The second round of the British Mathematical Olympiad was taken yesterday by the 100 or so top scoring eligible participants from the first round, as well as some open entries. Qualifying for BMO2 is worth celebrating in its own right. The goal of the setters is to find the sweet spot of difficult but stimulating for the eligible participants, which ultimately means it’s likely to be the most challenging exam many of the candidates sit while in high school, at least in mathematics.
I know that lots of students view BMO2 as something actively worth preparing for. As with everything, this is a good attitude in moderation. Part of the goal in writing about the questions at such length (and in particular not just presenting direct solutions) is because I think at this level it’s particularly easy to devote more time than needed to preparation, and use it poorly.
All these questions could be solved by able children. In fact, each could be solved by able children in less than an hour. You definitely count as an able child if you qualified or if your teacher allowed you to make an open entry! Others count too naturally. But most candidates won’t in fact solve all the questions, and many won’t solve any. And I think candidates often come up with the wrong reasons why they didn’t solve problems. “I didn’t know the right theorems” is very very rarely the reason. Olympiad problems have standard themes and recurring tropes, but the task is not to look at the problem and decide that it is an example of Olympiad technique #371. The task is actually to have as many ideas as possible, and eliminate the ones that don’t work as quickly as possible.
The best way to realise that an idea works is to solve the problem immediately after having the idea. For the majority of occasions when we’re not lucky enough for that to happen, the second-best way to realise that an idea works is to see that it makes the problem look a bit more like something familiar. Conversely, the best way to realise that an idea doesn’t work is to observe that if it worked it would solve a stronger but false problem too. (Eg Fermat’s Last Theorem *does* have solutions over the reals…) The second-best way to realise that an idea doesn’t work is to have the confidence that you’ve tried it enough and you’ve only made the problem harder, or less familiar.
Both of these second-best ideas do require a bit of experience, but I will try to explain why none of the ideas I needed for various solutions this year required any knowledge beyond the school syllabus, some similarities to recent BMOs, and a small bit of creativity.
As usual, the caveat that these are not really solutions, and certainly not official solutions, but they are close enough to spoil the problems for anyone who hasn’t tried them by themselves already. Of course, the copyright for the problems is held by BMOS, and reproduced here with permission.
I wrote this question. Perhaps as a focal point of the renaissance of my interest in geometry, or at least my interest in teaching geometry, I have quite a lot to say about the problem, its solutions, its origin story, the use of directed angles, the non-use of coordinate methods and so on. In an ideal world I would write a book about this sort of thing, and perhaps at some point I will.
I also wrote this problem, though I feel it’s only fair to show the version I submitted to the BMO committee. All the credit for the magical statement that appears above lies with them. There is a less magical origin story as well, but hopefully with some interesting combinatorial probability, which is postponed until the end of this post.One quick observation is that in my version Joe / Hatter gets to keep going forever. As we shall see, all the business happens in the first N steps, but a priori one doesn’t know that, and in my version it forces you to strategise slightly differently for Neel / Alice. In the competition version, we know Alice is done as soon as she visits a place for a second time, but not in the original. So in the original we only have to consider ‘avoid one place’ rather than the multiple possibilities now of ‘avoid one place’ or ‘visit a place again’.
But I think the best idea is to get Alice to avoid one particular place whenever possible. At all times she has two possible options for where to go next, lets say in the language of the original statement. We lose nothing by assuming , and certainly it would be ridiculous for Joe / Hatter ever to choose . The only time Alice’s strategy doesn’t work is when both of these are congruent to , which implies , and thus we must have . In other words, Alice’s strategy will always work if N is odd.
I think it’s really worth noticing that the previous argument is weak. We certainly did not show that N must be odd for Alice to win. We showed that Alice can avoid a congruence class modulo an odd integer. We didn’t really need that odd integer to be N for this to work. In particular, if N has an odd factor p (say a prime), then the same argument works to show that we can avoid visiting any site with label congruent to 1 modulo p.
It’s actually very slightly more complicated. In the original argument, we didn’t need to use any property of . But obviously here, if modulo p and , then certainly modulo p. So we have to prove instead that Alice can ensure she never ‘visits 1 modulo p for the first time’. Which is fine, by the same argument.
So, we’ve shown that Neel / Alice wins if N is odd, or has an odd factor. The only values that remain are powers of 2. I should confess that I was genuinely a little surprised that Joe / Hatter wins in the power of 2 case. You can find a construction fairly easily for N=2 and N=4, but I suspected that might be a facet of small numbers. Why? Because it still felt we could avoid a particular site. In order for Alice’s strategy to fail, we have to end up exactly opposite the particular site at exactly the time when the next , and so maybe we could try to avoid that second site as well, and so on backwards?
But that turned out to be a good example of something that got very complicated quite quickly with little insight. And, as discussed at the beginning, that’s often a sign in a competition problem that your idea isn’t so good. (Obviously, when composing a problem, that’s no guarantee at all. Sometimes things are true but no good ideas work.) So we want other ideas. Note that for N=4, the sequence (2,1,2) works for Joe / Hatter, because that forces Alice / Neel to visit either (0,2,1,3) or (0,2,3,1). In particular, this strategy gave Alice no control on the first step nor the last step, and the consequence is that we force her to visit the evens first, then transfer to an odd, and then force her to visit the other odd.
We might play around with N=8, or we might proceed directly to a general extension. If we have a Joe / Hatter strategy for N, then by doubling all the s, we have a strategy for 2N which visits all the even sites in the first N steps. But then we can move to an odd site eg by taking . Just as in the N=4 case, it doesn’t matter which odd site we start from, since if we again double all the s, we will visit all the other odd sites. This gives us an inductive construction of a strategy for powers of two. To check it’s understood, the sequence for N=8 is (4,2,4,1,4,2,4).
Although we don’t use it, note that this strategy takes Alice on a tour of sites described by decreasing order of largest power of two dividing the label of the site.