# BMO2 2019

The second round of the British Mathematical Olympiad was taken on Thursday 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 is because I think at this level it’s particularly easy to devote more time than needed to preparation, and use it poorly. This year time is tight at the end of semester, and so what follows is closer to a set of complete solutions than usual, for which apologies, although I hope it is still possible to get a sense of how one might have come across the solutions yourself. Of course, this means that what follows will certainly spoil the problems for anyone who hasn’t tried them by themselves already.

The copyright for the problems is held by BMOS, and reproduced here with permission.

Question 1

As if often the case in geometry questions, what you’ve been asked to prove here isn’t the most natural property of the configuration. A good first step would be to see if there are stronger statements which are true.You are asked to show that triangle BPE is isosceles, but you aren’t told which of the three vertices is the apex. In fact, the task is to show that $BP=EP$ or, alternatively, $\angle BEP=\angle PBE$. It’s not in general true that BE is equal to BP=EP. Unless you’re very unlucky, you can establish this from one diagram.

Now, you don’t immediately know whether it’s going to be easier to show that two lengths are equal, or that two angles are equal. However, you know that P lies on the perpendicular bisector of BC, hence BP=CP, which is a big clue. In particular, this means that P would be the centre of the circle through BCE. This clearly implies the given result, so deciding to prove this instead is a good strategy.

There are now a number of ways to prove this. Note that D lies on the altitude from A, and the feet of the perpendiculars from D to sides AB and AC are both present in the configuration so (just as for the orthocentre diagram) we can calculate most of the angles involving {A,B,C,D,E}.

For example, ABDE is cyclic, so $\angle BED=\angle BAD = 90-\hat{B}$, hence $\angle AEB=\hat{B},\,\angle EBA=\hat{C}$. This shows that AB is tangent to the circumcircle of BCE. But then the line L is a radius of this circle, and so its centre must be P, the unique point on L which is equidistant from B and C.

Alternatively, we could directly calculate $\angle BEC=180-\hat{B}$ and $\angle CBP=90-\hat{B}$. But BPC is isosceles so $\angle BPC=2\hat{B}$. In general, the converse of ‘angle at centre is twice angle at circumference’ does not hold, but when we know P is equidistant from B and C this does hold, and so the angle relations precisely confirm that P is the centre of the circle through BPE.

My intention had been that the triangle would be acute-angled, to reduce the number of diagram options based on the magnitude of $\hat{B}$. If pursuing this second approach, one would need to be careful to account for whether P is on the same side or the opposite side of BC to E. That said, unless you do something very exotic, it should be exactly the same argument or calculation, and such a case distinction probably isn’t very important.

Question 2

First, a short remark. As stated, if n=5, a piece could move 3 squares to the left then 4 squares up, by Pythagoras. Handling all such options is likely to be quite annoying, since some values of n can be written in this Pythagorean form, and others cannot. This brings us to some good general principles for olympiad problems which look like this one:

• A construction, when one exists, will probably be possible using simple versions of the allowed moves / structures.
• An argument why a construction is impossible should probably be based on ideas which treat the simple moves similarly to the more complicated moves.

The setup of the problem encourages you to think about dividing the board into $n^2$ sub-boards, each with dimensions $n\times n$. Continue reading