BMO1 2023

The first round of the British Mathematical Olympiad was marked in early December. Belatedly, here are some thoughts of the problems. These aren’t supposed to be official solutions, and some of them are not in fact solutions at all. Students reading this in future years are advised, as always, that such commentaries are normally more valuable after attempting and digesting the problem yourself first.

Note that the copyright to the problems is retained by BMO, and I’m reproducing here with permission. The paper can be found in its original format here.

Problem 1

I only have a few brief comments on this problem.

When you are asked to count some class of objects in a problem, it is common to make progress by rephrasing or recharacterising the objects you are trying to count. Often the recharacterised object may be more naturally suited to enumeration. As a slogan: do some thinking or apply some logic first, before starting counting.

However, when counting anything at all, make sure you read the question very carefully! There are some types of problem where misreading the problem will become obvious later. For example:

• In a ‘prove that X is true’ question, you might find examples which fit your misreading, but where X is clearly not true!
• In a geometry question, an accurate figure may be impossible if you assume the wrong pair of angles are equal, or you may end up proving something absurd, eg all triangles are equilateral.

But counting problems do not have this quality! Unlike a proof-based question, it always works as a problem to say count the number of […]. Whether it’s a good problem is another matter. Indeed, whether it is possible to get a nice answer may be hard to say. There are many natural combinatorial objects indexed by n which can’t be enumerated in closed-form (meaning as a single expression involving a combination of standard functions).

It would be a real shame to drop significant marks on this approachable P1 by instead answering (or wasting time struggling to answer) a different problem that carries no credit.

Problem 2

In the IMO programme, we discuss how the required conclusion of a problem might not be the most natural insight about the setup. Instead, the required conclusion might instead follow quickly from a much more natural fact which is not discussed at all in the problem statement. Especially in the context of geometry, this is often the crux of a good problem. Since we aren’t allowed to ask “say something interesting about this setup”, a well-structured problem might require the solver to investigate the setup and discover something interesting without prompting.

In this case, the investigation involves exploring the given recursion (for in terms of ) and noticing that it can be converted into a recursion for the increments in terms of . What makes this clever is that it works nicely but differently in the two cases. Note that because the conclusion involves ‘consecutive integers’ (ie where the increment is equal to 1), this is a clue that this is a good line of attack.

Specifically, we get or .

There has been no reference to or to the fact that the sequence is integer-valued, but now one can bring that back into focus. If , this means that for some . But are integers, and so the only option is .

Problem 3

This is an excellent BMO1 easy geometry problem, and should be near the top of the list for students in future years who are practising this subject while preparing.

This is a good exercise in avoiding mindless angle-chasing. In general, most geometry problems at this difficulty level can be solved just by identifying pairs of equal angles, possibly with an occasional use of something like or if there’s a clear geometrical meaning to those quantities.

One goal might be to prove carefully that triangle AXZ is isosceles with . This, combined with ABX being isosceles, shows that ABXZ is a kite, and so BZ is perpendicular to AX, which is the same line as AC. (*)

And with so many isosceles triangles given in the construction, as well as other pairs of angles coming from the cyclic quadrilaterals, this equality of angles in isosceles AXZ can be handled as shown below, or in many other orders.

The equal red angles and equal green angles immediately show that , since both are ‘180 minus red minus green’. Observe that because we’ve already handled a reduction to the problem at (*), this diagram didn’t actually need to have line BZ included, even though it’s part of the original conclusion!

A remark about diagrams: if you match angles efficiently eg as in the figure above, you will discover that triangle AXZ is isosceles even if it doesn’t look particularly isosceles on your figure. Similarly for CXY if you use that in your proof. However, if you draw an accurate figure using compasses, then depending on where the points end up, it may be very clear immediately from the diagram that these triangles are isosceles. This has good points and bad points:

• Good point: from an accurate figure, it’s more likely that you’ll have the thought process along the lines of “Oh, so AXBZ must be a kite, and that explains why those lines are perpendicular, so I should prove the kite result.”
• Bad point: it can be easy accidentally to assume one of these intermediate results before proving it. If you assume that CXY is isosceles, or something equivalent (like BCXY a kite, or X,Y are reflections in BC) this is likely to be viewed as a big hole in the proof, not much worse than assuming directly that AXZ is isosceles.

