be as high as 34 points (five problems), but I was just being silly.

## 3 thoughts on “IMO 2019 Aftermath”

1. Po-Shen Loh’s daughter went to mathcounts nationals(not in the top 560 as well as MPFG, correct?

Anyways, good job to the US team at IMO and thanks to all AoPS users for not believing my many troll comments about it(e.g. me posting IMO stole a problem from ASMP). Otherwise, there would have been complications…

Hopefully, I can make IMO at some point, even though that is nearly impossible given my current stats; making JMO at some point might be a more realistic goal.

Anyways, thank you Evan for another productive post and congratulations to djmathman for getting a problem on IMO.

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1. The “Geoff rule” is the colloquial name for the new policy instituted since IMO 2013 where problems 1, 2, 4, 5 are required to be different subjects (A/C/G/N).

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