EGMO 2019

Last week, we held our annual IMO training and selection camp in the lovely surroundings of Trinity College, Cambridge. Four of our students have subsequently spent this week in Kiev, for the ninth edition of the prestigious European Girls’ Mathematical Olympiad.

The UK team, none of whom had attended the competition before, and all of whom remain eligible to return at least once, did extremely well, placing fourth out of the official European countries, and earning three silver medals, and one gold medal, with complete or almost-complete solutions to all six problems between them. More details are available on the competition website.

In this post, I’m going to discuss some of the non-geometry problems. As a special treat, the discussion of Question 6 is led by Joe Benton, which is only fitting, since he wrote it. You can find the first day’s three problems here, and the second day’s problems here. The geometry problems are treated in a separate post.

Problem One

Triple equalities are hard to handle directly, so generally one starts with a single equality, for example the left hand one a^2b+c = b^2c+a, after noting that the setup is cyclic (but not symmetric) in the variables, under cycling a\to b \to c\to a.

Some quick notes:

  • The given equations are inhomogeneous, meaning that not every term has the same degree in terms of {a,b,c}. In complicated equations, normally we want to cancel terms, and we certainly can’t cancel terms with different degrees! So a good first idea is to find a way to make the equations homogeneous, for example using the condition, which is itself inhomogeneous. For example, one could replace c with c(ab+bc+ca), and it’s clear this might lead to some cancellation. In this case, the algebra is more straightforward if one rearranges and factorises first to obtain a(1-ab)=c(1-bc), from which the condition gives a(bc+ca)=c(ab+ac).
  • Shortly after this, we find a^2c=ac^2, which means a=c unless at least one of these terms is equal to zero.
  • This case distinction should not be brushed over lightly as a tiny detail. This is the sort of thing that can lose you significant marks, especially if the omitted case turns out to be more involved than the standard case.
  • This is a good example of planning your final write-up. The case distinction \{a,c\ne 0\}, \{a\text{ or }c=0\} is really clumsy. What about b? We have already said that the setup is cyclic in (a,b,c), and it’s annoying to throw this aspect away. It really would be much much to have the overall case distinction at the start of your argument, for example into \{a,b,c \ne 0\} and \{a\text{ or }b\text{ or }c=0\}, because then in the latter case you could assume that a=0 without loss of generality, then see what that implied for the other variables.
  • In this setting, one really should check that the claimed solutions satisfy both the condition and the triple-equality. This is a complicated family of simultaneous equations, and even if you’re lucky enough to have settled on an argument that is reversible, it’s much easier to demonstrate that everything is satisfied as desired than actually to argue that the argument is reversible.

Problem Two

Maybe I speak for a number of people when I say I’m a little tired of domino tiling problems, so will leave this one out.

Problem Five

I liked this problem. A lot. It did feel like there were potential time-wasting bear traps one might slip into, and perhaps these comments are relevant:

  • This feels like quite a natural problem to ask. In particular, the task specified by (A)+(B) is much more elegant than the RHS of (C). So unless you have immediate insight for the RHS of (C), it makes sense to start with the task, and aim for any bound similar to (C).
  • There are a lot of transformations one could make, for example, repeatedly removing n from one of the a_ns, or re-ordering them conveniently somehow, which shouldn’t affect any solution. You can also attack the RHS of (C) in a number of ways (perhaps with a case split for n odd or even?). As with involved but easy angle chases discussed in the companion post, it’s probably better to hold this in mind than instantly to do all of these, since it will just obscure the argument if you end up not using the result of the simplification!
  • One should always try small examples. Here, it’s worth trying examples large enough to get a sense of what’s happening, because I think the crucial observation is that (unless you were very very unlucky) there are lots of suitable sequences B=(b_i). (*)
  • Certainly the case where the (a_i)s themselves form a complete n-residue system is ‘easiest’ and certainly straightforward. One might say that the case where the (a_i) are equal (or at least congruent) is ‘hardest’ (in that (b_i-a_i) might have to take the largest values in some sense), but is also straightforward. There’s certainly the option of trying to rigorise this, but I don’t know whether this can be made to work.

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BMO1 2018

The first round of the British Mathematical Olympiad was sat yesterday. The paper can be found here, and video solutions here. Copyright for the questions is held by BMOS. They are reproduced here with permission.

I hope any students who sat the paper enjoyed at least some of the questions, and found it challenging! The following commentaries on the problems are not official solutions, and are not always full solutions at all, but contain significant steps of solutions, so would be best saved until after you have attempted the problems, if you are planning to do so. I’ve written quite a lot about Q5 because I found it hard (or at least time-consuming) and somewhat atypical, and I’ve written a lot about Q6 because there was a lot to say. I hope at least some of this is interesting to some readers of all levels of olympiad experience.

Question 1

A list of five two-digit positive integers is written in increasing order on a blackboard. Each of the five integers is a multiple of 3, and each digit {0,1,…,9} appears exactly once on the blackboard. In how many ways can this be done? (Note that a two-digit number cannot begin with zero.)

It’s a trope of BMO1 that the first question must be doable by some sort of exhaustive calculation or listing exercise. Of course, that is rarely the most efficient solution.

However, there is normally a trade-off between eliminating all listing, and reducing to a manageable task.

The key observation here is that writing the integers in increasing order is really just a way to indicate that order of the choices doesn’t matter. Even if that seems counter-intuitive. The question wants to know how many ways to choose these five numbers. The order of choice doesn’t matter since we’re going to put them in ascending order on the blackboard anyway.

You want to make your choices with as much independence as possible. So it would, for example, be a bad idea to choose the smallest number first. How many possibilities are there where the smallest number is 24? What about 42? What about 69? These are all different, and some are zero, so will make the computation very taxing.

However, you might notice that the digits {0,3,6,9} have to go together to form two numbers, and the rest have to pair up with one digit from {1,4,7} and one from {2,5,8}. You might know that an integer is divisible by 3 precisely if its digit sum is divisible by 3, but in this context you wouldn’t lose too much time by simply listing everything! These tasks are now completely separate, so you can take the number of ways to pair up {0,3,6,9} and multiply by the number of ways to pair up {1,4,7} and {2,5,8}. You need to take care over the ordering. It does (obviously) matter which is the first digit and which is the second digit in a number!

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