The configuration is intriguing at a broader level. If you extend BX to meet the circumcircle, you now have a cyclic hexagon where the diagonals meet at X and form three kites.

Problem 4

I submitted this problem, and a separate post is now available.

Problem 5

The wording of this problem follows a standard pattern, and the wording of a successful solution should also follow a standard template. If spending a few hours preparing for BMO1 or similar level competitions, it would be worth devoting an hour to problems of this sort, because they come up all the time.

A previous post has discussed the yellow and white houses from the MOG paper in 2018, which is similar in structure, but this P5 has more unusual constraints. Nonetheless, the principles of that post apply. A solution attempt must include a statement of the conjectured smallest number K of faults plus

• a construction X showing that it is possible to have K faults
• an argument explaining why K-1 is not possible.

Note that in general, an argument saying “start with X – you can’t remove any faults” will not be valid. After all, how do you know you haven’t missed some much better construction that is totally unrelated to X?

Whether it’s in a live competition or for practice, time spent investigating what K should be is essential. If you commit too early to a value of K that is too large, you certainly won’t be able to prove that K-1 is not possible!

Returning to the details of this question, the fact that the total number of dots is a multiple of 4 but not a multiple of 3 gives a clue that the optimal construction might be based on groups of 4. Probably many students will have found a repeating pattern of four colours that gives 250 faults in total. This is a plausible guess for the minimum.

There is one challenge in proving this, namely that the pattern RBBR has no faults within this group of four. However, it must create a fault at either end when you consider the next group of four on either side. So the key here is to index the faults by the left of the two affected dots (or, properly speaking, we should say the anticlockwise of the two affected dots, since they are on a circle). Then, one can justify, either by a reduction argument or by simply listing all possibilities, that each group of four indexes at least one fault, which proves that the minimum is 250 as required.

Problem 6

This year, more participants than usual scored close to full marks on BMO1, but I don’t think it is fair to say that this particular problem was too easy. The harder part of the problem (proving that it is necessarily true for n odd) is relatively short. That doesn’t mean that it’s obvious, and there are many ways to go astray. I personally spent 20 minutes on the n odd case pursuing ideas that are hard to articulate but certainly did not work. It just means that it’s relatively unlikely that a contestant had the main idea but ran out of time to write it.

It’s crucial to establish the correct answer by trying various constructions. It’s possible that the octagon n=8 was more instructive than the hexagon n=6. It’s crucial to notice that if you start with a regular k-gon (for any k>2), you can ‘decorate’ each side with congruent ‘shallow isosceles triangles’ to get a 2k-gon that satisfies the length condition but doesn’t have equal angles.

One advantage of taking the isosceles triangle to be very shallow is that we can claim that when is small enough, the 2k-gon is convex, and that the obtuse angle in the triangle is greater than the total obtuse angle at the vertices of the k-gon without formally justifying this with algebra.

Of course this approach does not make sense for constructing an n-gon when n is odd. But this is absolutely not a proof that there is no such construction for n odd. However, in the context of an olympiad problem at this level, it is reasonable evidence that the statement might be true for n odd.

The condition given about the lengths is a local condition, meaning that each condition only affects nearby vertices. When thinking about proving this result, one needs to consider whether we will find a local conclusion (eg two consecutive angles are equal), or a global conclusion (eg every angle is less than or equal to the ‘average angle’). This distinction is less clear when n is small, of course.

I won’t spoil the end of the problem, except to say that I wasted my 20 minutes thinking about global effects, whereas actually the main route involves a neat observation of the local situation.